Part IV · Quantum Mechanics — Chapter 4.5

The Spectral Theorem in Infinite Dimensions

An observable can have no eigenvectors at all. So "diagonalise it" has to be replaced by something that still says what the possible values are, and that replacement is what the rest of Part IV runs on.

Tools you'll need  — Chapter 0.5 above all: §6 for the finite-dimensional spectral theorem in both its forms, §6.4 for the projection form and the two promises made there, §7 for functions of an operator, and its closing paragraph, which this chapter finishes paying. Chapter 4.4 §3 for domains and the adjoint's domain, §4 for self-adjointness, §5.4 for the circle of momentum operators on an interval, and §6 for the extension count. Chapter 4.3 §4.3 for dominated convergence, §5.3 for the fact that a vector of $L^{2}$ has no value at a point, §6 for completeness, §7.3 for the four equivalent statements about a basis, and §8.5 for the warning this chapter answers. Chapter 0.9 §2.1 for the box limit and the manufacturing of the mode spacing, §2.3 for Plancherel and the unitarity of the Fourier transform, §3.1 for the derivative theorem, and §5 for the delta defined by what it does. Chapter 4.2 §4 for the second postulate, §5 for the Born rule, and §7 for time evolution and its generator.

1 · The half of the bill that is still unpaid

This section does three things and then gets out of the way. It reads back what Chapter 0.5 left outstanding and says which part of it belongs here. It states the difficulty in its sharpest form, which is an operator with no eigenvectors at all rather than an operator whose eigenvectors are hard to find. And it says, before any machinery arrives, what the three replacements are going to be, so that you can watch each one land. The part to hold on to is the second: the failure is not that a proof is missing, it is that the conclusion is false as stated.

1.1 · Which instalment this is

Chapter 0.5 ended by naming four steps in which its proofs had leaned on the dimension being finite, and it said the bill would come due in Chapters 4.4 and 4.5. Chapter 4.4 §1.2 took the four in order and settled two of them. The interchange of sums was paid outright, using a corollary of monotone convergence that Chapter 4.3 had already proved. The failure of injective implies surjective was worked in full, first as the shift operator and then, in that chapter's Worked example 3, as the reason a particle held on a half-line has no momentum observable.

Two remain, and Chapter 4.4 §1.3 assigned both of them here. They are the induction, whose very first step assumes an eigenvector exists, and rank–nullity, whose replacement is the observation that an image can be dense without being closed. Those look like two separate repairs and they are one, as §2 will show: coming arbitrarily close to every vector and reaching every vector are different achievements, and the gap between them is exactly where an operator keeps values that have no eigenvectors attached to them.

So this chapter is the second and last instalment of the bill Chapter 0.5 named. Counting instead from Chapter 4.3, which built the space rather than the operators on it, this is the third chapter of the run, and Chapter 4.4's closing brick counts it that way. Either count gives the same shape, and it is the shape Chapter 4.2's closing brick announced in advance: Chapter 0.4 built the space and Chapter 0.5 built the operators on it. Chapter 4.3 builds the space and Chapters 4.4 and 4.5 build the operators on it.

1.2 · The difficulty, in its sharpest form

Chapter 0.5 §6.3 proved its theorem by an induction whose first step was to produce one eigenvector, using the fact that a polynomial over $\C$ has a root. Here is what happens to that step. Take $\hat x$, multiplication by $x$, on $L^{2}(\R)$, and ask for a vector $\psi$ and a number $\lambda$ with $\hat x\psi=\lambda\psi$. Written out, that is the requirement

At every point where $x\ne\lambda$ the first factor is not zero, so $\psi(x)$ must be. The set where $x=\lambda$ is a single point, which has measure zero by Chapter 4.3 §2.4, so $\psi$ vanishes almost everywhere, and by the quotient of Chapter 4.3 §5.3 that means $\psi$ is the zero vector. There is no eigenvector for any $\lambda$ whatsoever, real or complex. This is the step where the finite-dimensional argument stops working, and it is worth slowing down over, because the usual reassurances do not apply: nothing here is a matter of the eigenvector being awkward or the proof being incomplete.

Momentum fails in the mirror-image way, and Chapter 0.9 §2.1 had already flagged it. The equation $-\ii\hbar\psi'=p\psi$ does have solutions, namely $\psi(x)=\ee^{\ii px/\hbar}$, and they are perfectly good functions. They are not vectors of $L^{2}(\R)$, because $\abs{\ee^{\ii px/\hbar}}^{2}=1$ and the integral of $1$ over the line is infinite. So for $\hat x$ the eigenvalue equation has only the zero solution, and for $\hat p$ it has solutions that live outside the space. Both operators are self-adjoint, by Chapter 4.4 §5.3 and its Worked example 1, and both are observables in the sense of Chapter 4.2's second postulate. Neither has a single eigenvector.

Say plainly what that costs. Chapter 0.5's conclusion was that a Hermitian operator has an orthonormal basis of eigenvectors, and Chapter 4.2 spent that conclusion everywhere: the possible outcomes of a measurement are the eigenvalues, the state expands in eigenstates, and the Born rule is a statement about projections onto eigenspaces. For position and momentum every one of those sentences is about an empty set. The theorem is not weakened here. It is inapplicable, which is a different and more serious thing, and Chapter 0.5's warning box used exactly that distinction about defective matrices.

1.3 · The three replacements, named in advance

What follows repairs this in three moves, and here they are as a list before any of them is built, because each is a weakening of something you already know and it helps to see which part is being given up.

• The eigenvalues become the spectrum. Instead of asking which $\lambda$ have eigenvectors, ask which $\lambda$ make $\hat A-\lambda$ fail to be invertible in a strong enough sense. In finite dimensions those two questions have the same answer, so nothing is being redefined. In infinite dimensions the second question has answers when the first has none, and the set of them is the spectrum.
• $A=UDU^{\dagger}$ becomes unitary equivalence to a multiplication operator. The finite-dimensional statement is that some unitary change of basis turns $A$ into a list of numbers attached to independent directions. The infinite-dimensional statement is that some unitary map turns $\hat A$ into multiplication by a real function on some space of square-integrable functions. A list of numbers is a function on a finite set, so this is the same sentence with a wider idea of what a list can be.
• The sum $\sum_k\lambda_kP_k$ becomes an integral $\int\lambda\,\dd P(\lambda)$. Chapter 0.5 §6.4 wrote the theorem a third way, as a weighted sum of projections onto eigenspaces, and said that this basis-independent form is the one that survives here. It does survive, with the sum replaced by an integral against a family of projections indexed by the real line. That family is what the physics is waiting for. The Born rule is a statement about projections, so until a projection exists for a continuous observable there is no probability to compute, and a position measurement has no theory of its outcome at all.

One of those three is quoted rather than proved, and it is the middle one. That is the single substantial mathematical flag of Part IV, and §3.2 states it with its hypotheses attached. The answer to the obvious objection is the point of §§4 and 5: the theorem is then verified by hand on $\hat x$, on $\hat p$ and on the oscillator Hamiltonian, which are the three operators everything later is built out of. You will finish this chapter holding a quoted theorem that you have checked yourself on the cases that matter, and §5.6 says exactly how far that checking reaches.

The three replacements are not where the chapter stops, and knowing now what they are for is what carries you through the checking. Once an observable has become multiplication by a real function, every function of that observable can be built the same way, and the one Part IV needs most is the exponential $\ee^{-\ii\hat Ht/\hbar}$. Section 9 shows that this operator is unitary, that evolving by $s$ and then by $t$ is evolving by $s+t$, and that differentiating it at the origin hands $\hat H$ back. That is Stone's theorem, and what it buys is that time evolution follows from self-adjointness instead of being postulated alongside it. Chapter 4.6 begins there.

2 · The spectrum, when there are no eigenvectors

Section 1 left a hole where the eigenvalues used to be, and this section fills it. We take the finite-dimensional definition of an eigenvalue, notice that it can be phrased in a way that never mentions an eigenvector, adopt that phrasing as the definition, and check that in finite dimensions it gives back exactly the old answer. Then we run it on position and momentum and get, in both cases, the whole real line. The result to carry forward is the last one: for a self-adjoint operator the spectrum is real and it has only two parts rather than three, and both of those facts are proved here rather than assumed.

2.1 · The definition an eigenvalue already had

In Chapter 0.5 a number $\lambda$ was an eigenvalue of $A$ when $Av=\lambda v$ for some $v\ne0$. That says $A-\lambda$ kills a non-zero vector, which is the same as saying $A-\lambda$ is not injective, which for a square matrix is the same as saying $A-\lambda$ is not invertible. The chain of equivalences repays a pause, because every link in it is a finite-dimensional fact. Killing nothing and reaching everything are the same condition only when there is a dimension count to convert one into the other, and Chapter 4.4 §1.2 showed that count failing in the flesh with the shift operator.

So the finite-dimensional definition can be read two ways, and the two readings come apart here. One of them mentions an eigenvector and the other does not, and we want the one that does not, because §1.2 has just shown there may be no eigenvector to mention. Taking invertibility as the primitive costs nothing in finite dimensions and buys everything in infinite ones.

2.2 · The definition, widened

Let $\hat A$ be an operator with a dense domain in a Hilbert space $H$, in the sense Chapter 4.4 §3 fixed. Since $\hat A$ acts only on $\operatorname{dom}(\hat A)$, so does $\hat A-\lambda$, and asking whether it is invertible means asking whether it maps that domain onto the whole space in a way that can be undone continuously. Write that out as the condition on $\lambda$:

The set $\rho(\hat A)$ of such $\lambda$ is the resolvent set, and $(\hat A-\lambda)^{-1}$ is the resolvent. Everything this chapter calls a value of an observable is defined by the failure of that condition. So the spectrum is what is left over:

Three demands are being made at once in (4.5.2) and each can fail on its own, which is why the spectrum splits into parts rather than being a single kind of thing. Notice also that boundedness of the inverse is a genuine third demand and not a convenience. Chapter 4.4 §2.1 measured what an unbounded operator does to a convergent sequence, and an unbounded inverse would mean that vectors $\hat A$ maps to nearly nothing are not nearly nothing themselves, which is precisely the situation an approximate measurement is in.

2.3 · Nothing has been redefined, only widened

Before using the new definition, confirm that it agrees with the old one where the old one applies, since otherwise the chapter would be changing the subject rather than continuing it. Let $A$ be an operator on a space of finite dimension $n$, so $\operatorname{dom}(A)$ is everything and every linear map is bounded by Chapter 0.6 §2. If $A-\lambda$ is injective then rank–nullity makes it surjective, and the inverse is a linear map on a finite-dimensional space and therefore bounded, so $\lambda$ is in the resolvent set. If $A-\lambda$ is not injective then $\lambda$ is an eigenvalue in the old sense. The two possibilities are exhaustive, so

Every step of that used finite dimension, which is the point: the new definition is the old one wherever the old one was available, and the two separate only where Chapter 0.5's proofs separate from their hypotheses. The word spectrum has been widened and nothing has been renamed.

2.4 · The three ways the condition can fail

Since (4.5.2) makes three demands, a $\lambda$ in the spectrum fails at least one of them, and sorting by which one fails first gives the standard division. Read the three as a sequence of increasingly mild failures.

• Point spectrum $\sigma_p$: $\hat A-\lambda$ is not one-to-one. Something non-zero is sent to zero, so there is an eigenvector, and $\lambda$ is an eigenvalue in Chapter 0.5's sense. This is the part that survives from finite dimensions unchanged.
• Continuous spectrum $\sigma_c$: $\hat A-\lambda$ is one-to-one, its image is dense, and its image is not the whole space. The inverse then exists on a dense set and is unbounded there. This is the case Chapter 4.4 §1.2 predicted when it said the replacement for rank–nullity is the distinction between a dense image and a closed one.
• Residual spectrum $\sigma_r$: $\hat A-\lambda$ is one-to-one and its image is not even dense. There is then a direction that no vector of the form $(\hat A-\lambda)\psi$ has any component along, which is a strong kind of failure.

One case is absent from that list, and here is why. A $\lambda$ could in principle leave $\hat A-\lambda$ one-to-one and onto while failing only the third demand, with an inverse that is defined everywhere and unbounded. For a closed operator that cannot happen. The inverse of a closed operator has a closed graph, and a closed map defined on the whole of a complete space is bounded, which is the closed graph theorem Chapter 4.4 §2.3 quotes. That mark is leaned on here rather than raised again, and §2.7 supplies the hypothesis by proving that a self-adjoint operator is closed.

With that case gone the three are mutually exclusive and together they are the whole spectrum. Section 2.7 will show that for a self-adjoint operator the third is always empty, so the physically relevant division is two-way: values that have states attached to them, and values that do not.

2.5 · Position: the spectrum is the whole real line

Now run the definition on the operator §1.2 showed has no eigenvectors. Take $\hat x$ on $L^{2}(\R)$, with $\operatorname{dom}(\hat x)=\{\psi:x\psi\in L^{2}\}$ from Chapter 4.4's Worked example 1. We want to know for which $\lambda$ the three demands hold, so start with the candidate inverse, which is forced: undoing multiplication by $x-\lambda$ can only be division by it.

Take $\lambda$ off the real axis first, since that is the easy case and it fixes the pattern. Writing $\lambda=a+\ii b$ with $b\ne0$, the modulus $\abs{x-\lambda}$ is at least $\abs b$ for every real $x$, so dividing by $x-\lambda$ multiplies the modulus of $\psi$ pointwise by at most $1/\abs b$. That makes (4.5.5) defined on every $\psi$ in the space and bounded, with $\norm{(\hat x-\lambda)^{-1}}\le1/\abs b$, and multiplying back by $x-\lambda$ returns $\psi$. All three demands hold, so no non-real number is in the spectrum.

Now take $\lambda$ real, and the useful thing to compute is not the inverse but how badly it fails to be bounded. Concentrate a unit vector on a shrinking interval to the right of $\lambda$, by setting $\psi_n=\sqrt n\,\mathbf 1_{[\lambda,\,\lambda+1/n]}$, which has $\norm{\psi_n}=1$ for every $n$. Apply $\hat x-\lambda$ to it and measure the result:

Unit vectors are being sent to vectors of length $1/n\sqrt3$, which tends to zero. Any inverse would have to send those short vectors back to unit ones, so its norm is at least $n\sqrt3$ for every $n$ and it cannot be bounded. Every real $\lambda$ is therefore in the spectrum, and by §1.2 none of them is an eigenvalue. Two further facts put them all in the continuous part, and the second is the one usually skipped. The image is dense: given any $\phi$, cut out a shrinking neighbourhood of $\lambda$ by setting $\phi_n=\phi\,\mathbf 1_{\{\abs{x-\lambda}\gt1/n\}}$. Then $\phi_n/(x-\lambda)$ is in the space, because the factor is bounded by $n$ there, and $\phi_n\to\phi$ in norm by dominated convergence with $\abs\phi^{2}$ as the dominating function, which is Chapter 4.3 §4.3. And the image is not the whole space, which the continuous case demands as well. Take $\phi=\mathbf 1_{[\lambda,\lambda+1]}$, a perfectly good vector. The only candidate for a $\psi$ with $(\hat x-\lambda)\psi=\phi$ is $\phi/(x-\lambda)$, and $\int_{\lambda}^{\lambda+1}(x-\lambda)^{-2}\dd x$ diverges, so that candidate is not in the space and $\phi$ is not in the image. So

Read that against what an instrument does. The possible readings of a position measurement are all the real numbers, which is the physically right answer, and the theory delivers it without producing a single state of definite position. The sequence $\psi_n$ is what stands in for the missing eigenvector, and it is not a formal device: $\psi_n$ is a state whose position is known to within $1/n$, which is what a real detector prepares.

2.6 · Momentum, by the same calculation in the other description

Momentum needs no separate argument, because Chapter 0.9 §2.3 supplies a unitary map that turns it into a multiplication operator, and the spectrum cannot notice a unitary change of description. Say why not, since the step is used repeatedly from here on. If $\hat W$ is unitary and $\hat B=\hat W\hat A\hat W^{-1}$, then $\hat B-\lambda=\hat W(\hat A-\lambda)\hat W^{-1}$, so one of the two is one-to-one, onto, or boundedly invertible exactly when the other is, since composing with a unitary map changes neither injectivity nor surjectivity nor any norm. Unitarily equivalent operators therefore have equal spectra, part by part.

Apply that with $\hat W$ the Fourier transform. Chapter 0.9 §3.1's derivative theorem says $\widetilde{\psi'}=\ii k\tilde\psi$, so transforming $\hat p=-\ii\hbar\,\dd/\dd x$ turns differentiation into multiplication:

In the transformed description $\hat p$ is multiplication by the real function $\hbar k$, which is $\hat x$ with $\hbar k$ in place of $x$. Section 2.5 therefore applies word for word, and the conclusion transfers back through the unitary:

That is the fact Chapter 0.9 §2.1 flagged and could not then justify, now derived. Momentum has every real number among its possible readings and not one momentum eigenstate in the space. It is also the first appearance of the pattern the whole chapter turns on, which is that a hard question about an operator became an easy question about a function once a unitary map had been found. Section 3 says that such a map always exists.

2.7 · Two facts about self-adjoint operators, proved

The definition in (4.5.2) allows complex $\lambda$, and an observable had better not have complex readings, so that has to be settled rather than assumed. It follows from self-adjointness together with completeness, and the argument also disposes of the residual spectrum. Start from the estimate that does the work. Let $\hat A$ be self-adjoint and let $\lambda=a+\ii b$ with $b\ne0$. Write $\hat B=\hat A-a$, which is symmetric because $\hat A$ is and $a$ is real. Now expand the norm for any $\psi$ in the domain:

The two middle terms cancelled because symmetry makes $\avg{\hat B\psi,\psi}$ and $\avg{\psi,\hat B\psi}$ the same number. Dropping the first term on the right gives $\norm{(\hat A-\lambda)\psi}\ge\abs b\,\norm\psi$, so $\hat A-\lambda$ is one-to-one and its inverse is bounded by $1/\abs b$ wherever it is defined. What remains is to show the inverse is defined everywhere, which is where completeness enters, and the grind box does it in two steps: the image is dense, and it is closed.

Grind box — the image of $\hat A-\lambda$ is the whole space when $\lambda$ is not real

First, the adjoint is closed. Suppose $u_n\in\operatorname{dom}(\hat A^{\dagger})$ with $u_n\to u$ and $\hat A^{\dagger}u_n\to w$. For every $v\in\operatorname{dom}(\hat A)$ the defining relation of Chapter 4.4 §3.3 gives $\avg{\hat A^{\dagger}u_n,v}=\avg{u_n,\hat Av}$. Both sides converge, because Cauchy–Schwarz makes the inner product continuous in each slot, and the limits are $\avg{w,v}=\avg{u,\hat Av}$. That is the statement that $u\in\operatorname{dom}(\hat A^{\dagger})$ with $\hat A^{\dagger}u=w$. So a convergent sequence in the domain whose images also converge has its limit in the domain, which is what closed means, and a self-adjoint operator is its own adjoint and therefore closed.

Second, the image is dense. Suppose it is not. Then there is $u\ne0$ orthogonal to it, so $\avg{u,(\hat A-\lambda)v}=0$ for every $v$ in the domain, which reads $\avg{u,\hat Av}=\avg{\bar\lambda u,v}$ and says $u\in\operatorname{dom}(\hat A^{\dagger})$ with $\hat A^{\dagger}u=\bar\lambda u$. Self-adjointness turns that into $\hat Au=\bar\lambda u$, so $\bar\lambda$ is an eigenvalue. But Chapter 4.4 §4.3 showed the eigenvalues of a symmetric operator are real, and $\bar\lambda$ is not. So no such $u$ exists.

Third, the image is closed. Let $y_n=(\hat A-\lambda)\psi_n$ converge to $y$. The estimate above applied to $\psi_n-\psi_m$ gives $\norm{\psi_n-\psi_m}\le\norm{y_n-y_m}/\abs b$, so $\psi_n$ is Cauchy and converges to some $\psi$ by completeness of the space, which is Chapter 4.3 §6.2. Then $\psi_n\to\psi$ and $\hat A\psi_n=y_n+\lambda\psi_n\to y+\lambda\psi$, so closedness puts $\psi$ in the domain with $\hat A\psi=y+\lambda\psi$, which is $(\hat A-\lambda)\psi=y$. So $y$ is in the image.

Dense and closed is everything, so $\hat A-\lambda$ maps its domain onto $H$, and with the bound already in hand all three demands of (4.5.2) hold.

That settles the first fact. Every later section leans on it, so it goes on its own line.

The second fact is the middle step of the grind box read again, and it costs nothing extra. If $\lambda$ is real and in the residual spectrum, then by definition the image of $\hat A-\lambda$ is not dense, so the same argument produces $u\ne0$ with $\hat Au=\lambda u$. That makes $\lambda$ an eigenvalue, so $\hat A-\lambda$ is not one-to-one, and the residual case required it to be. The two conditions contradict each other, so a self-adjoint operator has no residual spectrum. Every value an observable can return is therefore either an eigenvalue with a genuine state behind it, or a continuous-spectrum value with approximate states of the kind (4.5.6) exhibited, and there is no third possibility.

3 · The spectral theorem, in multiplication form

We now have a set of possible readings for an operator that has no eigenvectors, and no way yet to compute with it. This section supplies the statement that does the computing. It is the replacement for $A=UDU^{\dagger}$, it is quoted rather than proved, and the two sections after it check it by hand on the three operators the rest of Part IV rests on. Before the statement arrives, see what shape it has to have. The shape is forced by §2, and the theorem then looks inevitable rather than arbitrary.

3.1 · What a list of numbers has to become

Chapter 0.5's conclusion, stripped of its notation, was that a Hermitian operator is a list of numbers attached to independent directions, seen from a rotated angle. The rotation is $U$, the list is the diagonal of $D$, and the directions are the columns of $U$. Ask what each of those three pieces has to become here, and take them in the order that forces the answer.

The directions are the part that cannot survive, because §1.2 showed there may be none. So the theorem cannot be stated in terms of them, and whatever replaces the list has to be indexed by something other than a set of eigenvectors. The list itself is the piece to work on. A list of $n$ numbers is a function on the set $\{1,2,\dots,n\}$, and a diagonal matrix acts on a vector by multiplying its $j$th component by the $j$th entry of that function. Written that way, the operation $D$ performs is not "attach a number to each direction" but multiply by a function, and multiplication by a function makes sense on any set at all, finite or not.

That is the whole of the generalisation, and §2.6 has already shown one instance of it working. The Fourier transform turned $\hat p$ into multiplication by $\hbar k$, and $\hbar k$ is a function on the real line rather than a list on a finite set. Nothing about that calculation needed an eigenvector, and the spectrum came out of it immediately. The theorem says this is always available.

3.2 · The theorem, quoted with its hypotheses

3.3 · Read against $A=UDU^{\dagger}$, term by term

Put the statement side by side with the one it replaces, because every piece corresponds and the correspondence is what makes it recognisable rather than new. This is the infinite-dimensional reading of Chapter 0.5's $A=UDU^{\dagger}$, and the reading goes like this.

• $U$ becomes $\hat U$. In Chapter 0.5 the columns of $U$ were the eigenvectors, so $U$ was built out of the directions. Here $\hat U$ is a unitary map onto a space of functions, and it is not built out of anything: its existence is the theorem. Chapter 0.9 §2.3 already supplied one such map, the Fourier transform, and proved it unitary.
• $D$ becomes multiplication by $g$. A diagonal matrix multiplies the $j$th component by the $j$th entry, and multiplication by $g$ multiplies the value at $\xi$ by $g(\xi)$. The index set has changed from $\{1,\dots,n\}$ to $X$ and the operation has not changed at all.
• The diagonal entries become the values of $g$. Chapter 0.5's list $\lambda_1,\dots, \lambda_n$ was the spectrum. Here the spectrum is the set of values $g$ takes, in a sense §3.4 makes precise, and $g$ may take each of them on a set of positive measure rather than at a single index.
• $D$ real becomes $g$ real. Chapter 0.5 got real eigenvalues from Hermiticity in three lines. Here $g$ real is part of the statement, and §2.7 proved the consequence independently, which is a check on the two agreeing.
• The eigenvectors have no counterpart. This is the one entry with nothing on the right, and it is the honest reason a new theorem was needed rather than a new proof of the old one.

One more difference has to be named, because it is where the physics enters. In Chapter 0.5 the measure was invisible, since a finite list carries counting measure and nobody has to mention it. Here $\mu$ is part of the data, and §§4 and 5 will produce three operators whose measures are Lebesgue measure on the line, Lebesgue measure on the line again, and counting measure on the non-negative integers. The difference between a continuous spectrum and a discrete one is a difference of measure, and it is the only thing separating position from the oscillator.

3.4 · The spectrum, read off $g$

The theorem is only useful if the spectrum can be read out of $g$, so establish that now, since every later section quotes it. Multiplication by $g$ fails to be boundedly invertible at $\lambda$ exactly when $1/(g-\lambda)$ fails to be bounded, and by the argument of §2.5 that happens exactly when $g$ comes arbitrarily close to $\lambda$ on sets of positive measure. Give that set a name. The essential range of $g$ is

and the spectrum of $\hat A$ is exactly that set, since unitary equivalence preserves spectra by §2.6. The qualifier essential is doing real work rather than decorating: a value that $g$ takes only on a set of measure zero is invisible to the space, because Chapter 4.3 §5.3's quotient has already declared such a set to carry no vectors. The distinction is the same one that made a wavefunction have no value at a point, appearing again on the other side of the unitary map.

The point spectrum reads off equally directly. A vector $\Psi$ satisfies $g\Psi=\lambda\Psi$ when $(g-\lambda)\Psi=0$ almost everywhere, which forces $\Psi$ to vanish wherever $g\ne\lambda$. So $\lambda$ is an eigenvalue exactly when $g$ takes the value $\lambda$ on a set of positive measure, and the eigenspace is the set of functions supported there. For $\hat x$ the function is $g(x)=x$, which is constant on no set of positive measure at all, and §1.2's calculation is that observation made twice. Worked example 2 builds an operator where $g$ is constant on a half-line, and it has both kinds of spectrum at once.

a natural place to stop  ·  the theorem is on the table; what follows is three checks of it by hand

4 · Checked three times

A quoted theorem is worth what you can check of it, so this section and the next check it. The plan is simple and the order is chosen so that the cost rises: position needs no work at all, momentum needs one unitary map that Chapter 0.9 already built and proved unitary, and the oscillator needs a basis whose completeness is proved from scratch in §5. Two of the three land here, and the third is set up here and finished there. Two subsections interrupt that count and neither is a fourth verification: §4.2 exhibits one operator through two different triples, which §3.2 said to expect, and §4.4 puts the second verification against arithmetic. At the end of §5 the three are collected, and that collection is what turns the mark of §3.2 from a promise into an accounting.

4.1 · Verification 1: position, where there is nothing to do

The theorem asks for a measure space, a unitary map and a real function. For $\hat x$ on $L^{2}(\R,\dd x)$ every one of them is already sitting there, because the operator was defined as multiplication in the first place. Take $X=\R$ with Lebesgue measure, take $\hat U$ to be the identity, and take the function to be $g(x)=x$:

The domain condition of §3.2 asks that $g\cdot\hat U\psi=x\psi$ lie in the space. That is exactly $\operatorname{dom}(\hat x)$ as Chapter 4.4's Worked example 1 computed it, so nothing has to be adjusted. The essential range of $g(x)=x$ is the whole real line, since every interval around every real number has positive length, and that reproduces (4.5.7) from a one-line reading of $g$ instead of from the sequence argument of §2.5.

This verification is trivial, and its triviality is the content rather than an embarrassment. What §3.2 asserts is that every self-adjoint operator looks like this one. Position is the model that the general statement is copied from, and checking it amounts to noticing that the model is a case of the statement it models.

4.2 · The same operator, two different answers

Section 3.2 said the triple $(X,\mu,g)$ is not unique, and position is where that is easiest to see, so take a moment over it now rather than meeting it later as a surprise. Map the line onto a bounded interval by $x=\tan\theta$, and carry states across in the way that preserves the norm:

That map is unitary, by the substitution $x=\tan\theta$ with $\dd x=\sec^{2}\theta\,\dd\theta$, which turns $\int\abs{\hat V\psi}^{2}\dd\theta$ into $\int\abs\psi^{2}\dd x$. Conjugating $\hat x$ by it gives multiplication by $\tan\theta$ on $L^{2}\big((-\pi/2,\pi/2),\dd\theta\big)$, so the same operator now comes with a bounded interval, a different measure and a different function. The essential range of $\tan\theta$ over that interval is still the whole real line, so the spectrum is unchanged, as §2.6 requires. What the theorem fixes is the spectrum, not the triple that displays it, and a reader who expects $(X,\mu,g)$ to be canonical will be looking for something that is not there.

4.3 · Verification 2: momentum, on a unitary you already built

Momentum is the first case where the unitary map has to be found rather than noticed, and it has already been found. Chapter 0.9 §2.3 proved that the Fourier transform preserves the norm and read that back as the statement that it is unitary, and §2.6 above transformed $\hat p$ with it in one line. Collecting that as the theorem's data:

Check the domain as well, since §3.2 makes it part of the statement and Chapter 4.4 spent a whole chapter on the point that a domain is not an afterthought. The theorem's condition is that $\hbar k\tilde\psi$ lie in $L^{2}(\dd k)$, and Chapter 4.4 §5.3 defined $\operatorname{dom}(\hat p)$ as the functions that are integrals of an $L^{2}$ derivative. Those are the same set, and Plancherel is what says so: $\psi'\in L^{2}$ transforms to $\ii k\tilde\psi\in L^{2}$, and the norms agree. So the description of the domain that took most of a page in Chapter 4.4 becomes, in the transformed picture, the single requirement that $\tilde\psi$ decay fast enough for $k\tilde\psi$ to stay square integrable.

Notice two things about how cheap that was. The whole verification is one line of algebra applied to a theorem you proved in Chapter 0.9, and the measure space is again the real line with Lebesgue measure, which is why position and momentum have identical spectra despite being different operators. Chapter 4.3 §8.5 had already said that the two are one Hilbert space seen through a unitary map; here that statement acquires its consequence for the values.

4.4 · The second verification, checked against arithmetic

Run the identity in (4.5.15) numerically, because it is the one place in this chapter where a claim about an unbounded operator can be put against a number. Work in units where $\hbar=1$ and lengths are measured in units of the width of the test state, and take a grid of $4096$ points across a box of length $40$. The test state is a Gaussian of unit width centred at $x=1$ and given a boost of $k=3$, which is $\psi(x)\propto\ee^{-(x-1)^{2}/2}\ee^{3\ii x}$, normalised on the grid.

Now compute $\hat p\psi$ two ways that share nothing. The first is the theorem's way: transform, multiply by $\hbar k$, transform back. The second is the definition's way: approximate $-\ii\hbar\,\dd\psi/\dd x$ by the centred difference $-\ii\hbar\big[\psi(x+h)-\psi(x-h)\big]/2h$. The two answers agree to a relative discrepancy of

and the size of that number is less informative than the way it moves. Halving the grid spacing divides it by four every time, running $3.13\times10^{-3}$, $7.82\times10^{-4}$, $1.96\times10^{-4}$, $4.89\times10^{-5}$ over $1024$, $2048$, $4096$ and $8192$ points. A centred difference has an error proportional to the square of the spacing, so that factor of four identifies the entire discrepancy as the finite difference's error, with nothing left over to attribute to the transform. The expectation values say the same thing more bluntly: multiplying by $\hbar k$ in the transformed picture returns $\avg{\hat p}=3.0000000000$, which is the boost that was put in, while the centred difference returns $2.9994993505$.

4.5 · Verification 3: the shape, and the one thing missing

The third operator is the harmonic oscillator Hamiltonian, and it is the case the rest of Part IV uses most, so it gets done properly. Write it with the particle's mass $m$ and the angular frequency $\omega$:

This one is different in kind from the first two, and the difference is the whole reason the theorem has to allow an arbitrary measure. Position and momentum have no eigenvectors, so their measure spaces had to be continuous. The oscillator does have eigenvectors, and they are the Hermite functions, so its measure space is going to be a set of isolated points carrying counting measure. If the theorem could not accommodate that, it would not cover the operator whose spectrum every physics course computes first.

The unitary map is the one Chapter 0.5 would have used: send a state to its list of coefficients against the eigenfunctions. What makes that map unitary is Parseval, and what makes Parseval available is that the eigenfunctions form a genuine orthonormal basis rather than merely an orthonormal set. Chapter 4.3 §7.3 is explicit that those are different claims, and it proved they are equivalent to two others, one of which is checkable. So the verification reduces to a single statement about the Hermite functions, and that statement is what §5 proves.

5 · The Hermite functions are complete

This section builds the one ingredient §4.5 was missing, and it builds it rather than quoting it, because Chapter 4.3's closing brick promised in writing that this chapter would. The route is short to describe. Define the Hermite functions by a generating function, and get two recurrences and a differential equation out of it. Use those to show the functions are eigenfunctions of $\hat H_{\text{osc}}$ and that they are orthonormal. Then prove the claim that costs something, which is that no non-zero vector is orthogonal to all of them. The proof of that last step uses dominated convergence and Plancherel and nothing else. No complex analysis is required and nothing here carries a mark.

5.1 · The functions, and the variable that removes the constants

Constants clutter this calculation and there is a length in the problem that removes all of them, so introduce it first. The oscillator has one length that can be built out of $\hbar$, $m$ and $\omega$, namely $x_0=\sqrt{\hbar/m\omega}$. Writing $\xi=x/x_0$ turns (4.5.17) into

One notational borrowing, since that letter is already in use. From §3.2 onward $\xi$ has been the generic point of the abstract measure space $X$, and for the whole of §5 it is the oscillator's dimensionless coordinate instead. It goes back to the abstract meaning at §6.1.

Both terms carry the same factor because that is what $x_0$ was chosen to arrange, and every statement below is about the bracket. Now define the polynomials that will appear. Rather than write them down and check their properties one at a time, define all of them at once by a single function whose expansion produces them. That is the device Chapter 0.9 §7.2 used for probability distributions:

The left-hand side is an entire function of $t$ for each fixed $\xi$, so the expansion exists and defines each $H_n$ as a polynomial in $\xi$. The first few are $H_0=1$, $H_1=2\xi$ and $H_2=4\xi^{2}-2$, read off by expanding. The Hermite functions are these polynomials damped by a Gaussian and normalised. The constant is chosen now and justified in §5.3:

These are real, they decay faster than any exponential, and they are square integrable, so each one is a genuine vector of $L^{2}(\R)$ in a way the momentum eigenfunctions of §1.2 were not. That is the first thing to notice about the oscillator: unlike position and momentum, it has candidate eigenvectors that actually live in the space.

5.2 · They are the eigenfunctions, and the eigenvalues are what you expect

We want $\hat H_{\text{osc}}h_n=(n+\half)\hbar\omega\,h_n$, and by (4.5.18) that is the single statement $-h_n''+\xi^{2}h_n=(2n+1)h_n$. Everything needed comes out of (4.5.19) by differentiating it once in each variable, and the grind box does that. What comes out is two recurrences and, from them, a differential equation for $H_n$:

Grind box — the two recurrences and the equation, from the generating function

Write $G(\xi,t)=\ee^{2\xi t-t^{2}}$. Differentiating in $t$ gives $\partial_tG=(2\xi-2t)G$, and expanding both sides using (4.5.19),

$$\sum_{n\ge1}H_n\frac{t^{n-1}}{(n-1)!} \;=\; 2\xi\sum_{n\ge0}H_n\frac{t^{n}}{n!} \;-\; 2\sum_{n\ge0}H_n\frac{t^{n+1}}{n!}.$$

Comparing the coefficient of $t^{n}$ on the two sides gives $H_{n+1}/n!=2\xi H_n/n!-2H_{n-1}/(n-1)!$, and multiplying through by $n!$ is the first recurrence.

Differentiating in $\xi$ instead gives $\partial_\xi G=2tG$, so

$$\sum_{n\ge0}H_n'\frac{t^{n}}{n!} \;=\; 2\sum_{n\ge0}H_n\frac{t^{n+1}}{n!},$$

and the coefficient of $t^{n}$ reads $H_n'/n!=2H_{n-1}/(n-1)!$, which is the second recurrence.

Now eliminate. Substituting $2nH_{n-1}=H_n'$ into the first recurrence gives $H_{n+1}=2\xi H_n-H_n'$. Differentiate that in $\xi$ and use the second recurrence at index $n+1$, which says $H_{n+1}'=2(n+1)H_n$:

$$2(n+1)H_n \;=\; 2H_n+2\xi H_n'-H_n'',$$

and cancelling $2H_n$ from both sides leaves $H_n''-2\xi H_n'+2nH_n=0$.

Finally carry that to $u_n=H_n\ee^{-\xi^{2}/2}$. Differentiating twice, $u_n''=\big(H_n''-2\xi H_n'-H_n+\xi^{2}H_n\big)\ee^{-\xi^{2}/2}$, and the equation replaces $H_n''-2\xi H_n'$ by $-2nH_n$, leaving $u_n''=(\xi^{2}-2n-1)u_n$.

The last line of the grind box rearranges to $-u_n''+\xi^{2}u_n=(2n+1)u_n$, and $h_n$ is $u_n$ times a constant, so the same holds for it. Putting that into (4.5.18),

So the oscillator has eigenvectors, they are in the space, and the eigenvalues are the equally spaced ladder every physics course quotes. The half in the ground-state energy has arrived from nothing more than the recurrences. No physical argument produced it, and Chapter 4.8 will produce it a second way and get the same number.

5.3 · They are orthonormal

Orthonormality also comes out of (4.5.19), by multiplying two copies of it together and integrating. Take the generating function at two parameters $s$ and $t$, multiply, and include the Gaussian weight, completing the square in the exponent so that Chapter 0.2's Gaussian integral applies:

Now expand the two ends in powers and compare. The left-hand side, with each exponential replaced by its Hermite series, is $\sum_{m,n}\frac{s^{m}t^{n}}{m!\,n!}\int H_mH_n\ee^{-\xi^{2}}\dd\xi$. The right-hand side is $\sqrt\pi\sum_n(2st)^{n}/n!$, which carries only terms with $m=n$. Two power series in two variables agree only coefficient by coefficient, so

The normalising constant in (4.5.20) was chosen to be exactly the square root of that number, which is why the second equality is immediate. So the Hermite functions are an orthonormal set. Chapter 4.3 §7.3 is emphatic that this is a weaker claim than being a basis, and it named the four statements that would upgrade it. The next subsection proves one of them.

5.4 · No vector is orthogonal to all of them

Of Chapter 4.3 §7.3's four equivalent statements, the one that can be checked here is (d): the only vector orthogonal to every $h_n$ is the zero vector. Suppose then that $f\in L^{2}(\R)$ satisfies $\avg{h_n,f}=0$ for every $n$, and follow the consequences.

First convert the hypothesis into a statement about powers. The recurrence in (4.5.21) shows $H_n$ has degree exactly $n$, since it builds $H_{n+1}$ by multiplying $H_n$ by $2\xi$ and subtracting something of lower degree. So the span of $H_0,\dots,H_N$ is every polynomial of degree at most $N$, and each power $\xi^{j}$ is a finite combination of Hermite polynomials. The $h_n$ are real, so $\avg{h_n,f}=\int h_nf$, and the hypothesis therefore says

Write $q=f\,\ee^{-\xi^{2}/2}$, which is in $L^{2}$ because $\abs q\le\abs f$ and in $L^{1}$ by Cauchy–Schwarz, since $\int\abs f\ee^{-\xi^{2}/2}\le\norm f\,\norm{\ee^{-\xi^{2}/2}}$ and both factors are finite. Our goal is to show its Fourier transform vanishes identically, so write the transform out and replace $\ee^{-\ii k\xi}$ by its power series, which converges for every $\xi$:

where every integral in the sum vanishes by (4.5.25). The only step that needs an argument is the interchange of the sum with the integral, and it is dominated convergence, which is exactly the theorem Chapter 4.3 §4.3 proved and exactly the reason this proof belongs after that chapter rather than before it. The partial sums of the exponential series satisfy $\abs{\sum_{j\le J}(-\ii k\xi)^{j}/j!}\le\ee^{\abs{k\xi}}$ for every $J$, so every partial sum of the integrand is dominated by

the bound again being Cauchy–Schwarz, and the remaining integral being finite because completing the square turns it into $2\ee^{k^{2}}\int_0^{\infty}\ee^{-(\xi-\abs k)^{2}}\dd\xi$. So a single integrable function dominates the whole sequence of partial sums, for each fixed $k$, and the interchange is licensed.

The rest is one line of Chapter 0.9. Plancherel says the transform preserves the norm, so $\norm q=\norm{\tilde q}=0$, which makes $q$ the zero vector, so $q=0$ almost everywhere. Since $\ee^{-\xi^{2}/2}$ is never zero, $f=0$ almost everywhere, which is to say $f$ is the zero vector of $L^{2}$. Statement (d) holds, and Chapter 4.3 §7.3's equivalence upgrades it to statement (b): every $f$ in the space equals $\sum_n\avg{h_n,f}h_n$, with the series converging in norm. The Hermite functions are an orthonormal basis of $L^{2}(\R)$. $\blacksquare$

Chapter 4.3's closing brick said this proof would be statement (d) of its §7.3 run on an integral that only its §4.3 licenses, and that is precisely what happened. Nothing else was used, and in particular nothing about analytic functions of a complex variable, which this book has not built and does not build until Chapter 5.4.

5.5 · Verification 3, completed

With a genuine basis in hand the third verification is immediate. Send each state to its list of coefficients against the Hermite functions. That is Chapter 0.5's move, made with an infinite list:

That map is unitary onto $\ell^{2}$, and both halves of the claim are already proved. It preserves the norm by Parseval, which is statement (c) of Chapter 4.3 §7.3, and it hits every square-summable list, which is that chapter's §7.2. Now identify the measure space. A sequence space is a space of functions on the set $\{0,1,2,\dots\}$, and the sum in (4.5.28) is the integral of $\abs{c}^{2}$ against the measure that gives each point weight one. So $\ell^{2}$ is not merely like an $L^{2}$ space, it is one:

Multiplication by $g$ is what $\hat H_{\text{osc}}$ becomes, by (4.5.22) applied term by term, and the domain condition of §3.2 reads $\sum_n(n+\half)^{2}\hbar^{2}\omega^{2}\abs{c_n}^{2}\lt\infty$, with the constants written out even though they cannot affect whether the sum converges. Every point of $X$ has measure one, hence positive measure, so §3.4 makes every value of $g$ an eigenvalue and there is nothing else in the essential range. The spectrum is pure point, and it is the ladder.

One honest remark about what has been fixed and by what. The differential expression in (4.5.17) is a formula, and Chapter 4.4 §7 showed at length that a formula does not determine an operator. What determines this one is (4.5.29): the operator is multiplication by a real function in the $\hat U$ picture, which makes it self-adjoint by construction, and its action agrees with the formula on every $h_n$ and hence on every finite combination of them. That is the operator the rest of Part IV means by $\hat H_{\text{osc}}$, and saying so here saves an argument in Chapter 4.8.

5.6 · The three verifications, collected

Put the three side by side, because the comparison is the point of having done them.

• $\hat x$. $X$ is the line, $\mu$ is Lebesgue measure, $\hat U$ is the identity, and $g(x)=x$. Spectrum $\R$, purely continuous, no eigenvectors.
• $\hat p$. $X$ is the line, $\mu$ is Lebesgue measure, $\hat U$ is the Fourier transform of Chapter 0.9, and $g(k)=\hbar k$. Spectrum $\R$, purely continuous, no eigenvectors.
• $\hat H_{\text{osc}}$. $X$ is the non-negative integers, $\mu$ is counting measure, $\hat U$ is expansion in Hermite functions, and $g(n)=(n+\half)\hbar\omega$. Spectrum a ladder, purely point, an orthonormal basis of eigenvectors.

Now say what that buys, because it is the reason the mark in §3.2 is defensible and it must not be passed over. You are holding a theorem this book quotes rather than proves, and you have personally checked it in every case this book applies it to. Then say exactly what that covers, since it is easy to overstate. The theorem is verified here for these three operators and for nothing else. Every later use in Part IV is one of three kinds. It is a function of one of the three, built by §6.6's functional calculus, which never leaves the multiplication picture. Or it is an operator whose own chapter exhibits a complete orthonormal family of eigenfunctions, supplying the triple directly in the way §5.5 supplied it for the oscillator. Or it is $\hat p^{2}/2m+V(\hat x)$, whose self-adjointness is a separate question about $V$ that Chapter 4.6 §4.4 answers case by case.

That third kind is the one to be careful about, because there is a shortcut here and it does not work. Unitary equivalence to a multiplication operator is not inherited by sums. For unbounded self-adjoint $\hat A$ and $\hat B$ the sum on $\operatorname{dom}(\hat A)\cap\operatorname{dom}(\hat B)$ need not be self-adjoint, need not be essentially self-adjoint, and need not even be densely defined. So "built out of operators that have been verified" certifies nothing at all. Adding $V(\hat x)$ to $\hat p^{2}/2m$ produces a fourth operator whose admissibility has to be established rather than inherited, and §8.5 below is the standing warning against assuming otherwise. That is the standard the book set for itself when it decided what a mark means, and here it is met rather than approximated.

The three also make one structural point that a single verification could not. The operators differ in their measure and in nothing else that matters: continuous spectrum and discrete spectrum are Lebesgue measure and counting measure, seen through the same statement. Chapter 0.5 never had to mention a measure because a finite list carries only one, and that silence is exactly what had to be broken to get here.

5.7 · The numerical confirmation

Parseval in the Hermite basis is the statement to check, since the whole of §5.5 rests on it, and there is a test function whose coefficients come out in closed form. Take the ground state displaced sideways by $a$, meaning $\psi_a(\xi)=\pi^{-1/4}\ee^{-(\xi-a)^{2}/2}$, and pair it against (4.5.19). Completing the square in the exponent gives $\int G(\xi,t)\ee^{-\xi^{2}/2}\psi_a\,\dd\xi=\pi^{1/4}\ee^{-a^{2}/4}\ee^{at}$. Matching powers of $t$ against the Hermite series then produces every coefficient at once:

Those squared coefficients are the Poisson probabilities with mean $a^{2}/2$, and the mathematics is identical rather than analogous: the same expression, the same normalisation, the same parameter in the same place. So Parseval's claim that they sum to one is, in this instance, the statement that a Poisson distribution is a distribution. That is a strong check, because it is a closed form to compare arithmetic against rather than a number produced by the same arithmetic.

Run it at $a=2$, so the mean is $2$. Computing the coefficients numerically on a grid of $4096$ points across $\abs\xi\le12$ reproduces $c_0,\dots,c_5=0.3678794$, $0.5202601$, $0.5202601$, $0.4247906$, $0.3003723$, $0.1899721$. Those match (4.5.30) to sixteen digits. The truncated sum is

That remainder is the exact Poisson tail summed in arithmetic of higher precision, not the difference of the printed sum from one, which is contaminated by rounding in the twelfth figure. The orthonormality of (4.5.24) can be checked on the same grid by forming the whole Gram matrix $\avg{h_m,h_n}$ for $m,n\lt40$, and its largest departure from the identity is $4.0\times10^{-15}$. Read that last figure for what it is. It is the floor of double-precision arithmetic rather than anything about the Hermite functions, and it moves by a factor of ten either way with the quadrature rule and the exact placement of the grid points. What is worth knowing is how the departure grows once the grid is too narrow, and that part is truncation and is implementation-independent: at half-width $12$ it is still at the rounding floor, at $11$ it is $6.34\times10^{-11}$, at $10$ it is $8.16\times10^{-6}$ and at $9$ it is $2.17\times10^{-2}$. The reason is visible in (4.5.22). The state $h_{39}$ has energy $39.5\hbar\omega$, so its classical turning point sits at $\xi=\sqrt{79}=8.89$, and a grid that stops at $9$ is cutting the function off where it is still large.

a natural place to stop  ·  the checking is finished; next comes the projection a continuous observable needs before the Born rule can give it a probability at all

6 · The projection-valued measure, and the integral that replaces the sum

Chapter 0.5 wrote its theorem three ways, and only two of them have been dealt with. The eigenvector form is gone, the matrix form $A=UDU^{\dagger}$ has become §3.2, and the third form, the weighted sum of projections, is what this section builds. It gets a section of its own because it is the form the physics is written in: the Born rule is a statement about projections, and until a projection exists for a continuous observable there is no probability to compute. By the end you will have $\hat A=\int\lambda\,\dd P(\lambda)$, the Born rule for a continuous variable, and the recipe for building functions of an operator that §9 needs.

6.1 · Projections, read off the multiplication form

Multiplication operators come with an obvious supply of projections, because multiplying by the indicator function of a set is a projection: doing it twice is the same as doing it once, and it is real, so it is its own adjoint. Transport that supply back through $\hat U$. For each Borel subset $E$ of the real line, take the part of the measure space where $g$ lands inside $E$. From here $\xi$ is the generic point of $X$ again, as it was in §3.2, and §5's oscillator coordinate is finished with. Define

Each $P(E)$ is an orthogonal projection on $H$, since conjugating by a unitary preserves both $P^{2}=P$ and $P^{\dagger}=P$. Each is defined on the whole space and bounded by one, because multiplying by an indicator never increases a modulus. The family $E\mapsto P(E)$ is the projection-valued measure of $\hat A$. It deserves the word measure because it satisfies the axioms Chapter 4.3 §2.1 laid down, with projections in place of numbers:

the last for any countable family of disjoint sets, with the series converging in norm. All four are the corresponding statements about indicator functions, carried across the unitary. The first three are pointwise identities, and the fourth is dominated convergence one more time: the partial sums of $\sum_j\mathbf 1_{E_j}$ converge pointwise to $\mathbf 1_{\bigcup E_j}$ and are dominated by it, so the images converge in $L^{2}$ by Chapter 4.3 §4.3.

The definite article in the projection-valued measure is doing physics, and it has to be paid for, which is why §3.2's mark has two clauses rather than one. The definition (4.5.32) runs through a particular $(X,\mu,\hat U,g)$, and §4.2 has already displayed $\hat x$ through two of them. The mark's second clause is what says the resulting $P$ comes out the same either way. Without it, the probability §6.5 assigns to an interval would depend on whether the calculation was done on the line or on $(-\pi/2,\pi/2)$ through §4.2's map, and there is no reading of the Born rule under which that could be allowed. Uniqueness of the spectral measure is a real theorem rather than a formality, this book quotes it rather than proving it, and §3.2 is where the quotation is recorded.

Two special values are worth noticing at once. Taking $E$ to be a single point $\{\lambda\}$ gives the projection onto the eigenspace at $\lambda$, by §3.4, so $P(\{\lambda\})=0$ exactly when $\lambda$ is not an eigenvalue. And taking $E$ to be the complement of the spectrum gives zero, since $g$ never lands outside its essential range on a set of positive measure. So the whole of $P$ lives on $\sigma(\hat A)$, which is the sense in which the spectrum is the complete list of readings.

6.2 · The measure a state induces

A projection-valued measure is not yet a probability, and the step that makes one is to pair it with a state. Fix a unit vector $\psi$ and define, for each Borel set $E$,

where the middle equality is $P(E)^{2}=P(E)$ together with $P(E)^{\dagger}=P(E)$, and the right one is (4.5.32) written out. Read the right-hand expression and $\mu_\psi$ is visibly an ordinary measure in Chapter 4.3's sense: it is non-negative, it is countably additive because the integral is, its total mass is $\int\abs\Psi^{2}\dd\mu=\norm\psi^{2}=1$, and by the end of §6.1 it assigns zero to everything outside the spectrum. So each state turns the observable into a probability distribution on the real line, concentrated on the possible readings, and that distribution is an object Chapter 4.3 knows how to integrate against.

6.3 · The integral that replaces the sum

Now compute the expectation value of $\hat A$ in the state $\psi$ and watch $\mu_\psi$ appear. Work in the multiplication picture, where the operator is nothing but a factor:

The last equality is the only one doing work, and it is the standard argument for pushing a measure forward along a map, run entirely inside Chapter 4.3. For an indicator $\mathbf 1_E$ the two sides agree by the definition (4.5.34), so they agree for simple functions by linearity. They then agree for every non-negative measurable function by monotone convergence, which is Chapter 4.3 §4.1, and for a general one by splitting into positive and negative parts as Chapter 4.3 §3 does. Nothing beyond that chapter is used.

That identity is what licenses writing the operator itself as an integral. Define $\int\lambda\,\dd P(\lambda)$ to be the operator whose expectation value in every state is the right-hand side of (4.5.35), on the set of states for which the second moment is finite. That does pick out one operator rather than a family, and here is the reason. On a complex space the expectation values determine the operator, by the move Chapter 4.2 §7.2 made for a unitary: expand $\avg{u+v,\hat A(u+v)}$, then run the same line with $v$ replaced by $\ii v$, and every $\avg{u,\hat Av}$ is recovered. Then $\hat A$ is that operator, and the two labels below say what the integral runs over and what is being integrated against:

That is Chapter 0.5's third form, and it now exists for an operator with no eigenvectors at all. Nothing on the right-hand side mentions one. The sum over a finite list of eigenvalues has become an integral over the spectrum, and the projections onto eigenspaces have become the $P(E)$ of §6.1, which exist whether or not any state sits behind a value. That is the promise Chapter 0.5 §6.4 made about this form, and §6.4 below collects it against the words that chapter used.

The domain is not an extra stipulation, and that has to be checked rather than asserted, because Chapter 4.4 spent a chapter establishing that domains are forced. The condition $\int\lambda^{2}\dd\mu_\psi\lt\infty$ is, by the same pushforward argument, $\int_X g^{2}\abs\Psi^{2}\dd\mu\lt\infty$, which says $g\Psi\in L^{2}$, which is exactly the domain condition §3.2 attached to the multiplication form. So the integral form carries its own domain and it is the same one.

6.4 · Chapter 0.5's sum, recovered

Take a finite-dimensional space and watch (4.5.36) collapse. There $\sigma(A)$ is a finite set of eigenvalues by (4.5.4), and each $P(\{\lambda_k\})$ is the projection $P_k$ onto the eigenspace $E_{\lambda_k}$ of Chapter 0.5 §6.4. The measure $\mu_\psi$ is then a sum of atoms, $\mu_\psi=\sum_k\norm{P_k\psi}^{2}\delta_{\lambda_k}$, so an integral against it is a finite sum, and (4.5.36) reads

which is Chapter 0.5 §6.4 verbatim, with its two side conditions arriving as the third and second identities of (4.5.33). That chapter said of exactly this formula that "this is the basis-independent form, and it is the one that survives to infinite dimensions in Chapter 4.5". It survives. The sense in which it survives is that the sum over a finite list of atoms has become an integral over a measure that may have no atoms at all.

The second promise made in the same place is longer and it is worth reading back before answering it. Chapter 0.5's closing warning box said that in infinite dimensions "the spectrum becomes continuous, the sum $\sum_k\lambda_kP_k$ becomes an integral $\int\lambda\,\dd P(\lambda)$ over projection-valued measures, and completeness must be re-proved rather than assumed. Chapter 4.5 pays this bill in full." Take the three clauses in order. The spectrum became continuous in §2.5 and §2.6, for position and for momentum, and the widened definition that made the word available is §2.2. The sum became the integral in (4.5.36), over the projection-valued measure of §6.1. And completeness was re-proved rather than assumed, three separate times: completeness of the space in Chapter 4.3 §6.2, completeness of the Fourier basis in Chapter 4.3 §8.3, and completeness of the Hermite basis in §5.4 above. That is the bill, paid clause by clause.

6.5 · The Born rule for a continuous variable

Chapter 4.2's third postulate said that the probability of the outcome $\lambda_k$ in the state $\psi$ is $\norm{P_k\ket\psi}^{2}$, and it was stated for an observable with eigenvalues and eigenspaces. Position has neither, so the postulate as written says nothing about the one measurement every introductory treatment performs first. The repair needs no new assumption, only the projection that §6.1 supplies: the probability that a measurement of $\hat A$ returns a value in the set $E$ is $\norm{P(E)\psi}^{2}$, which is $\mu_\psi(E)$. Setting $E=\{\lambda_k\}$ gives back the original statement word for word, so this is the same postulate with a wider notion of outcome and not a second one.

Now run it on position, where (4.5.13) makes $\hat U$ the identity and $P([a,b])$ multiplication by $\mathbf 1_{[a,b]}$:

That closes a loop the reader will have felt open since Chapter 4.3. That chapter's §5.3 warned that a vector of $L^{2}$ has no value at any point, so $\abs{\psi(x_0)}^{2}$ is not a probability of anything, and it said the Born rule for a continuous variable would have to be stated over a region instead. Here is why it is stated that way, and it is not a convention adopted for safety. The projection $P(\{x_0\})$ onto a single point is the zero operator, because a point has measure zero, so the probability of finding the particle exactly at $x_0$ is zero for every state and every point. The integral over an interval is the only quantity the theory produces, and $\abs\psi^{2}$ is a density rather than a probability.

6.6 · Functions of an operator

One more construction comes free, and §9 needs it. Chapter 0.5 §7 defined $f(A)$ in finite dimensions by diagonalising and applying $f$ to each diagonal entry, which in the present language is applying $f$ to the multiplying function. So define

for any measurable $f$, with the second expression meaning what §6.3 made it mean. If $f$ is bounded on the spectrum then $f(\hat A)$ is defined on the whole space with $\norm{f(\hat A)}\le\sup_{\lambda\in\sigma(\hat A)}\abs{f(\lambda)}$, since multiplying by a bounded function does exactly that to a norm. Three properties follow by composing functions pointwise, and they are the ones every later use needs: $(fh)(\hat A)=f(\hat A)h(\hat A)$, the map $f\mapsto f(\hat A)$ is linear, and $\overline{f}(\hat A)=f(\hat A)^{\dagger}$. In particular $f$ real makes $f(\hat A)$ self-adjoint, and $\abs f=1$ makes $f(\hat A)$ unitary, which is the observation §9 is built on.

7 · What $\ket x$ and $\ket p$ actually mean

Every physics text writes $\ket x$ and $\ket p$, inserts $\int\ket x\bra x\dd x=\hat I$ between two operators, and writes $\avg{x|p}=\ee^{\ii px/\hbar}/\sqrt{2\pi\hbar}$. Chapter 4.3 §8.5 explicitly declined to say what those symbols are and sent the question here. This section answers it. The answer is a procedure rather than a definition: put the system in a box, or on a lattice, do the computation where every object is a genuine vector, and then take a limit in which the spacing is manufactured into the sums. The procedure always works, it is what every one of those equations abbreviates, and by the end of the section you will also know exactly which of the comfortable statements about $\ket x$ and $\ket p$ are false.

7.1 · Neither object is a vector, and both failures are already proved

Start by fixing what is wrong, since both halves have been established elsewhere and it is worth having them together. A state of definite momentum would be $\ee^{\ii px/\hbar}$, and Chapter 0.9 §2.1 pointed out that $\int\abs{\ee^{\ii px/\hbar}}^{2}\dd x$ diverges, so it is not in $L^{2}(\R)$. A state of definite position would be concentrated at a point, and Chapter 4.3 §5.3 showed that such a thing is the zero vector, because a point is a null set. So one object is too big for the space and the other is too small, and §1.2 turned both failures into the statement that neither operator has an eigenvector.

What makes this urgent rather than pedantic is that the notation works. Calculations performed with $\ket x$ and $\ket p$ give right answers, reliably, and have done since the 1920s. Something is being abbreviated, and the job here is to say what.

7.2 · Momentum, in a box: box normalisation worked in full

The move is to change the problem to one where the object exists, and Chapter 4.4 §5.4 has already built the replacement. It goes by the name box normalisation, and this subsection works it in full so that you have a procedure rather than a slogan. One notice about the label first. Everything from here to §7.5 is written in the wavenumber $k$ rather than the momentum $p$, because the box's modes are labelled by $k$ and the constants only get in the way. Section 7.6 puts the $\hbar$ back and turns $\ket k$ into the $\ket p$ this section was opened with.

Put the particle on the interval $[-L/2,L/2]$ with the periodic condition $\psi(L/2)=\psi(-L/2)$, which is that section's $\hat p_\theta$ at $\theta=0$ with the interval recentred, and which that section proved self-adjoint. This is not the operator §3.2 used as its counterexample. There the wavefunction was pinned to zero at both ends, and the result was symmetric with no eigenvalues at all. Here the two ends are joined rather than pinned, which is the difference Chapter 4.4 §5.4 turned into a self-adjointness question and settled. So on this interval momentum does have eigenvectors, and they are the Fourier modes:

Everything about this situation is legitimate and nothing is being smuggled. The $u_n$ satisfy the periodic condition, they are orthonormal by direct integration, and Chapter 4.3 §8.3 proved they are a genuine orthonormal basis. That chapter proved it on $[-\pi,\pi]$, and the interval here is a unitary rescaling of that one, which carries a basis to a basis. So every statement of Chapter 4.3 §7.3 applies to them. In particular any state expands as $\psi=\sum_nc_nu_n$ with $c_n=\avg{u_n,\psi}$, and $\sum_n\abs{c_n}^{2}=\norm\psi^{2}$.

Now prepare for the limit, and the preparation is Chapter 0.9 §2.1's move made once more. The mode spacing is $\Delta k=2\pi/L$, and a sum over $n$ becomes an integral over $k$ only if every term carries a factor of $\Delta k$. So rescale the basis vectors by whatever makes that happen, which means dividing by $\sqrt{\Delta k}$:

These rescaled vectors are still perfectly good vectors of the box's Hilbert space. Their lengths grow as $L$ grows, which is the first sign of what goes wrong in the limit. Watch what the rescaling does to the three statements that matter. The coefficient becomes the Fourier transform of Chapter 0.9, exactly and with no approximation:

The resolution of the identity becomes an integral. Chapter 4.3 §7.3's expansion statement, written in the Dirac notation Chapter 0.5 §2.2 installed, is $\sum_n\ket{u_n}\bra{u_n}=\hat I$. Each $\ket{u_n}\bra{u_n}$ is $\Delta k\ket{k_n}\bra{k_n}$ by (4.5.41), so the sum is already a Riemann sum:

And Parseval becomes Plancherel, since $\sum_n\abs{c_n}^{2}=\sum_n\Delta k\,\abs{\avg{k_n|\psi}}^{2}$ is a Riemann sum for $\int\abs{\tilde\psi}^{2}\dd k$. Three familiar identities, and every one of them is exactly true at finite $L$ before any limit is taken.

Now be exact about the three arrows themselves, because §7.3 sets a standard just below and this subsection has to meet it too. What is proved above is everything at finite $L$: the $u_n$ are an orthonormal basis, (4.5.42) is an identity, and the resolution of the identity holds in the box. The arrows are not proved here, and it is worth saying why not. Each of them compares objects living in different spaces. The space $L^{2}[-L/2,L/2]$ changes as $L$ grows, $\psi$ lives on the whole line, and the labels $k_n$ move as well. No topology has been named in which such a comparison is a limit, and none is named here.

What makes the arrows safe to write is that their destinations are already known from the other side. Chapter 0.9 §2.1 built $\tilde\psi$ as exactly this Riemann-sum limit, and Chapter 0.9 §2.3 proved Plancherel outright, without a box anywhere in the argument. So the box does not earn those identities and this subsection does not claim it does. What the box supplies is the thing that was actually wanted, which is a finite, legitimate statement that each of them abbreviates. Section 7.3's matching claim is different in kind and is proved there in four lines, because its objects all sit in one fixed space and the limit is an ordinary one.

The orthogonality relation is the one that needs care, because $\delta_{mn}/\Delta k$ has no limit as a number. It has a limit as an instruction, which is what Chapter 0.9 §5.1 insisted a delta always was. Test it the only way it is ever used, against a coefficient function $f$, remembering that a sum over modes carries $\Delta k$:

and the left-hand side is the Riemann sum for $\int f(k)\avg{k'|k}\,\dd k$. So in the limit the symbol $\avg{k'|k}$ is whatever reproduces $f(k')$ when integrated against $f$, and that is precisely Chapter 0.9 §5.1's definition of $\delta(k-k')$. The equation $\avg{k'|k}=\delta(k-k')$ is therefore true, and true as a statement about what happens under an integral sign rather than as a statement about the value of an inner product.

7.3 · Position, on a lattice

Position needs the same treatment with a different replacement, and the reason a box does nothing for it takes one line. Confining the line to an interval leaves position as multiplication by $x$, so its spectrum shrinks to that interval and stays continuous, with no eigenvector anywhere in it. Problem 1(c) runs that calculation. Bounding the space achieves nothing, because what has to be made discrete is the space itself rather than its extent. So chop the line into cells $C_j=[ja,(j+1)a)$ of width $a$ and take the normalised indicators $e_j=a^{-1/2}\mathbf 1_{C_j}$, which are orthonormal because the cells are disjoint. They do not span the space, since their closed span is the functions that are constant on every cell, so the missing step is that the projection onto that span tends to the identity as $a\to0$.

That is worth proving rather than waving through, and it takes four lines. For a continuous $\psi$ vanishing outside a bounded set, the cell-average approximation differs from $\psi$ by at most the oscillation of $\psi$ over one cell, which tends to zero uniformly, so it converges in norm. Chapter 4.3 §8.1 proved such functions are dense in $L^{2}$. And each projection has norm one, so for a general $\psi$ choose a continuous $\phi$ within $\epsilon$, note that the projection moves $\phi$ by less than $\epsilon$ once $a$ is small, and the triangle inequality bounds the total by $3\epsilon$. So the lattice recovers everything in the limit.

Now rescale as before. The spacing is $a$ rather than $\Delta k$, so divide by $\sqrt a$:

Here $\Pi_a$ is the projection of the previous paragraph, so the middle statement is exactly true at every $a$ and becomes the identity only in the limit. The rescaled object $\ket{x_j}$ is the cell indicator divided by the cell width, which is a bump of unit area narrowing as $a\to0$. That is exactly Chapter 0.9 §5.2's picture of the delta as a limit of narrowing bumps, arriving here from a different direction, and $\avg{x_i|x_j}=\delta_{ij}/a$ tends to $\delta(x-y)$ by the same test as (4.5.44). What is being written $\ket x$ is a bump, not a point, and the width is going to zero afterwards rather than being zero to begin with.

The pairing with a state also behaves. For continuous $\psi$ the number $\avg{x_j|\psi}=a^{-1}\int_{C_j}\psi$ is the average of $\psi$ over the cell, which tends to $\psi(x)$ as the cell shrinks onto $x$. So $\avg{x|\psi}=\psi(x)$ is true in the sense that the cell average converges to the value, and no other sense is available, because Chapter 4.3 §5.3 established that a vector of $L^{2}$ has no values at points to converge to. For a general state what actually gets used is the norm statement of the previous paragraph, and that holds without any continuity.

7.4 · The procedure, stated once

Both calculations are the same three steps, and stating them once is the point of having done both, because from here on any manipulation with these symbols can be checked against the list rather than trusted.

• Replace the continuum by something with a genuine orthonormal basis, carrying a spacing parameter. A periodic box for momentum, which is where the name box normalisation comes from, and a lattice for position. Chapter 4.4 §5.4 supplies the self-adjoint operator in the first case and the indicator functions supply the basis in the second.
• Do the computation there. Every step is Chapter 4.3 §7.3 applied to an honest orthonormal basis, every object is a vector of a Hilbert space, and nothing needs excusing.
• Rescale so that each term carries the spacing, then let the spacing go to zero. Sums become integrals as Riemann sums do, a Kronecker delta divided by the spacing becomes a Dirac delta in Chapter 0.9 §5.1's sense, and the answer is the equation everyone writes.

Every equation containing $\ket x$ or $\ket k$ is an abbreviation for the output of that procedure. The abbreviation is safe because the procedure always terminates and always gives the same answer, and it is worth using because writing the box out each time would triple the length of every calculation in Part IV. What it is not is a statement about vectors of $L^{2}(\R)$, and §7.6 says which comfortable readings are therefore false.

7.5 · The rigged Hilbert space

There is a framework in which $\ket x$ and $\ket k$ are objects in their own right rather than abbreviations, and it is stated here, because it explains what kind of thing they are and because you will meet its name elsewhere. The book quotes it and does not build it.

7.6 · What has been closed, and what has not

Chapter 0.9 §3.2 finished its account of why Fourier analysis works with a confession, and the confession named this chapter: "Chapter 0.5's spectral theorem was proved in finite dimensions, and the argument above applies it to an operator on a function space, where the eigenfunctions are not even in the space. That gap is real. Chapter 4.5 closes it, and the price is a genuinely more careful theory." The gap is now closed, and it is worth saying exactly which gap that was and what the price turned out to be.

The gap was that the manipulations of Chapter 0.9 treated $\ee^{\ii kx}$ as an eigenbasis of the derivative when those functions are not in $L^{2}(\R)$. That chapter's §2.1 named the same thing from the other side, saying of a basis labelled by a real number rather than by an integer that "that one change is the difficulty Chapter 4.5 has to work to make legitimate." Section 7.2 supplies the finite statement standing behind each of them: the expansion, the resolution of the identity, the Plancherel identity and the continuum normalisation are each exactly true in a box, and the notation records what the box gives in the limit. The price is the procedure itself, which is that these are limits and not identities between vectors. The word closes applies to the gap Chapter 0.9 named and to nothing wider. The general theory of distributions, in which objects like $\delta'$ and the Fourier transform of a tempered distribution are built rather than approached through limits, is Chapter 5.4's, and this section claims none of it.

Chapter 4.3 §8.5's warning also stands, unamended, and it is worth restating because the notation invites forgetting it. Three things remain false. The families $\{\ket x\}$ and $\{\ket p\}$ are not orthonormal bases in Chapter 4.3 §7.3's sense, because that chapter's §7.4 proved every orthonormal basis of this space is countable and the points of a line are not countable. The function $\ee^{\ii kx}$ is not in $L^{2}(\R)$ and no limit makes it so. And a state concentrated at a point is the zero vector, not a small vector. What is true is what §7.4 says: a procedure exists, it is finite at every stage, and the notation records its output.

One piece of bookkeeping remains, and Chapter 4.6 needs it. Everything above was in the wavenumber $k$, and Part IV writes momentum $p=\hbar k$ with the constants visible. Since $\avg{k'|k}=\delta(k-k')$ and $\delta(k-k')=\hbar\,\delta(p-p')$, the momentum kets have to be rescaled by $\sqrt\hbar$ to keep the normalisation. That gives

Worked example 3 derives the last of those from the box and checks that it reproduces the momentum-space wavefunction $\hat\psi(p)$ that Chapter 4.3 §8.5 wrote with the $\sqrt\hbar$ in place. That agreement is not automatic and it is the reason the symmetric Fourier convention was chosen in Chapter 0.9 §2.2.

a natural place to stop  ·  the notation is legitimised; what follows is a checklist and one theorem

8 · What is now safe to do

The rest of Part IV performs four manipulations constantly, and until this chapter none of them was licensed for an operator without eigenvectors. This section is the licence, written as a checklist with the condition attached to each item, because a condition stated once at the point of use is worth more than a theorem stated once and forgotten. The fourth item is Chapter 4.4's rather than this chapter's, and it is on the list because it is the one that fails most often in practice. The third depends on §9, which has not been proved yet, and the entry says so.

8.1 · Insert a resolution of the identity

Writing $\hat I$ as a sum or an integral of projections and slipping it between two operators is the commonest move in the subject. It is legitimate in exactly three situations and there is no fourth. If $\{e_n\}$ is a genuine orthonormal basis, then $\sum_n\ket{e_n}\bra{e_n}=\hat I$ is statement (b) of Chapter 4.3 §7.3, and the word basis is carrying the whole weight: an orthonormal set that is merely orthonormal gives an inequality rather than an identity, by Bessel. If the observable is self-adjoint, then $\int\dd P(\lambda)=P(\R)=\hat I$ is (4.5.33), whether or not there are any eigenvectors. And if the family is $\{\ket x\}$ or $\{\ket k\}$, the identity is (4.5.43) or (4.5.45), meaning the limit of a box or lattice statement in §7.4's sense.

8.2 · Expand a state in eigenstates

This is legitimate exactly when the spectrum is pure point and the eigenvectors are complete, and that is a genuine restriction rather than a formality. It holds for $\hat H_{\text{osc}}$, by §5.5, and it will hold for angular momentum in Chapter 4.11, in each case because a complete orthonormal family of eigenvectors gets exhibited. For hydrogen in Chapter 4.13 it holds on the bound-state subspace and not on the whole space, for the reason given at the end of this paragraph. It fails for $\hat x$ and for $\hat p$, where §1.2 showed there is nothing to expand in, and the replacement is §7.4's procedure rather than a different expansion. It also fails, more quietly, for a Hamiltonian with both a discrete and a continuous part, where an expansion in bound states alone misses the scattering states entirely. Chapter 4.7 meets that case at a barrier, and hydrogen is an instance of it, since its spectrum carries the whole of $[0,\infty)$ above the bound states.

8.3 · Write a function of an operator, including the exponential

For any self-adjoint $\hat A$ and any measurable $f$, the operator $f(\hat A)$ of (4.5.39) exists, and if $f$ is bounded on the spectrum then $f(\hat A)$ is bounded and defined on the whole space. The case the book uses most is $f_t(\lambda)=\ee^{-\ii\lambda t/\hbar}$, which has modulus one on the real line and therefore gives a bounded operator for every real $t$. That $\ee^{-\ii\hat Ht/\hbar}$ is not merely bounded but unitary, and that it is what Chapter 4.2 §7 meant by time evolution, is the content of §9 below, so this entry is provisional until then and complete afterwards.

8.4 · Integrate by parts and drop the boundary term

This one is Chapter 4.4's and it is included because it is the item that fails silently. The boundary term of Chapter 0.2 §3.2 may be dropped when both functions lie in the domain of the operator, and the whole of Chapter 4.4 §5 is the demonstration that "both functions are nice" is not the same condition. On the line the term vanishes for every pair in the domain. On an interval it vanishes only for the pairs a chosen phase permits. On a half-line it cannot be made to vanish at all, and the operator is not an observable. The rule to carry is that the boundary term is where the domain lives, so dropping it is a claim about domains and must be checked as one.

8.5 · What is still not safe

Four things need naming, since a checklist that only permits is a checklist that gets misread.

• Multiplying two unbounded operators without checking domains. The product $\hat A\hat B$ is defined only where $\hat B\psi$ lands in $\operatorname{dom}(\hat A)$, and Chapter 4.4's Worked example 2 showed that even squaring a self-adjoint operator changes the domain in a way that matters physically.
• Adding two self-adjoint operators and expecting a third. On $\operatorname{dom}(\hat A)\cap\operatorname{dom}(\hat B)$ the sum need not be self-adjoint, need not be essentially self-adjoint, and need not even be densely defined. Nothing in §3.2 is inherited by sums, which is why §5.6 refuses to certify a Hamiltonian by the operators it was assembled from and why Chapter 4.6 §4.4 has to check $\hat p^{2}/2m+V(\hat x)$ case by case.
• Assuming a symmetric operator has a spectral theorem. It does not, and §3.2 named the counterexample. Every result in this chapter has self-adjoint in its hypothesis, which is a condition about domains and nothing else.
• Treating $\ket x$ as a vector. It has no norm, it cannot be normalised, and $\avg{x|x}$ is not a number. Every appearance of it is §7.4's procedure in shorthand.

9 · Stone's theorem

Chapter 4.2 §7.3 derived, from unitarity and the group law alone, that time evolution has the form $\ee^{-\ii\hat Gt/\hbar}$ for some Hermitian $\hat G$ with the dimensions of energy. It did that by differentiating the group law, and it said in place that the rigorous infinite-dimensional version of that step is Stone's theorem, quoted here. This section supplies it. One direction is three lines from §6.6 and is proved. The other is the hard one and is quoted. Together they say that "time evolution is unitary" and "the Hamiltonian is self-adjoint" are the same statement, which is the sentence Chapter 0.5 §7 has been pointing at since Part 0.

9.1 · The forward direction, built

One letter is about to carry two jobs, so separate them before anything is computed with it. The $\hat U$ of §3.2 is the spectral unitary. It is fixed once and for all by $\hat H$, it never carries an argument, and it is the map (4.5.39) conjugates by. The $\hat U(t)$ this section builds is the evolution family, a different operator for each $t$. The argument is the only thing separating them, so read each $\hat U$ below by whether it has one.

Let $\hat H$ be self-adjoint, and define $\hat U(t)=\ee^{-\ii\hat Ht/\hbar}$ by (4.5.39), which is legitimate because $\lambda\mapsto\ee^{-\ii\lambda t/\hbar}$ is measurable and bounded. Then work on the far side of the spectral unitary, in §3.2's multiplication picture, where $\hat H$ is multiplication by $g$. A state $\psi$ appears there as $\Psi=\hat U\psi$, which is the notation §6.2 set up, and $\hat U(t)$ appears as multiplication by $\ee^{-\ii g(\xi)t/\hbar}$. So every claim becomes a claim about a function of modulus one. Three things follow immediately.

It is unitary, because multiplying by a function of modulus one leaves $\int\abs\Psi^{2}\dd\mu$ unchanged and is undone by multiplying by the conjugate. It satisfies the group law, because $\ee^{-\ii g s/\hbar}\ee^{-\ii g t/\hbar}=\ee^{-\ii g(s+t)/\hbar}$ pointwise, and $\hat U(0)=\hat I$. And it is strongly continuous, meaning $\hat U(t)\psi\to\psi$ for every state as $t\to0$, which is the one that needs an argument:

because the integrand tends to zero pointwise and is dominated by $4\abs\Psi^{2}$, which is integrable. That is dominated convergence, Chapter 4.3 §4.3, used for the fourth time in this chapter. Note what was not claimed: $\norm{\hat U(t)-\hat I}$ does not tend to zero unless $\hat H$ is bounded, so the continuity is state by state and not uniform. Since every observable in this book except spin is unbounded, that distinction is the normal case rather than the exception.

9.2 · The generator, recovered by differentiating

The remaining half of the forward direction is that $\hat H$ can be read back off the flow, which is what makes it the generator rather than merely an ingredient. Take $\psi\in\operatorname{dom}(\hat H)$ and form the difference quotient in the multiplication picture, since that turns an operator limit into a limit of an integral:

The integrand tends to zero pointwise, because that is the derivative of $\ee^{-\ii gt/\hbar}$ at $t=0$ computed for each fixed $\xi$. And it is dominated: the elementary bound $\abs{\ee^{\ii\theta}-1}\le\abs\theta$ makes the first term at most $\abs g/\hbar$ in modulus, so the whole bracket is at most $2\abs g/\hbar$, and $\int g^{2}\abs\Psi^{2}\dd\mu$ is finite precisely because $\psi$ is in the domain. Dominated convergence again, and the limit is zero. So

with the derivative taken in the norm of the space rather than pointwise in $x$. That is the forward direction complete: a self-adjoint operator generates a strongly continuous one-parameter unitary group, and it is recoverable from that group by differentiating at the origin. Every step used §6.6 and Chapter 4.3 §4.3 and nothing else.

9.3 · The converse

9.4 · What the two halves say together

Read the two directions as one statement and it is a dictionary entry rather than a theorem. A strongly continuous one-parameter unitary group and a self-adjoint operator are the same object described twice. Chapter 4.2 §7 assumed the first, from linearity and conservation of probability, and deduced the second by differentiating the group law formally. Stone is what makes that differentiation legitimate in a space where the generator is unbounded and defined on a proper subset, which is the only case that arises.

Three consequences follow, and Part IV spends all three.

• Probability is conserved as a theorem. $\norm{\hat U(t)\psi}=\norm\psi$ for every $t$, because $\hat U(t)$ is unitary, so a state that starts normalised stays normalised for all time and nothing has to be imposed to keep it so.
• Self-adjointness is a physical requirement, not a technical one. Chapter 4.4 §5.5 showed that momentum on a half-line is symmetric with no self-adjoint extension, and its Worked example 3 showed the corresponding flow runs forwards only. Stone explains why those two facts are the same fact: no self-adjoint generator means no two-sided unitary group, and a system whose evolution cannot be run backwards is not a system this framework describes.
• The Hamiltonian is what a clock reads off. Given the flow, (4.5.49) extracts $\hat H$ uniquely, so there is no freedom to choose a different generator for the same evolution. Chapter 4.2's fifth postulate is then the single remaining physical statement, that this uniquely determined operator is the energy.

A note on signs, since the exponential now carries one. This book writes $\hat U(t)=\ee^{-\ii\hat Ht/\hbar}$, with the minus sign, matching $\ii\hbar\,\partial_t\ket\psi=\hat H\ket\psi$. Chapter 4.6 §2 states the convention and what it buys. Some engineering literature uses the opposite time convention throughout, and a formula copied across from it will differ by the sign of every $\ii$.

10 · Worked examples

11 · Your turn