States of Certain Energy · Quantum Mechanics (Phillips)

A complex time factor turns a standing wave into something that rotates without moving. That single change of the string’s cosine into an exponential is where the chapter turns.

§4.2 solved a vibrating string. This section runs the identical procedure on the Schrödinger equation — same separation, same eigenvalue problem, same superposition waiting at the end.

One thing changes, and everything in the chapter follows from it: the Schrödinger equation is first order in time where the wave equation was second. That single difference turns $\cos\omega t$ into $e^{-iEt/\hbar}$ , and a wave that flexes into one that merely rotates.

Separating the Schrödinger equation

The equation, from Eq. (2.17), is

i\hbar\frac{\partial\Psi}{\partial t} = \left[-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf r)\right]\Psi\tag{4.17}

and we look for the same kind of solution as before — a fixed spatial shape with a common time factor:

\Psi(\mathbf r, t) = \psi(\mathbf r)\,T(t)\tag{4.18}

Substituting and separating gives

\frac{i\hbar}{T}\frac{\mathrm dT}{\mathrm dt} = \frac{1}{\psi}\left[-\frac{\hbar^2}{2m}\nabla^2\psi + V(\mathbf r)\psi\right]\tag{4.19}

The argument is word for word §4.2’s: a function of $t$ alone equals a function of $\mathbf r$ alone for all $\mathbf r$ and $t$ , so both must be the same constant. Phillips calls it $E$ and remarks, dryly, that some readers may have guessed its meaning.

The time factor — where the chapter turns

\frac{\mathrm dT}{\mathrm dt} = \left(\frac{-iE}{\hbar}\right)T\tag{4.20}

T(t) = A\,e^{-iEt/\hbar}\tag{4.21}

Compare this with the string, because it is the entire point

The string’s time factor was $T(t) = A\cos\omega t + B\sin\omega t$ — a real function that passes through zero twice a cycle. That is why the string in §4.2’s widget visibly flexes: every point’s displacement shrinks to nothing and grows back.

The quantum time factor is $e^{-iEt/\hbar}$ , and $|e^{-iEt/\hbar}| = 1$ exactly, for all $t$ . It never shrinks. It never passes through zero. All it does is rotate in the complex plane.

The difference traces to one structural fact: the wave equation is second order in time, giving $\mathrm d^2T/\mathrm dt^2 = -\omega^2T$ and real oscillation; the Schrödinger equation is first order, giving $\mathrm dT/\mathrm dt \propto -iT$ and pure rotation. §2.1 showed that first-order-in-time is forced if plane waves are to satisfy $E = p^2/2m$ — so the rotation is not a modelling choice either.

The energy eigenvalue equation

The spatial half is

\left[-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf r)\right]\psi(\mathbf r) = E\psi(\mathbf r)\tag{4.22}

which, using $\hat H$ , is just

\hat H\psi(\mathbf r) = E\psi(\mathbf r)\tag{4.23}

This is the energy eigenvalue equation, $\psi$ is the eigenfunction of $\hat H$ belonging to the eigenvalue $E$ , and — the name you will see everywhere else — it is also called the time-independent Schrödinger equation. In practice there are many eigenvalues and many eigenfunctions.

Combining the eigenfunction with the time factor gives a special solution of the Schrödinger equation:

\Psi(\mathbf r, t) = \psi(\mathbf r)\,e^{-iEt/\hbar}\tag{4.24}

Engineer's bridge — you have already solved this numerically

Equation (4.23) is a matrix eigenvalue problem wearing a hat.

§0.3 built $\hat H$ as an explicit matrix — the tridiagonal $[1, -2, 1]/\delta x^2$ stencil for $\nabla^2$ , plus $V$ on the diagonal — and handed it to an eigensolver. Everything on this page is the continuous version of that:

Eq. (4.23)	`qm.ts`
$\hat H$	a symmetric tridiagonal matrix
$\psi(\mathbf r)$	an eigenvector, one component per grid point
$E$	its eigenvalue
“there are many eigenvalues”	an $N\times N$ matrix has $N$ of them
$\hat H\psi = E\psi$	`eigh(T, k, dx)`

PotentialExplorer has been solving this equation live since chapter 0. What §4.3 supplies is the meaning: the eigenvector is a state, and the eigenvalue is the energy you will measure.

Where it breaks. A finite matrix has finitely many eigenvalues and a bounded spectrum; the real $\hat H$ has infinitely many, and for some potentials a continuum as well — scattering states, which cannot be normalized and so are not eigenvectors in any ordinary sense. Footnote 1 of §4.5 flags this and defers it to chapter 7, where Dirac delta normalization handles it. A grid quietly hides the whole issue, which is a good reason to distrust numerics as proof even while trusting them as intuition.

Proving ΔE = 0

Phillips now proves the chapter’s opening claim, and the proof is short enough to hold in your head.

The energy uncertainty is defined exactly as position and momentum were in Eq. (3.33):

\Delta E = \sqrt{\langle E^2\rangle - \langle E\rangle^2}\tag{4.25}

with both expectation values given by sandwich integrals — §3.5’s recipe with $\hat H$ as the filling:

\langle E\rangle = \int\Psi^*(\mathbf r,t)\,\hat H\,\Psi(\mathbf r,t)\,\mathrm d^3\mathbf r\tag{4.26}

\langle E^2\rangle = \int\Psi^*(\mathbf r,t)\,\hat H^2\,\Psi(\mathbf r,t)\,\mathrm d^3\mathbf r\tag{4.27}

step 1 of 4

Four lines. Notice that the time factor is never used — it cancels immediately, which is itself the reason the result is so clean.

1Ψ(r,t) is ψ(r) times a constant-in-space factor, so applying Ĥ (which only differentiates in SPACE) passes straight through it. Ψ is therefore an eigenfunction of Ĥ too, with the same eigenvalue.
$\hat{H} Ψ (r, t) = E Ψ (r, t)$
Equation (4.28).
2Put that into Eq. (4.26). E is a number, so it comes straight out of the integral — and what remains is the normalization integral, which is 1.
$⟨ E ⟩ = \int Ψ^{*} \hat{H} Ψ d^{3} r = E \int Ψ^{*} Ψ d^{3} r = E$
3Apply Ĥ twice: the first application gives EΨ, and the second gives E times EΨ. So Ψ is an eigenfunction of Ĥ² with eigenvalue E².
$\hat{H}^{2} Ψ (r, t) = E^{2} Ψ (r, t)$
Equation (4.29).
4The same argument on Eq. (4.27) gives ⟨E²⟩ = E². Substitute both into Eq. (4.25).
$Δ E = E^{2} - E^{2} = 0$
Not small. Not approximately zero. Exactly zero — an energy measurement on this state is CERTAIN to return E.

An eigenfunction of the Hamiltonian always describes a state of definite energy.

Why this is stronger than it looks

Compare with position and momentum. For every state in chapter 3, $\Delta x$ and $\Delta p$ were both non-zero — the Gaussian minimised their product at $\hbar/2$ but neither factor ever vanished. Uncertainty seemed unavoidable.

Energy is different. There exist states — infinitely many of them — for which the energy is perfectly certain. Measure it a thousand times and get $E$ a thousand times.

The reason is that these states are constructed as eigenfunctions of the very operator being measured. That is the general pattern, and chapter 7 makes it a theorem: an observable has a definite value in exactly those states that are eigenfunctions of its operator. Position and momentum have no normalizable eigenfunctions, which is precisely why $\Delta x$ and $\Delta p$ can never be zero for a real state.

So “certain energy” is not a special indulgence. It is what happens whenever a state is an eigenfunction of the thing you are measuring — and for $\hat H$ , unlike $\hat x$ and $\hat p$ , such states are honest, normalizable wave functions.

A state that rotates without moving

Put a number on it — how fast does an electron's phase rotate?

The rotation rate is $\omega = E/\hbar$ . For the ground state of a 1 nm box, §3.4 computed $E_1 = 0.376$ eV, so

\omega = \frac{E_1}{\hbar} = \frac{0.376 \times 1.602\times10^{-19}}{1.055\times10^{-34}} \approx 5.7\times10^{14}\ \mathrm{s^{-1}}

a period of about 11 femtoseconds. The wave function completes roughly 90 trillion rotations per second.

And nothing observable happens during any of them. Every measurable quantity is unchanged from one rotation to the next, and would be unchanged after a billion years. This is the sharpest illustration of why $\Psi$ is not a physical wave: something is racing round the complex plane at $10^{14}$ Hz, and no experiment can detect it.

What can be detected is a difference of two such rotations — and that is exactly what §4.6 is about.

Reproduce it in NumPy

Three claims, tested rather than restated: that Ĥψ/ψ is a constant only for an eigenfunction, that ΔE is exactly zero for one and non-zero for a mixture, and that the time factor leaves every observable untouched. Natural units ħ = m = a = 1.

import numpy as np

hbar = m = a = 1.0
N = 4000
x = np.linspace(0, a, N)
dx = x[1] - x[0]

def H(psi):                      # -hbar^2/2m d2/dx2, with V = 0 inside the box
    d2 = np.zeros_like(psi)
    d2[1:-1] = (psi[2:] - 2*psi[1:-1] + psi[:-2])/dx**2
    return -hbar**2/(2*m)*d2

box = lambda n: np.sqrt(2/a)*np.sin(n*np.pi*x/a)
E_exact = lambda n: n**2*np.pi**2*hbar**2/(2*m*a**2)

def ratio(psi):
    """H psi / psi, sampled only where psi is safely away from its nodes --
    at a node the ratio is 0/0 and any number the grid returns is noise."""
    keep = np.abs(psi) > 0.1*np.abs(psi).max()
    keep[:2] = keep[-2:] = False          # the stencil is one-sided at the walls
    return H(psi)[keep]/psi[keep]         # index BEFORE dividing, so no 0/0 arises

print("an eigenfunction is the function H gives back unchanged")
print("=" * 62)
print("  H psi / psi, sampled away from the nodes:")
for n in (1, 2, 3):
    r = ratio(box(n))
    print(f"    n={n}:  min {r.min():9.4f}   max {r.max():9.4f}   "
          f"spread {r.max()-r.min():.1e}   E_n = {E_exact(n):.4f}")
mix = (box(1) + box(2))/np.sqrt(2)
r = ratio(mix)
print(f"    mix:   min {r.min():9.4f}   max {r.max():9.4f}   "
      f"spread {r.max()-r.min():.1e}   <- NOT an eigenfunction")

print("\nDelta E = 0 for an eigenstate, by the sandwich integrals (4.26) and (4.27)")
def stats(psi):
    psi = psi/np.sqrt(np.trapezoid(psi**2, x))
    E1 = np.trapezoid(psi*H(psi), x)
    E2 = np.trapezoid(psi*H(H(psi)), x)
    return E1, E2, np.sqrt(max(0.0, E2 - E1**2))
print(f"  {'state':16} {'<E>':>10} {'<E^2>':>12} {'<E>^2':>12} {'Delta E':>11}")
for n in (1, 2, 3):
    E1, E2, dE = stats(box(n))
    print(f"  {'box n='+str(n):16} {E1:10.4f} {E2:12.4f} {E1**2:12.4f} {dE:11.2e}")
E1, E2, dE = stats(mix)
print(f"  {'equal mix 1+2':16} {E1:10.4f} {E2:12.4f} {E1**2:12.4f} {dE:11.4f}")
print(f"  the mix should give <E> = (E1+E2)/2 = {(E_exact(1)+E_exact(2))/2:.4f}"
      f" and Delta E = |E2-E1|/2 = {abs(E_exact(2)-E_exact(1))/2:.4f}")

print("\nthe time factor has modulus one, so |Psi|^2 never moves")
psi = box(2)
for t in (0.0, 0.31, 1.7, 12.0):
    Psi = psi*np.exp(-1j*E_exact(2)*t/hbar)
    print(f"  t = {t:5.2f}:  max|Psi|^2 = {np.abs(Psi).max()**2:.6f}   "
          f"integral |Psi|^2 = {np.trapezoid(np.abs(Psi)**2, x):.6f}   "
          f"arg Psi at peak = {np.angle(Psi[np.argmax(np.abs(psi))]):+.4f} rad")
print("  the phase advances; nothing observable does.")

printsan eigenfunction is the function H gives back unchanged ============================================================== H psi / psi, sampled away from the nodes: n=1: min 4.9348 max 4.9348 spread 1.0e-07 E_n = 4.9348 n=2: min 19.7392 max 19.7392 spread 2.5e-07 E_n = 19.7392 n=3: min 44.4132 max 44.4132 spread 5.1e-07 E_n = 44.4132 mix: min -56.8571 max 86.1359 spread 1.4e+02 <- NOT an eigenfunction Delta E = 0 for an eigenstate, by the sandwich integrals (4.26) and (4.27) state <E> <E^2> <E>^2 Delta E box n=1 4.9348 24.3523 24.3523 0.00e+00 box n=2 19.7392 389.6362 389.6362 0.00e+00 box n=3 44.4132 1972.5323 1972.5323 3.75e-06 equal mix 1+2 12.3370 206.9942 152.2017 7.4022 the mix should give <E> = (E1+E2)/2 = 12.3370 and Delta E = |E2-E1|/2 = 7.4022 the time factor has modulus one, so |Psi|^2 never moves t = 0.00: max|Psi|^2 = 2.000000 integral |Psi|^2 = 1.000000 arg Psi at peak = +0.0000 rad t = 0.31: max|Psi|^2 = 2.000000 integral |Psi|^2 = 1.000000 arg Psi at peak = +0.1640 rad t = 1.70: max|Psi|^2 = 2.000000 integral |Psi|^2 = 1.000000 arg Psi at peak = -2.1407 rad t = 12.00: max|Psi|^2 = 2.000000 integral |Psi|^2 = 1.000000 arg Psi at peak = +1.8905 rad the phase advances; nothing observable does.

Where this leaves us

We have a recipe for states of definite energy: solve $\hat H\psi = E\psi$ , attach $e^{-iEt/\hbar}$ , and the result has $\Delta E = 0$ with no observable time dependence at all.

What we have not done is solve that equation for any actual potential. §4.4 does it for the box — and because the eigenvalue problem turns out to be identical to the string’s from §4.2, the answer is already sitting in Fig. 4.1. What is new is what the eigenvalues mean: not frequencies now, but energies, and discrete ones.

Check yourself

0 / 7 answered

1.The string's time factor was $cos ω t$ ; the quantum one is $e^{- i E t /ℏ}$ . What structural fact causes the difference?
2.What makes $ψ$ an eigenfunction of $\hat{H}$ rather than just some function?
3.In the $Δ E = 0$ proof, the time factor is never used. Why not?
4. $Δ x$ and $Δ p$ are never zero for a real state, yet $Δ E$ can be exactly zero. What explains the asymmetry?
5.An electron in a 1 nm box has $ω = E_{1} /ℏ \approx 5.7 \times 1 0^{14} s^{- 1}$ — about 90 trillion rotations a second. What is observable about that?
6.How does Eq. (4.23) relate to what `PotentialExplorer` has been doing since chapter 0?
7.Where does the matrix picture of $\hat{H}$ genuinely mislead?