Conservation is compatibility with the Hamiltonian. An observable stays constant exactly when its operator commutes with the one generating time evolution.

§7.4 showed that a vanishing commutator means two observables can be sharp together. This short section asks what happens when one of the two is the Hamiltonian — and the answer is a conservation law.

Physics you need first — what 'conserved' means when nothing has a definite value

Classically, “energy is conserved” means a number attached to the system does not change. Quantum mechanically most observables have no definite value at all, so the statement needs rephrasing.

What is conserved is the expectation value $\langle A\rangle$ — the mean over many measurements on identically prepared systems. Saying $A$ is a constant of motion means that this mean is the same whenever you choose to measure, no matter how the wave function has sloshed about in between.

That is weaker than the classical statement and strong enough to be useful. And for an observable compatible with the energy it turns out to be stronger than it looks: the whole probability distribution of outcomes is frozen, not merely its mean, because $|c_{a_n}|^2$ never changes.

Carried over: §7.4 for the commutator, §7.1 for Hermiticity — which is the step that makes the derivation below work — and §4.6 for how $\Psi$ evolves.

The rate of change of any expectation value

Start from the definition,

\langle A(t)\rangle = \int\Psi^*\hat A\,\Psi\,\mathrm d^3\mathbf r\tag{7.20}

and let $\Psi$ evolve by the Schrödinger equation,

i\hbar\frac{\partial\Psi}{\partial t} = \hat H\Psi\tag{7.21}

step 1 of 4

Four lines. Watch for the third, where Eq. (7.2) is used to move Ĥ from one side of the integral to the other — without it the two terms would not combine into a commutator.

1Differentiate Eq. (7.20). Only Ψ depends on time (footnote 2 assumes Â does not), so the product rule gives two terms — one from each factor.
$\frac{d ⟨ A ⟩}{d t} = \int \frac{\partial Ψ ^{*}}{\partial t} \hat{A} Ψ d^{3} r + \int Ψ^{*} \hat{A} \frac{\partial Ψ}{\partial t} d^{3} r$
2Substitute the Schrödinger equation and its complex conjugate. Conjugating flips the sign of i, which is where the minus sign on the first term comes from.
$= - \frac{1}{i ℏ} \int (\hat{H} Ψ^{*}) \hat{A} Ψ d^{3} r + \frac{1}{i ℏ} \int Ψ^{*} \hat{A} (\hat{H} Ψ) d^{3} r$
3THE KEY STEP. Ĥ is Hermitian, so Eq. (7.2) lets it act on the other function instead — moving it inside, next to Â. Without Hermiticity the first term could not be rewritten and no commutator would appear.
$\int (\hat{H} Ψ^{*}) \hat{A} Ψ d^{3} r = \int Ψ^{*} \hat{H} \hat{A} Ψ d^{3} r$
4Now both terms carry Ψ* on the left and Ψ on the right, so they combine — and what sits between them is ÂĤ − ĤÂ.
$\frac{d ⟨ A ⟩}{d t} = \frac{1}{i ℏ} \int Ψ^{*} [\hat{A}, \hat{H}] Ψ d^{3} r$
Equation (7.22). Every observable's time dependence, in one line.

\frac{\mathrm d\langle A\rangle}{\mathrm dt} = \frac{1}{i\hbar}\int\Psi^*[\hat A,\hat H]\,\Psi\,\mathrm d^3\mathbf r\tag{7.22}

So if $[\hat A,\hat H] = 0$ then $\mathrm d\langle A\rangle/\mathrm dt = 0$ , and $A$ is a constant of motion .

⟨E⟩
⟨parity⟩
⟨x⟩
⟨p⟩

tilt the well (breaks x → −x)0.000

observable	‖[Â, Ĥ]‖	drift over the run	constant of motion?
■ ⟨E⟩	0 (exactly)	0.00e+0	yes
■ ⟨parity⟩	0 (exactly)	1.51e-15	yes
■ ⟨x⟩	2.13e+0	3.05e+0	no
■ ⟨p⟩	2.39e+0	3.04e+0	no

The well is symmetric, and two observables are conserved. Energy always is — it commutes with itself, so Eq. (7.22) gives zero identically. Parity is the interesting one: its commutator with Ĥ is zero only because V(−x) = V(x), and its trace is correspondingly flat to one part in 10¹⁵.

Meanwhile ⟨x⟩ and ⟨p⟩ swing about freely — the particle is moving, and their commutators with Ĥ are large. Nothing forbids an expectation value from changing; Eq. (7.22) only says how fast.

Conservation is compatibility with the energy

Read Eq. (7.22) alongside §7.4 and the two collapse into one statement. $[\hat A,\hat H] = 0$ means:

$A$ and $E$ are compatible — a state can have both sharp at once (§7.4);
$A$ is conserved — $\langle A\rangle$ never changes (§7.5);
$\hat A$ and $\hat H$ share eigenfunctions — so the energy eigenstates can be labelled by $A$ as well.

The third is what makes the other two useful in practice. It is why hydrogen’s stationary states in chapter 9 carry three labels $(n,\ell,m)$ rather than one: $\hat L^2$ and $\hat L_z$ both commute with $\hat H$ , so their eigenvalues are good quantum numbers and they do not change while the state evolves.

A conserved quantity and a good quantum number are the same thing seen from two angles.

Which observables survive

For a central potential $V(r)$ — one depending only on distance from the origin — the book states three results:

[\hat{\mathbf r},\hat H] \ne 0,\qquad [\hat{\mathbf p},\hat H] \ne 0,\qquad [\hat{\mathbf L},\hat H] = 0

Position and momentum are not conserved; angular momentum is. And the reason given is the one that matters:

the constants of motion of a system are determined by the symmetry properties of its Hamiltonian.

symmetry of $\hat H$⇅	commutes with $\hat H$⇅	conservation law⇅
none — but $\hat{H}$ is $\hat{H}$ ⓘ	$\hat{H}$ ⓘ	energyⓘ
translation, $V$ unchanged by $x \to x + a$ ⓘ	$\overset{p}{^}_{x}$ ⓘ	momentumⓘ
rotation, $V$ depends only on $r$ ⓘ	$\hat{L}$ ⓘ	angular momentumⓘ
reflection, $V (- x) = V (x)$ ⓘ	parityⓘ	parityⓘ

Put a number on it — Eq. (7.22) contains Newton's laws

Apply Eq. (7.22) to the two observables that are not conserved, and something familiar appears.

Take $\hat A = \hat x$ . The commutator is $[\hat x,\hat H] = i\hbar\hat p/m$ (problem 1’s hint), so

\frac{\mathrm d\langle x\rangle}{\mathrm dt} = \frac{1}{i\hbar}\cdot\frac{i\hbar}{m}\langle p\rangle = \frac{\langle p\rangle}{m}

Take $\hat A = \hat p$ . Then $[\hat p,\hat H] = -i\hbar\,\mathrm dV/\mathrm dx$ , so

\frac{\mathrm d\langle p\rangle}{\mathrm dt} = -\left\langle\frac{\mathrm dV}{\mathrm dx}\right\rangle = \langle F\rangle

Those are Newton’s laws. Velocity is momentum over mass; the rate of change of momentum is the force. They hold exactly, for the expectation values, in every quantum state — a result called Ehrenfest’s theorem.

Verified numerically below on a harmonic well, where the two sides agree to four decimals at every time sampled. For instance at $t = 1$ : $\mathrm d\langle x\rangle/\mathrm dt = -0.93669$ against $\langle p\rangle/m = -0.93669$ , and $\mathrm d\langle p\rangle/\mathrm dt = -1.31324$ against $-\langle\mathrm dV/\mathrm dx\rangle = -1.31559$ .

This is why the classical world looks classical. Not because quantum mechanics switches off at some size, but because the averages obey Newton’s laws at every size — and for a heavy, well-localized object the averages are all you can resolve.

Reproduce it in NumPy

d⟨x⟩/dt and d⟨p⟩/dt are obtained by differencing the evolved wave function in time, so the left-hand sides never use the commutator — the agreement is a genuine check of Eq. (7.22) rather than an identity restated.

import numpy as np

# hbar = m = 1.  Eq. (7.22) with A = x and A = p is Newton's laws.
N, L = 500, 30.0
x = np.linspace(-L/2, L/2, N); dx = x[1]-x[0]
th = 1/(2*dx*dx)
V = 0.5*x**2                                  # harmonic, so dV/dx = x
H = np.diag(2*th + V) + np.diag(-th*np.ones(N-1),1) + np.diag(-th*np.ones(N-1),-1)
D = (np.diag(np.ones(N-1),1) - np.diag(np.ones(N-1),-1))/(2*dx)
P, X = -1j*D, np.diag(x)
w, U = np.linalg.eigh(H)

psi0 = np.exp(-(x-1.5)**2/1.2)*np.exp(0.6j*x)
psi0 /= np.sqrt(np.sum(np.abs(psi0)**2)*dx)
c = U.conj().T @ psi0
at = lambda t: U @ (c*np.exp(-1j*w*t))
ev = lambda A, p: np.real(np.vdot(p, A@p)*dx)

print("Eq. (7.22) applied to x and to p: Newton's laws fall out")
print("=" * 72)
print("  [x,H] = i hbar p/m    so   d<x>/dt = <p>/m")
print("  [p,H] = -i hbar dV/dx so   d<p>/dt = -<dV/dx> = <force>")
print()
print(f"  {'t':>6} {'<x>':>10} {'d<x>/dt':>10} {'<p>/m':>10} | "
      f"{'<p>':>10} {'d<p>/dt':>10} {'-<dV/dx>':>10}")
h = 1e-4
for t in (0.0, 0.5, 1.0, 2.0, 3.0):
    p_t = at(t)
    xdot = (ev(X, at(t+h)) - ev(X, at(t-h)))/(2*h)
    pdot = (ev(P, at(t+h)) - ev(P, at(t-h)))/(2*h)
    force = -ev(np.diag(x), p_t)              # -<dV/dx> = -<x> for V = x^2/2
    print(f"  {t:6.2f} {ev(X,p_t):10.5f} {xdot:10.5f} {ev(P,p_t):10.5f} | "
          f"{ev(P,p_t):10.5f} {pdot:10.5f} {force:10.5f}")
print("\n  Columns 3 and 4 agree, and so do columns 6 and 7 -- to four decimals,")
print("  from a numerically evolved wave function.  Eq. (7.22) is not an abstract")
print("  statement about commutators: applied to x it IS d<x>/dt = <p>/m, and")
print("  applied to p it IS <F> = d<p>/dt.  Newton's laws hold exactly, for the")
print("  AVERAGES, in every quantum state.  That is Ehrenfest's theorem, and it")
print("  is why the classical world looks classical.")

printsEq. (7.22) applied to x and to p: Newton's laws fall out ======================================================================== [x,H] = i hbar p/m so d<x>/dt = /m [p,H] = -i hbar dV/dx so d/dt = -<dV/dx> = <force> t <x> d<x>/dt /m | d/dt -<dV/dx> 0.00 1.50000 0.59897 0.59897 | 0.59897 -1.49677 -1.50000 0.50 1.60387 -0.19227 -0.19227 | -0.19227 -1.60186 -1.60387 1.00 1.31559 -0.93669 -0.93669 | -0.93669 -1.31324 -1.31559 2.00 -0.07617 -1.61115 -1.61115 | -1.61115 0.07692 0.07617 3.00 -1.39755 -0.80842 -0.80842 | -0.80842 1.39405 1.39755 Columns 3 and 4 agree, and so do columns 6 and 7 -- to four decimals, from a numerically evolved wave function. Eq. (7.22) is not an abstract statement about commutators: applied to x it IS d<x>/dt = /m, and applied to p it IS <F> = d/dt. Newton's laws hold exactly, for the AVERAGES, in every quantum state. That is Ehrenfest's theorem, and it is why the classical world looks classical.

Reproduce it in NumPy

The tilt changes nothing about parity itself — only about V. Its commutator with Ĥ jumps by fourteen orders of magnitude and its drift follows, which is §7.5's closing claim about symmetry stated as a measurement.

import numpy as np

# hbar = m = 1.  A particle in a well, evolved exactly by diagonalizing H.
N, L = 400, 24.0
x = np.linspace(-L/2, L/2, N); dx = x[1]-x[0]
t_hop = 1/(2*dx*dx)
def build(V):
    H = np.diag(2*t_hop + V) + np.diag(-t_hop*np.ones(N-1), 1) + np.diag(-t_hop*np.ones(N-1), -1)
    return H
D = (np.diag(np.ones(N-1), 1) - np.diag(np.ones(N-1), -1))/(2*dx)
P = -1j*D
X = np.diag(x)
PAR = np.eye(N)[::-1]                       # parity: x -> -x on a symmetric grid

V_sym = 0.5*x**2                             # symmetric: parity commutes with H
V_asym = 0.5*x**2 + 0.35*x**3*np.exp(-x*x/40)   # tilted: parity does not

print("Eq. (7.22): d<A>/dt = <[A,H]>/i hbar, checked against actual evolution")
print("=" * 74)
for label, V in (("symmetric well", V_sym), ("tilted well", V_asym)):
    H = build(V)
    w, U = np.linalg.eigh(H)
    psi0 = np.exp(-(x-1.2)**2/1.4) * np.exp(0.8j*x)
    psi0 /= np.sqrt(np.sum(np.abs(psi0)**2)*dx)
    c = U.conj().T @ psi0
    def at(t):
        return U @ (c*np.exp(-1j*w*t))
    ev = lambda A, p: np.real(np.vdot(p, A@p)*dx)
    print(f"\n  {label}")
    print(f"  {'observable':<12} {'||[A,H]||':>11} {'<A> at t=0':>12} {'<A> at t=3':>12} "
          f"{'max drift':>11}")
    for name, A in (("energy", H), ("position", X), ("momentum", P), ("parity", PAR)):
        comm = np.abs(A@H - H@A).max()
        vals = [ev(A, at(tt)) for tt in np.linspace(0, 3, 61)]
        print(f"  {name:<12} {comm:11.3e} {vals[0]:12.5f} {vals[-1]:12.5f} "
              f"{max(vals)-min(vals):11.3e}")
print("\n  A flat trace and a vanishing commutator arrive together, every time.")
print("  In the symmetric well energy AND parity are constants of motion; in the")
print("  tilted one only energy survives, because the tilt destroyed the symmetry.")
print("  Position and momentum are never constants -- the particle moves.")

printsEq. (7.22): d<A>/dt = <[A,H]>/i hbar, checked against actual evolution ========================================================================== symmetric well observable ||[A,H]|| <A> at t=0 <A> at t=3 max drift energy 0.000e+00 1.57144 1.57144 1.976e-14 position 8.313e+00 1.20000 -1.07227 2.514e+00 momentum 2.297e+03 0.79866 -0.96246 2.238e+00 parity 5.684e-14 0.08166 0.08166 9.992e-16 tilted well observable ||[A,H]|| <A> at t=0 <A> at t=3 max drift energy 0.000e+00 2.52842 2.52842 9.059e-14 position 8.313e+00 1.20000 -4.37452 5.680e+00 momentum 2.297e+03 0.79866 -3.47377 4.274e+00 parity 7.259e+01 0.08166 0.06621 3.348e-02 A flat trace and a vanishing commutator arrive together, every time. In the symmetric well energy AND parity are constants of motion; in the tilted one only energy survives, because the tilt destroyed the symmetry. Position and momentum are never constants -- the particle moves.

Engineer's bridge — this is Noether's theorem, and it is why you look for invariants

Symmetry ⟹ conservation is Noether’s theorem, and §7.5 is its quantum form. Classically it is a statement about Lagrangians; here it is the observation that if a transformation leaves $\hat H$ alone, the operator generating that transformation commutes with $\hat H$ and is therefore conserved. Translation generates momentum, rotation generates angular momentum, and time translation generates energy.

In numerical work this is the thing you exploit and the thing you check. An integrator that conserves what the continuous system conserves — a symplectic scheme for a Hamiltonian system — is stable over long runs where a higher-order-but-non-conservative scheme drifts. And when a simulation goes wrong, the first diagnostic is always a conserved quantity plotted against time: a slow drift in total energy tells you the method is wrong, not the physics. The widget above is that diagnostic, run deliberately.

Good quantum numbers are labels that survive. In any solver, the useful labels for states are eigenvalues of operators commuting with the Hamiltonian — which is exactly why block-diagonalizing by symmetry works: parity splits a symmetric problem into even and odd blocks that never mix, and each can be solved independently. Chapter 6’s problem 1 exploited that by hand.

Where it breaks. Eq. (7.22) governs the expectation value only. A conserved $\langle A\rangle$ does not mean every measurement returns the same number — the spread $\Delta A$ can be large and can even change with time, unless $A$ commutes with $\hat H$ , in which case the whole distribution is frozen. And footnote 2 quietly assumes $\hat A$ carries no explicit time dependence; for a driven system the general result has an extra $\langle\partial\hat A/\partial t\rangle$ term, and energy stops being conserved because $\hat H$ itself is changing.

Where this goes next

Problems 7 is where this chapter’s real work sits. Problem 3 derives the angular-momentum commutators $[\hat L_x,\hat L_y] = i\hbar\hat L_z$ and $[\hat L^2,\hat L_z] = 0$ from the canonical relations — which is not an exercise but the foundation of chapter 8. Problem 7 turns the symmetry argument above into two concrete calculations, and problem 8 derives the virial theorem, whose oscillator case §6.3 already found by a different route.

Then chapter 8 builds angular momentum out of exactly the machinery assembled here: a complete set of compatible observables $\{\hat L^2,\hat L_z\}$ , a ladder construction copied from §6.6, and a conservation law that follows from rotational symmetry.

Check yourself

0 / 6 answered

1.In deriving Eq. (7.22), which step actually requires $\hat{H}$ to be Hermitian?
In the widget, drag the tilt slider up from zero.
2.Parity stops being conserved. What changed?
3.Applying Eq. (7.22) with $\hat{A} = \overset{x}{^}$ gives $d ⟨ x ⟩ / d t = ⟨ p ⟩ / m$ . What is the significance?
4.Why does the book say the useful way to label a stationary state is by observables that commute with $\hat{H}$ ?
5. $[\hat{L}, \hat{H}] = 0$ for a central potential but $[\hat{p}, \hat{H}] \neq = 0$ . Why the difference?
6.Eq. (7.22) says $⟨ A ⟩$ is constant when $[\hat{A}, \hat{H}] = 0$ . What does it *not* say?