Part IV · Quantum Mechanics — Chapter 4.6

The Schrödinger Equation

Two things are already on the table: a flow that conserves probability, and a theorem saying that such a flow has a self-adjoint generator. The equation is what you get by differentiating. The only physics added anywhere in this chapter is one choice of operator inside the derivation and one measurement where the numbers arrive, and both are marked in place.

Tools you'll need  — Chapter 4.5 above all: §9 for Stone's theorem in both directions, §3 for the spectral theorem in multiplication form, §6.6 for functions of an operator, §7 for what $\ket x$ and $\ket p$ mean, and §8 for the checklist of what is now safe to write down. Chapter 4.2 §7 for the argument that evolution is unitary and has a Hermitian generator, §7.4 for the postulate that the generator is the energy, and §8 for the canonical commutator. Chapter 4.4 §2 for unbounded operators, §3 for domains, §4 for the difference between symmetric and self-adjoint, and §5.4 for momentum on a ring. Chapter 4.3 §6 for completeness of $L^{2}$ and §7.3 for what a basis gives you. Chapter 0.9 §2.3 for Plancherel, §3.1 for the derivative theorem, and §6 for the bandwidth theorem. Chapter 0.7 §6 for the continuity equation and §7.5 for the Laplacian. Chapter 0.2 §4 for the Gaussian integral. Chapter 1.3 §2.2 for the classical Hamiltonian and §6.2 for the fundamental Poisson brackets. Chapter 0.5 §7 for functions of an operator in finite dimensions. Chapter 0.8 §7.6 for the continuum limit of a chain.

1 · What time evolution has to be

Here is where this section is going. We collect the properties that any rule for moving a quantum state forward in time is obliged to have, and the point worth watching is that none of them is a matter of taste. Two of them were derived in Chapter 4.2 from conservation of probability, and the third is the mathematical shadow of the physical statement that a state does not jump. At the end we will have exactly three conditions, they will be exactly the three conditions Stone's theorem takes as input, and one of them will be marked as the one a later chapter has to give up.

1.1 · Linear, and preserving the length of every state

Write $\hat U(t)$ for the map carrying the state at time $0$ to the state at time $t$, so that $\ket{\psi(t)}=\hat U(t)\ket{\psi(0)}$. Chapter 4.2 §7.1 required two things of it and derived the rest. The first is linearity, which is forced by the superposition postulate. If a system can be prepared in either of two states it can be prepared in any combination of them, and evolution has to carry the combination to the combination of the evolved states. The second is that total probability stays at one, which by Chapter 4.2 §5.2 is the statement $\norm{\psi(t)}=1$ at every time.

Those two together already fix a great deal, and Chapter 4.2 §7.2 spent three lines showing it. Preserving all lengths looks weaker than preserving all inner products, and it is not, because the missing information can be extracted from lengths by expanding $\norm{u+v}^{2}$ and then running the same line with $v$ replaced by $\ii v$. The conclusion was

That much is Chapter 0.5's promise collected, in the words it used. That chapter wrote that "$U(t)$ must be unitary because $\norm\psi^{2}=1$ is a total probability and probabilities have to keep summing to one." One gap in that sentence has to be named now rather than later, because Chapter 4.4 §1.2 made a point of it. In infinite dimensions (4.6.1) makes $\hat U(t)$ an isometry, and an isometry need not be onto: the shift that sends $(a_1,a_2,\dots)$ to $(0,a_1,a_2,\dots)$ preserves every length and still misses a whole direction, since its range is everything whose first coordinate is zero. Unitary means isometric and onto. Section 1.3 supplies the missing half from the group law, and until then the honest word is isometry.

1.2 · There is no third option

Before going on, ask what the alternative would have looked like. A factor $\lambda$ multiplying a state has modulus less than one, equal to one, or greater than one, and those three are the whole list. What this subsection shows is that only the middle one survives. Suppose $\hat U$ is an isometry and $\hat U\psi=\lambda\psi$ for some non-zero $\psi$. Then $\norm\psi=\norm{\hat U\psi}=\abs\lambda\,\norm\psi$, and dividing by $\norm\psi$ leaves $\abs\lambda=1$. That is the whole argument. Chapter 0.5 made it in finite dimensions, in its Problem 2, and stated the consequence in exactly these terms: "if $\abs\lambda\lt1$ the state would fade away and probability would leak out of the universe."

What Chapter 4.5 adds is that the argument does not depend on there being any eigenvectors to argue about. An operator on an infinite-dimensional space may have none at all, and Chapter 4.5 §2 showed that position and momentum are both in that situation. The one-line calculation above never needed the spectral theorem: it needed a vector, and if there is no such vector the conclusion is carried instead by the multiplication form of Chapter 4.5 §3, under which a unitary becomes multiplication by a function of modulus one. Either way the possible phases are on the unit circle and nothing else is available. A factor of modulus less than one is decay, a factor of modulus greater than one is growth, and probability permits neither.

1.3 · The group law, and what it assumes

Now bring in the fact that $t$ is a continuous parameter rather than a single instant. Take a system whose physical arrangement is not itself changing, with no field being switched on and no apparatus being moved. For such a system, waiting $s$ and then waiting $t$ has to be the same as waiting $s+t$, because nothing distinguishes the intermediate instant from any other. Chapter 4.2 §7.3 wrote that as

Two things follow immediately and both matter. Setting $s=-t$ gives $\hat U(t)\hat U(-t)=\hat I$, so every $\hat U(t)$ has a two-sided inverse and is therefore onto. That closes the gap §1.1 left open, and from here unitary is the right word rather than isometry. And the group law is where the words no memory enter the theory: the state at time $t+s$ depends on the state at time $s$ and on nothing that happened before it.

The assumption behind (4.6.2) deserves to be said out loud, since a whole later chapter is built on its failure. It assumes the Hamiltonian does not depend on time. Drive the system, with a laser pulse or a swept magnetic field, and the evolution from $s$ to $t$ depends on both times and not only on their difference. Then $\hat U(t)\hat U(s)\ne\hat U(t+s)$, the exponential of this chapter is wrong, and Chapter 4.17 §3 replaces it with a time-ordered series.

1.4 · Strong continuity, which is weaker than it looks

One condition remains, and it is the one that is easy to state carelessly. We want to say that the state moves continuously, meaning that a system left alone for a very short time is in very nearly the state it started in. Written as a limit that is

and the name for it is strong continuity. The word strong is unfortunate, because the condition is the weak one. It is continuity state by state, and the tempting alternative is to ask for $\norm{\hat U(t)-\hat I}\to0$, which is continuity of the operator itself and is a genuinely stronger demand.

Asking for the stronger version would be a mistake, and Chapter 4.5 §9.3 says exactly why. A one-parameter unitary group that is continuous in the operator norm has a bounded generator. Every Hamiltonian in this book except the one for spin is unbounded, by the argument of Chapter 4.4 §2, so the norm-continuous theory would describe nothing we want to describe. The weaker condition is the one that has the theorems, and it is also the one with physical content, since what an experiment tests is what happens to a state rather than what happens to an operator.

Continuity at $t=0$ is enough to give continuity everywhere, and the group law is what does it. Slide by $t$ and use the fact that $\hat U(t)$ preserves lengths:

So the flow of a single state is a continuous curve in the space for all time, and the only thing that had to be checked was its behaviour at the origin. Register that pattern, because §3.1 runs the same move again: a group law turns a statement at one point into a statement everywhere.

1.5 · The three conditions, and which one goes

Collect them, because from here on this list is the input and nothing else is.

• Each $\hat U(t)$ is unitary. From linearity and conservation of probability, by Chapter 4.2 §7.2, with the onto half supplied by the group law.
• The family obeys the group law $\hat U(t)\hat U(s)=\hat U(t+s)$, with $\hat U(0)=\hat I$. From the physical arrangement not changing during the evolution.
• The family is strongly continuous, in the sense of (4.6.3). From the state not jumping.

Each of the first two fails somewhere in this book, and knowing where is worth more than knowing that they hold. Unitarity is what fails on a half-line: Chapter 4.4 §5.5 found that momentum on $[0,\infty)$ has no self-adjoint extension, and its Worked example 3 showed the corresponding flow runs forwards only, as a family of isometries with no inverses. The group law is what fails for a driven system, as §1.3 said, and Chapter 4.17 §3 is where that case is handled. Strong continuity is never given up anywhere in this book.

2 · Stone, and the generator

Section 1 produced a list of three conditions. This section hands that list to a theorem, gets an operator back, and then does two pieces of bookkeeping that the rest of Part IV depends on. The theorem is Stone's, proved forward and quoted backward in Chapter 4.5 §9. The first piece of bookkeeping is the identification of the operator it produces with the energy, which is the single physical assumption in the whole chapter. The second is the sign convention, which two written sentences of this book name this section for, and which is stated here once and then used without comment for the rest of Part IV.

2.1 · The theorem, in the form we need it

Stone's theorem is an equivalence between an object with a physical description and an object with an algebraic one. In the direction we need it reads as follows, and Chapter 4.5 §9.3 is where it is quoted with its hypotheses and its proof named.

Let $\{\hat U(t)\}_{t\in\R}$ satisfy §1.5's three conditions. Then there is exactly one self-adjoint operator $\hat G$ with $\hat U(t)=\ee^{-\ii\hat Gt/\hbar}$ for every real $t$. Its domain is not an extra stipulation. It is the set on which the difference quotient converges:

Three features of that statement carry the chapter, and it leans on all three. The operator is produced rather than assumed, so nothing has to be guessed and then checked. It is unique, so there is no freedom to describe the same evolution with a different generator, and a measurement of the flow is therefore a measurement of $\hat G$. And it is self-adjoint rather than merely symmetric, which is the distinction Chapter 4.4 §4 built and the reason Chapter 4.5 can apply the spectral theorem to it later without further apology. Unitarity of each $\hat U(t)$ is precisely what delivers that stronger word.

The converse half of Stone is the hard half and this book does not prove it. It is marked at Chapter 4.5 §9.3, with its hypotheses and its four-page proof named, and it is cited here rather than marked again, since a result carries its flag once and at the place it is quoted. What Chapter 4.5 §9.1 did prove is the direction running the other way, that a self-adjoint operator generates a strongly continuous unitary group and can be recovered from it by differentiating at the origin. The two halves together are what make time evolution is unitary and the Hamiltonian is self-adjoint the same sentence.

2.2 · Why $\hbar$ appears at all

Nothing so far has said why the constant in the exponent is $\hbar$, and the answer is that it is fixed by dimensions once you decide what $\hat G$ is going to be called. The exponent has to be a pure number, and $t$ carries a time, so $\hat G$ divided by the constant must carry one over a time. Write the constant as $\hbar$ and $\hat G$ acquires the dimensions of $\hbar$ over time, which is action over time, which is energy. Chapter 4.1 §5.7 already found that the new constant of quantum mechanics has the dimensions of action, and this is where that fact is spent: it is the only constant available that converts the rate at which a phase turns into an energy.

Read backwards, the same sentence is Chapter 4.1's $E=h\nu$ in operator form. A state whose phase turns at angular frequency $\omega$ has $\hat G$ equal to $\hbar\omega$ on it, and the constant of proportionality between a frequency and an energy is the constant Planck had to introduce to fit a spectrum. Nothing new is being assumed. The dimensional argument fixes the constant up to a pure number, and Chapter 4.1's measurement fixes the number.

2.3 · The identification, and what it buys

Everything to this point is forced. What is not forced is that $\hat G$ is the energy, meaning the same quantity a calorimeter measures, the same quantity Chapter 1.3 §2.2 built as the Legendre transform of the Lagrangian. Chapter 4.2 §7.4 stated that as its fifth postulate, writing the operator $\hat H$ and calling it the Hamiltonian, and it remains a postulate here. Half of it was derived and half of it was not, and the division is exact. That the generator exists, is unique, is self-adjoint and has the dimensions of energy is §2.1 and §2.2. That it is the energy is an identification with experiments behind it.

Here is what that identification buys, since a postulate that bought nothing would have no reason to be made, and this one buys a great deal.

• The generator becomes computable. Without the identification, $\hat G$ is whatever operator the flow happens to have, and there is no way to write it down before measuring the flow. With it, the classical energy of the system tells you what to write, which is the subject of §4 and the reason a quantum problem can be posed before it is solved.
• The spectrum of the generator becomes a measurable list. Chapter 4.2's Born rule says the possible values of an observable are the points of its spectrum, so the possible energies are the spectrum of the operator that generates time. A spectrometer is then an instrument that measures the generator of time evolution, which is a strange sentence and a true one.
• Frequencies become energy differences. Section 9 shows that a superposition of two states of definite energy has a probability density beating at $(E_2-E_1)/\hbar$, so a spectral line is a difference of two eigenvalues of $\hat H$. That is the sharpest experimental test the identification has, and it is the one that was available in 1913.
• Conservation of energy becomes a triviality of the algebra. Chapter 4.2 §7.5 showed that an observable is conserved exactly when it commutes with $\hat H$, and $\hat H$ commutes with itself. So a system whose arrangement does not change conserves its energy, and the proof is one line rather than a theorem.

2.4 · The sign convention, stated once and loudly

This book writes $\hat U(t)=\ee^{-\ii\hat Ht/\hbar}$, with the minus sign, so that a state of definite energy $E$ carries the time factor $\ee^{-\ii Et/\hbar}$ and a wave of definite momentum $p$ moving to the right is $\ee^{\ii(px-Et)/\hbar}$. That choice propagates into every formula in Part IV and into every formula in Parts V and VI, so the next two paragraphs separate what in it is a convention from what is not, because the two are usually run together.

What is genuinely conventional is a single global decision, which of the two square roots of $-1$ gets the name $\ii$. Nothing in mathematics distinguishes them and nothing in physics does either. Replace $\ii$ by $-\ii$ everywhere at once and you obtain a description in which every wavefunction is the complex conjugate of the one here, every predicted probability is identical because probabilities are moduli squared, and no experiment could tell the two apart.

What is not conventional is the relative sign, and this is the part that gets copied across wrongly. Once $\ii$ has a name, Chapter 4.2's canonical commutator $[\hat x,\hat p]=\ii\hbar$ fixes the sign of the momentum operator, as §5.2 checks by direct computation, and it comes out as $\hat p=-\ii\hbar\,\partial_x$ rather than $+\ii\hbar\,\partial_x$. Given that, the sign in the exponent is fixed by the requirement that a particle with positive momentum move in the positive direction. Take the free evolution of $\ee^{\ii px/\hbar}$ and ask where the crest goes. With the minus sign the phase is $(px-Et)/\hbar$ and the crest moves to the right, which is the direction a particle of positive momentum goes. With the opposite sign it would move left. Only the direction is being used here, and not the speed, which is $E/p=p/2m$ rather than $p/m$; §10.4 is where that mismatch is taken up. So the pair of signs is locked together and only their common flip is free.

The practical consequence is a warning. A good deal of engineering literature, in signal processing and in optics, writes time dependence as $\ee^{+\ii\omega t}$ and often calls the imaginary unit $j$. That is the conjugate convention, consistently applied. A formula lifted from it into this book will differ by the sign of every $\ii$. That is invisible in anything real and fatal in anything complex, such as an impedance, a scattering amplitude, a transmission coefficient's phase, or the sense in which a circular polarisation rotates.

2.5 · What the exponential of an unbounded operator means

One question has been left hanging and Chapter 4.3 named this chapter for it. The symbol $\ee^{-\ii\hat Ht/\hbar}$ has to be given a meaning, because $\hat H$ is unbounded and the power series $\sum_m(-\ii t/\hbar)^m\hat H^m/m!$ makes no sense applied to a state that is not in the domain of every power of $\hat H$. The general definition is Chapter 4.5 §6.6's functional calculus. For a self-adjoint $\hat A$ and a measurable $f$ the operator $f(\hat A)$ exists, and here we take $f_t(\lambda)=\ee^{-\ii\lambda t/\hbar}$, which has modulus one on the whole real line. The result is an operator that is bounded and defined on every state. That is the definition Chapter 4.5 §8.3 licensed and it is the one in force.

There is a second and more concrete situation, and it is the one every computation from Chapter 4.7 onwards actually uses. Suppose $\hat H$ has an orthonormal basis of eigenvectors $\{\ket n\}$ with $\hat H\ket n=E_n\ket n$, which is the case for a particle in a box, for the oscillator, and for the bound states of hydrogen. Expand the initial state as $\ket{\psi(0)}=\sum_nc_n\ket n$ and the exponential acts term by term:

The right-hand equality is why the series has a sum at all. Its partial sums form a Cauchy sequence, because the tails of $\sum\abs{c_n}^{2}$ go to zero, and a Cauchy sequence has a limit in the space because $L^{2}$ is complete, which Chapter 4.3 §6 proved and this book has spent elsewhere. Chapter 4.3's closing brick said this chapter would need completeness to make $\ee^{-\ii\hat Ht/\hbar}$ map states to states, and this is the place: without it the sum would be a formal expression rather than a state, and the evolved system would not be anywhere.

3 · The equation

All the work is done and this section collects the payment. We differentiate the flow, read the result, and then spend the rest of the section on what the equation does not say, which is the part that causes trouble later. By the end the equation will be on the table together with an exact statement of which states satisfy it, and the two questions it leaves open will be named and handed to §4 and §5.

3.1 · Differentiating the flow

We want the rate of change of the state, so take a state in the domain of the generator and differentiate $\hat U(t)\psi$ with respect to $t$. The group law does the work, in the same move §1.4 used: split the increment off the front and let the flow carry the rest.

Every step there is legitimate, and each of the three reasons has to be named. The middle equality is the group law together with the fact that $\hat U(t)$ is bounded, so it passes through the limit. The last equality is Chapter 4.5 §9.2, which is the half of Stone this book proves rather than quotes: the difference quotient converges in norm precisely on $\operatorname{dom}(\hat H)$, and its limit there is $-\ii\hat H\psi/\hbar$. Finally, $\hat U(t)$ and $\hat H$ commute, because both are functions of the same self-adjoint operator in the sense of Chapter 4.5 §6.6, so $\hat U(t)\hat H\psi=\hat H\hat U(t)\psi$. That closes the bookkeeping.

So multiply by $\ii\hbar$ and write $\ket{\psi(t)}=\hat U(t)\ket{\psi(0)}$. What is left is the equation this chapter is named after:

Take the two sides in turn. On the left is the rate at which the state turns, with $\ii\hbar$ converting that rate into an energy, which is the job §2.2 gave it. On the right is the energy operator applied to the state. The equation sets the two equal, so what a state does next is fixed by its own energy and by nothing else.

The box itself is not new on the page. Chapter 4.2 §7.4 wrote it in finite dimensions, with the generator postulated, and said in place that the infinite-dimensional step needed care. This time it has been earned. Stone's theorem produced $\hat H$ rather than anyone assuming it, and the derivative above converges in the norm of the space, on a domain handed over with the operator. Its solution is the exponential we started from and now know how to read:

Those two lines are Chapter 0.1's first forward pointer, paid in the form it was written. That chapter said, in its opening pages and before any machinery existed, that the fundamental law of quantum mechanics would turn out to be the first of them, with the second following. Chapter 1.1 made the same promise from the other side, naming the Hamiltonian as the operator that generates time evolution through (4.6.8) and pointing at Chapter 1.3's Legendre transform for its classical ancestor.

3.2 · What went into it

The reason to run the argument in this order is that the ingredient list is short enough to print, and printing it is the difference between a derivation and a postulate presented as one. Four statements went in, and only the last is physics beyond bookkeeping.

• Superposition and the Born rule, which make evolution linear and norm-preserving, hence unitary. Chapter 4.2's P1 and P3.
• The system is not being interfered with, which makes the family obey the group law.
• The state does not jump, which makes the family strongly continuous.
• The generator is the energy, which is Chapter 4.2's P5 and is an identification rather than a theorem.

Stone's theorem is what converts the first three into the existence of a self-adjoint $\hat H$, and the differentiation above is what converts that existence into a differential equation. The debt Chapter 4.2 §7 left at the same box, named there, is now paid, and (4.6.8) is a consequence of the four items above rather than a fifth item alongside them.

3.3 · Which states satisfy it

The equation holds for states in $\operatorname{dom}(\hat H)$ and for no others, and that is not a technicality to be waved past. Chapter 4.4 spent a whole chapter showing that a domain is part of an operator rather than an afterthought, and its §7 showed that for a particle in a box the domain is where the physics of the walls lives. Two facts keep the restriction from being a nuisance.

The first is that the flow preserves the domain. Since $\hat U(t)$ and $\hat H$ commute, a state that starts in $\operatorname{dom}(\hat H)$ stays there for all time, so the equation holds along the whole trajectory once it holds at the start. The second is that the domain is dense, so every state is a limit of states the equation applies to. For a state outside the domain the derivative in (4.6.8) does not exist, and the correct statement of its evolution is the integrated form (4.6.9), which is defined for every state without exception. That is the sense in which the exponential is the more fundamental of the two expressions, and the differential equation is the version you can compute with.

3.4 · What the equation is not

a natural place to stop  ·  the equation is on the table; what follows is the one physical choice it needs, and what it looks like when a state is a function of position

4 · Which operator is the generator

Equation (4.6.8) is complete and useless until somebody says what $\hat H$ is, and this section says it. The point of the section is not the formula, which you have seen since school, but its status. It is chosen rather than derived, the choice is made here in the open, and the rest of the section is about what makes a choice like that defensible and where it is known to fail. By the end $\hat H$ will be a specific operator with a specific domain, and §5 will be free to ask what it looks like when a state is a function of position.

4.1 · The question the equation leaves open

Stone's theorem produces the generator of whatever flow you hand it. That is a strength when you already have the flow and a weakness when you do not, and in practice you never do: nobody measures $\hat U(t)$ and then differentiates it. What a physicist has in advance is a description of the system, meaning its masses and the forces acting in it, and what is needed is a rule taking that description to an operator. Classical mechanics has such a rule, since Chapter 1.3 §2.2 built the Hamiltonian from the Lagrangian by a Legendre transform and Chapter 1.3 §2.3 said exactly when it comes out as kinetic plus potential energy. The rule adopted here is to carry that classical answer across.

4.2 · The identification

4.3 · What makes a choice like this defensible

An identification that cannot be derived can still be tested, and the test is the whole of the rest of Part IV. Three kinds of evidence have to be kept apart, since they are usually run together into a single gesture at "it works".

The first is that the classical limit comes out right, and comes out right as a theorem rather than as a hope. Chapter 4.9 derives Ehrenfest's relations, which say that the expectations $\avg{\hat{\vv x}}$ and $\avg{\hat{\vv p}}$ obey equations of exactly Hamilton's form, so a narrow packet in a slowly varying potential moves along a classical trajectory. Section 8 below supplies the piece of that argument which belongs in this chapter.

The second is quantitative agreement in cases with no adjustable parameters. Chapter 4.13 solves the hydrogen atom with this Hamiltonian and the Coulomb potential and gets the Rydberg constant, and Chapter 4.16 gets the fine structure. Neither calculation has a knob in it.

The third is that the failures are known, named and fixed by adding physics rather than by adjusting the rule. A charged particle in a magnetic field needs $\hat{\vv p}$ replaced by $\hat{\vv p}-q\vv A(\hat{\vv x})$, which does mix position and momentum and therefore does carry an ordering question, and Chapter 6.3 shows where that replacement comes from. Spin is not in the expression at all and has to be adjoined, in Chapter 4.12. Relativistic corrections are outside its reach altogether and are Chapter 5.5's. A rule whose failures are catalogued is in better standing than one whose successes are advertised.

4.4 · The domain, which is not automatic

One more thing has to be true before the operator in the box is admissible at all, and it is the condition Chapters 4.4 and 4.5 exist to enforce. The sum has to be self-adjoint, not merely symmetric, because Stone's theorem has self-adjointness in its hypothesis and the spectral theorem has it in its hypothesis, and Chapter 4.4 §4 showed the two words come apart the moment a domain is restricted. Self-adjointness of $\hat{\vv p}^{2}/2m+V(\hat{\vv x})$ is a genuine question about $V$, with a genuine answer that depends on how singular the potential is and on what happens at the edges of the region.

This book does not quote a general theorem for it. It checks the cases it uses, which is the standard Chapter 4.5 §5.6 set when it verified the spectral theorem three times rather than trusting it once. The free operator $\hat{\vv p}^{2}/2m$ is self-adjoint on the line, by Chapter 4.5 §4.3's verification, since the Fourier transform makes it multiplication by $\hbar^{2}k^{2}/2m$. A bounded potential can be added to that without damage, and the reason is one line rather than a claim. If $\hat B$ is bounded, self-adjoint and defined on the whole space, then $(\hat A+\hat B)^{\dagger}=\hat A^{\dagger}+\hat B$ on $\operatorname{dom}(\hat A)$, since Chapter 4.4 §3.3's defining relation splits across the sum and $\hat B$ imposes no condition of its own. The sum therefore has $\hat A$'s domain and is self-adjoint whenever $\hat A$ is. For a particle confined to an interval the question becomes the choice of boundary condition, which Chapter 4.4 §7 answered by counting: there is a four-parameter family of self-adjoint Hamiltonians for one box, and picking one is picking a wall. Chapter 4.7 takes exactly that route. The oscillator is settled by Chapter 4.5 §5, which exhibits a complete orthonormal family of eigenfunctions and builds the operator out of them.

Hydrogen is not settled that way, and saying so plainly here does more than a reassurance would. The Coulomb Hamiltonian is $\hat{\vv p}^{\,2}/2m-e^{2}/4\pi\epsilon_{0}r$, which §6 will write as $-\tfrac{\hbar^{2}}{2m}\nabla^{2}-e^{2}/4\pi\epsilon_{0}r$ once $\hat{\vv p}$ has been turned into a derivative. It has a ladder of negative eigenvalues accumulating at zero and, above them, the whole of $[0,\infty)$ as continuous spectrum. Its bound states span the point-spectrum subspace and are not a basis of $L^{2}(\R^{3})$, which closes the oscillator's route before it opens. That is the case §9.3 below takes up, on the failing side of the condition stated there. The bounded-potential clause above does not reach the case either, since $1/r$ is unbounded near the origin. Self-adjointness for the Coulomb Hamiltonian is a different theorem, it is not in this book, and it has to be established before an eigenfunction is looked for rather than read off one afterwards. Chapter 4.13 carries a mark for the theorem it uses.

One habit of thought has to be refused before it takes hold, since it is the shortcut this section exists to close. Unitary equivalence to a multiplication operator is not inherited by sums, so no Hamiltonian is certified by being assembled out of operators that have been verified, and Chapter 4.5 §5.6 says the same in its own terms. The list above is not of that shape in any case. The box Hamiltonian is a self-adjoint extension selected from a four-parameter family rather than a sum, and $V(\hat{\vv x})$ is a function of $\hat{\vv x}$ through Chapter 4.5 §6.6's functional calculus rather than a product of two operators. Every entry is checked on its own, and that is what checking case by case means.

5 · Momentum in the position representation

The Hamiltonian of §4 is written in terms of two operators whose commutator is known and whose form is not. That is enough to do algebra with and not enough to compute anything, so this section converts them into objects that act on functions. We fix what "the position representation" means, write down the momentum operator in it, check it against the commutator by direct computation, and then ask the question that has to be asked about any such formula: whether it is the only one. The answer is that it is not the only one and that all the others are copies of it, with the hypotheses of that statement mattering more than the statement.

5.1 · What the position representation is

Chapter 4.5 §7 supplied the meaning of the symbol $\ket x$, and the meaning is a procedure rather than a vector. Put the particle on a lattice of spacing $a$, where the indicator function of each cell is an honest vector and the family of them is an honest orthonormal basis. Do the computation there, rescale so that each term carries the spacing, and let the spacing go to zero. What comes out is that $\avg{x|\psi}$ is the limit of the average of $\psi$ over a cell shrinking onto $x$, and that $\int\ket x\bra x\dd x=\hat I$ is the limit of a true statement at finite spacing. Working in the position representation means writing every state as the function $\psi(\vv x)=\avg{\vv x|\psi}$ and every operator by what it does to that function.

Position itself is easy in this description, since $\hat{\vv x}$ acts by multiplying $\psi(\vv x)$ by $\vv x$, which is where Chapter 4.5 §4.1 started. What has to be worked out is momentum, and the requirement it has to meet is Chapter 4.2's canonical commutator.

5.2 · The operator, and the commutator checked

The realisation this book uses, in one dimension first so that the check fits on a line, is

That is Chapter 1.3's promise collected in the words it used, that "$\hat p=-\ii\hbar\, \partial/\partial q$ is the standard realisation" of the operator conjugate to position, and Chapter 4.2 §8's promise that this chapter would supply it. What has to be verified is that it meets the commutator, and the verification is worth doing rather than asserting, because it is where the sign of §2.4 is fixed. Apply the commutator to an arbitrary $\psi$ and use the product rule on the second term:

The two derivative terms cancel and what survives is the term where the derivative landed on $x$ itself, so $[\hat x,\hat p]=\ii\hbar\hat I$ as required. Notice that the opposite sign would have given $-\ii\hbar$, which is why §2.4 said the sign of the momentum operator is not a free choice once the imaginary unit has a name. In three dimensions the same line with $\partial_i$ and $x_j$ produces $\ii\hbar\delta_{ij}$, since $\partial_ix_j=\delta_{ij}$, and the other two relations of Chapter 4.2 §8 hold because partial derivatives commute with each other and coordinates commute with each other.

A formula is not yet an operator, and Chapter 4.4 spent a chapter on the difference. Equation (4.6.10) needs a domain, and which domain depends on where the particle lives. On the whole line the operator with the natural domain is self-adjoint, which Chapter 4.4 §5.3 established by showing the boundary term vanishes for every pair in the domain. On an interval it is self-adjoint only after a phase is chosen at the boundary, and Chapter 4.4 §5.4 exhibited the whole circle of choices. On a half-line no choice works at all, by Chapter 4.4 §5.5, and the same fact reappeared in §1.5 above as a flow that runs forwards only. The formula is the easy part.

5.3 · A realisation, not the realisation

Nothing above says (4.6.10) is the only operator meeting the requirement, and it is not. Take any smooth real function $\chi(x)$ and conjugate by the unitary that multiplies a state by $\ee^{\ii\chi(x)}$. Running the derivative through that factor gives

and this operator satisfies the same commutator with $\hat x$, because the extra term is a function of $x$ alone and commutes with $\hat x$. So there is a whole family of realisations, one for each $\chi$, and every member of the family reproduces every prediction of every other, since they differ by conjugation with a unitary and Chapter 4.2 §5 makes all probabilities inner products. Physicists meet this family again as the freedom to change the phase of a wavefunction pointwise while changing the momentum operator to match, which is the seed of the gauge principle of Chapter 6.3.

The question that matters is whether every realisation is of this kind, meaning a copy of the standard one obtained by conjugating with some unitary. If the answer were no, then the commutator would not determine the physics and Chapter 4.2's postulate would be badly underdetermined. The answer is yes, with hypotheses, and the hypotheses are the interesting part.

5.4 · The uniqueness theorem, and the hypothesis that fails later

With the theorem in hand, the status of (4.6.10) is exactly right. It is a choice of coordinates on the space of states rather than an additional physical assumption. Chapter 4.2 §8.4 said in advance that the realisation cannot be dodged, and the reason is that the commutator forces the space to be infinite-dimensional, so something concrete has to be written down before any number can be computed. This is that something, and the theorem says any other honest choice is the same choice in different clothes.

5.5 · The other representation, and why moving between them is free

The same argument run with the roles exchanged gives the momentum representation, in which a state is the function $\tilde\psi(p)=\avg{p|\psi}$, momentum acts by multiplying by $p$, and position acts as $\ii\hbar\,\partial/\partial p$. Chapter 4.5 §7.6 fixed the constants relating the two descriptions, finishing with $\avg{x|p}=\ee^{\ii px/\hbar}/\sqrt{2\pi\hbar}$, so the map between them is Chapter 0.9's Fourier transform with $k=p/\hbar$.

What makes that map safe to use without a second thought is the property Chapter 0.9 §2.3 proved and said this chapter would need. The symmetric convention was chosen there for one reason, that it makes the transform unitary, and Plancherel is the statement $\int\abs\psi^{2}\dd x=\int\abs{\tilde\psi}^{2}\dd p$. So a state normalised in position is normalised in momentum, and the total probability is one in both descriptions without any rescaling. Chapter 0.9's closing paragraph said that this single line is why a wavefunction can be moved freely between the two representations without the probabilities ceasing to add to one. That is precisely how it gets used here and in every chapter after it.

Chapter 4.3 §8.5 put the limits of that statement on record and they carry forward here unamended. What is true is that $L^{2}$ in the variable $x$ and $L^{2}$ in the variable $p$ are the same Hilbert space with a unitary map between the two descriptions, and that is enough for every computation in Chapters 4.6 to 4.10. What is not true is that $\{\ket x\}$ or $\{\ket p\}$ is an orthonormal basis in Chapter 4.3 §7.3's sense, and every equation containing either symbol is shorthand for Chapter 4.5 §7.4's procedure.

6 · The equation as a partial differential equation

Three separate things are now on the table and this section is where they meet. Section 3 gave the equation of motion in the abstract, §4 said which operator generates it, and §5 said what that operator does to a function of position. Putting them together turns an equation about vectors in a Hilbert space into a partial differential equation for a function of space and time, which is the form every calculation in the next four chapters starts from. The section is short because the work is already done, and its one job is to make sure you can read the result term by term.

6.1 · The three pieces assembled

Only one computation is needed and it is squaring the momentum operator. Apply (4.6.10) twice, and the two factors of $-\ii\hbar$ multiply to give $-\hbar^{2}$:

Now substitute that into §4's Hamiltonian, put the Hamiltonian into (4.6.8), and write $\psi(\vv x,t)$ for the state as a function of position. The result is the equation the chapter is named after, in the form in which it is used:

Read across it once. The left-hand side is a rate, with $\ii\hbar$ converting the rate at which a phase turns into an energy, exactly as §2.2 arranged. The two terms on the right are the two terms of the classical energy, each converted into an instruction about the function: the first measures how sharply $\psi$ bends, the second measures where $\psi$ is sitting in the landscape $V$. That is the whole content, and every problem in Chapters 4.7, 4.8 and 4.13 is this equation with a different $V$.

6.2 · Why the kinetic term is a Laplacian

The appearance of $\nabla^{2}$ deserves a sentence of explanation rather than being left as the output of an algebraic substitution, and Chapter 0.7 §7.5 named this chapter for it. The clean reason lives in the momentum representation. There kinetic energy is multiplication by $\vv p^{\,2}/2m$, which is as unmysterious as a classical formula can be, and Chapter 0.9 §3.1's derivative theorem says that multiplication by $\ii k$ in the transform is differentiation in position. Multiplying by $k^{2}$ is therefore differentiating twice and changing the sign, which is what (4.6.13) reports.

So the Laplacian is not an ingredient of quantum mechanics that has to be motivated separately. It is what "kinetic energy" looks like after the Fourier transform has been applied to it, and Chapter 0.9 §3.2 already said that the transform is the spectral theorem for the derivative. Chapter 0.7's closing list sent the Laplacian to three destinations and this is the quantum one. The reason that puts a Laplacian into Poisson's equation is this same reason seen from the other side of the transform. The only rotationally symmetric quadratic in the momentum is $\vv p^{\,2}$, and $\nabla^{2}$ is what multiplication by it becomes.

6.3 · Curvature costs energy, and that is the whole of confinement

There is a way of reading (4.6.14) that turns out to explain most of the qualitative behaviour in the next two chapters, so install it now. Rearrange the kinetic term as an average: for a normalised state the expected kinetic energy is

the second form following from an integration by parts whose boundary term vanishes for a state in the domain, which is Chapter 4.5 §8.4's fourth licence used exactly as that section says it may be. Written that way the kinetic energy is manifestly non-negative and it is a measure of how fast $\psi$ varies from place to place. A state squeezed into a small region has to rise and fall over that region, so $\nabla\psi$ is large, so $\avg T$ is large.

That single observation is the reason confinement costs energy, and it is where the ground-state energy of a box comes from in Chapter 4.7. Having it in this form makes the result a statement about the equation rather than an appeal to a principle. The lowest energy a confined particle can have is set by how gently a function can be made to vanish at two walls a fixed distance apart, and that is a question about calculus.

6.4 · The form used for the rest of Part IV

Most of what follows happens in one dimension, where the equation reads

and Chapter 4.7 solves it for a well and a barrier, Chapter 4.8 for a parabola, and Chapter 4.13 returns to three dimensions for the Coulomb potential. Nothing about the reduction to one dimension is an approximation. It is the exact equation for a particle whose motion in the other two directions has been separated off, and Chapter 4.13 shows how that separation works when the potential is spherically symmetric.

7 · What the $\ii$ is doing

Chapter 0.7 §7.5 put the diffusion equation and the equation of §6 side by side, observed that one is the other with an imaginary coefficient, and said that this chapter would explain what the imaginary unit is doing there. This section explains it. The claim being made is a strong one, that the $\ii$ is the entire difference between an evolution that forgets and an evolution that conserves probability, and by the end of the section it will be a computation rather than a slogan.

7.1 · The two equations

Take the free case, $V=0$, so that the comparison is between two equations with the same terms. Divide (4.6.14) through by $\ii\hbar$ and set it beside the diffusion equation Chapter 0.7 built for a concentration $C$:

They are the same equation with $D$ replaced by $\ii\hbar/2m$. Everything structural is shared: first order in time, second order in space, linear, with constant coefficients. So whatever separates them has to be extractable from the single fact that one coefficient is real and positive and the other is purely imaginary.

7.2 · One mode, and the whole difference

The fastest way to see it is to solve both equations for a single Fourier mode, which is legitimate because Chapter 0.9 §3.2 showed that the transform diagonalises every derivative. Put $\ee^{\ii\vv k\cdot\vv x}$ into each and read off the time dependence of its coefficient:

Compare the moduli, which is where the two part company completely. The diffusion factor is a real number less than one, decreasing, and decreasing faster the larger $k$ is, so fine structure dies first and the profile smooths towards a constant. The quantum factor has modulus exactly one at every $t$ and every $k$, so no mode ever loses any of its size: what changes is only the phase, and the phase advances at a rate proportional to $k^{2}$.

That is the difference stated once, and both halves of §1 are visible in it. A factor of modulus less than one is the case §1.2 ruled out, since it is probability leaking away. A factor of modulus exactly one is what unitarity demands, and Chapter 0.9 §2.3's Plancherel theorem promotes the statement about each mode into the statement about the whole state: if no coefficient changes size then $\int\abs\psi^{2}$ does not change either. Chapter 0.7 said this chapter would say what the $\ii$ is, and (4.6.18) is the answer. The $\ii$ turns a real exponential, which shrinks, into a complex exponential, which turns.

7.3 · Which of them can be run backwards

The same line settles reversibility, which is physics here rather than a curiosity. Run the diffusion equation backwards by taking $t$ negative, and the factor $\ee^{-Dk^{2}t}$ becomes $\ee^{+Dk^{2}\abs t}$, which grows without bound and grows fastest for the finest structure. Any error in the data at short wavelengths is amplified enormously, which is why reconstructing a past concentration profile from a present one is a hard and unstable problem in practice, and an ill-posed one in principle.

Now run the quantum equation backwards. The factor becomes $\ee^{+\ii\hbar k^{2}\abs t/2m}$, of modulus one, and nothing grows. That is the concrete content of $\hat U(t)^{-1}=\hat U(-t)$ from §1.3, and it says the equation loses nothing: the state at any one time contains the state at every other time. An arrow of time is not present in (4.6.14) and cannot be extracted from it, and that fact is the one to hold on to when Chapter 4.20 asks where the apparent irreversibility of measurement comes from.

7.4 · What this is not: the wave equation with an $\ii$

One comparison in circulation has to be corrected, because Chapter 0.8's closing list makes it in passing. The Schrödinger equation is not the wave equation with an $\ii$ inserted. Chapter 0.8 §7.6's wave equation is second order in time, so it needs a displacement and a velocity to start it, and (4.6.14) is first order and needs one function. Inserting an $\ii$ into $\partial_t^{2}u=v^{2}\nabla^{2}u$ produces neither equation.

The accurate statement is Chapter 0.7's, and it is the one made above: the equation that becomes the Schrödinger equation when its coefficient is multiplied by $\ii$ is the diffusion equation. What the wave equation and (4.6.14) genuinely share is the Laplacian on the right and therefore the mechanism by which spatial structure drives change in time. What they do not share is the order in time, and the difference shows up in everything: a wave has a single propagation speed and a quantum packet does not, which is the dispersion §10 computes.

7.5 · The substitution that converts one into the other

Since the two equations differ by a factor of $\ii$ in one coefficient, there is a substitution taking either to the other, and it has a name and a use. Set $t=-\ii\tau$ in (4.6.17). The chain rule gives $\partial_t=\ii\,\partial_\tau$, so the quantum equation becomes $\partial_\tau\psi=(\hbar/2m)\nabla^{2}\psi$, which is diffusion with $D=\hbar/2m$. The substitution is called Wick rotation, Chapter 0.7 named it, and Chapter 5.6 uses it to turn quantum field theory into statistical mechanics.

It has a practical face as well, and Problem 4 works it out. In imaginary time every mode decays at a rate proportional to its energy, so the component with the lowest energy is the one that survives longest. Evolving an arbitrary state in imaginary time and renormalising it therefore drives the state towards the ground state. That is how a large class of numerical methods find ground states, and it is the same fact read as an algorithm rather than as a theorem.

a natural place to stop  ·  the equation is written and read; what follows is the conservation law that turns "the state stays normalised" into a statement about a local flow

8 · The probability current

This is the centre of the chapter. Section 1 required that total probability stay at one, and Stone's theorem delivered that requirement as a property of the flow, so in one sense there is nothing left to prove. In another sense there is everything left to prove, because a global statement about a number is much weaker than the physical claim anyone actually wants, which is that probability does not vanish here and reappear there. What this section builds is the local version: a current density $\vv J$, made out of $\psi$, satisfying Chapter 0.7's continuity equation exactly. By the end, "the wavefunction stays normalised" will be a theorem with a mechanism attached rather than a condition imposed by hand.

8.1 · The statement worth proving

Chapter 0.7 §6.1 was emphatic about the difference, and the emphasis carries straight over here. Global conservation says the total in the universe does not change, and that would permit probability to disappear in one region and appear simultaneously in another. Local conservation forbids it: to leave a region the stuff has to cross the boundary, and to get anywhere it has to pass through the places in between. Chapter 0.7 stated the general form,

and said that a conservation law is a continuity equation, with the global version its integrated consequence rather than its definition. Its §6.2 then listed probability in quantum mechanics among the things that equation governs, and named this chapter for it. What has to be produced here is the pair $(\rho,\vv J)$ for a quantum particle, with $\rho$ the probability density the Born rule already fixed. Nothing may be chosen to make the equation come out: $\rho=\abs\psi^{2}$ is not negotiable, so either a $\vv J$ exists that fits it or the claim of local conservation is false.

8.2 · The derivation, with every step shown

Start from the density and differentiate it, using the product rule on $\rho=\psi^{*}\psi$:

Both time derivatives are supplied by the equation of motion, so the next step is to get them from (4.6.14). Divide that equation by $\ii\hbar$ and then take the complex conjugate of the result. The pair we need is

The conjugation used one property of $V$ and it is the property the whole result rests on: $V$ is real, so $V^{*}=V$ and the potential terms come back with only their sign of $\ii$ flipped. Section 8.5 returns to what happens when that fails. Substituting both into (4.6.20) and collecting terms,

The second bracket is zero. That cancellation is the reason the potential never appears in the final answer, and it happened because the two terms differed only by which of $\psi$ and $\psi^{*}$ carried the conjugate, while $V$ carried none. What remains is a difference of two Laplacian terms, and the move that turns it into a divergence is an identity worth checking rather than quoting. Expand $\nabla\cdot(\psi^{*}\nabla\psi)$ by the product rule, do the same with the roles exchanged, and subtract:

The cross terms are equal and cancel, which is why the combination with the minus sign is the one that works. Feeding (4.6.23) back into (4.6.22) puts the whole right-hand side under a divergence, and moving it to the other side puts the result in Chapter 0.7's form:

The last equality is a piece of arithmetic about complex numbers: the bracket is $z-z^{*}$ with $z=\psi^{*}\nabla\psi$, which is $2\ii\operatorname{Im}z$, and the factor of $2\ii$ cancels against the one in front. The symbol $\vv J$ is Chapter 0.7's own, chosen so that this is visibly the same equation as the one for charge and for mass rather than a relative of it.

8.3 · Reading the current

Equation (4.6.24) is more informative than it looks, and the way to see what it says is to split $\psi$ into a modulus and a phase. Write $\psi=R\,\ee^{\ii S/\hbar}$ with $R$ and $S$ real, which is always possible and puts $S$ in units of action. Then $\psi^{*}\nabla\psi=R\nabla R+\tfrac{\ii}{\hbar}R^{2}\nabla S$, and taking the imaginary part kills the first term entirely:

So the current has exactly the classical form, density times velocity, with the velocity read off the gradient of the phase. Two consequences follow at once and both are used constantly. A wavefunction that is real, or real up to a constant phase, has $\nabla S=0$ and therefore carries no current at all, however sharply peaked it is: the shape of $\abs\psi$ says nothing about the flow. And the current is not something added to the theory alongside $\psi$; it is the phase, which is the part of $\psi$ the Born rule discards.

One check confirms that $\vv J$ deserves its name. Integrate it over all space and compare with the mean momentum. Since $\avg{\hat{\vv p}}=-\ii\hbar\int\psi^{*}\nabla\psi$ is real, it equals $\hbar\int\operatorname{Im}(\psi^{*}\nabla\psi)$, and therefore

The total probability current is the mean momentum over the mass, which is what "current" ought to mean and is not something we arranged.

8.4 · Normalisation is now a theorem

The global statement follows from the local one in the way Chapter 0.7 §6.1 described. Integrate over a large ball and apply the divergence theorem:

the limit vanishing for the reason Chapter 4.4 §5.3 gave in one dimension. For a state in the domain both $\psi$ and $\nabla\psi$ are in $L^{2}$, so Cauchy–Schwarz makes $\vv J$ integrable over the whole of space. Integrating the flux over all radii therefore gives a finite answer, so the flux cannot be held away from zero, and it vanishes along a sequence of radii running to infinity, which is all the limit above needs. What is not available is the pointwise version: a function of $L^{2}$ need not tend to zero at infinity at all, and the argument above is built to avoid needing it to. Hence $\norm{\psi(t)}$ does not change. That is the same conclusion §1 obtained from unitarity, and having it twice is not redundancy. The first derivation says the number is constant. This one says why, by exhibiting the flow that carries probability from place to place and showing it never leaks out the sides.

Chapter 1.3 §5 is the classical statement of the same idea, and the ledger sends it here. Liouville's theorem says the density of representative points in phase space is carried along by the classical flow without being compressed or created, which makes classical probability conserved as an identity rather than an assumption. Unitarity is the Hilbert-space version of that sentence, and (4.6.24) is the local form of the quantum half. The comparison stops one term short. Liouville's density lives on phase space, while $\rho=\abs\psi^{2}$ lives on position alone and answers to that density integrated over the momenta. Section 5.4's pair leaves no quantum joint density of position and momentum for $\vv J$ to be the current of. In both theories the conservation of probability is a property of the equations of motion rather than an extra postulate about them.

8.5 · Where the reality of the potential was used, and what breaks without it

Section 8.2 pointed at one step, and this subsection isolates it, since it is the whole physical content of "the Hamiltonian is self-adjoint" written out in position space. Suppose $V=V_{R}-\ii\Gamma/2$ with $\Gamma$ real and positive, which is a common way of modelling a state that decays out of the system. Then the potential terms in (4.6.22) no longer cancel, and repeating the calculation gives

Probability now drains away everywhere at a fixed per-unit-time rate, which is first-order kinetics with rate constant $\Gamma/\hbar$ and half-life $\hbar\ln2/\Gamma$. Nothing about that is wrong as a piece of modelling, and Chapter 4.17 uses exactly this device for a state that decays. What it is not is a description of a closed system: the operator with a complex potential is not self-adjoint, so §1's argument does not apply to it, and the probability that vanishes has gone into degrees of freedom the model left out. Worked example 3 does the calculation in full. The lesson to carry is that self-adjointness of $\hat H$ and conservation of probability are the same requirement seen from two sides, and that the reality of $V$ is where that requirement is enforced in practice.

8.6 · What the current is for

Two later chapters need $\vv J$, and which parts of it they need is what §8.3 was building for.

Chapter 4.7 uses it to define transmission and reflection for a particle meeting a step or a barrier. The honest definition of a transmission probability is a ratio of fluxes rather than a ratio of squared amplitudes, because the transmitted and incident waves can have different wavenumbers when the potential differs on the two sides, and (4.6.25) is what makes the difference. Problem 2 computes the current for a superposition of a right-moving and a left-moving wave, which is the piece that chapter starts from.

Chapter 4.10 uses it to say what the classical limit is. Substituting $\psi=R\ee^{\ii S/\hbar}$ into (4.6.14) and separating real and imaginary parts gives two real equations. The first is the Hamilton–Jacobi equation of Chapter 1.3 §8.2 with a single extra term carrying all the $\hbar$, and the second is (4.6.24) again, written with the same $\vv J$. That the amplitude equation of the classical limit is this chapter's continuity equation is the reason the phase $S$ deserves to be called an action.

8.7 · Ehrenfest, stated here and proved later

Section 4.3 named the classical limit as the first evidence for §4's identification and said this section would supply the piece of that argument belonging in this chapter. Here it is, with its proof elsewhere. For a particle in a potential the expectation values obey

which are Hamilton's equations of Chapter 1.3 §3 with every quantity replaced by its average. The first of them is (4.6.26) and the continuity equation one integration by parts apart, and the second needs machinery this chapter does not have. Chapter 4.9 proves both from the Heisenberg equation, which is the uniform route to statements of this kind, and it also supplies the caveat that decides how much they are worth: $\avg{\nabla V}$ is not $\nabla V(\avg{\vv x})$ unless $V$ is at most quadratic. Chapter 1.1 §4.4 promised these relations and warned in advance that they are a derived statement about averages rather than a fundamental law, and that warning is the right one to carry into Chapter 4.9.

9 · Stationary states, and the time-independent equation

This section is short and it is the hinge on which the next two chapters turn. The equation of §6 is a partial differential equation in four variables, and nobody solves those directly. What is solved instead is an eigenvalue problem in the spatial variables alone, and this section is the three lines that reduce one to the other. It also says exactly what is being claimed when the reduction is made, since a superposition of solutions is a solution and the eigenvalue problem finds only the special ones.

9.1 · Separating the variables

We are looking for the solutions on which the time dependence is as simple as possible, so try one in which time and position appear in separate factors, $\psi(\vv x,t)=f(t)\,u(\vv x)$. Substitute that into (4.6.14) and divide through by $f u$. The variables separate completely:

The left side depends only on $t$ and the right only on $\vv x$, so both are equal to the same constant, which has the dimensions of energy and is called $E$. Read off the two halves. The one for $f$ is first order with solution $f=\ee^{-\ii Et/\hbar}$, and the one for $u$ is the time-independent Schrödinger equation:

So the separable solutions are exactly the eigenvectors of the Hamiltonian, each carrying its own phase factor $\ee^{-\ii Et/\hbar}$ whose rate is its own eigenvalue over $\hbar$. Chapter 0.8 §3 said that a linear differential equation is an eigenvalue problem and named this chapter as one of the places the observation would be cashed; (4.6.31) is the cashing. Chapter 0.8's closing list also promised that the eigenvectors of $\hat H$ would be the stationary states here, and that Chapters 4.7 and 4.8 would be the work of finding them, for a well, for a barrier and for the oscillator. That division is the right one: this chapter defines them and those chapters find them.

9.2 · Why they are called stationary

The name is not decoration and it says exactly what is constant. For a separable solution the density is

with no time in it, because the phase factor has modulus one. The same cancellation happens for the expectation of any observable that does not itself depend on time, since the two phase factors in $\avg{\psi,\hat A\psi}$ are conjugates of each other. So nothing measurable about a stationary state ever changes, which is why an atom left alone in an energy eigenstate sits there indefinitely.

Two cautions belong with the name. A stationary state is not a state in which nothing is moving: §8.3 showed that the current is set by the gradient of the phase, and Worked example 2 exhibits a stationary state with a constant non-zero current running through it. And the state is stationary only because $\hat H$ has no time dependence, which is §1.3's assumption again and is exactly what Chapter 4.17 removes.

9.3 · When every solution is a superposition of these

Separation of variables produces special solutions, and it is a fair question why they are worth anything, since almost no state is one of them. The answer is linearity together with Chapter 4.5, and it carries a condition that has to be stated alongside it rather than after it. The equation is linear, so any combination of solutions solves it. If the eigenvectors of $\hat H$ form a complete orthonormal family, which by Chapter 4.5 §8.2 is the statement that the spectrum is pure point, then every state is such a combination. That gives the general solution in one line, which is (4.6.6) written out:

Every solvable problem in Chapters 4.7 and 4.8 is that line plus a diagonalisation. Find the eigenvalues and eigenfunctions, project the initial state onto them, attach a phase to each, and add. The only work is the first step, and that is why those chapters look like exercises in solving ordinary differential equations rather than exercises in evolution.

Hydrogen is the case §4.4 sent here, and it sits on the other side of the condition. Chapter 4.13 finds the bound states of the Coulomb Hamiltonian, and those are complete on the point-spectrum subspace and not on $L^{2}(\R^{3})$. So the sum above reproduces a bound state of hydrogen exactly and misses everything above the ionisation threshold, and finding the bound states is not the same achievement as diagonalising the operator.

What the condition excludes is the part to carry forward. It holds for the oscillator and for the bound states of a well, and it fails for a Hamiltonian with a continuous part. There an expansion in bound states alone misses the scattering states entirely, and the general solution needs an integral alongside the sum, which is Chapter 4.5 §6's $\int\lambda\,\dd P(\lambda)$ doing the work the sum cannot. Chapter 4.7 meets that case at a step and a barrier, hydrogen meets it above threshold, and §10 below meets the extreme version of it, a free particle with no bound states at all.

9.4 · Two levels, and where a spectral line comes from

The cheapest non-stationary state is a superposition of two stationary ones, and writing it down shows what §2.3 claimed the identification of $\hat H$ with the energy buys. Take $\psi=c_1u_1\ee^{-\ii E_1t/\hbar}+c_2u_2\ee^{-\ii E_2t/\hbar}$ with $u_1,u_2$ orthonormal. Squaring gives

and the cross term oscillates at the angular frequency $\omega_{21}=(E_2-E_1)/\hbar$. The two squared terms are static and the interference between them is not, so the density beats at a frequency fixed entirely by the difference of two eigenvalues of $\hat H$. That is the Bohr frequency condition, $\Delta E=\hbar\omega$, and it is the reason a spectrum is a set of lines rather than a continuum. It is also the sharpest test the fifth postulate has. The same $\hat H$ whose eigenvalues are supposed to be the energies is the one generating the time dependence, so the frequency a spectrometer measures checks both readings against each other at once. Chapter 4.17 turns this observation into transition rates.

a natural place to stop  ·  the machinery is finished; what follows is the one problem that can be solved on the spot, and what it says about a moving particle

10 · A free packet: group velocity, and spreading

Everything in this chapter so far has been structural. This section solves something. The free particle is the only problem in Part IV whose time evolution can be written in closed form with no special functions. It is done in full here for three reasons. It shows what "the particle moves" means when the state is not a point. It produces the one quantitative prediction about a lone particle that this chapter can make. And it is the machinery Chapter 4.2 §10.3 named when that section took a shortcut with neutrino oscillations. By the end we will have a formula for the width of a packet at any time and a number for how long an electron stays where you put it. Two things then follow the numbers. The measurement they rest on is marked where it arrives, in §10.6, and §§10.7 and 10.8 check the formula against a direct integration of the equation.

10.1 · Why the free particle needs a packet at all

Set $V=0$ and the time-independent equation (4.6.31) becomes $-\hbar^{2}u''/2m=Eu$, whose solutions are $\ee^{\pm\ii kx}$ with $E=\hbar^{2}k^{2}/2m$. Those are not states. Chapter 0.9 §2.1 pointed out that $\int\abs{\ee^{\ii kx}}^{2}\dd x$ diverges, and Chapter 4.5 §2 turned that into the statement that momentum has no eigenvectors in the space at all. So §9.3's recipe of expanding in eigenfunctions and attaching phases cannot be applied literally here, since there is nothing in the space to expand in.

What replaces it is the continuous version of the same idea, and Chapter 4.5 §6 built it: the sum over eigenvalues becomes an integral against the spectral measure, which for the free particle is Chapter 0.9's Fourier transform. A genuine state is a superposition of plane waves over a range of $k$, which is a wave packet, and the packet is normalisable even though none of its ingredients is. Chapter 4.3's Problem 3 drew a travelling bump escaping to infinity and said this chapter would have to handle exactly that for a free packet, with Chapter 4.7 taking the unbound states of a genuine potential. This is that handling.

10.2 · The packet, and its two widths

We want a state that is narrow in space and moving, so take a Gaussian centred at the origin and multiply it by a plane wave:

The normalisation is Chapter 0.2 §4's Gaussian integral and nothing more, and the shape of the density is the point Chapter 0.2's Worked example 2 made in advance. Squaring $\psi$ squares the Gaussian and leaves the phase factor alone, so $\abs\psi^{2}\propto\ee^{-x^{2}/2\sigma_0^{2}}$, which in that chapter's notation is $\ee^{-ax^{2}}$ becoming $\ee^{-2ax^{2}}$, the same shape with $a$ doubled. The probability density is a normal distribution of standard deviation $\sigma_0$, so $\Delta x=\sigma_0$.

Now transform, since the momentum width is what will drive everything that follows. Chapter 0.2 §4.3 computed the Gaussian with a linear term in the exponent, which is what a transform supplies, and that chapter's Problem 4 did this transform in full. With $k=p/\hbar$ the result is another Gaussian, centred on $k_0=p_0/\hbar$:

Multiply the two widths and the $\sigma_0$ cancels, leaving $\Delta x\,\Delta p=\hbar/2$ for every $\sigma_0$. Chapter 0.9 §6 proved that no state does better than $\hbar/2$ and that Gaussians are the only ones achieving it, so this packet sits exactly on the bound. Chapter 0.2 said in advance that quantum mechanics would contribute exactly one physical identification to that line, $b=p/\hbar$, and that with it the bandwidth theorem reads $\Delta x\,\Delta p\ge\hbar/2$. The identification is (4.6.10), made in §5, and §10.6 says what stands behind it. The general inequality for an arbitrary pair of observables is Chapter 4.9's, and it needs nothing from this chapter.

10.3 · Evolving it

The evolution is trivial in the transform and awkward in position, which is the whole reason for transforming. Each mode carries the phase §7.2 computed, so $\tilde\psi(k,t)=\tilde\psi(k,0)\,\ee^{-\ii\hbar k^{2}t/2m}$, and the state in position is the inverse transform of that. The integral is a Gaussian with a complex coefficient, and completing the square does it.

The integral, completed square by square

Write $\alpha=\sigma_0^{2}+\ii\hbar t/2m$ and $v_g=\hbar k_0/m$, and put $q=k-k_0$. The exponent of the integrand is

$$-\sigma_0^{2}q^{2}+\ii\Big[(q+k_0)x-\frac{\hbar t}{2m}(q+k_0)^{2}\Big] \;=\; -\alpha q^{2}+\ii q\,(x-v_gt)+\ii\Big(k_0x-\frac{\hbar k_0^{2}t}{2m}\Big),$$

the $q^{2}$ terms combining into $\alpha$ and the $q^{1}$ terms into the displacement $\xi\equiv x-v_gt$. So the whole $x$ and $t$ dependence sits in the single integral $F(\xi)=\int\ee^{-\alpha q^{2}}\ee^{\ii q\xi}\dd q$, and we want that as a function of $\xi$. Rather than evaluate it, differentiate it, which is Chapter 0.2 §4.4's move and is licensed by the same domination. Then use $q\,\ee^{-\alpha q^{2}}=-\frac{1}{2\alpha}\partial_q\ee^{-\alpha q^{2}}$ and integrate by parts, the boundary term vanishing because $\operatorname{Re}\alpha=\sigma_0^{2}\gt0$ makes the integrand decay:

$$F'(\xi) \;=\; \int \ii q\,\ee^{-\alpha q^{2}}\ee^{\ii q\xi}\dd q \;=\; -\frac{\ii}{2\alpha}\int\Big(\partial_q\ee^{-\alpha q^{2}}\Big)\ee^{\ii q\xi}\dd q \;=\; -\frac{\xi}{2\alpha}\,F(\xi).$$

That is separable, so $F(\xi)=F(0)\,\ee^{-\xi^{2}/4\alpha}$, and no complex Gaussian had to be evaluated. Assembling, with $C(t)$ standing for everything independent of $x$,

$$\psi(x,t) \;=\; C(t)\; \ee^{-\xi^{2}/4\alpha}\;\ee^{\ii(k_0x-\hbar k_0^{2}t/2m)}, \qquad \xi=x-v_gt.$$

Now take the modulus squared. Since $\abs{\ee^{z}}^{2}=\ee^{2\operatorname{Re}z}$ and $\operatorname{Re}(1/\alpha)=\sigma_0^{2}/\abs\alpha^{2}$, the exponent becomes $-\xi^{2}\sigma_0^{2}/2\abs\alpha^{2}$, and writing $\abs\alpha^{2}=\sigma_0^{4}+(\hbar t/2m)^{2}=\sigma_0^{2}\sigma(t)^{2}$ turns that into $-\xi^{2}/2\sigma(t)^{2}$. The density is therefore a Gaussian of standard deviation $\sigma(t)$ centred at $v_gt$, and the constant $\abs{C(t)}^{2}$ in front is fixed without any further work: §8 proves the norm is preserved, so it has to be whatever normalises a Gaussian of that width, which is $(2\pi\sigma(t)^{2})^{-1/2}$.

The density is what the grind box assembles, and it is a Gaussian at every time:

10.4 · Reading the answer

Two things are in that formula and they are independent of each other. The centre moves at $v_g=p_0/m$, which is the classical velocity of a particle of momentum $p_0$, and the width grows. Take the velocity first, since it carries a subtlety.

The packet's centre moves at the group velocity $v_g=\dd\omega/\dd k$ evaluated at $k_0$, where $\omega(k)=E(k)/\hbar=\hbar k^{2}/2m$ is the rate at which the mode of wavenumber $k$ turns. Differentiating gives $\hbar k_0/m=p_0/m$. The individual crests inside the packet move at the phase velocity $\omega/k=\hbar k_0/2m$, which is half as fast. Relative to the envelope, then, each crest drifts backwards at $v_p-v_g=-\hbar k_0/2m$. So a crest is continually born at the leading edge of the packet and dies at the trailing edge, while the packet as a whole outruns every one of them. Get that direction the wrong way round and the picture you are carrying is a water wave: in deep water $v_p=2v_g$, so crests there run forward through the group, and a free particle does the reverse. The quantum factor of two, $v_p=v_g/2$, is a standing reminder that the wave and the particle are different objects: the particle's velocity is the group velocity, and no crest ever travels at it.

Now the width. The formula $\sigma(t)^{2}=\sigma_0^{2}+(\hbar t/2m\sigma_0)^{2}$ has the shape of variances adding, and the second contribution is exactly what a spread in velocity produces. The momentum width is $\Delta p=\hbar/2\sigma_0$ from (4.6.36), so the velocity width is $\Delta v=\hbar/2m\sigma_0$, and a spread of velocities $\Delta v$ carries a spread of positions $\Delta v\cdot t$ after time $t$. The formula is therefore $\sigma(t)^{2}=\sigma_0^{2}+(\Delta v\,t)^{2}$, which is what a cloud of classical particles with a distribution of velocities and no interactions would do.

The quantum content is not the spreading. It is the fact that $\Delta v$ cannot be made small without making $\sigma_0$ large, because their product is fixed at $\hbar/2m$. So the packet cannot be made both narrow and slow to spread, and minimising $\sigma(t)$ at a chosen time by choosing $\sigma_0$ gives $\sigma_0^{2}=\hbar t/2m$ and $\sigma(t)^{2}=\hbar t/m$. That is the best any state can do, and the best gets worse with time.

10.5 · Numbers

Put an electron in it. Localise one to $\sigma_0=1\ \mathrm{nm}$, which is a few atomic diameters, and the velocity spread that comes with it is

which is faster than an artillery shell and follows from nothing but having pinned the electron down to a nanometre. The width doubles when $(\Delta v\,t)^{2}=3\sigma_0^{2}$, that is at $t=\sqrt3\,\sigma_0/\Delta v=3.0\times10^{-14}\ \mathrm{s}$, and the packet reaches a micron across in $1.7\times10^{-11}\ \mathrm{s}$. An electron does not stay where you put it for any length of time worth naming.

Run the same arithmetic on something heavier and the effect vanishes, which is the reassurance the formula owes us. A grain of dust of mass $10^{-9}\ \mathrm{kg}$ localised to a micron has $\Delta v=5.3\times10^{-20}\ \mathrm{m\,s^{-1}}$, and its width doubles in $3.3\times10^{13}$ seconds, which is about a million years. The spreading is real for both and the mass in the denominator is what decides whether it matters, which is the shape every classical limit in Chapter 4.10 will have.

10.6 · The experimental input

10.7 · The equation, integrated numerically

The figure below integrates (4.6.16) directly, by the split-operator method. Advance the potential term for half a step by multiplying by a phase, advance the kinetic term for a full step by multiplying by a phase in the Fourier transform, then finish the potential half step. Every factor has modulus one, so the method conserves the norm by construction rather than by luck, which makes it a fair test of the physics rather than of the arithmetic.

free step barrier oscillator

initial wavenumber k₀

2.00

run reset

t = 0.000 · norm = 1.000000000000 · |norm − 1| = 2.22e-16

width = 1.00000000 · formula = 1.00000000 · relative difference = 3.55e-14 · centre = -8.000000 against -8.000000

10.8 · The numerical confirmation

The same method, run offline at higher resolution, is what the claims above were checked against. On a $4096$-point grid spanning $\abs x\le100$ with $\Delta t=0.002$, in units $\hbar=m=1$, a packet with $\sigma_0=1$ and $k_0=2$ was propagated for $10$ time units, and the three figures that follow are all from that single run. The norm stayed within $4.5\times10^{-13}$ of one, the measured centre agreed with $v_gt=2t$ to $1.9\times10^{-12}$, and the measured width agreed with (4.6.37) to a relative error of $7.4\times10^{-14}$. That is the formula confirmed to the precision the arithmetic allows rather than to the precision of the physics.

The oscillator run is the sharper test, because there the answer is known independently. Starting the same shape displaced to $x_0=2$ in a potential $V=x^{2}/2$, with the width chosen as $\sigma_0=1/\sqrt2$ so that it matches the ground state, the centre tracked $2\cos t$ with a maximum departure of $2.6\times10^{-6}$, and the width held its starting value $1/\sqrt2$ to $3.5\times10^{-7}$. That departure of the centre is the integrator's and not the physics': halving the time step to $0.001$ reduced it to $6.6\times10^{-7}$, and doubling it to $0.004$ raised it to $1.1\times10^{-5}$, which is the factor-of-four scaling a second-order splitting is supposed to have. Chapter 4.9 explains why the mean position of a packet in a quadratic potential follows the classical trajectory exactly, and Chapter 4.8 explains why this particular width does not change.

11 · Worked examples

12 · Your turn