Part IV · Quantum Mechanics — Chapter 4.9

Commutators, Uncertainty, and Symmetry

The uncertainty relation was proved in Part 0, with no physics anywhere in it, and it has been waiting three parts for one substitution. This chapter makes the substitution in a line. Then it spends the rest of its length on the fact that the same commutator which bounds a product of spreads is also what moves every observable in time and what generates every continuous symmetry. Very little here is new. Almost all of it is one object, met from three sides.

Tools you'll need  — Chapter 0.5 above all: §1.4 for Cauchy–Schwarz and its equality condition, §7.1 for the exponential of $\ii$ times a Hermitian operator, and §8 for commuting operators and simultaneous diagonalisation. Chapter 0.9 §6.4 for the bandwidth theorem and §6.5 for the Gaussian that saturates it. Chapter 4.2 §4.3 for compatible observables and quantum numbers, §5.3 for the variance written as a length, §7.2 and §7.3 for the argument from unitarity to a Hermitian generator, §7.5 for the first-order action of a generator, §8 for the canonical commutator, and §9 for the tensor product. Chapter 4.6 §2 for $\hat U(t)$ and Stone's theorem in the form used here, §5 for momentum in the position representation, §8.7 for the Ehrenfest relations as stated there, and §9.2 for what a stationary state is. Chapter 1.3 §2.1 for what a conjugate momentum is, §6.1 for the classical equation of motion, and §7 for the flow an observable generates. Chapter 1.4 §3 for the three classical symmetries and §7.3 for the algebra of the rotation charges. Chapter 0.4 §4 for change of basis and the similarity transformation. Chapter 4.4 §3 for domains and §5.4 for momentum on a ring. Chapter 4.5 §6 for the projection-valued measure and §9 for Stone. Chapter 4.3 §8.1 for the density of continuous functions in $L^{2}$.

1 · One substitution: $p=\hbar k$

You have seen the answer before you have seen this section. Chapter 4.6 §10.2 multiplied one Gaussian packet's two widths together and got $\hbar/2$, and Chapter 4.8 §6.1 sat the oscillator ground state on the same number. Both were particular states, taken one shape at a time. What is still owed is the bound for every state, and it costs one multiplication, because Chapter 0.9 did the hard part with no physics in it at all.

1.1 · What was proved in Part 0, and what was missing from it

Chapter 0.9 §6.4 proved the bandwidth theorem: for any normalised $f$ with finite widths, the spread of $f$ across positions and the spread of its Fourier transform across wavenumbers satisfy $\Delta x\,\Delta k\ge\tfrac12$. The proof used the definition of the transform, Plancherel, the derivative theorem, Cauchy–Schwarz and one integration by parts, and Chapter 0.9 listed the ingredients in a bulleted list precisely because the shortness of the list was the point. No particle appears in it, no measurement, no observer, and no $\hbar$. A radio engineer shaping a pulse obeys it, and so does a seismic trace.

What was missing was a reason to care, in physics. The theorem constrains a wave, and a wave has a wavenumber. Nothing in Part 0 said that a particle has one.

1.2 · The substitution, and the line it takes

Chapter 4.6 supplied the missing sentence, and supplied it twice over. Its §5.5 showed that the map between the position and momentum representations of a state is Chapter 0.9's Fourier transform with $k=p/\hbar$, and that the map is unitary, so a state normalised in one description is normalised in the other. The density of the measured momentum is therefore the density of the wavenumber re-expressed under $p=\hbar k$. Its §10.6 marked de Broglie's relation as the one experimental input of that chapter, and that mark stands there rather than being raised again here. Those are the two facts §10.2 used on its Gaussian, and neither of them mentions the Gaussian. Since $k\mapsto\hbar k$ is linear, and a linear change of variable multiplies a standard deviation by its slope, we have $\Delta p=\hbar\,\Delta k$ for any state with both widths finite. So multiply Chapter 0.9's inequality through by $\hbar$:

That is the whole of it. The inequality was never quantum. What is quantum, and what is strange, is the claim that a particle's momentum is the wavenumber of a wave. Grant that one claim and the Heisenberg uncertainty principle for position and momentum stops being an additional mystery and becomes arithmetic you did in Part 0.

Now read the display as a prohibition with a scale on it, because the scale is what makes it a statement about the world. Confine an electron to a region the width of an atom, $\Delta x=10^{-10}\ \mathrm{m}$, and (4.9.1) forces $\Delta p\ge5.27\times10^{-25}\ \mathrm{kg\,m\,s^{-1}}$. That is a spread in speed of $5.8\times10^{5}\ \mathrm{m\,s^{-1}}$, and an energy of order $(\Delta p)^{2}/2m_{\mathrm e}=0.95\ \mathrm{eV}$. The electronvolt is the scale of atomic binding energies, and this is where that scale comes from: confinement to the size of an atom, priced by (4.9.1). Chapter 4.7 §3.5 read a confined particle's ground-state energy in exactly that way and got the same order back.

It is worth saying what has not been proved by that line, because §2 exists to prove it. Equation (4.9.1) is a statement about one pair of quantities, obtained through a Fourier transform, and the Fourier transform was available only because position and momentum are related in that particular way. Two arbitrary observables are not each other's transform, and for them the argument above says nothing at all.

1.3 · Every conjugate pair, and where the dimensions came from

Before leaving, one more paragraph, because it removes a puzzle that would otherwise sit in your way for the rest of Part IV. Chapter 1.3 §2.1 was careful to say that the canonical momentum is not $m\vv v$, and that for the pendulum with coordinate $\theta$ it comes out as $p_{\theta}=m\ell^{2}\dot\theta$, an angular momentum measured in $\mathrm{J\,s}$ rather than in $\mathrm{kg\,m\,s^{-1}}$. It then said what $p_{i}$ really is: whatever pairs with $q^{i}$ so that $p_{i}\,\dd q^{i}$ has the dimensions of action.

Chapter 4.1 §5.7 found that the new constant of quantum mechanics has the dimensions of action, and Chapter 4.6 §2.2 spent that fact to fix the $\hbar$ in the exponent. Put the two together and the bound $\Delta q\,\Delta p\ge\hbar/2$ is dimensionally consistent for every conjugate pair and not only for a Cartesian one, because both sides carry an action either way. An angle paired with an angular momentum works as well as a length paired with a linear momentum.

Whether the inequality holds for those other pairs is a separate question, and a more delicate one than it looks. The angle is the case where the general theorem's fine print does real work, and Worked example 2 is where that is shown rather than described.

2 · The general relation, and what it is not

Here is where this section is going. We want the same bound for an arbitrary pair of observables, with the right-hand side telling us which pair we are dealing with. The whole derivation is Chapter 0.5's Cauchy–Schwarz applied to two vectors that Chapter 4.2 has already built, and it takes three steps. Then we spend rather longer on what the result means, because the inequality is short and the number of things it is routinely taken to say is not.

2.1 · The two vectors, already in hand

Chapter 4.2 §5.3 derived the variance of an observable from the Born rule and then wrote it in a second form:

It said in place why the second form was the one this chapter would need. The spread of an observable is the length of a vector, and Cauchy–Schwarz is a statement about lengths. So name the two vectors, which is the only preparation the argument requires:

By (4.9.2) these have lengths $\norm f=\Delta A$ and $\norm g=\Delta B$. Chapter 0.5's insight box wrote down this pair three parts ago and said that nothing would be added to it here except the meaning of the symbols.

2.2 · Cauchy–Schwarz, once

The inequality of Chapter 0.5 §1.4 says $\abs{\avg{f,g}}^{2}\le\avg{f,f}\avg{g,g}$ for any two vectors of an inner-product space. Apply it to (4.9.3) and take the square root of both sides, which is legitimate because both sides are non-negative:

The uncertainty relation is now proved. Everything remaining is the identification of the right-hand side, which is where the commutator will appear, and it appears because we are about to ask which part of $\avg{f,g}$ survives when the two observables are swapped.

2.3 · Splitting the overlap into two real numbers

We want $\avg{f,g}$ expressed in the observables rather than in the vectors, so move the first operator across the inner product using its Hermiticity. Writing $a=\avg{\hat A}$ and $b=\avg{\hat B}$, both real by Chapter 4.2 §4.1, and abbreviating $\Delta\hat A=\hat A-a$ and $\Delta\hat B=\hat B-b$,

Two symbols differing by one hat are now in play, and they are not each other's value, so fix them apart before they appear in the same display. $\Delta\hat A$ is the operator $\hat A-a$, and its expectation is zero by construction. $\Delta A$ is a length, the norm of $\Delta\hat A\ket\psi$, and it is the standard deviation the Conventions paragraph fixed. The hat here separates an operator from a number built out of it, not from its own expectation.

That product of two Hermitian operators is not itself Hermitian, and the standard repair is to split it into a part that is and a part that is not. Any product splits that way, by adding and subtracting half of the reversed product:

where $\{\hat X,\hat Y\}=\hat X\hat Y+\hat Y\hat X$ is the anticommutator, and the constants $a$ and $b$ have dropped out of the commutator because a number commutes with everything. Now read off the character of each piece. The anticommutator of two Hermitian operators is Hermitian, since $(\hat X\hat Y+\hat Y\hat X)^{\dagger}=\hat Y\hat X+\hat X\hat Y$, so its expectation is real. The commutator is anti-Hermitian, since $[\hat X,\hat Y]^{\dagger}=\hat Y\hat X-\hat X\hat Y =-[\hat X,\hat Y]$, so its expectation is purely imaginary. The two terms of (4.9.6) are therefore the real and imaginary parts of one complex number, and a complex number's modulus squared is the sum of the squares of its parts:

Both terms are non-negative, so dropping the first can only weaken the statement. Drop it, feed what is left into (4.9.4), and take the square root:

Three steps, and the third of them was arithmetic on a complex number. Set $\hat A=\hat x$ and $\hat B=\hat p$ and use Chapter 4.2 §8's $[\hat x,\hat p]=\ii\hbar$: the right-hand side becomes $\tfrac12\abs{\ii\hbar}=\hbar/2$ in every state whatsoever, and (4.9.1) comes back. The two derivations share no step. Section 1 went through the Fourier transform and de Broglie's relation, this one through Cauchy–Schwarz and one postulated commutator, and they agree. That agreement is worth more than either route alone.

2.4 · The sharper form, which cost nothing

Look again at what was thrown away. Keeping the anticommutator term in (4.9.7) gives a stronger inequality for the same work:

The discarded quantity $\tfrac12\avg{\{\Delta\hat A,\Delta\hat B\}}$ is the covariance of the two observables, in exactly the sense used of any two random quantities, symmetrised because the operators do not commute. So (4.9.8) is what you get by pretending the two are uncorrelated when they need not be. For position and momentum in a state with no correlation between them the two forms agree, and Problem 1 exhibits a perfectly ordinary Gaussian in which they do not. The geometry of that difference is drawn in the figure at the end of this section, with two caveats and a pair of numerical checks standing between here and there.

2.5 · The hypothesis, which is not decoration

Every line above assumed it was legal to move an operator across an inner product. Chapter 4.4 spent a whole chapter on when that is legal, and the answer was not "always". Written out, the derivation needs $\Delta\hat B\ket\psi$ to lie in the domain of $\hat A$ and $\Delta\hat A\ket\psi$ to lie in the domain of $\hat B$, so that (4.9.5) is an identity rather than a hope. Call it the domain hypothesis. For a bounded operator it costs nothing, since Chapter 4.4 §2.1 showed bounded and continuous to be the same condition and a continuous operator on a dense domain extends by continuity to the whole space. For unbounded ones it is a genuine condition.

Something the domain hypothesis quietly requires is easier to miss, because it looks like a formality and is not. Both sides of (4.9.8) assume that $\Delta A$ and $\Delta B$ exist, and a normalised state is under no obligation to supply them. Chapter 4.3 §5.5 gave the structural reason, that $L^{2}(\R)$ does not sit inside $L^{1}(\R)$, and said the physical reading belonged in this chapter. Here it is. Normalisation is a statement about $\int\abs\psi^{2}$ and guarantees nothing about $\int x\abs\psi^{2}$, so a perfectly legitimate state can have no mean position and no $\Delta x$ at all, as that chapter's Worked example 3 exhibits. For such a state the relation is true and empty, in exactly the way Chapter 0.9 §6.5's grind box said the bandwidth theorem is true and empty when a width is infinite. Nothing has gone wrong. The inequality compares two numbers, and it says nothing in a state where one of them is not a number.

The domain hypothesis itself fails in the most quoted example after position and momentum. Take an angle and the angular momentum conjugate to it. The naive reading of §1.3 would give $\Delta\varphi\,\Delta L_{z}\ge\hbar/2$. That statement is false, and Worked example 2 exhibits a state in which the left-hand side is exactly zero, identifies the line of the derivation that breaks, and computes the boundary term the breakage leaves behind. The repair is to replace the angle by a bounded periodic function of it, after which the hypothesis holds and the theorem applies with no modification. Chapter 4.4 §5.4 built the operator involved, so the machinery for the repair is already on the shelf.

2.6 · What the relation does not say

2.7 · The relation, measured

Two checks, neither of which uses the derivation above, and the first of which is the one to remember. A third, on angular momentum, waits until §6.3, where the commutator it needs has been built.

The oscillator. Chapter 4.8 obtains $\avg{\hat x^{2}}$ and $\avg{\hat p^{2}}$ for the energy eigenstates by algebra, giving an uncertainty product of $(n+\half)\hbar$. Computing the same two integrals numerically for the Hermite functions at forty-digit working precision returns $\Delta x\,\Delta p=0.500000000000$ for the ground state and $3.500000000000$ for $n=3$ in units where $\hbar=m=\omega=1$, printed to twelve figures, and departing from $\tfrac12\hbar$ and $\tfrac72\hbar$ by less than $10^{-40}$, which is the quadrature's error at that working precision rather than a property of the states. Read those two numbers side by side. The ground state sits exactly on the floor set by (4.9.1), and the $n=3$ state sits seven times above it. The relation is a floor and not a prediction. It says how small the product cannot be, and it says nothing whatever about how large it is in any particular state. Chapter 4.7 §3.5 made the same point from the other end before this theorem existed, reading the infinite well's ground-state energy as an uncertainty and getting $\Delta x\,\Delta p=0.567862\,\hbar$ against a floor of $\hbar/2$. An instance computed before the theorem is worth more than the theorem alone, and both of those instances were.

Random pairs. Two hundred thousand random Hermitian pairs and random states in dimensions two to five, all built independently of the derivation, satisfy both (4.9.8) and (4.9.9) in every trial. The smallest slack in the sharper form (4.9.9) is $-3.0\times10^{-14}$, which is roundoff and means the sharper form is attained; the smallest slack in (4.9.8) over the same trials is $7.8\times10^{-10}$. The identity (4.9.7), on which the whole split rests, holds over twenty thousand random cases with a worst residual of $2.5\times10^{-14}$.

a natural place to stop  ·  the bound is proved and spent; what follows is the commutator when it vanishes, and then when it moves things

3 · Compatible observables, and a complete set

Section 2 asked what happens when a commutator is large. This section asks what happens when it is zero, and then answers a question Chapter 4.2 §4.3 raised, named this section for, and deliberately left open. The answer is a criterion, and the criterion is what Chapters 4.11 and 4.13 will apply every time they label a state.

3.1 · The two halves, now both proved

Chapter 0.5 §8 proved that two Hermitian operators admit a common orthonormal eigenbasis exactly when they commute, and Chapter 4.2 §4.3 renamed both halves. If $[\hat A,\hat B]=0$ the two are compatible: there are states in which both quantities have definite values, and enough of them to span the space. If $[\hat A,\hat B]\neq0$ there is no common eigenbasis, hence no state at all in which both are sharp.

That second half is the qualitative content of the uncertainty principle: a non-zero commutator rules out every state in which both are sharp, without saying by how much. Equation (4.9.8) is now the quantitative version, and the two fit together with one caution worth fixing now. A non-zero commutator forbids a common eigenbasis outright. The inequality, by contrast, is evaluated in a particular state, and $\avg{[\hat A,\hat B]}$ can vanish in a state even when the operator $[\hat A,\hat B]$ does not. In that state the bound has fallen silent while the prohibition still stands, and §6.3 puts a number on exactly that case.

3.2 · The question Chapter 4.2 left here

Chapter 4.2 §4.3 gave the definition this part of the book runs on. A set $\{\hat A_{1},\dots,\hat A_{m}\}$ of mutually commuting observables is a complete set of commuting observables when their common eigenspaces are all one-dimensional, so that the list of eigenvalues $(a_{1},\dots,a_{m})$ fixes the state up to phase. That list is what the word quantum numbers means, and writing a hydrogen state as $\ket{n,\ell,m_{\ell}}$ is naming the eigenvalues of three commuting operators and nothing more.

What it left open was how one knows a set is complete. It is a fair question and it is not answered by the definition, because the definition quantifies over eigenspaces you would have to find first. What you want is a test you can run on the operators.

3.3 · Maximality is the same condition

Here is the test, and its usefulness is that it never mentions eigenspaces. One phrase in it does all the work, so fix its meaning before the statement arrives. A function of a commuting set means what Chapter 0.5 §7 meant for a single operator, extended from one operator to a family: choose a number $c(a_{1},\dots,a_{m})$ for each tuple of joint eigenvalues, and let the operator multiply by that number on the corresponding common eigenspace, so that $\hat C=\sum c(a_{1},\dots,a_{m})\hat P_{(a)}$ with $\hat P_{(a)}$ the projection onto it. Nothing else counts as a function of the set.

Theorem. A set of mutually commuting observables on a finite-dimensional space has all its common eigenspaces one-dimensional if and only if every observable that commutes with all of them is a function of them.

A set with that property is called maximal: you cannot add a genuinely new commuting observable, because anything you might add is already there. Both directions are short, and both use the same fact about a common eigenspace.

One-dimensional common eigenspaces imply maximality. Let $\hat C$ commute with every $\hat A_{i}$, and let $\ket v$ be a common eigenvector with eigenvalues $(a_{1},\dots,a_{m})$. Then $\hat A_{i}\hat C\ket v=\hat C\hat A_{i}\ket v=a_{i}\hat C\ket v$, so $\hat C\ket v$ sits in the same common eigenspace. That eigenspace is one-dimensional and contains $\ket v$, so $\hat C\ket v$ is a multiple of $\ket v$. Every common eigenvector is therefore an eigenvector of $\hat C$, and since those eigenvectors form a basis by Chapter 0.5 §8, we may define $c$ on the tuples and write $\hat C=\sum c(a_{1},\dots,a_{m})\hat P_{(a)}$, which is a function of the $\hat A_{i}$ in exactly the sense just fixed.

A common eigenspace of dimension two or more destroys maximality. Let $W$ be such an eigenspace and pick a unit vector $\ket u$ inside it. Take $\hat C=\ket u\bra u$. It commutes with each $\hat A_{i}$, because $\hat A_{i}\ket u\bra u=a_{i}\ket u\bra u$ and, using Hermiticity, $\ket u\bra u\hat A_{i}=a_{i}\ket u\bra u$ as well. But any function of the $\hat A_{i}$ acts on all of $W$ as a single multiple of the identity, since $W$ carries one tuple of eigenvalues, whereas $\hat C$ annihilates the part of $W$ orthogonal to $\ket u$, which is not empty. So $\hat C$ commutes with everything in the set and is not a function of the set. $\blacksquare$

3.4 · How you show it in practice: exhibit the count

What §3.3 buys is not the working test. It makes the definition safe, by showing that the property you cannot check without finding eigenspaces is the same property as one you can state about the operators alone, so nothing turns on which of the two anyone happens to mean. The working test comes from the definition and the basis, and needs neither direction of the theorem. Since the joint eigenvectors form a basis, a set is complete exactly when the number of distinct eigenvalue tuples equals the dimension of the space. So you do not test maximality directly. You list the tuples and you count.

• Angular momentum. On a single multiplet of dimension $2j+1$, the set $\{\hat J^{2},\hat J_{z}\}$ produces the tuples $(j(j+1)\hbar^{2},m\hbar)$ for $m=-j,\dots,+j$, which is $2j+1$ of them. The count matches and the set is complete. Chapter 4.11 constructs the multiplet and then makes exactly this remark.
• Hydrogen. The set $\{\hat H,\hat L^{2},\hat L_{z}\}$ produces, at fixed principal quantum number $n$, the tuples with $\ell=0,\dots,n-1$ and $m_{\ell}=-\ell,\dots,+\ell$, so the count is $\sum_{\ell=0}^{n-1}(2\ell+1)$. That sum telescopes to $n^{2}$, since consecutive squares differ by consecutive odd numbers. Chapter 4.13 computes the dimension of the $n$-th energy level and finds $n^{2}$, at which point the count matches and the set is complete; Chapter 4.14 explains why the degeneracy is that number and not something else. What belongs here is the criterion those chapters apply.
• A set that fails. Problem 4 builds two qubits and shows that the total of the two "which-state" observables is not by itself complete. It then exhibits an operator commuting with that total which is not a function of it, and completes the set in two different ways whose members do not commute with each other. That last point is the entire structure of Chapter 4.12's coupled and uncoupled bases, met early and with no angular momentum in it.

3.5 · What the criterion becomes in infinite dimensions

Name the difficulty rather than stepping over it. The proof in §3.3 used a basis of eigenvectors, and Chapter 4.5 §2 showed that an observable in infinite dimensions may have no eigenvectors at all. The replacement is the projection-valued measure of Chapter 4.5 §6. Commuting observables have commuting spectral measures, and the joint measure assigns a projection to each region of the joint spectrum. The set is complete when that joint measure has multiplicity one, meaning that no region can be split further by any projection commuting with all of them, which is the one-dimensional condition with "dimension" replaced by "cannot be cut".

This book needs the finite-dimensional statement and needs it repeatedly, because every use in Chapters 4.11 to 4.16 is a count inside a finite-dimensional eigenspace of $\hat H$. The infinite-dimensional version is stated so the definition is not silently changed later, and the momentum components of a free particle in three dimensions are the standard example: one joint generalised eigenstate per momentum vector, nothing left to cut, and Chapter 4.5 §7 is where such objects were given their meaning.

One separate caution is worth parking here before §4 begins, since it belongs to the criterion rather than to infinite dimensions and Chapter 4.2 §4.3 stated it with a figure to go with it. The theorem promises that a common eigenbasis exists. It does not promise that a numerical eigensolver hands you that one, because inside a degenerate eigenspace the solver returned some basis rather than the basis the second observable prefers. Chapter 4.13's degeneracies are exactly where that bites.

What you now hold is a test you can run on paper: list the tuples, count them, compare with the dimension. Chapters 4.11 and 4.13 run it on angular momentum and on hydrogen, and Problem 4 runs it on two qubits, on a set that passes and a set that fails. The chapter now changes subject from a single instant to time.

4 · The Heisenberg picture, and the Heisenberg equation

Everything so far has been about a single instant. This section puts the time back in, and the first thing to settle is where the time is allowed to sit. The answer is that there is a choice, that the choice is a change of basis rather than a change of physics, and that making it produces an equation of motion for observables which is the exact quantum copy of the one Chapter 1.3 derived classically. That equation is what §5 and §6 both run on.

4.1 · Where the time can be put

Every prediction the theory makes is an expectation value. Chapter 4.2 §5.3 derived $\avg{\hat A}=\bra\psi\hat A\ket\psi$ from the Born rule, and Chapter 4.2 §5 showed that a probability is such a number with $\hat A$ a projection, so there is nothing else to compute. Take one at time $t$ and substitute the evolution $\ket{\psi(t)}=\hat U(t)\ket{\psi(0)}$ that Chapter 4.6 §2 produced:

Now look at the middle of that expression and notice that the brackets can be grouped two ways. Group the $\hat U$'s with the state on either side and you have the picture used since Chapter 4.6: the state moves and the operator sits still. Group them with the operator instead and you have a different bookkeeping with the same value, in which the state sits still and the operator moves. Define

and (4.9.10) reads $\avg{\hat A}(t)=\bra{\psi_{H}}\hat A_{H}(t)\ket{\psi_{H}}$. This is the Heisenberg picture, and the one before it is the Schrödinger picture. Since every prediction of the theory is one of these numbers and the number has not changed, the two pictures agree on everything measurable by construction, and not as a theorem that could have come out otherwise.

4.2 · It is a change of basis, and the domain the basis does not carry

Look at the form of (4.9.11) rather than at what it does. Chapter 0.4 §4.2 derived the transformation of a matrix under a change of basis and got $A'=P^{-1}AP$, calling two matrices related that way similar and summarising the content as: similar matrices are the same map, seen twice. Here $P=\hat U(t)$, and $\hat U^{-1}=\hat U^{\dagger}$ because $\hat U$ is unitary. So (4.9.11) is a similarity transformation, and as algebra it is nothing else. The Heisenberg picture is the Schrödinger picture written in a basis that rotates along with the state.

Chapter 0.4's warning box listed the consequences of confusing a map with its matrix, and one of them was "why the Schrödinger and Heisenberg pictures look like different physics instead of different bases". Collect it here, and pay for it, because in infinite dimensions the algebra leaves something out and Chapters 4.4 and 4.5 were spent on exactly that thing. An operator is a formula together with a domain, and Chapter 4.4 §3.1 fixed that two operators are the same operator only when the formulae agree and the domains do. Conjugation carries the domain along with the formula, so $\operatorname{dom}\big(\hat A_{H}(t)\big)=\hat U^{\dagger}(t)\operatorname{dom}(\hat A)$, which for an unbounded $\hat A$ is a different subspace at every time. That is what makes $\hat A_{H}(t)$ a genuinely different operator at each instant rather than one operator wearing a rotating coat, and it is why the differentiation of §4.3 is a strong derivative taken on vectors that stay inside the moving domain rather than a derivative in the operator norm. In finite dimensions the slogan costs a sentence about similar matrices. Here it costs the domain as well, and the domain is what the algebra is silent about.

The claim has consequences you can check rather than admire, and the one that matters is the list of readings. Chapter 4.5 §2.1 took invertibility rather than an eigenvector as the primitive, exactly so that this kind of question could be asked about $\hat x$ and $\hat H$, which have no determinant, no trace and in general no eigenvectors. Use it here. Since $\hat A_{H}-\lambda=\hat U^{\dagger}(\hat A-\lambda)\hat U$, the operator $\hat U^{\dagger}(\hat A-\lambda)^{-1}\hat U$ is a bounded everywhere-defined inverse of the left-hand side exactly when $(\hat A-\lambda)^{-1}$ is one for $\hat A-\lambda$. The two resolvent sets are therefore the same set, so the spectrum of $\hat A_{H}(t)$ equals the spectrum of $\hat A$ at every time. The list of values a measurement of $A$ can return does not move. If it did, the two pictures would disagree about what an apparatus can read, and no amount of algebra would repair that.

4.3 · Differentiating, and the equation that falls out

We have $\hat A_{H}(t)$ as an explicit formula in $t$, and what we want is its rate of change, so differentiate (4.9.11) with the product rule. Chapter 4.6 §3.1 supplies the derivative of the evolution operator, $\dv{\hat U}{t}=-\tfrac{\ii}{\hbar}\hat H\hat U$, and taking the adjoint of that gives $\dv{\hat U^{\dagger}}{t}=+\tfrac{\ii}{\hbar}\hat U^{\dagger}\hat H$ since $\hat H$ is self-adjoint. Allowing $\hat A$ its own explicit time dependence as well, the three terms are

The first two terms combine, and the operations have to go in one particular order. Insert $\hat U\hat U^{\dagger}=\hat I$ between the operator and the Hamiltonian in each term first, which converts each factor separately into its Heisenberg form and leaves $\tfrac{\ii}{\hbar}\big(\hat H_{H}\hat A_{H}-\hat A_{H}\hat H_{H}\big)$. Only now is there a commutator to read off, and it is $\tfrac{\ii}{\hbar}\big[\hat H_{H},\hat A_{H}\big]$. Two sign changes then bring that to the form written below and they cancel: reversing the bracket into $\big[\hat A_{H},\hat H_{H}\big]$ contributes one, and $\tfrac{\ii}{\hbar}=-\tfrac{1}{\ii\hbar}$ contributes the other. For a Hamiltonian with no explicit time dependence $\hat H$ commutes with $\hat U$, so $\hat H_{H}=\hat H$ and no subscript is needed on it. What is left is the Heisenberg equation:

The two terms answer to different things and it is worth keeping them apart from the start. The second is present only if the observable was defined with a time in it, as $\hat x\cos\omega t$ is, and it has nothing to do with the dynamics. The first is the dynamics, entirely, and it is a commutator.

4.4 · Beside the classical equation, term by term

Now set that against what Chapter 1.3 §6.1 derived, before any quantum mechanics existed in this book, for the rate of change of an arbitrary function on phase space:

Chapter 1.3 called this the equation of motion for every observable of every Hamiltonian system, and noted that it contains Hamilton's equations as the two special cases $f=q^{i}$ and $f=p_{i}$. Compare the two equations piece by piece:

• The left-hand sides are the same, with a function replaced by an operator.
• The explicit-time terms are the same, with the same meaning and the same irrelevance to the dynamics.
• The Hamiltonian appears in the same slot in both.
• The only difference anywhere is which bracket. The Poisson bracket $\{\;,\;\}$ has become $\tfrac{1}{\ii\hbar}[\;,\;]$.

That substitution is not being introduced here. Chapter 1.3 §6.4 wrote it down and Chapter 1.3's closing brick sent the bracket to this chapter to be replaced by $\tfrac{1}{\ii\hbar}[\;,\;]$, while Chapter 4.2 §7.5 derived the same replacement from the first-order expansion of a unitary. What is new is the equation itself, obtained from $\hat U(t)$ and the product rule rather than by analogy. Chapter 1.3 also warned that the substitution cannot be extended to every pair of classical observables at once, and that warning is still standing; Chapter 4.10 §8 is where it is proved.

4.5 · Conservation, in one line, and stronger than the classical version

Set $\pdv{\hat A}{t}=0$ in (4.9.13) and the left-hand side vanishes exactly when the commutator does:

Chapter 1.3 obtained the same statement with the same shape, and there too conservation stopped being something you established by solving the equations of motion and became something you established by computing one bracket. The quantum version says slightly more than the classical one, and the extra is worth having. What is constant is the operator $\hat A_{H}$, not merely its expectation. To see how much that buys, run it through the projections rather than the moments, since constant moments do not by themselves pin a distribution down for an unbounded observable. Because $[\hat A,\hat H]=0$ makes $\hat A$'s spectral projections of Chapter 4.5 §6 commute with $\hat U(t)$, the probability $\norm{P(E)\ket{\psi(t)}}^{2}$ of a reading landing in any region $E$ is the same at every time. The entire probability distribution of the measured values of $A$ is frozen, not merely its mean. Chapter 4.6's Worked example 1 found precisely this for the momentum of a free particle and obtained it there by a special argument. Here it is the general case.

4.6 · Two commutators the rest of the chapter needs

Before using (4.9.13) on anything, we need to be able to compute the commutator of a Hamiltonian with something. One rule does almost all of the work. Expanding both sides and cancelling the two middle terms gives, for any three operators,

which is the product rule with the ordering respected. Since the commutator with a fixed operator therefore behaves like a derivative, everything below is one application of it. With $[\hat x,\hat p]=\ii\hbar$ from Chapter 4.2 §8, applying (4.9.16) to $\hat p\hat p$ gives

The other one needs the position representation, and Chapter 4.6 §5 established that there $\hat p=-\ii\hbar\,\partial_{x}$ and $V(\hat x)$ is multiplication by $V(x)$. Acting on any differentiable $\psi$ in the domain, the product rule gives $\hat p(V\psi)=-\ii\hbar V'\psi-\ii\hbar V\psi'$ while $V\hat p\psi=-\ii\hbar V\psi'$, and the second terms cancel:

Both are verified symbolically. Now feed the Hamiltonian $\hat H=\hat p^{2}/2m+V(\hat x)$ of Chapter 4.6 §4 into (4.9.13) twice, once with $\hat A=\hat x$ and once with $\hat A=\hat p$. Since $\hat x$ commutes with $V(\hat x)$ and $\hat p$ commutes with $\hat p^{2}$, only one term survives in each case, and the factors of $\ii\hbar$ cancel:

Those are Hamilton's equations of Chapter 1.3 §3 with hats on everything, and they are exact operator identities with no approximation anywhere in them. Section 5 takes their expectation values and Problem 2 solves them in closed form for the oscillator.

4.7 · Which picture is used where

Neither picture is more correct, so the choice is made on convenience, and the convenience runs in opposite directions in the two halves of this book. Everything in Part IV is easier in the Schrödinger picture, because a bound-state problem is an eigenvalue problem for a fixed operator and Chapters 4.7 and 4.8 want the operator to hold still while they diagonalise it.

Part V reverses that. A field is an operator attached to each point of spacetime, and its time dependence has to sit in the operator rather than in the state. Otherwise the time coordinate is treated differently from the three spatial ones, and the relativity of Part II is invisible in the notation. Chapters 5.2 and 5.3 are written in the Heisenberg picture throughout for that reason, and (4.9.13) is the equation they solve. Building it here costs one section; discovering it there would cost a chapter.

a natural place to stop  ·  the equation of motion is built; what follows is what it says about averages, and then about symmetries

5 · Ehrenfest, and the potentials for which it is exact

Chapter 4.6 §8.7 wrote down two relations and left the second unproved, for want of machinery it did not have. The machinery is (4.9.13), and with it both relations take two lines. The section is not really about deriving them. It is about the caveat that comes with them, which decides how much they are worth and which is the reason the harmonic oscillator occupies the position in physics that it does.

5.1 · Both relations, from one equation

Take the expectation of (4.9.13) in the Heisenberg state, which is fixed, so the derivative passes straight through the bracket. Translated back into the Schrödinger picture, where it is more familiar, the statement is that for any observable

Everything needed is already in (4.9.19). Neither $\hat x$ nor $\hat p$ carries an explicit time, so the last term is absent. Taking expectations of those two operator identities gives Ehrenfest's relations, written here in three dimensions because the promotion is free and componentwise: §4.6's two commutators used one axis at a time, $[\hat x_{i},\hat p_{j}]=\ii\hbar\delta_{ij}$ keeps different axes from mixing, and $V'$ becomes $\nabla V$ one component at a time.

Those are the two relations Chapter 4.6 §8.7 stated and deferred, and the route to them here is uniform: one equation of motion, two choices of observable. Chapter 4.6's Worked example 1 obtained the first of them for a free particle by a special argument in three lines, and its agreement with the general case is a check on both.

5.2 · The warning Chapter 1.1 issued, and why it stands

Chapter 1.1 §4.4 argued that force is not a fundamental concept in quantum mechanics: there is no force operator, nobody lists its eigenvalues, and it appears in no commutation relation. It then named (4.9.21) as the closest thing to a force that quantum mechanics contains, and it added a warning at the same time, which is worth quoting because softening it would spoil the point. Read carefully, it said, this "is not a fundamental law but a derived statement about expectation values, in which the potential $V$ is the primitive object and the force is what you get by differentiating it".

Everything in this section confirms that reading. The potential entered through $\hat H$, which is where Chapter 4.6 §4 put it. The gradient appeared in (4.9.18) as the by-product of a commutator, and the object it acts on is an operator, not a trajectory. Nothing anywhere obeys Newton's second law. Two averages obey an equation that looks like it.

5.3 · The step that is not there

5.4 · At most quadratic, and nothing else

To see how big the gap is, expand $V'$ in a Taylor series about the mean position, which Chapter 0.3 licenses for a smooth potential, and take the expectation term by term. The first-order term vanishes because $\avg{\hat x-\avg{\hat x}}=0$ by construction, so the first surviving correction is the second-order one:

Read the correction term. It is controlled by the third derivative of the potential and by the square of the packet width, which by (4.9.1) cannot be sent to zero at finite momentum spread. Two consequences follow, and they are the point of the section.

The gap vanishes identically when $V'''\equiv0$, which is to say when $V$ is a polynomial of degree at most two. Then $V'$ is an affine function, the average of an affine function is the affine function of the average, and every term after the first in (4.9.22) is absent rather than small. Statement (b) becomes exactly true, for every state, however wide, however skewed, and however far from classical it looks. That is the quadratics and nothing else: the free particle, a uniform field, the harmonic oscillator, and the upside-down oscillator of an unstable equilibrium, where the coefficient is negative and the classical reading is exact all the same.

The gap does not vanish otherwise. If $V'''$ is non-zero somewhere then a packet sitting there has a discrepancy of order $\half V'''(\Delta x)^{2}$, and since $\Delta x$ cannot be sent to zero without sending $\Delta p$ to infinity, there is no state that escapes it. Ehrenfest's theorem survives untouched, because it was statement (a). What fails is the classical reading.

One more thing needs saying, because the exactness in the quadratic case is weaker than it sounds. An oscillator energy eigenstate has $\avg{\hat x}=0$ at every time, and it satisfies the classical equation with both sides zero forever. Nothing about it resembles a swinging pendulum. Exact-in-the-mean is a statement about one number extracted from the state, and a state can satisfy it while looking nothing like a trajectory. The state that does look like a trajectory is Chapter 4.8 §7's coherent state, which keeps its width while its centre traces the classical orbit, and Chapter 4.10 takes up the question of when a general system admits anything of the kind.

5.5 · The two cases, measured against each other

The way to settle this is to integrate the Schrödinger equation twice with everything held fixed except the potential, and watch the residuals. The integrator is the split-operator scheme of Chapter 4.6 §10.7, in which every factor has modulus one, so the norm is conserved by construction and the readouts test the physics rather than the arithmetic. Units are $\hbar=m=1$, the grid is $4096$ points on $[-20,20]$, the initial state is a Gaussian of width $0.6$ released from rest at $x=1.3$, and the run is to $t=6$ with $\dd t=10^{-3}$. The two potentials are $V=x^{2}/2$ and $V=x^{4}/4$, chosen so that the first has $V'''=0$ and the second has $V'''=6x$. The time derivatives in the table are fourth-order centred differences of the recorded series, which is the one detail the first two rows depend on and the one a reader reproducing the table would otherwise have to guess.

largest residual over the run	$V=x^{2}/2$	$V=x^{4}/4$
$\abs{\dv{}{t}\avg{\hat x}-\avg{\hat p}}$	$2.2\times10^{-7}$	$1.0\times10^{-6}$
$\abs{\dv{}{t}\avg{\hat p}+\avg{V'(\hat x)}}$	$1.1\times10^{-7}$	$3.1\times10^{-6}$
$\abs{\avg{V'(\hat x)}-V'(\avg{\hat x})}$	$0$, bit for bit	$1.665772$
$\abs{\dv{}{t}\avg{\hat p}+V'(\avg{\hat x})}$	$1.1\times10^{-7}$	$1.665771$

Read the table by rows and then by columns. The first two rows are Ehrenfest's relations, and they hold in both potentials with residuals of order $10^{-6}$ or smaller against quantities of order one. Those residuals belong to the integrator rather than to the theorem, and here is how you know: halving $\dd t$ divides them by four, three times running, which is a second-order error in the step size doing what second-order errors do. That test fixes the order and not the source, and the source here is the splitting's own second-order error seen through the difference stencil. The theorem is exact and the arithmetic is not.

The third row is the substitution the warning box refused to make. In the quadratic potential $\avg{V'(\hat x)}$ and $V'(\avg{\hat x})$ are the same floating-point number at every one of the six thousand steps, because $V'$ is the identity there and averaging commutes with doing nothing. In the quartic potential the two differ by as much as $1.67$ against forces of order one, with a mean gap of $0.62$ over the run. The fourth row is the consequence: the centre of the quartic packet is not obeying Newton's second law, it is missing it by an amount comparable to the force itself, and that number does not move when $\dd t$ is halved because it is physics rather than error. Meanwhile the quadratic run's centre follows $\avg{\hat x}(t)=1.3\cos t$ to $2.6\times10^{-7}$ throughout, although the packet is not a coherent state and its width is breathing the whole time.

a natural place to stop  ·  the commutator has bounded and it has moved things; what remains is what it generates

6 · Symmetries, generators, and conserved quantities

Chapter 1.4 ended Part I with a claim it called the real destination: a conserved quantity and the symmetry it comes from are one object under two names. Chapter 4.2 §7.5 restated that claim with operators in place of functions and expanded a unitary to first order to show the commutator appearing where the Poisson bracket had been. What it could not do was supply the generator's existence, because in infinite dimensions differentiating a one-parameter family of unitaries is not a formal manoeuvre, and it said so at the time. That link now exists, so the correspondence Chapter 4.2 could only state can be assembled here, and then run on the three cases Chapter 1.4 §3 ran classically.

6.1 · The chain, and which links were missing

A symmetry is a transformation of states that leaves every probability alone. Follow that through the machinery already built, one link at a time. Every link below was built somewhere else, so this subsection is where they are put in order rather than where anything is computed. The computing is in §6.2 and §6.3, on the three cases. The second link is the one Chapter 4.2 could not supply, and it is the reason this section can exist at all.

• Symmetry gives unitary. A map preserving all probabilities preserves all norms, and Chapter 4.2 §7.2 showed that a linear norm-preserving map preserves every inner product and is unitary. Linearity is assumed rather than derived at this link, exactly as Chapter 4.2 §7.1 assumed it, and every family in this section is visibly linear, so the assumption is discharged case by case instead of in general.
• A continuous family gives a self-adjoint generator. If the transformations form a strongly continuous one-parameter group, Stone's theorem produces a unique self-adjoint $\hat G$ with $\hat U_{\epsilon}=\ee^{-\ii\epsilon\hat G/\hbar}$. That is the theorem Chapter 4.5 §9 proved forward and quoted backward, and Chapter 4.6 §2 used it on time. This is the link Chapter 4.2 could not supply, and the reason it could not is that Chapter 0.5 §7.1's finite-dimensional argument assumes a bounded operator with an eigenbasis, which the generators here are not and do not have.
• The generator is an observable. Self-adjoint is exactly the condition Chapter 4.2's second postulate requires, in the corrected form Chapter 4.4 §4.2 gave it.
• The symmetry acting on an observable is a commutator. Chapter 4.2 §7.5 expanded $\hat U^{\dagger}_{\epsilon}\hat f\hat U_{\epsilon}$ to first order and obtained $\delta\hat f=\epsilon\cdot\tfrac{1}{\ii\hbar}[\hat f,\hat G]$.
• Symmetry of the dynamics means conservation. The family is a symmetry of the system when it leaves $\hat H$ alone, which by the previous line means $[\hat H,\hat G]=0$. By (4.9.15) that is precisely the statement that $\hat G$ is conserved.

Read the last two links in the other order and the converse is the same equation. If $\hat G$ is conserved then it commutes with $\hat H$, so the family it generates leaves $\hat H$ alone and is a symmetry. A conserved quantity is the generator of its symmetry, which is what Chapter 1.4 §7 proved classically, repeated here with the generator's existence supplied rather than assumed. The three cases below are Chapter 1.4 §3's three, and the third of them is already done.

6.2 · Translation, and momentum

Define the family that slides a wavefunction along, writing it $\hat U_{a}$ in the notation of §6.1, so that $(\hat U_{a}\psi)(x)=\psi(x-a)$. It is unitary because $\int\abs{\psi(x-a)}^{2}\dd x=\int\abs{\psi}^{2}\dd x$, Lebesgue measure being translation-invariant by Chapter 4.3 §2.1, and it is onto because $\hat U_{-a}$ undoes it. The group law holds because sliding twice is sliding once by the sum. Strong continuity takes one line: for a continuous $\psi$ of compact support, $\norm{\hat U_{a}\psi-\psi}\to0$ by uniform continuity, and Chapter 4.3 §8.1 proved such functions dense, so the general case follows from $\norm{\hat U_{a}}=1$.

Stone therefore hands us a generator, and to find it we differentiate the family at the origin, which is what Stone's formula for the domain in Chapter 4.6 §2.1 instructs. Since $\hat U_{a}=\ee^{-\ii a\hat G/\hbar}$ gives $\dv{\hat U_{a}}{a}\big|_{0}=-\tfrac{\ii}{\hbar}\hat G$, and differentiating $\psi(x-a)$ at $a=0$ gives $-\psi'(x)$, the two must agree:

Momentum is the generator of translations, with the operator Chapter 4.6 §5 identified arriving here for a second and independent reason. Chapter 1.4 §7.1 computed $\{x,p_{x}\}=1$ and read the same sentence off it classically.

Now close the loop with a computation rather than a slogan. By (4.9.18), $[\hat p,\hat H]=[\hat p,V(\hat x)]=-\ii\hbar V'(\hat x)$, which is the zero operator exactly when $V$ is constant. So momentum is conserved precisely when the potential is unchanged by translation, and the two statements are one commutator apart. Chapter 1.4 §3.2 reached the same conclusion from the action, and the two derivations have no step in common.

6.3 · Rotation, and the commutator Chapter 4.11 is built on

Do the same in three dimensions with rotations about the $z$-axis. The family is $(\hat R(\theta)\psi)(\vv r)=\psi(R^{-1}(\theta)\vv r)$, unitary because a rotation preserves volume, and differentiating at $\theta=0$ gives, with $R^{-1}(\theta)\vv r=(x\cos\theta+y\sin\theta,\, -x\sin\theta+y\cos\theta,\,z)$,

Angular momentum is the generator of rotations, matching Chapter 1.4 §7.1's classical computation with $L_{z}=xp_{y}-yp_{x}$, which also rotated the momentum as a bonus. The operator is the classical expression with hats, and the ordering ambiguity that usually attends such a substitution is absent here because $\hat x$ and $\hat p_{y}$ commute.

What that costs is one line and what it buys is the whole of Chapter 4.11, so compute the thing it buys. Rotations about different axes do not commute, and the algebraic residue of that geometric fact is the commutator of the generators. Take $\hat L_{x}=\hat y\hat p_{z}-\hat z\hat p_{y}$ and $\hat L_{y}=\hat z\hat p_{x}-\hat x\hat p_{z}$ and expand with (4.9.16). Of the four terms, two vanish outright because they contain no conjugate pair. The survivors are the first and the last, $[\hat y\hat p_{z},\hat z\hat p_{x}]=\hat y[\hat p_{z},\hat z]\hat p_{x}=-\ii\hbar\,\hat y\hat p_{x}$ and $[\hat z\hat p_{y},\hat x\hat p_{z}]=\hat x[\hat z,\hat p_{z}]\hat p_{y} =+\ii\hbar\,\hat x\hat p_{y}$. Both enter the sum with a plus sign, the first because it carried no minus sign in the expansion at all and the last because the two minus signs it carried multiply together. Adding them,

with the other two relations following by the cyclic relabelling $x\to y\to z\to x$. Verified in the Weyl algebra: normal-ordering the products using $[\hat x_{i},\hat p_{j}]=\ii\hbar\delta_{ij}$ and nothing else returns zero for all three differences.

That verification deserves a sentence of its own, because of what it does not use. Chapter 4.2 §8 postulated the operator commutator for the single pair $(\hat q,\hat p)$ and was explicit that the general correspondence between Poisson brackets and commutators is not derivable and is in fact false. Equation (4.9.25) is what the correspondence predicts for a pair it was never postulated for, since Chapter 1.4 §7.3 computed $\{L_{i},L_{j}\}=\varepsilon_{ijk}L_{k}$ classically, and applying $\{\;,\;\}\mapsto\tfrac{1}{\ii\hbar}[\;,\;]$ to that turns it into (4.9.25) exactly. The prediction is correct, and it was obtained here from the one postulated pair by algebra alone. Chapter 4.10 §8 explains why such agreement cannot be made universal.

Two consequences, and then the hand-off. First, feeding (4.9.25) into (4.9.8) gives $\Delta L_{x}\,\Delta L_{y}\ge\tfrac{\hbar}{2}\abs{\avg{\hat L_{z}}}$, so no state has all three components sharp unless all three vanish, which is what Chapter 1.4's grind box anticipated when it noted that $\{L_{x},L_{y}\}\neq0$ would survive quantisation.

Second, the check §2.7 held back. It was the third of that section's three, and the commutator it needed now exists. On the three-dimensional representation Chapter 4.11 builds from (4.9.25), the state with the largest $\hat L_{z}$ gives $\Delta L_{x}\,\Delta L_{y}=0.500000000000000\,\hbar^{2}$ against a bound $\tfrac{\hbar}{2}\abs{\avg{\hat L_{z}}}=0.500000000000000\,\hbar^{2}$, so the relation is saturated to fifteen figures. The middle state of the same set gives $\Delta L_{x}\,\Delta L_{y}=1.0\,\hbar^{2}$ against a bound of exactly zero, since $\avg{\hat L_{z}}=0$ there. That second case is the one to keep. A bound that vanishes has not announced that both quantities are sharp. It has fallen silent, while (4.9.25) still forbids a common eigenvector, which is §3.1's distinction made numerical.

And now the hand-off: equation (4.9.25) is the entire input to Chapter 4.11. That chapter takes it as the definition of what an angular momentum is, forgets where it came from, and extracts the whole spectrum from it by the ladder move Chapter 4.8 ran on the oscillator. Nothing about rotations in space is used again.

6.4 · Time, which was done first

The third case needs no work, because Chapter 4.6 did it before the pattern was visible. Stone applied to time evolution produced $\hat H$ as the generator, so the Hamiltonian generates time translation, and $[\hat H,\hat H]=0$ makes energy conserved by (4.9.15) with no calculation at all. Chapter 1.4 §3.1 obtained conservation of energy from invariance of the action under a shift of the time origin, and Chapter 1.4 §7.1's third example observed that time evolution is the transformation generated by the energy, so it is not a different kind of process from a symmetry. All three of Chapter 1.4's classics have now been repeated with operators, and the table below is that chapter's §3 in quantum form.

symmetry	unitary family	generator	conserved when
translation in space	$\psi(x)\mapsto\psi(x-a)$	$\hat p$	$V$ is translation-invariant
rotation	$\psi(\vv r)\mapsto\psi(R^{-1}\vv r)$	$\hat L_{z}$	$V$ depends on $r$ alone
translation in time	$\ket\psi\mapsto\hat U(t)\ket\psi$	$\hat H$	always, for a closed system

6.5 · A symmetry with no generator of its own

The chain of §6.1 needs a continuous family, and not every symmetry sits inside one that tells you anything. Parity, built in Chapter 4.7 §2 and written $\hat\Pi$, is the map $\psi(x)\mapsto\psi(-x)$. It is unitary, it is self-adjoint, and it commutes with $\hat H$ whenever $V$ is even. What it is not is a member of a family that could produce something new. Geometrically the reflection $\vv r\mapsto-\vv r$ has determinant $-1$, so no continuous family of spatial transformations reaches it from the identity. Algebraically a family can be manufactured, since the projections onto the even and the odd functions are functions of $\hat\Pi$ and exponentiating the odd one runs from $\hat I$ to $\hat\Pi$. But Stone applied to that family returns a generator which is itself a function of $\hat\Pi$, so it is no observable that was not already parity, and nothing has been generated. What survives is the conclusion without the machinery: parity commutes with $\hat H$, so by (4.9.15) it is conserved, and since $\hat\Pi^{2}=\hat I$ leaves it only two eigenvalues, what is conserved is a label taking two values rather than a quantity taking a continuum of them.

Chapter 1.4 §5 recorded the classical version of the same limitation and called it an honest exception, and this is what it looks like on this side. Noether's theorem needs a continuous symmetry that reaches the identity through the transformations themselves, and so does anything Stone can usefully hand back. A discrete symmetry produces a conserved quantum number with no independent generator behind it. Chapters 4.17 and 4.18 use exactly that, since a selection rule is a statement about a discrete label rather than about a continuous charge.

6.6 · What the commutator has not told us

Take stock, because the chapter has said three things about one object and it is worth being explicit that a fourth is missing.

The commutator of two observables bounds how sharp they can be at once ((4.9.8)), and it vanishes exactly when they can be sharp together (§3). The commutator with the Hamiltonian moves every observable in time ((4.9.13)), and it vanishes exactly when the observable is conserved ((4.9.15)). The commutator with any self-adjoint operator generates the symmetry that operator belongs to (§6.1). Three jobs, one object, and the classical shadow of all three is the Poisson bracket of Chapter 1.3.

What none of it says is what happens when $\hbar$ is small compared with the action in play. Every equation in this chapter carries $\hbar$ in a place where setting it to zero either destroys the statement or makes it vacuous: (4.9.8) becomes the empty assertion that a product of spreads is non-negative, and (4.9.13) becomes $0/0$. Something more careful is needed, and it is a genuine limit rather than a substitution.

That is Chapter 4.10, and it comes with a warning worth meeting before the proof. The correspondence between classical and quantum observables looks, from (4.9.13) and Chapter 4.2 §7.5, like a dictionary that should extend to everything. It does not. Chapter 4.10 §8 proves that no consistent dictionary exists taking every classical observable to an operator while turning every Poisson bracket into the corresponding commutator. Chapter 4.2 kept its distance from that general claim twice, in §7.5 when it separated the derived algebraic statement from the undeliverable general one and in §8 when it postulated the correspondence for a single pair rather than for all of them. Now that you have watched the correspondence succeed for angular momentum in (4.9.25), the claim that it must eventually fail is worth arriving with rather than being told afterwards.

7 · Worked examples

8 · Your turn