Part IV · Quantum Mechanics — Chapter 4.4

Domains, and the Adjoint's Domain

In infinite dimensions an operator is not a formula. It is a formula together with a domain, the domain is forced rather than chosen, and the word "Hermitian" turns out not to have been enough.

Tools you'll need  — Chapter 4.3 above all: §5 for $L^{2}$ and the quotient that repaired the third inner-product axiom, §5.4 for Cauchy–Schwarz carried over intact, §6 for completeness, §7.4 for the dense set of step functions, and §8 for the Fourier basis on an interval. Chapter 0.5 §4 for the definition of the adjoint, which this chapter reads more carefully than that chapter needed to, §6 for the spectral theorem in finite dimensions, and its closing paragraph, which is the debt §1 collects. Chapter 0.2 §3.2 for integration by parts and for the sentence that sent its boundary term here. Chapter 0.6 §2 for the lemma that every linear map on $\R^{n}$ is bounded, and for the clause naming the derivative as the operator that breaks it. Chapter 4.2 §4 for the second postulate and the warning attached to it, §7 for time evolution, and §8 for the three lines that force infinite dimension. Chapter 0.9 §5 for what integration by parts means when the derivative is taken away from the function and given to the test function instead.

1 · The four places Chapter 0.5 used finite dimension

Here is where this section is going. Chapter 0.5 finished by naming four steps in which its proofs leaned on the dimension being finite, and it said the bill would come due in this chapter and the next. We read that sentence back and take the four in order. By the end you will know which one belongs here, which belong to Chapter 4.5, and why the repair could not be done in a single chapter. The fourth item is the one to watch, because it is the only one with no finite-dimensional shadow at all.

1.1 · The sentence, read back

Chapter 0.5's last paragraph is short enough to quote whole, and it is worth having in front of you rather than in memory:

"One thing remains outstanding, and it is worth naming so that you notice when it is paid. Every proof in this chapter used finite dimension: in the induction, in rank–nullity, in the interchange of sums, in the claim that an injective map is surjective. Quantum mechanics happens in infinite dimensions. Chapters 4.4 and 4.5 are where the bill comes due."

Chapter 4.2 §8 then made the second sentence unavoidable rather than atmospheric. If $[\hat x,\hat p]=\ii\hbar\hat I$ held on a space of finite dimension $n$, taking the trace of both sides would give $0$ on the left, because the trace of a commutator of matrices vanishes, and $\ii\hbar n$ on the right. Three lines, no physics beyond the commutator itself, and quantum mechanics is infinite-dimensional before a single measurement has been described. Chapter 4.3 built the space that follows from that. This chapter and the next have to rebuild the operators.

1.2 · The four, one at a time

Take them in Chapter 0.5's own order, because the order is roughly the order of increasing damage.

The induction. Chapter 0.5 §6.3 proved the spectral theorem by finding one eigenvector, restricting the operator to the orthogonal complement of that eigenvector, and repeating on a space one dimension smaller. The recursion halts because the dimension is a positive integer and cannot decrease forever. Strip out finiteness and two things break rather than one. The recursion no longer halts, which is the visible failure. The first step also fails, which is the serious one: there need not be any eigenvector to start from. The operator "multiply by $x$" on $L^{2}(\R)$ has none at all, since $x\psi(x)=\lambda\psi(x)$ forces $\psi$ to vanish wherever $x\ne\lambda$, and a function supported at one point is the zero vector of Chapter 4.3 §5.3. That failure is Chapter 4.5's, and it is why that chapter has to widen the word spectrum before it can state a theorem at all.

Rank–nullity. The identity $\dim\ker\hat A+\dim\operatorname{im}\hat A=\dim V$ is not so much false in infinite dimensions as empty, because both sides can be infinite and an equation between two infinities constrains nothing. What that identity was for was converting a statement about the kernel into a statement about the image, and the replacement is not a theorem about dimensions at all. It is the observation that an image can be dense without being closed, so that coming arbitrarily close to every vector and reaching every vector are two different achievements, and counting cannot tell them apart. Chapter 4.5 is where that becomes the difference between the parts of the spectrum, because $\hat A-\lambda$ can be one-to-one with a dense image and still have no bounded inverse.

The interchange of sums. Chapter 0.5 rearranged double sums freely, which finitely many terms always permit. The tempting move is to hand this item to Chapter 4.3's convergence theorems and walk away, but those license the exchange of a limit with an integral, whereas rearranging $\sum_j\sum_k a_{jk}$ exchanges two sums, and an exchange of two sums is an exchange of two integrals in disguise. Chapter 4.3 §4.1 is explicit about that one: exchanging two integrals is Fubini's theorem, and it stays where Chapter 0.2 §4.2 marked it. What pays this item without raising anything new is the corollary the same section proves for non-negative terms, $\int\sum_k h_k\,\dd\mu=\sum_k\int h_k\,\dd\mu$. Apply it to the functions $h_k=\sum_j a_{jk}\mathbf 1_{[j,j+1)}$, whose integral is $\sum_j a_{jk}$, and the two sides of the corollary are the two orders of summation. That settles it whenever every $a_{jk}\ge0$, and a double sum that converges absolutely follows by splitting it into positive and negative parts the way Chapter 4.3 §3 splits a function before integrating it. This item is paid, and the material that pays it was in place before this chapter opened.

Injective does not imply surjective. A one-to-one linear map of a finite-dimensional space to itself is onto, and Chapter 0.5's grind box on the converse of $\hat U=\ee^{\ii\hat A}$ used exactly that, in the words "being injective on a finite-dimensional space it is onto, hence unitary on $W$". It is false here, and the counterexample is small enough to hold in one hand. Take an orthonormal basis $e_1,e_2,e_3,\ldots$ of the space, which Chapter 4.3 §7.4 says can always be listed, and let $\hat S$ move every basis vector one place along that list:

Nothing non-zero is sent to zero, so $\hat S$ is injective, and by Parseval it preserves the norm exactly, since the same coefficients appear in the answer. Yet $e_1$ is not in its image, because every vector $\hat S\psi$ has coefficient zero against $e_1$. So $\hat S$ is one-to-one and not onto, and in finite dimensions no such map exists. Hold on to this one. It returns in §5.5 wearing physical clothes, as the reason a particle confined to a half-line has no momentum observable, and again in Worked example 3, where $\hat S$ turns out to be a translation with nowhere to come from.

1.3 · Which chapter pays which

The division follows from the four items and is worth stating before you meet it rather than after. This chapter is about the first half of the word "self-adjoint": what an operator is allowed to act on, and what its adjoint is therefore allowed to act on. That is the machinery the induction's first step will need, and it is a prerequisite for the spectrum rather than a part of it. Chapter 4.5 is about the values. It widens spectrum so that an operator with no eigenvectors still has one, states the theorem that replaces $\hat A=\hat U\hat D\hat U^{\dagger}$, and checks it in every case this book uses.

Neither half stands alone. You cannot ask what the spectrum of an operator is until you have said which operator, domain included, and Chapter 4.5's theorem has self-adjoint in its hypothesis, which is a statement about domains and nothing else. So the bill Chapter 0.5 named is settled across two chapters, in that order, for that reason.

2 · Bounded, unbounded, and why the derivative cannot be tamed

Section 1 closed Chapter 0.5's account. The debt this section collects is a different one, owed to Chapter 0.6, which proved that every linear map on $\R^{n}$ is bounded and named the derivative as the map that breaks the proof once the dimension is infinite. The plan for this section is to make one word precise and then use it to prove something that sounds like an inconvenience and is really a structural fact. We define what it means for an operator to be bounded and show that it is the same thing as being continuous. We then measure the derivative and find that it is neither. Finally we quote one theorem and get from it, in two lines, the statement that decides the shape of this whole chapter: an unbounded operator that we want to be an observable cannot be defined on the whole space, no matter how cleverly we try.

2.1 · The operator norm, and what it measures

Chapter 0.6 §2 proved that every linear map on $\R^{n}$ satisfies $\abs{\hat A\vv v}\le C\abs{\vv v}$ for some constant depending only on $\hat A$, and the proof wrote $\vv v$ in a basis and used Cauchy–Schwarz on the finitely many coefficients. The property that proof establishes has a name and a number attached. For a linear $\hat A$ defined on a subspace $\mathcal{D}$ of a Hilbert space, put

which is the largest factor by which $\hat A$ can stretch anything. Call $\hat A$ bounded when this number is finite and unbounded when it is not. The supremum in (4.4.2) is over a set of non-negative real numbers, so it always exists in the extended sense, and the only question is whether it is a number.

Before using the word it is worth knowing what it means, because "bounded" sounds like a technical convenience and it is not one. It is continuity, under another name. One direction takes a line: if $\norm{\hat A}$ is finite then

so nearby inputs have nearby outputs, with a uniform rate. The converse takes three lines and is worth having because it is what makes the word carry weight. Suppose $\norm{\hat A}$ is infinite. Then for each $n$ there is a unit vector $\psi_n$ in the domain with $\norm{\hat A\psi_n}\ge n$. Scale it down by setting $\phi_n=\psi_n/\sqrt n$. Now $\norm{\phi_n}=1/\sqrt n\to0$, so the inputs march into the origin, while $\norm{\hat A\phi_n}\ge n/\sqrt n=\sqrt n\to\infty$, so the outputs run away. An unbounded operator is therefore discontinuous, and discontinuous at every point rather than at some awkward ones, since linearity carries the failure at the origin everywhere else.

That is the honest content of the word. An unbounded observable is one whose value cannot be controlled by controlling the state to within a small error in the norm of the space, and the next subsection shows that momentum is one.

2.2 · The derivative is unbounded, exhibited

Chapter 0.6 §2 named the offender in advance, saying that "in infinite dimensions (Chapter 4.4) linear maps can be unbounded, $\dv{}{x}$ being the standard offender, and a great deal of quantum mechanics' technical difficulty descends from exactly that." The exhibit takes one family of functions. Work on $L^{2}[0,L]$ and take the exponential modes, with $kL$ a multiple of $2\pi$, which are the basis Chapter 4.3 §8 built on $[-\pi,\pi]$ with the interval rescaled:

Read the three parts of (4.4.4) together and the conclusion is immediate. The norm of $\psi_k$ does not depend on $k$ at all, while the norm of $\hat p\psi_k$ is $\hbar\abs{k}\sqrt L$, so the ratio in (4.4.2) is $\hbar\abs k$ and $k$ runs over $2\pi\mathbb{Z}/L$ without limit. The operator norm of $\hat p$ is infinite. Every one of these functions is smooth, periodic and as well behaved as a function can be, so the failure is not caused by anything pathological in the inputs.

The same conclusion holds on the whole line, and it needs one extra step because the modes themselves are not square-integrable there. Fix any smooth $g$ that vanishes outside a bounded interval and has $\norm g=1$, and set $\psi_k=\ee^{\ii kx}g(x)$. The modulus of $\psi_k$ is the modulus of $g$, so $\norm{\psi_k}=1$ for every $k$. Differentiating the product gives $\hat p\psi_k=\hbar k\psi_k-\ii\hbar\ee^{\ii kx}g'$, and the triangle inequality then bounds the second term away rather than adding to it:

So $\hat p$ is unbounded on $L^{2}(\R)$ too, and the reason is not subtle. The number $\hbar k$ in (4.4.4) is a momentum, the operator norm of $\hat p$ would be the largest momentum a state can have, and there is no largest momentum. Unboundedness is what a physically unbounded quantity looks like when you write it as an operator. The same argument, run on multiplication by $x$, says the same thing about position, and Worked example 1 runs it.

2.3 · Unboundedness cannot be designed away

The natural response to §2.2 is to look for a better formulation. The hope would be that momentum has been written down clumsily, and that some equivalent operator, defined on every state, does the same physical job while staying continuous. This subsection closes that door, and closes it completely, using one theorem that this book quotes rather than proves.

Now the two lines. Let $\hat A$ be defined on all of $H$ and satisfy $\avg{\hat Au,v}=\avg{u,\hat Av}$ for every pair of vectors, which is Chapter 0.5's Hermitian condition with nothing added. Call an operator satisfying that condition symmetric. The word is worth having now, before it is needed in anger: §4.1 states it again once an operator has a domain to state it on, and the whole force of this chapter is in the difference between the two statements. We check the closed-graph condition. Suppose $\psi_n\to\psi$ and $\hat A\psi_n\to\phi$. For an arbitrary $v$ in the space, move $\hat A$ across, take the limit, and move it back:

where each limit passes through the inner product because Cauchy–Schwarz makes the inner product continuous in each slot, as Chapter 4.3 §5.4 transferred it. So $\avg{v,\phi-\hat A\psi}=0$ for every $v$, and choosing $v=\phi-\hat A\psi$ gives $\phi=\hat A\psi$. The closed-graph condition holds, the quoted theorem applies, and $\hat A$ is bounded. That is Hellinger–Toeplitz, and it is worth stating on its own line because the rest of the chapter is a consequence of it.

a natural place to stop  ·  the restriction is compulsory; what follows is the vocabulary for it

3 · Domains, and the adjoint's domain

What follows is the vocabulary the rest of the chapter uses, and it is short. An operator is promoted from a formula to a pair, a formula and a set of vectors it acts on. We fix the domain of $\hat p$ and check that it is dense, saying what "dense" is for. Then we re-read Chapter 0.5's definition of the adjoint with that in hand, and find that the definition hands the adjoint a domain of its own, determined but not chosen, with no reason to agree with the one we started from. The last subsection draws the consequence that makes the whole of §5 predictable in advance.

3.1 · An operator is a formula and a domain

From here on, an operator on a Hilbert space $H$ means two things given together: a subspace $\operatorname{dom}(\hat A)\subseteq H$, called its domain, and a linear map from that subspace into $H$. Two operators are the same operator when the formulae agree and the domains agree. If the domains differ, they are different operators, even when the formula is the same symbols in the same order.

That convention looks pedantic on first meeting and it is the entire subject of the chapter. In finite dimensions it costs nothing, because a matrix acts on everything and there is only one domain available, which is why Chapter 0.5 never mentioned the word. Here the choice is real, the choice changes the physics, and §7 will show two operators with the same formula, both perfectly legitimate, whose energy levels are different numbers.

When $\hat B$ has a larger domain than $\hat A$ and agrees with $\hat A$ everywhere on $\operatorname{dom}(\hat A)$, we say $\hat B$ extends $\hat A$. That word does the work in §6, where the question is how many self-adjoint extensions a given operator has.

3.2 · The domain of $\hat p$, and what "dense" is for

Momentum needs a domain, and §2.3 says the domain cannot be everything. The natural choice is the largest set on which the formula makes sense and lands back in the space. Take

The phrase "is the integral of a function" is the fundamental theorem of Chapter 0.2 §2 used as a definition rather than as a conclusion, and it is what lets $\psi$ be non-smooth while still having a derivative that Chapter 4.3's integral can handle. A function with a jump in it fails the condition, because no integral has a jump. The condition $\psi'\in L^{2}$ is what makes $\hat p\psi$ a vector of the space rather than merely a function.

One property of this set matters more than any other, and it is the reason the next subsection works at all. The domain is dense: every vector of $L^{2}$ is a limit of vectors in it. On an interval this is one sentence, since the modes of (4.4.4) all lie in the domain and Chapter 4.3 §8 proved that finite combinations of them reach everything. On the whole line the argument is Chapter 4.3's as well, and it is already written down there. That chapter's §7.4 showed that step functions are dense, and its §8.1 then replaced the vertical sides of an indicator by straight ramps of width $\delta$, at a cost of at most $2\delta$ in squared distance. Those trapezoids are exactly what is wanted here: each is continuous, each is the integral of its own step-function derivative, and that derivative is bounded and supported on a bounded set, hence in $L^{2}$. So every trapezoid is in (4.4.7), and density transfers along Chapter 4.3's own chain.

Now the reason to care. Density is not a technical hygiene condition. It is what makes the adjoint exist as a function of $u$ at all, and the next subsection is where you can watch it do that job.

3.3 · The adjoint, with a domain of its own

Chapter 0.5 §4 defined the adjoint by the relation $\avg{\hat A^{\dagger}u,v}=\avg{u,\hat Av}$, and in finite dimensions that relation determines a matrix, namely the conjugate transpose, with nothing further to say. Read the same relation here, where $\hat A$ acts only on its own domain, and it stops being a formula for $\hat A^{\dagger}$ and becomes a demand that a given $u$ may or may not be able to meet:

The two labels under (4.4.8) are the whole of the difficulty. What is being demanded is a single vector $\hat A^{\dagger}u$ whose inner product against every $v$ in the domain reproduces those computable numbers, and sometimes no vector in the space does that job. So the definition sorts the vectors of $H$ into those for which the demand can be met and those for which it cannot, and admitting the first sort and refusing the second is what defines $\operatorname{dom}(\hat A^{\dagger})$:

with $\hat A^{\dagger}u$ defined to be that $w$. Here is where density earns its keep. If two vectors $w_1$ and $w_2$ both satisfied (4.4.9), then $\avg{w_1-w_2,v}=0$ for every $v$ in the domain, and because the domain is dense and the inner product is continuous, the same holds for every $v$ in the whole space. Taking $v=w_1-w_2$ gives $w_1=w_2$. Without density there would be a direction the test vectors never probe, $w$ could be shifted along it freely, and $\hat A^{\dagger}u$ would not be a well-defined vector. That is the whole job of the word dense, and it is done here and nowhere else.

This is the sentence the chapter turns on, so it is worth saying without any equation attached. The definition of the adjoint determines its domain. Nobody chooses it, and there is no reason whatever for it to coincide with the domain we started from. In finite dimensions the two coincide because both are the whole space and there is nothing to compare. Here they are two different subspaces produced by two different considerations, and the rest of the chapter is the study of when they happen to agree.

3.4 · The two domains move in opposite directions

One structural fact makes everything in §§5 to 7 predictable before you compute anything, and it follows from (4.4.9) by reading the words rather than the symbols. Suppose $\hat B$ extends $\hat A$, so that $\operatorname{dom}(\hat A)\subseteq\operatorname{dom}(\hat B)$. The condition on $u$ in (4.4.9) then has to hold for more vectors $v$, which is a harder test, so fewer vectors $u$ pass it. In symbols,

Enlarging an operator shrinks its adjoint, and the two move towards each other. That single observation is why the family of self-adjoint operators built from a given formula is a family at all, rather than a single answer or none: you start with a domain too small and an adjoint domain too big, and you enlarge the first, which shrinks the second, until they meet. Whether they can be made to meet, and in how many ways, is the question §6 answers and §5 answers first by hand.

4 · Symmetric is not self-adjoint

The two conditions in the title are the same condition in finite dimensions, and Chapter 4.2 §4 used the finite-dimensional word. This section separates them, says what each one buys, corrects the second postulate as that chapter announced would happen here, and is honest about what the difference is not: it is invisible to every test you would naturally run, including the test that measured values come out real. Section 5 then makes the difference bite on an operator you can check by hand.

4.1 · Two conditions, and how far apart they are

Section 2.3 has met the first of the two words already, for an operator that acts on everything. With domains in hand it can be said in the form the rest of the chapter needs. Call $\hat A$ symmetric when moving it across the inner product changes nothing, for vectors it is allowed to act on:

Compare (4.4.11) with the definition of the adjoint in (4.4.9) and read off what symmetry says in that language. It says that every $u$ in $\operatorname{dom}(\hat A)$ passes the test defining $\operatorname{dom}(\hat A^{\dagger})$, with $w=\hat Au$ as the witness. So symmetry is exactly the statement that

In the vocabulary of §3.1, a symmetric operator is one whose adjoint extends it. Call $\hat A$ self-adjoint when the containment in (4.4.12) is an equality, so that $\hat A^{\dagger}=\hat A$ with the domains included and not merely the formulae. The gap between the two words is therefore a gap between two subspaces, and measuring it on one operator after another is what the rest of the chapter does. Each measurement is of a stated operator, since the gap belongs to the domain and not to the formula. Section 5.3 measures it for momentum on the line, on the largest domain the formula allows, and finds no gap at all. Section 5.4 measures it on a bounded interval, for momentum pinned to zero at both ends, and finds a gap of two complex dimensions. It then closes that gap in a circle of ways, and each of them is an operator with no gap left. Section 5.5 measures it on a half-line, for the largest symmetric domain available there, finds a gap of exactly one complex dimension, and shows that nothing closes it.

In finite dimensions this distinction cannot be drawn, because both domains are the whole space and (4.4.12) is an equality the moment it is a containment. Chapter 0.5 was therefore right to use one word, and Chapter 4.2 inherited it correctly for the finite-dimensional systems that chapter's examples used.

4.2 · The second postulate, corrected

Chapter 4.2 §4 stated P2 as an observable is a Hermitian operator, and attached a warning saying the word would be sharpened here, in this section, for a reason that is physical rather than pedantic. The correction is now available and it is one line.

4.3 · What the difference is not visible to

It is worth naming the difficulty out loud here, because the difference between the two words is invisible to every test a physicist would think to apply, and a reader who does not know that will reasonably suspect the distinction of being empty.

Consider the property that made Hermitian operators attractive in the first place. If $\hat A$ is symmetric and $\hat A\psi=\lambda\psi$ for a non-zero $\psi$ in the domain, then (4.4.11) with $u=v=\psi$ gives $\lambda\norm\psi^{2}=\overline\lambda\norm\psi^{2}$, so $\lambda$ is real. Chapter 0.5 §6.1's argument transfers with no change and needs only symmetry. Symmetric operators also have orthogonal eigenvectors for distinct eigenvalues, by Chapter 0.5 §6.2's argument, again unchanged. And every expectation value $\avg{\psi,\hat A\psi}$ is real for the same reason. Chapter 4.2 §4.1 collected three theorems as the payoff of P2, and the first two of them are already true for a merely symmetric operator.

The third one is where the words part company, and it is the one you cannot check by making a measurement. Chapter 4.2 §4.1's third theorem was every state is a superposition of outcomes, and its proof there was Chapter 0.5 §6.3's orthonormal basis of eigenvectors. That is the spectral theorem, and the spectral theorem is exactly what symmetry fails to buy. Two structures the rest of Part IV stands on go with it.

• The spectral theorem. Chapter 4.5's hypothesis is self-adjointness, not symmetry, and there are symmetric operators for which the conclusion is false. Without it there is no resolution of the identity, so no way to write a state as a superposition of the outcomes of a measurement, so no Born rule for that quantity.
• Stone's theorem. Chapter 4.5 §9 gives $\ee^{-\ii\hat Ht/\hbar}$ as a unitary flow exactly when $\hat H$ is self-adjoint. A merely symmetric Hamiltonian does not generate one, so the state after a finite time is undefined rather than wrong.

Notice the shape of that. The two facts a reader would test by hand, on a single measurement, survive the weaker condition. The one that fails is a statement about all states at once, and no single measurement can see it. That is why the next section works entirely by hand: the only way to believe the distinction is to watch it happen on an operator you can differentiate.

a natural place to stop  ·  the vocabulary is built; what follows is one operator on three intervals

5 · Three intervals, one operator, worked by hand

This is the section the chapter exists for, and everything in it is integration by parts. We redo Chapter 0.2's calculation with the boundary term kept instead of waved through, identify what the adjoint of $\hat p$ is when no boundary condition has been imposed at all, and then ask on which functions the boundary term vanishes. Three intervals give three different answers. On the line there is nothing to impose. On a bounded interval there is exactly a circle of possible impositions, and you will see the whole circle. On a half-line the boundary term cannot be killed without killing the operator, and the conclusion is that momentum there is not an observable. No classification theorem is used anywhere in this section; §6 supplies one afterwards and it will have to agree with what you find here.

5.1 · The boundary term, kept this time

Chapter 0.2 §3.2 listed the places integration by parts does structural work, and one of the entries was a promise about this chapter: "An operator gets an adjoint. In Chapter 4.4, $\int \psi^{*}\,\partial_x\phi\,\dd x = -\int(\partial_x\psi)^{*}\phi\,\dd x$ when boundary terms vanish. The minus sign is why $\hat p = -\ii\hbar\,\partial_x$ is Hermitian and $\partial_x$ alone is not." That chapter needed the identity with the boundary term gone. Here the boundary term is the whole content, so we keep it.

Work on an interval $(a,b)$, where $a$ may be $-\infty$ and $b$ may be $+\infty$. For $u$ and $v$ with $u',v'\in L^{2}$, Chapter 0.2's identity reads $\int_a^b\overline u\,v'=[\overline u v]_a^b-\int_a^b\overline{u'}v$. We want that in terms of $\hat p$ rather than of $\dd/\dd x$, so multiply through by $-\ii\hbar$. The left side becomes $\avg{u,\hat pv}$. On the right the integral becomes $+\ii\hbar\int\overline{u'}v$, which is $\avg{\hat pu,v}$, because conjugating the constant flips its sign: $\overline{-\ii\hbar u'}=+\ii\hbar\,\overline{u'}$. Moving that term across leaves

One point about the hypotheses is worth settling here rather than leaving it to be noticed. Chapter 0.2 proved integration by parts for continuously differentiable functions, and the members of (4.4.7) need not be differentiable everywhere. The extension is a single step: write each function as the integral of its derivative, substitute, and split the resulting square of integration into its two triangles, whereupon each side of the identity turns into the other. That step is Fubini's theorem, the standing mark of Chapter 0.2 §4.2, and it is the same interchange the grind box of §5.2 names below. Nothing else is added.

Read (4.4.13) as the definition of symmetry rather than as an identity about integrals, and the section is already organised. By (4.4.11), an operator with the formula $-\ii\hbar\,\dd/\dd x$ is symmetric on a given domain precisely when the right-hand side vanishes for every pair $u,v$ drawn from that domain. Everything below is that one question, asked on three intervals.

5.2 · What the adjoint is before any condition is imposed

Before choosing domains it is worth knowing the answer to a question that does not depend on the choice: which $u$ satisfy (4.4.9), given that the domain of $\hat p$ contains at least the functions that die away before either end of the interval is reached? Every domain considered below contains those, whatever it does at the ends, so the answer is the same in all three cases, and it is the most permissive one available.

Write $\mathcal{C}$ for the members of (4.4.7) that vanish outside a closed subinterval of the open interval $(a,b)$. If $\mathcal{C}\subseteq\operatorname{dom}(\hat p)$, then every $u$ in $\operatorname{dom}(\hat p^{\dagger})$ is an integral of a function $u'\in L^{2}$, and $\hat p^{\dagger}u=-\ii\hbar u'$. No condition at the endpoints comes out of this step. The grind box proves it, and the one part of it that is not integration by parts is the fact that a function whose derivative vanishes in this weak sense is constant, which is done there in full.

That leaves one further requirement, and it is the whole of the three subsections below. Testing against $\mathcal{C}$ alone cannot see the ends of the interval, so it cannot decide which of those $u$ are admitted. What decides it is that the defining relation must hold against every $v$ in the domain rather than only against those that die away first, and the difference between the two demands is precisely the boundary term of (4.4.13). So the adjoint's domain is the maximal set cut down by the condition that the boundary term vanish, and the three intervals cut it down by different amounts.

Grind box — identifying the adjoint of the derivative, once, for all three intervals

Recall $\mathcal{C}$: the continuous functions that vanish outside a closed subinterval of $(a,b)$ and are integrals of a derivative lying in $L^{2}$. It contains §3.2's trapezoids, so it is dense in $L^{2}$ over every bounded subinterval by the chain Chapter 4.3 §8.1 built, and it is contained in every domain this section considers, since its members vanish at both ends. Let $u\in\operatorname{dom}(\hat p^{\dagger})$ with $w=\hat p^{\dagger}u$, so $\avg{w,v}=\avg{u,-\ii\hbar v'}$ for all $v\in\operatorname{dom}(\hat p)$ and in particular for all $v\in\mathcal{C}$. Put $h=\ii w/\hbar$, which lies in $L^{2}$. Dividing the relation by $-\ii\hbar$ and conjugating the constant gives

$$\int_a^b \overline{u}\,v' \;=\; -\int_a^b\overline{h}\,v \qquad\text{for all } v\in\mathcal{C}.$$

Fix $x_0\in(a,b)$ and set $H(x)=\int_{x_0}^{x}h$, which is finite for $x$ in $(a,b)$ because $L^{2}$ functions are integrable on bounded subintervals by Cauchy–Schwarz. Interchanging the two integrations then gives $\int\overline H v'=-\int\overline h v$ for the same $v$. That interchange is Fubini's theorem, and it is worth naming rather than performing quietly. Chapter 4.3 §4.1 says in as many words that it settles the exchange of a sum with an integral and leaves the exchange of two integrals where Chapter 0.2 §4.2 marked it, so we are spending that standing mark here rather than raising a new one. The version needed is the undemanding one: the region is a bounded triangle on which the modulus of the integrand has finite integral, which is the non-negative case Chapter 0.2 noted holds with no extra hypotheses. Subtracting,

$$\int_a^b \overline{(u-H)}\;v' \;=\;0 \qquad\text{for all } v\in\mathcal{C}.$$

The constancy step. Fix $\eta\in\mathcal{C}$ with $\int\eta=1$. Given any $\varphi\in\mathcal{C}$, put $c=\int\varphi$ and $\chi(x)=\int_a^{x}(\varphi-c\,\eta)$. Then $\chi'=\varphi-c\eta$, and $\chi$ vanishes outside a closed subinterval of $(a,b)$, because the total integral of $\varphi-c\eta$ is zero, so $\chi$ returns to zero beyond both supports. Hence $\chi\in\mathcal{C}$ and

$$\int\overline{(u-H)}\,\varphi \;=\; \int\overline{(u-H)}\,(c\,\eta+\chi') \;=\; c\int\overline{(u-H)}\,\eta \;=\; \gamma\int\varphi ,$$

where $\gamma=\int\overline{(u-H)}\eta$ is one fixed number, the middle step being the displayed identity applied to $v=\chi$. So $\int\overline{(u-H)}\varphi=\int\gamma\varphi$ for every $\varphi\in\mathcal{C}$, and by the density noted at the top, $\overline{u-H}=\gamma$ almost everywhere. Therefore $u=H+\overline\gamma$.

Conclusion. $u$ is an integral of $h$ plus a constant, so $u$ is the integral of a function, that function is $h\in L^{2}$, and $\hat p^{\dagger}u=w=-\ii\hbar h=-\ii\hbar u'$. No endpoint condition was used and none was produced, so nothing constrains $u$ at $a$ or $b$ beyond membership of $L^{2}$. Conversely every such $u$ is in $\operatorname{dom}(\hat p^{\dagger})$ when the boundary term of (4.4.13) vanishes against the domain, which is checked case by case below. $\blacksquare$

Two remarks. The argument used only test functions supported strictly inside the interval, which is why it is indifferent to what happens at the ends and why one proof covers all three cases. And it is the honest content of the phrase "integrate by parts and drop the boundary term": the dropping is legitimate against test functions that vanish there, and against nothing else without a reason.

5.3 · The line: nothing to impose

Take $(a,b)=(-\infty,\infty)$ and the domain (4.4.7), which is the largest one the formula admits. The boundary term in (4.4.13) is a limit at each end, and the claim is that both limits are zero for every pair in the domain, with no condition imposed at all.

Here is the argument, and after (4.4.13) it uses Cauchy–Schwarz twice and nothing else. Put $F=\overline u v$. Read (4.4.13) on the bounded interval $[0,X]$ and divide by $\ii\hbar$: it says exactly that $F(X)-F(0)=\int_0^{X}\big(\overline{u'}v+\overline uv'\big)$. Now bound the two pieces. Both $u$ and $v$ are in $L^{2}$, so $F$ is in $L^{1}$ by Cauchy–Schwarz, meaning $\int\abs F$ is finite. The integrand above is a sum of two products of $L^{2}$ functions, so it is in $L^{1}$ for the same reason, and its integral over $[0,X]$ therefore converges as $X\to\infty$. So $F(X)$ has a finite limit. A function with a finite limit at infinity whose modulus has a finite integral must have that limit equal to zero, since a non-zero limit would make $\int\abs F$ diverge. So $F(X)\to0$, and the same at $-\infty$.

The boundary term therefore vanishes identically and $\hat p$ is symmetric on the domain (4.4.7). Now compare that domain with the adjoint's. Section 5.2 says every $u$ in the adjoint's domain is an integral of a function in $L^{2}$, which is membership of (4.4.7), and the further requirement it left open is that the boundary term vanish, which the paragraph above has just shown it does for every such $u$. Nothing is cut down. The two domains are the same set, so

There is nothing to choose, no parameter, and no family. The momentum of a particle on a line is an observable, uniquely, and the reason it looked as though no work was needed is that in this one case no work is needed. Keep the feeling of that, because the next two cases are the same operator and the same calculation with different endpoints, and they do not come out this way.

5.4 · The interval: a circle of momentum operators

Now take $(a,b)=(0,L)$ with $L$ finite. The boundary term in (4.4.13) is no longer a limit, it is two numbers, and symmetry on a domain $\mathcal D$ is the requirement

Start with the choice almost everyone makes first, which is to pin the wavefunction to zero at both ends: $\mathcal D_{\min}=\{v : v'\in L^{2},\ v(0)=v(L)=0\}$. Condition (4.4.15) holds, since both sides are zero, so this operator is symmetric. Now ask what its adjoint is. By §5.2 the adjoint's domain is every $u$ with $u'\in L^{2}$ and no endpoint condition, because the boundary term is already zero for every such $u$ once $v$ is pinned. So $\operatorname{dom}(\hat p^{\dagger})$ is strictly larger than $\mathcal D_{\min}$, by every function that fails to vanish at an end.

That is the promised example, and it is worth pausing on. Take the claims one at a time. The operator $-\ii\hbar\,\dd/\dd x$ with the wavefunction pinned at both ends is symmetric, by the paragraph above. Its expectation values are real, by §4.3, and so would any eigenvalue be. It passes that eigenvalue test vacuously, because $-\ii\hbar\psi'=p\psi$ forces $\psi=\ee^{\ii px/\hbar}$, which never vanishes and so cannot be pinned at either end, leaving the operator with no eigenvalues at all. And it is not self-adjoint. The gap between the two domains is exactly two complex dimensions, because what an element of the larger domain may do, and an element of the smaller one may not, is take a non-zero value at each of the two ends. Chapter 4.2 §4's warning box was pointing at this operator.

So pinning both ends is too strong. The question is which domains are not, and (4.4.15) answers it in two lines. Set $u=v$ in it and take moduli:

Two cases follow, and there are no others. Either every $v$ in $\mathcal D$ has $v(0)=0$, and then (4.4.16) forces $v(L)=0$ as well, so $\mathcal D$ is $\mathcal D_{\min}$ or a subspace of it. The first was dealt with in the paragraph above, and a subspace of it has an adjoint at least as large again, by (4.4.10), so it fails for the same reason and worse. Or some $v_0\in\mathcal D$ has $v_0(0)\ne0$, and we may scale it so that $v_0(0)=1$, whereupon (4.4.16) makes $v_0(L)$ a number of modulus one, say $\ee^{\ii\theta}$. Put $u=v_0$ in (4.4.15) and every other $v$ in the domain is caught by it:

So a single phase, fixed once for the whole domain, is the only symmetric possibility beyond the pinned one. Define $\hat p_\theta$ to be $-\ii\hbar\,\dd/\dd x$ on $\mathcal D_\theta=\{v: v'\in L^{2},\ v(L)=\ee^{\ii\theta}v(0)\}$.

Each of these is self-adjoint, and the check is the same equation read in the other direction. Let $u$ be in the adjoint's domain, which by §5.2 means $u'\in L^{2}$ with nothing at the ends, and impose (4.4.15) against every $v\in\mathcal D_\theta$. Substituting $v(L)=\ee^{\ii\theta}v(0)$ turns it into $\big(\overline{u(L)}\ee^{\ii\theta}-\overline{u(0)}\big)v(0)=0$, and there are functions in $\mathcal D_\theta$ with $v(0)\ne0$, for instance $\ee^{\ii\theta x/L}$. So the bracket vanishes, which conjugates to $u(L)=\ee^{\ii\theta}u(0)$: the adjoint's domain is $\mathcal D_\theta$ again. The two domains coincide and $\hat p_\theta$ is self-adjoint.

One step of the case analysis is still open, and it is the step the first case took explicitly. Condition (4.4.17) puts $\mathcal D$ inside $\mathcal D_\theta$ without saying the two are equal, and a subspace of a symmetric domain is symmetric as well, so containment on its own settles nothing. But the operator on a proper subspace cannot be self-adjoint. If $\mathcal D$ is strictly inside $\mathcal D_\theta$, then $\hat p_\theta$ extends $-\ii\hbar\,\dd/\dd x$ on $\mathcal D$, so (4.4.10) makes the adjoint's domain contain $\operatorname{dom}(\hat p_\theta^{\dagger})=\mathcal D_\theta$, which is strictly larger than $\mathcal D$. So $\mathcal D=\mathcal D_\theta$ exactly, and the phases are the whole list.

Since $\theta$ and $\theta+2\pi$ give the same condition, the family is a circle. Momentum on a bounded interval is not one observable but a circle of them, and they are genuinely different operators, because their eigenvalues differ. Solving $-\ii\hbar\psi'=p\psi$ gives $\psi=\ee^{\ii px/\hbar}$, and the condition in (4.4.17) reads $\ee^{\ii pL/\hbar}=\ee^{\ii\theta}$, so

A whole ladder of allowed momenta, sliding rigidly as $\theta$ turns, and returning to itself after one full turn. Choosing $\theta$ is choosing what happens at the wall, and §7 will show the same phenomenon for energy with more room in it.

5.5 · The half-line: the term that cannot be killed

Take $(a,b)=(0,\infty)$, which is the case Chapter 4.2 §4 named twice. The end at infinity is handled by §5.3's argument word for word, so the boundary term of (4.4.13) is one number:

Ask the same question as before: for which domains does this vanish on every pair? Set $u=v$ in (4.4.19) and the answer arrives at once, because $\abs{v(0)}^{2}=0$ has only one solution. Every $v$ in a symmetric domain must satisfy $v(0)=0$. There is no phase to choose, because there is only one endpoint and a phase relates two.

So the largest symmetric domain available is $\mathcal D_{0}=\{v:v'\in L^{2},\ v(0)=0\}$, and we can ask whether that one is self-adjoint. Section 5.2 gives its adjoint's domain as every $u$ with $u'\in L^{2}$ and no condition at zero, since (4.4.19) already vanishes once $v(0)=0$. That is strictly larger: the function $u(x)=\ee^{-x/a}$, for any length $a$, has $u'\in L^{2}$ and $u(0)=1$, so it sits in the adjoint's domain and not in $\mathcal D_0$. The operator is symmetric and not self-adjoint.

Now put the two paragraphs together, which is the step that makes this different from §5.4. In that section the repair was to enlarge the domain until the two met. Here enlarging is not available: any enlargement contains a function with $v(0)\ne0$, and the first paragraph shows that such a domain is not symmetric, so it is not a candidate at all. The domain cannot grow and the adjoint's domain cannot shrink. They never meet.

The reason for the asymmetry deserves a sentence, because it is Chapter 0.5's fourth item returning under a different name. On a bounded interval, whatever leaves one end can be made to arrive at the other, and $\theta$ records how much phase it picks up on the way. On a half-line there is a place to go and nowhere to come from. That is exactly the shift $\hat S$ of (4.4.1): norm-preserving, one-to-one, and not onto. Worked example 3 makes the identification precise by exponentiating $\hat p$ and finding that the flow it generates runs in one direction only.

a natural place to stop  ·  the answers are in hand; what follows is the theorem that counts them

6 · Counting the extensions: deficiency indices

Section 5 produced three answers by hand: a unique operator, a circle of them, and none. This section quotes the theorem that predicts which of those three outcomes occurs, without solving a boundary-value problem, and reduces the prediction to counting the square-integrable solutions of one first-order equation. It has to reproduce §5 exactly three times, and it does.

6.1 · The question, in a form a theorem can answer

Section 3.4 said what the general situation looks like. A symmetric operator sits inside its adjoint, enlarging the operator shrinks the adjoint, and a self-adjoint extension is a place where the two meet. So the size of the gap between $\operatorname{dom}(\hat A)$ and $\operatorname{dom}(\hat A^{\dagger})$ is what decides everything, and the useful way to measure that gap is to look at what $\hat A^{\dagger}$ can do that $\hat A$ cannot.

The measurement von Neumann found is this. A symmetric operator has real eigenvalues, by §4.3, so it can have no eigenvector with a strictly complex eigenvalue. Its adjoint has a larger domain and is under no such restriction, so it can. Counting the eigenvectors of $\hat A^{\dagger}$ at one complex eigenvalue above the axis and at its mirror image below measures exactly how much room there is between the operator and its adjoint, and that count is what the theorem is stated in terms of.

6.2 · The classification, quoted

6.3 · The three cases, counted

Say first which operator is being extended in each case, since the theorem counts the extensions of a given one. On the line it is $\hat p$ on (4.4.7). On the interval it is $\hat p$ on $\mathcal D_{\min}$, the pinned domain, which is the smallest sensible starting point and the one whose adjoint §5.4 computed. On the half-line it is $\hat p$ on $\mathcal D_{0}$. In all three the adjoint acts by the same formula on the maximal domain, so the calculation is the same one three times over.

Take $\lambda$ to be a fixed positive momentum $p_0$, so that the units work. The equation $\hat p^{\dagger}f=\pm\ii p_0 f$ is $-\ii\hbar f'=\pm\ii p_0f$, so $f'=\mp p_0f/\hbar$ and

each unique up to a constant, since a first-order equation has a one-dimensional solution space. An exponential has its derivative proportional to itself, so being in the domain of $\hat p^{\dagger}$ at all reduces to the single question of whether the exponential is square-integrable. Hence $n_\pm$ is $1$ when the corresponding exponential lies in $L^{2}$ of the interval and $0$ when it does not, and that is the entire computation. Take the three intervals in the order §5 took them.

• The line. $\ee^{-p_0x/\hbar}$ blows up as $x\to-\infty$ and $\ee^{+p_0x/\hbar}$ blows up as $x\to+\infty$, so neither is in $L^{2}(\R)$ and the indices are $(0,0)$. Exactly one self-adjoint extension. Section 5.3 found exactly one.
• The interval. Both exponentials are continuous on a bounded interval, so both are in $L^{2}[0,L]$, and the indices are $(1,1)$. The unitary maps between two one-dimensional spaces are multiplication by a phase, so the family is $U(1)$, a circle, with $1^{2}=1$ real parameter. Section 5.4 found a circle, parametrised by one phase.
• The half-line. $\ee^{-p_0x/\hbar}$ decays and is in $L^{2}[0,\infty)$, while $\ee^{+p_0x/\hbar}$ grows and is not. The indices are $(1,0)$, they are unequal, and there are no self-adjoint extensions. Section 5.5 found none.

Three cases, three agreements, and the agreement is not a coincidence of arithmetic. Look at the totals. The sum $n_++n_-$ comes out $0$, $2$ and $1$, which is the number of independent pieces of boundary data the interval carries, because each piece is one thing the boundary term of (4.4.13) has to cancel. For a first derivative that data is one number at each finite end, the value of the function there, which is why the totals here also read as a count of finite endpoints. Take the count of data as the rule and the count of endpoints as a coincidence of first order, because §7 has the same two finite ends and gets $(2,2)$: a second derivative reads a value and a slope at each of them. Self-adjoint extensions exist when the data can be paired off against itself, which needs an even total split evenly, and that is $n_+=n_-$. The half-line has one datum and no partner for it. What the theorem adds is that this survives into situations where the hand calculation would be much longer, and §7 is the first of those.

a natural place to stop  ·  the first derivative is finished; what follows is the second, with the machinery put to work

7 · The particle in a box has four parameters, not one

Everything so far has been about a first derivative. Energy is a second derivative, and this section runs the whole machine on it. The boundary term acquires a second pair of terms, the deficiency indices come out $(2,2)$ rather than $(1,1)$, and the family of self-adjoint Hamiltonians for a particle in a box is therefore a $U(2)$: four real parameters, not one. We then take a one-parameter slice of that family, solve it, and find that different members have genuinely different energy levels. That is the point of the section. The boundary condition is not bookkeeping and it is not a choice of convenience. It is physics, and it is measurable.

7.1 · The boundary form for a second derivative

Take $\hat H=-\dfrac{\hbar^{2}}{2m}\,\dvn{2}{}{x}$ on $L^{2}[0,L]$, which is Chapter 4.2's free Hamiltonian confined to a box. Integrating by parts twice moves both derivatives across and leaves two boundary terms rather than one:

The structure is the same as (4.4.13) and the arithmetic is one step longer. Symmetry on a domain is again the demand that the right-hand side of (4.4.21) vanish for every pair drawn from it, but now the data at each end is a pair of numbers, the value and the slope, so there is more to arrange and more ways to arrange it.

7.2 · The count, and the number it produces

Rather than classify by hand, use §6.2, which is exactly the situation it was quoted for. As §6.3 insisted, the theorem counts the extensions of a given operator, so the first job is to say which operator is being extended. It is the one whose domain kills every boundary number there is:

with $\hat H_{00}$ the formula $-\tfrac{\hbar^{2}}{2m}\dvn{2}{}{x}$ on that domain. All four numbers are killed and not two, which is more than §5.4's $\mathcal D_{\min}$ asked: that domain pinned the two values and left the slopes free. The distinction decides what is being counted. Pinning the values alone, with $\psi''$ in $L^{2}$, is the Dirichlet domain, and the list below puts Dirichlet at a single point of the family we are about to count, whereas the operator a family is counted from has to be one that every member of the family extends.

Two hypotheses of §6.2 have to be in place before it can be used. $\hat H_{00}$ is symmetric, because every term in the bracket of (4.4.21) carries a factor that vanishes at each end. Its domain is dense, by §3.2's chain with one adjustment: replace each straight ramp there by the ramp $\sin^{2}(\pi t/2\delta)$ across the same width $\delta$, which meets the flat parts on either side with zero slope and has a bounded second derivative, so the approximating functions lie in (4.4.22) and are no further away in $L^{2}$ than the straight ones were. The adjoint of $\hat H_{00}$ is then §5.2's answer run twice: the formula on the maximal domain, with no endpoint condition surviving, because the test functions that argument uses vanish identically near both ends and so constrain nothing there. So we solve $\hat H_{00}^{\dagger}f=\pm\ii E_0f$ for a fixed positive energy $E_0$, with no boundary conditions imposed:

What matters about (4.4.23) is not the value of $\mu_\pm$ but the order of the equation. A second-order linear equation with constant coefficients has a two-dimensional solution space, spanned by $\ee^{\mu x}$ and $\ee^{-\mu x}$, and every one of those solutions is continuous on the bounded interval $[0,L]$ and therefore square-integrable on it, whatever complex number $\mu$ happens to be. Both deficiency spaces are two-dimensional and the indices are $(2,2)$.

That is where §6.3's rule comes good, and the two versions of it are worth reading against each other here. Two finite ends, two pieces of boundary data at each, four pieces in all, split evenly into $(2,2)$. Counting data gives the right answer and counting endpoints gives two, which is why §6.3 asked you to keep the first version and treat the second as an accident of first order.

Feed that into §6.2 and the answer is a family of self-adjoint Hamiltonians in one-to-one correspondence with the unitary maps of one two-dimensional space onto another, which is $U(2)$:

The real dimension of $U(n)$ is $n^{2}$, and §6.2 quoted that along with the theorem. At $n=1$ you can check it on sight, a $1\times1$ unitary being a phase and nothing else, but $n=2$ is not visible that way, so here is the count. A $2\times2$ unitary matrix is one whose two columns are orthonormal. The first column is a unit vector in $\C^{2}$: four real numbers with one real equation on them, so three real parameters. The second column has to be a unit vector orthogonal to the first, and the vectors orthogonal to a given non-zero vector in $\C^{2}$ form a line of one complex dimension, on which the unit vectors differ by a phase. One more real parameter, and three and one make four.

Four real parameters. That is worth stating loudly, because the number one is easy to arrive at and wrong. Momentum on an interval gave a single phase in §5.4, the ladder in (4.4.18) slid rigidly as that phase turned, and it is natural to carry the picture across. Energy is a second derivative, the boundary data at each end is a value and a slope, and the family is four-dimensional. Some familiar conditions and where they sit in it:

• Dirichlet, $\psi(0)=\psi(L)=0$, which is the textbook infinite well. One point of $U(2)$.
• Neumann, $\psi'(0)=\psi'(L)=0$. A different point.
• Periodic, $\psi(L)=\psi(0)$ and $\psi'(L)=\psi'(0)$, which is a particle on a ring, and antiperiodic, the same with both signs reversed. Two more.
• The Robin family, $\psi'(0)=\alpha\,\psi(0)$ and $\psi'(L)=-\alpha\,\psi(L)$ with $\alpha$ real, a curve through the first two: $\alpha=0$ is Neumann and the limit of large $\abs\alpha$ is Dirichlet. This is the slice §7.3 works.

Each entry satisfies (4.4.21) directly, and you can check any of them in a line by substituting the two conditions into the bracket. The Robin case is done in full below, including the step from symmetric to self-adjoint, because it is the one whose spectrum we want.

7.3 · The Robin family, worked

Fix a real number $\alpha$ with the units of an inverse length and take

Check symmetry first. Substitute the two conditions into the bracket of (4.4.21) and look at each end separately. At $x=L$ the bracket is $\overline{u(L)}\,(-\alpha v(L))-\overline{(-\alpha u(L))}\,v(L)$, which is $-\alpha\overline{u(L)}v(L)+\alpha\overline{u(L)}v(L)=0$ because $\alpha$ is real and comes out of the conjugate unchanged. At $x=0$ the same cancellation happens with the sign the other way. Each end contributes zero on its own, so $\hat H_\alpha$ is symmetric.

Self-adjointness is the same computation read backwards, and it starts from §7.2 rather than from nothing. A function with all four boundary numbers zero satisfies the two Robin conditions, so $\mathcal{D}_{00}\subseteq\operatorname{dom}(\hat H_\alpha)$, and (4.4.10) then puts $\operatorname{dom}(\hat H_\alpha^{\dagger})$ inside $\operatorname{dom}(\hat H_{00}^{\dagger})$, which §7.2 identified as the maximal domain. So let $u$ have $u''\in L^{2}$ and no endpoint condition, and demand that the bracket vanish against every $v\in\operatorname{dom}(\hat H_\alpha)$. Substituting the conditions on $v$ collects the bracket into

The two numbers $v(0)$ and $v(L)$ can be prescribed independently inside the domain, because for any pair of target values there is a cubic polynomial matching those values and the two slopes the conditions then demand, and a cubic has exactly four coefficients to do it with. So both brackets in (4.4.26) vanish separately, and conjugating them returns $u'(L)=-\alpha u(L)$ and $u'(0)=\alpha u(0)$. Those are the conditions defining (4.4.25), so the adjoint's domain is the domain, and $\hat H_\alpha$ is self-adjoint for every real $\alpha$.

7.4 · Different extensions, different spectra

Now solve it, because the whole claim of this section is that the choice is physical, and the way to demonstrate that is to show the choice moving the energy levels. Write $E=\hbar^{2}k^{2}/2m$ for a positive energy, so that $\psi''=-k^{2}\psi$ and $\psi=A\cos kx+B\sin kx$. The condition at $x=0$ reads $Bk=\alpha A$, which is solved by $A=k$ and $B=\alpha$ up to scale. Imposing the condition at $x=L$ on that function and collecting terms gives one equation in one unknown:

That is a transcendental equation and it has no closed-form solution, which is itself informative: the clean $n^{2}$ spectrum of the textbook box is a property of one point of the family and not of the formula. Two limits are worth reading off before solving it numerically. At $\alpha=0$ the equation collapses to $\sin kL=0$, so $k=n\pi/L$ and $E_n=n^{2}\pi^{2}\hbar^{2}/2mL^{2}$, which is the Neumann ladder. It runs from $n=0$, and the $n=0$ member is the constant function, which the scaling $A=k$ above discards and which the function $\Delta(E)$ defined below puts back. As $\abs\alpha\to\infty$ the condition $\psi'(0)=\alpha\psi(0)$ forces $\psi(0)\to0$ with the slope finite, so the limit is the Dirichlet ladder, the same numbers with $n=0$ removed and the lowest level at $\pi^{2}\hbar^{2}/2mL^{2}$.

Negative energies have to be looked for separately, and they are where the family does something the textbook box never does. Put $E=-\hbar^{2}\kappa^{2}/2m$ with $\kappa\gt0$, repeat the two steps with hyperbolic functions in place of trigonometric ones, and the condition becomes $(\alpha^{2}+\kappa^{2})\sinh\kappa L+2\alpha\kappa\cosh\kappa L=0$, which needs $\alpha\lt0$ to have any solution at all. Both cases are captured by one function of the energy, the characteristic function of this boundary-value problem, which shares its name with Chapter 0.9's characteristic function of a distribution and nothing else. It is obtained by dividing (4.4.27) by $k$ and letting $k$ be imaginary:

where $s(q)=\sin(\sqrt q\,L)/\sqrt q$ and $c(q)=\cos(\sqrt q\,L)$, both of which are power series in $q$ with infinite radius of convergence and are therefore perfectly well defined for $q$ negative, where they become $\sinh$ and $\cosh$. The energies are the zeros of $\Delta$, and because $\Delta$ depends smoothly on $\alpha$ as well as on $E$, the levels slide continuously as $\alpha$ is turned.

One value of $\Delta$ can be read without any computation and it predicts the whole qualitative story. At $q=0$ we have $s=L$ and $c=1$, so

which vanishes at $\alpha=0$ and at $\alpha=-2/L$ and nowhere else. Those are the only two settings at which a level sits exactly at zero energy, so they are the only two at which a level can pass from one side of zero to the other. Solving the equation confirms that nothing else happens: below zero there are no states for $\alpha\gt0$, one state for $-2/L\lt\alpha\lt0$, and two for $\alpha\lt-2/L$. A negative $\alpha$ makes each wall attractive, and the two thresholds in (4.4.29) are where the first and then the second state bound to a wall drops below the bottom of the positive ladder.

Solving (4.4.28) by Newton's method gives the numbers. Energies are in units of $\hbar^{2}/2mL^{2}$, and the last column is the Dirichlet limit for comparison.

level	$\alpha L=-4$	$\alpha L=0$	$\alpha L=+5$	$\alpha L\to\infty$
$n=0$	$-17.06249$	$0$	$5.218729$	$\pi^{2}=9.869604$
$n=1$	$-14.66902$	$\pi^{2}=9.869604$	$22.66987$	$4\pi^{2}=39.47842$
$n=2$	$24.18110$	$4\pi^{2}=39.47842$	$55.70646$	$9\pi^{2}=88.82644$
$n=3$	$73.09505$	$9\pi^{2}=88.82644$	$106.6389$	$16\pi^{2}=157.9137$

Read the middle column first. At $\alpha=0$ the roots of (4.4.28) come out at $n^{2}\pi^{2}$ to every digit printed, which is the check that the numerical method is solving the equation we wrote down rather than a neighbour of it. Then read across. The three columns are three different sets of numbers. Same particle, same mass, same box, same formula for the energy, and three different spectra, because three different self-adjoint operators were selected out of the same $U(2)$.

boundary parameter αL

alpha L = 0.000

Neumann second threshold towards Dirichlet

levels: E0 = 0.000000 E1 = 9.869604 E2 = 39.478418 E3 = 88.826440

n^2 pi^2: 0.000000 9.869604 39.478418 88.826440 largest relative departure from n^2 pi^2 = 0.000e+0 (at alpha = 0 the closed form is n^2 pi^2 hbar^2 / 2 m L^2 and this reads zero to the last stored digit)

levels strictly below E = 0: 0 Delta(0) = alpha (alpha L + 2) = 0.000000 -- it vanishes only at alpha L = 0 and alpha L = -2, and those are the only two settings at which the count above changes

The physical reading is the point of the whole chapter and it is worth stating without hedging. The formula $-\hbar^{2}\psi''/2m$ does not determine a Hamiltonian for a particle in a box. It determines a four-parameter family of them, all self-adjoint, all legitimate observables, with different energy levels. Choosing among them is not a mathematical formality performed before the physics starts. It is the physics of what the wall is made of, and any experiment that measures the levels is measuring which member of the family the wall implements.

8 · Worked examples

9 · Your turn