Stationary States & Superposition

4 hours ~9 min read

Stationary States & Superposition

The time-dependent Schrödinger equation looks fearsome — a complex PDE — but for time-independent potentials it surrenders to the oldest trick in mathematical physics: separation of variables. The reward is the time-independent equation, whose solutions — stationary states — are the atoms of quantum dynamics. Individually they do nothing at all, like a cat at 2 p.m.; superpose two of them, and the relative phase makes everything move.

Learning Objectives

After this lesson you will be able to:

  1. Carry out separation of variables on the Schrödinger equation and obtain the time-independent Schrödinger equation H^ψ=Eψ\hat H\psi = E\psi.
  2. Prove that the separation constant EE must be real and must exceed VminV_{\min}.
  3. Derive the three defining properties of stationary states: frozen Ψ2|\Psi|^2, constant expectation values, and definite energy.
  4. Expand an initial state in energy eigenstates via orthonormality and Fourier's trick, interpreting cn2|c_n|^2 as energy-measurement probabilities.
  5. Derive the two-state sloshing of the infinite square well and explain why all quantum dynamics is evolving relative phase.

Intuition

Faced with itΨ=H^Ψi\hbar\,\partial_t\Psi = \hat H\Psi, ask the physicist's first question: are there solutions with trivial time dependence? Yes — products ψ(x)eiEt/\psi(x)\,e^{-iEt/\hbar}, whose time dependence is a pure phase. The Born rule sees only Ψ2|\Psi|^2, so the phase cancels: the probability density is frozen. These are the stationary states, one per allowed energy.

That sounds like a dead end — a theory of motion whose basic solutions never move. The escape is linearity: the general solution is a superposition of stationary states, each spinning its phase at its own rate En/E_n/\hbar. In a superposition the phases no longer cancel in Ψ2|\Psi|^2; they beat against each other, and the beats are the dynamics. An electron sloshing in a well, a vibrating bond, a qubit mid-gate — all of it is relative phase between energy eigenstates. The strategy for every problem in Courses P.5–P.6 follows: solve the time-independent equation once for the catalog {ψn,En}\{\psi_n, E_n\}, then assemble any motion from them.


Theory

Separation of variables

Let V=V(x)V = V(x) and try a product solution Ψ(x,t)=ψ(x)φ(t)\Psi(x,t) = \psi(x)\,\varphi(t). Insert into itΨ=22mx2Ψ+VΨi\hbar\,\partial_t\Psi = -\frac{\hbar^2}{2m}\partial_x^2\Psi + V\Psi and divide by ψφ\psi\varphi:

i1φdφdt=22m1ψd2ψdx2+V(x).i\hbar\,\frac{1}{\varphi}\frac{d\varphi}{dt} = -\frac{\hbar^2}{2m}\frac{1}{\psi}\frac{d^2\psi}{dx^2} + V(x).

The left side depends only on tt, the right only on xx; a function of tt can equal a function of xx only if both equal the same constant, EE (the name is not innocent). The PDE splits into two ODEs.

Time part. iφ˙=Eφi\hbar\,\dot\varphi = E\varphi, so

φ(t)=eiEt/\varphi(t) = e^{-iEt/\hbar}

(the integration constant is absorbed into ψ\psi): every separable solution wears the same uniform, a phase rotating at ω=E/\omega = E/\hbar — Planck–Einstein again, now as an output.

Space part. The remainder is the time-independent Schrödinger equation (TISE):

  H^ψ=Eψ,H^=22md2dx2+V(x)  \boxed{\;\hat H\psi = E\psi, \qquad \hat H = -\frac{\hbar^2}{2m}\frac{d^2}{dx^2} + V(x)\;}

— an eigenvalue problem for the Hamiltonian H^\hat H: allowed separation constants are eigenvalues, stationary profiles are eigenfunctions. For bound problems, the admissibility conditions of P.4.1 (normalizability, continuity) select a discrete set {ψn,En}\{\psi_n, E_n\}: quantization emerges from a boundary-value problem, not from an extra postulate.

EE must be real, and E>VminE > V_{\min}

Real. If E=E0+iΓE = E_0 + i\Gamma with Γ0\Gamma \neq 0, then φ(t)2=e2Γt/|\varphi(t)|^2 = e^{2\Gamma t/\hbar} and Ψ2dx=e2Γt/ψ2dx\int|\Psi|^2dx = e^{2\Gamma t/\hbar}\int|\psi|^2dx grows or decays — contradicting conservation of probability (P.4.1). Normalizability for all time forces ERE \in \mathbb{R}. (Term 1 language: H^\hat H is Hermitian, so its eigenvalues are real — 0.1.5.)

E>VminE > V_{\min}. Rearrange the TISE:

d2ψdx2=2m2(V(x)E)ψ.\frac{d^2\psi}{dx^2} = \frac{2m}{\hbar^2}\big(V(x) - E\big)\,\psi.

If E<VminV(x)E < V_{\min} \le V(x) everywhere, then VE>0V - E > 0 everywhere, so ψ\psi'' has the same sign as ψ\psi at every point: where ψ>0\psi>0 it curves up, away from the axis; where ψ<0\psi<0 it curves down, also away. Such a function bends away from zero and cannot decay to it at both ends — it blows up at ±\pm\infty and is never normalizable. Physically: T0\langle T\rangle \ge 0, so you cannot have less energy than the least potential energy. (EE may still be negative if Vmin<0V_{\min}<0 — the bound states of wells, P.5.2.)

Properties of stationary states

Let Ψn=ψn(x)eiEnt/\Psi_n = \psi_n(x)e^{-iE_nt/\hbar}, ψn\psi_n normalized. Three properties, one line each:

1. Frozen density. Ψn2=ψne+iEnt/ψneiEnt/=ψn2|\Psi_n|^2 = \psi_n^*e^{+iE_nt/\hbar}\psi_ne^{-iE_nt/\hbar} = |\psi_n|^2 — time-independent; hence stationary.

2. Every expectation value constant. For any Q^=Q(x^,p^)\hat Q = Q(\hat x,\hat p) with no explicit tt, the phase passes through Q^\hat Q and cancels:

Q=ψne+iEnt/Q^ψneiEnt/dx=ψnQ^ψndx.\langle Q\rangle = \int \psi_n^*e^{+iE_nt/\hbar}\,\hat Q\,\psi_ne^{-iE_nt/\hbar}dx = \int\psi_n^*\hat Q\,\psi_n\,dx.

Nothing measurable ever changes.

3. Definite energy. Using H^ψn=Enψn\hat H\psi_n = E_n\psi_n: H=ψnH^ψndx=En\langle H\rangle = \int\psi_n^*\hat H\psi_ndx = E_n, and H2=ψnH^(H^ψn)dx=En2\langle H^2\rangle = \int\psi_n^*\hat H(\hat H\psi_n)dx = E_n^2, so

σH2=H2H2=0:\sigma_H^2 = \langle H^2\rangle - \langle H\rangle^2 = 0:

an energy measurement returns EnE_n with certainty. Stationary states are states of definite energy.

Caution. A single stationary state predicts nothing moving. Not "slowly" — every probability and every expectation value of every observable is constant. If your system visibly does anything, it is not in a stationary state. Motion — including every interference trick quantum computing plays — lives in superpositions and their evolving relative phases. Stationary states are the alphabet; dynamics is the poetry.

Vocabulary: eigensolutions and degeneracy

Fix the words now, because the rest of the term uses them constantly. A problem whose solutions exist only for a specific set of allowed values of some parameter — here the energy — is an eigenvalue problem; the allowed values are the eigenvalues, and the particular function attached to each is its eigenfunction. Together they form an eigensolution. Since our parameter is an energy, the EnE_n are eigenenergies and the ψn\psi_n are energy eigenfunctions. "Stationary state", "energy eigenstate", and "eigensolution of H^\hat H" all name the same object.

One eigenvalue need not have only one eigenfunction. When two or more linearly independent eigenfunctions share a single eigenvalue, the eigenvalue is called degenerate, and the number of independent eigenfunctions belonging to it is its degeneracy. Nothing forbids it, and symmetry tends to cause it: the 1D problems of Course P.5 are all non-degenerate (a second-order ODE with two decaying-tail boundary conditions has one solution per energy), but the moment a problem gains a symmetry direction, degeneracy appears — a particle in a cube has Enx2+ny2+nz2E \propto n_x^2 + n_y^2 + n_z^2, so (2,1,1)(2,1,1), (1,2,1)(1,2,1) and (1,1,2)(1,1,2) share one energy, a three-fold degeneracy. The hydrogen atom's n2n^2-fold degeneracy in \ell and mm (P.6.2) is the headline example, and lifting degeneracies with fields is how spectroscopy reads atoms apart (P.6.3).

Degeneracy matters practically for the expansion below: within a degenerate set, any linear combination is still an eigenfunction of the same energy, so the basis is not unique — you may (and should) choose combinations adapted to another symmetry, exactly the freedom exploited when a degenerate subspace is diagonalized in 0.1.5.

The general solution: superposition of stationary states

Linearity lets us add separable solutions, and the spectral theorem guarantees every solution arises this way:

Ψ(x,t)=ncnψn(x)eiEnt/,cnC.\Psi(x,t) = \sum_n c_n\,\psi_n(x)\,e^{-iE_nt/\hbar}, \qquad c_n \in \mathbb{C}.

This is the master formula of wave mechanics: solve the TISE once, then evolution is "attach the phase wiggles."

Orthonormality and completeness. The eigenfunctions can be chosen with

ψmψndx=δmn,\int_{-\infty}^{\infty}\psi_m^*\psi_n\,dx = \delta_{mn},

and they are complete: any admissible f(x)f(x) expands as f=ncnψnf = \sum_nc_n\psi_n. This is the spectral theorem of 0.1.5 Eigenvalues & the Spectral Theorem promoted to infinite dimensions: H^=nEnψnψn\hat H = \sum_n E_n\lvert\psi_n\rangle\langle\psi_n\rvert.

Fourier's trick. Multiply Ψ(x,0)=ncnψn\Psi(x,0) = \sum_nc_n\psi_n by ψm\psi_m^* and integrate; orthonormality collapses the sum:

ψmΨ(x,0)dx=ncnδmn=cmcn=ψn(x)Ψ(x,0)dx.\int\psi_m^*\,\Psi(x,0)\,dx = \sum_n c_n\,\delta_{mn} = c_m \quad\Longrightarrow\quad c_n = \int_{-\infty}^{\infty}\psi_n^*(x)\,\Psi(x,0)\,dx.

Exactly the Fourier-coefficient recipe of 0.3.2 — a Fourier series is an eigenfunction expansion (for the infinite well, literally a sine series).

What the cnc_n mean. Insert the expansion into Ψ2dx\int|\Psi|^2dx and H\langle H\rangle; the cross terms cmcnei(EmEn)t/ψmψndxc_m^*c_ne^{i(E_m-E_n)t/\hbar}\int\psi_m^*\psi_ndx die by orthonormality:

1=ncn2,H=ncn2En,1 = \sum_n|c_n|^2, \qquad \langle H\rangle = \sum_n|c_n|^2E_n,

both time-independent (energy is conserved). The reading, confirmed by 1.1.2: cn2|c_n|^2 is the probability that an energy measurement returns EnE_n — the Born rule in the energy basis, with ncn2=1\sum_n|c_n|^2 = 1 its consistency condition.

Dynamics from interference: the sloshing well

To watch phases do something, take the cleanest arena: the infinite square well on [0,L][0,L], whose stationary states (derived in full in P.5.2) are

ψn(x)=2LsinnπxL,En=n2π222mL2,n=1,2,3,\psi_n(x) = \sqrt{\frac{2}{L}}\,\sin\frac{n\pi x}{L}, \qquad E_n = \frac{n^2\pi^2\hbar^2}{2mL^2}, \qquad n = 1,2,3,\dots

Prepare Ψ=12(ψ1eiE1t/+ψ2eiE2t/)\Psi = \frac{1}{\sqrt2}\big(\psi_1e^{-iE_1t/\hbar} + \psi_2e^{-iE_2t/\hbar}\big) and compute the density (the ψn\psi_n are real):

Ψ2=12(ψ12+ψ22+ψ1ψ2(ei(E1E2)t/+ei(E1E2)t/)),|\Psi|^2 = \frac12\Big(\psi_1^2 + \psi_2^2 + \psi_1\psi_2\big(e^{i(E_1-E_2)t/\hbar} + e^{-i(E_1-E_2)t/\hbar}\big)\Big),

and since the bracket is 2cos((E2E1)t/)2\cos\big((E_2-E_1)t/\hbar\big),

  Ψ(x,t)2=12(ψ12+ψ22)+ψ1(x)ψ2(x)cos(ω21t),ω21E2E1  \boxed{\;|\Psi(x,t)|^2 = \frac12\big(\psi_1^2 + \psi_2^2\big) + \psi_1(x)\psi_2(x)\cos(\omega_{21}t), \qquad \omega_{21} \equiv \frac{E_2 - E_1}{\hbar}\;}

The cross term — pure interference — oscillates at the Bohr frequency ω21\omega_{21}. Since ψ1ψ2>0\psi_1\psi_2 > 0 on the left half of the well and <0< 0 on the right, the probability lump sloshes left–right with period T=2π/(E2E1)T = 2\pi\hbar/(E_2-E_1). Note what mattered: not the absolute phases (a global eiγe^{i\gamma} changes nothing) but the relative phase (E2E1)t/(E_2-E_1)t/\hbar. All quantum dynamics is relative phase in a superposition. This is also the frequency an atom radiates: ω21=E2E1\hbar\omega_{21} = E_2 - E_1, the Bohr condition of P.2.2.

Connections. The phases assemble into the evolution operator U(t)=eiH^t/=neiEnt/ψnψnU(t) = e^{-i\hat Ht/\hbar} = \sum_ne^{-iE_nt/\hbar}\lvert\psi_n\rangle\langle\psi_n\rvert — exactly the unitary of 1.1.3 The Evolution Postulate; what we found by separating variables, Term 1 postulates directly. And a stationary state's eiEnt/e^{-iE_nt/\hbar} alone is a global phase — unobservable; between two components it is a relative phase — the whole show: precisely the distinction of 1.1.1, now with a clock attached.


Worked Examples

Example 1 — Energy statistics of a three-term superposition

A particle in the infinite well is prepared in Ψ(x,0)=16ψ1+i3ψ2+12ψ3\Psi(x,0) = \frac{1}{\sqrt6}\psi_1 + \frac{i}{\sqrt3}\psi_2 + \frac{1}{\sqrt2}\psi_3. Check normalization; find the energy-measurement outcomes, probabilities, and H\langle H\rangle.

cn2=16+13+12=1\sum|c_n|^2 = \frac16 + \frac13 + \frac12 = 1. ✓ Outcomes are the eigenvalues present: P(E1)=16P(E_1) = \frac16, P(E2)=13P(E_2) = \frac13, P(E3)=12P(E_3) = \frac12 (the ii in c2c_2 never enters — only c22|c_2|^2 does). With En=n2E1E_n = n^2E_1:

H=16E1+13(4E1)+12(9E1)=6E1.\langle H\rangle = \tfrac16E_1 + \tfrac13(4E_1) + \tfrac12(9E_1) = 6\,E_1.

Note 6E16E_1 is not an eigenvalue: the average need not be a possible outcome. And all these numbers are constants of the motion — measure at any tt, same statistics.

Example 2 — The sloshing amplitude x(t)\langle x\rangle(t)

For the two-state superposition above, compute x(t)\langle x\rangle(t).

x=120Lx(ψ12+ψ22)dx+cos(ω21t)0Lxψ1ψ2dx.\langle x\rangle = \frac12\int_0^L x\big(\psi_1^2 + \psi_2^2\big)dx + \cos(\omega_{21}t)\int_0^L x\,\psi_1\psi_2\,dx.

Each ψn2|\psi_n|^2 is symmetric about L/2L/2, so the first term is L/2L/2. For the cross integral, use 2sinAsinB=cos(AB)cos(A+B)2\sin A\sin B = \cos(A-B) - \cos(A+B) and 0LxcosnπxLdx=L2n2π2((1)n1)\int_0^L x\cos\frac{n\pi x}{L}dx = \frac{L^2}{n^2\pi^2}\big((-1)^n - 1\big):

0Lx2LsinπxLsin2πxLdx=1L(2L2π2+2L29π2)=16L9π2.\int_0^L x\,\frac2L\sin\frac{\pi x}{L}\sin\frac{2\pi x}{L}dx = \frac1L\left(-\frac{2L^2}{\pi^2} + \frac{2L^2}{9\pi^2}\right) = -\frac{16L}{9\pi^2}.

Therefore

x(t)=L216L9π2cos(ω21t)L20.18Lcos(ω21t):\langle x\rangle(t) = \frac L2 - \frac{16L}{9\pi^2}\cos(\omega_{21}t) \approx \frac L2 - 0.18\,L\cos(\omega_{21}t):

the mean position swings across the well at the Bohr frequency ω21=3π2/2mL2\omega_{21} = 3\pi^2\hbar/2mL^2. Each ingredient state alone predicts a frozen x=L/2\langle x\rangle = L/2; the superposition moves.


Hands-on (Python)

import numpy as np
import matplotlib.pyplot as plt

# --- Two-state sloshing in the infinite square well --------------------------
# Numerical values hbar = m = L = 1 (physics keeps ħ explicit; arrays don't care).
hbar, m, L = 1.0, 1.0, 1.0
x = np.linspace(0, L, 800)

psi_n = lambda n, x: np.sqrt(2/L) * np.sin(n*np.pi*x/L)   # √(2/L) sin(nπx/L)
E_n   = lambda n: n**2 * np.pi**2 * hbar**2 / (2*m*L**2)  # n²π²ħ²/2mL²

w21 = (E_n(2) - E_n(1)) / hbar                # Bohr frequency
T = 2*np.pi / w21                             # sloshing period
Psi = lambda x, t: (psi_n(1,x)*np.exp(-1j*E_n(1)*t/hbar)
                  + psi_n(2,x)*np.exp(-1j*E_n(2)*t/hbar)) / np.sqrt(2)

for frac in [0, 0.25, 0.5, 0.75]:             # snapshots across one period
    plt.plot(x, np.abs(Psi(x, frac*T))**2, label=f"t = {frac:.2f} T")
plt.xlabel("x/L"); plt.ylabel(r"$|\Psi|^2$"); plt.legend(); plt.show()
# Expected: lump on the left at t=0, symmetric at T/4 and 3T/4, on the right at T/2.

t = np.linspace(0, 2*T, 400)
x_mean = [np.trapz(x*np.abs(Psi(x, tt))**2, x) for tt in t]
plt.plot(t/T, x_mean); plt.axhline(0.5, ls="--", c="gray")
plt.xlabel("t/T"); plt.ylabel(r"$\langle x\rangle$"); plt.show()
# Expected: <x>(t) = 0.5 - (16/9π²)cos(ω21 t) ≈ 0.5 - 0.180 cos(ω21 t):
# a clean cosine at exactly ω21 = (E2 - E1)/ħ.
# --- Expand a triangular initial state in well eigenstates -------------------
tri = np.where(x < L/2, x, L - x)             # triangle peaked at L/2
tri = tri / np.sqrt(np.trapz(tri**2, x))      # normalize

N = 40                                        # c_n by numerical quadrature
c = np.array([np.trapz(psi_n(n, x)*tri, x) for n in range(1, N+1)])
print(np.cumsum(np.abs(c)**2)[[0, 2, 9, 39]])
# Expected: ≈ [0.9855 0.9977 0.9998 1.0000] — Σ|c_n|² → 1 (completeness);
# even n vanish by symmetry and |c_n|² ~ 1/n⁴, so convergence is fast.

def Psi_rec(x, t):                            # reconstruct Σ c_n ψ_n e^{-iE_n t/ħ}
    return sum(c[n-1]*psi_n(n, x)*np.exp(-1j*E_n(n)*t/hbar) for n in range(1, N+1))

T1 = 2*np.pi*hbar/E_n(1)                      # fundamental period
plt.plot(x, tri**2, "k--", label=r"exact $|\Psi(x,0)|^2$")
for tt, lab in [(0, "t=0"), (0.1*T1, "t=0.1$T_1$"), (0.25*T1, "t=0.25$T_1$")]:
    plt.plot(x, np.abs(Psi_rec(x, tt))**2, label=lab)
plt.xlabel("x/L"); plt.legend(); plt.show()
# Expected: at t=0 the reconstruction overlays the triangle (dashed); later
# snapshots show the peak melting into interference wiggles — dynamics generated
# purely by the relative phases e^{-iE_n t/ħ}.

Exercises

E1 (easy). Verify by substitution that Ψn=ψn(x)eiEnt/\Psi_n = \psi_n(x)e^{-iE_nt/\hbar} solves the time-dependent Schrödinger equation whenever H^ψn=Enψn\hat H\psi_n = E_n\psi_n.

Solution

Left side: $i\hbar,\partial_t\Psi_n = i\hbar(-iE_n/\hbar)\psi_ne^{-iE_nt/\hbar} = E_n\psi_ne^{-iE_nt/\hbar}.Rightside:. Right side: \hat H\Psi_n = (\hat H\psi_n)e^{-iE_nt/\hbar} = E_n\psi_ne^{-iE_nt/\hbar}(thephaseisconstantin (the phase is constant in x,soitpassesthrough, so it passes through \hat H$). Equal. ✓

E2 (easy). For Ψ(x,0)=15ψ1+25ψ4\Psi(x,0) = \frac{1}{\sqrt5}\psi_1 + \frac{2}{\sqrt5}\psi_4 in the infinite well: which energies can be measured, with what probabilities? What is H\langle H\rangle in units of E1E_1? Do the answers change at later times?

Solution

E1E_1 with P=1/5P = 1/5; E4=16E1E_4 = 16E_1 with P=4/5P = 4/5. H=15E1+4516E1=13E1\langle H\rangle = \frac15E_1 + \frac45\cdot16E_1 = 13\,E_1. No change: evolution sends cncneiEnt/c_n \to c_ne^{-iE_nt/\hbar}, so every cn2|c_n|^2 is constant — energy statistics are conserved.

E3 (medium). Prove that eigenfunctions of H^\hat H with distinct energies are orthogonal: EmEnψmψndx=0E_m \neq E_n \Rightarrow \int\psi_m^*\psi_n\,dx = 0. (Hint: evaluate ψm(H^ψn)dx(H^ψm)ψndx\int\psi_m^*(\hat H\psi_n)dx - \int(\hat H\psi_m)^*\psi_ndx two ways.)

Solution

Call the difference DD. The VV terms cancel (VV real); the kinetic terms give $-\frac{\hbar^2}{2m}\int(\psi_m^\psi_n'' - \psi_m''^\psi_n)dx = -\frac{\hbar^2}{2m}\big[\psi_m^\psi_n' - \psi_m'^\psi_n\big]_{-\infty}^{\infty} = 0$ after two integrations by parts (boundary terms vanish for normalizable states). So D=0D = 0. But evaluating directly with the eigenvalue equations (and real eigenvalues, proved in Theory): D=(EnEm)ψmψndxD = (E_n - E_m)\int\psi_m^*\psi_ndx. Since EmEnE_m \neq E_n, the integral vanishes — Hermiticity at work, as in 0.1.5.

E4 (medium). For $\Psi = \frac{1}{\sqrt2}\big(\psi_1e^{-iE_1t/\hbar} + e^{i\alpha}\psi_2e^{-iE_2t/\hbar}\big)withconstant with constant \alpha,show, show |\Psi|^2$ has the same sloshing form as in Theory but time-shifted. What does this say about α\alpha versus a global phase?

Solution

The cross term becomes ψ1ψ2cos(ω21tα)\psi_1\psi_2\cos(\omega_{21}t - \alpha): the relative phase α\alpha shifts the starting position of the slosh (α=π\alpha = \pi starts the lump on the right) but not the frequency. A global phase eiγΨe^{i\gamma}\Psi cancels in Ψ2|\Psi|^2 entirely. Relative phase: physical initial condition. Global phase: nothing — the verdict of 1.1.1.

E5 (hard). Show that in the infinite well any state revives exactly: Ψ(x,t+Trev)=Ψ(x,t)\Psi(x, t + T_{\mathrm{rev}}) = \Psi(x,t) with Trev=4mL2/πT_{\mathrm{rev}} = 4mL^2/\pi\hbar. Why does no such universal revival occur for a generic potential?

Solution

En=n2E1E_n = n^2E_1 with E1=π22/2mL2E_1 = \pi^2\hbar^2/2mL^2. After time TT, the nn-th term gains ein2E1T/e^{-in^2E_1T/\hbar}. Choose E1T/=2πE_1T/\hbar = 2\pi, i.e. Trev=2π/E1=4mL2/πT_{\mathrm{rev}} = 2\pi\hbar/E_1 = 4mL^2/\pi\hbar: then e2πin2=1e^{-2\pi in^2} = 1 for every integer nn, so every coefficient returns exactly (sub-multiples of TrevT_{\mathrm{rev}} give nontrivial global phases and fractional revivals). The magic: all EnE_n are integer multiples of one quantum E1E_1, so all Bohr frequencies are commensurate. A generic potential has incommensurate level spacings — the relative phases never realign, and the state never exactly recurs.


Checkpoint

  1. Walk through separation of variables: what plays the role of the separation constant, and which two ODEs result?
  2. Why must EE be real, and why must E>VminE > V_{\min}?
  3. Name and derive (one line each) the three defining properties of a stationary state.
  4. Given Ψ(x,0)\Psi(x,0), how do you find the cnc_n, and what do cn2|c_n|^2, cn2\sum|c_n|^2, and cn2En\sum|c_n|^2E_n mean?
  5. Nothing about a stationary state changes in time — so where does quantum dynamics come from?
Answers
  1. Try Ψ=ψ(x)φ(t)\Psi = \psi(x)\varphi(t); dividing by ψφ\psi\varphi separates tt from xx, each side equaling a constant EE. Results: iφ˙=Eφi\hbar\dot\varphi = E\varphi (so φ=eiEt/\varphi = e^{-iEt/\hbar}) and the TISE H^ψ=Eψ\hat H\psi = E\psi.
  2. Complex EE makes Ψ2dxe2Im(E)t/\int|\Psi|^2dx \propto e^{2\,\mathrm{Im}(E)t/\hbar}, violating conservation of probability. If E<VminE < V_{\min}, ψ\psi'' and ψ\psi share a sign everywhere, so ψ\psi curves away from the axis and cannot be normalizable.
  3. (i) Ψn2=ψn2|\Psi_n|^2 = |\psi_n|^2 — the phases cancel; (ii) Q\langle Q\rangle constant — the phase passes through Q^\hat Q and cancels; (iii) H=En\langle H\rangle = E_n, H2=En2\langle H^2\rangle = E_n^2, so σH=0\sigma_H = 0 — definite energy.
  4. Fourier's trick: cn=ψnΨ(x,0)dxc_n = \int\psi_n^*\Psi(x,0)dx (orthonormality projects each coefficient out). cn2|c_n|^2 = probability of measuring EnE_n; cn2=1\sum|c_n|^2 = 1 = normalization; cn2En=H\sum|c_n|^2E_n = \langle H\rangle — all constant in time.
  5. From superposition: components cnψneiEnt/c_n\psi_ne^{-iE_nt/\hbar} accumulate relative phases at the Bohr frequencies (EmEn)/(E_m - E_n)/\hbar, which survive in the cross terms of Ψ2|\Psi|^2. All quantum dynamics is evolving relative phase.

Further Reading

  • [Gri] Griffiths & Schroeter, §2.1 — stationary states; the source of this lesson's structure.
  • [Gri] Griffiths & Schroeter, §2.2 — the infinite square well and its superposition dynamics.
  • [Sha] Shankar, §4.3 & Ch. 5 — the energy basis and the propagator U(t)U(t), bridging to Term 1.
  • [ER] Eisberg & Resnick, §5.5–5.7 — separation of variables and the meaning of eigenvalues.
  • [Pre] Preskill, Ph219, Ch. 2 — Hamiltonian evolution stated axiomatically, for contrast.

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