The expansion here is an ordinary Fourier series. Everything quantum on this page is in what the coefficients are taken to mean, not in how they are obtained.

Everything so far has had a definite energy. These two sections build states that do not, and then show what that costs — or rather, what it buys: the ability to change.

4.5 States of Uncertain Energy

§4.3 showed that a state of sharply defined energy $E_n$ is

\Psi_n(\mathbf r,t) = \psi_n(\mathbf r)\,e^{-iE_nt/\hbar}\tag{4.45}

The claim of this section is that a state of uncertain energy is

\Psi(\mathbf r,t) = \sum_{n=1,2,\dots}c_n\psi_n(\mathbf r)\,e^{-iE_nt/\hbar}\tag{4.46}

and making that precise needs two ideas: the mathematical one of a complete set of basis functions, and the physical one of an energy probability amplitude.

Basis functions

The Schrödinger equation is a homogeneous linear PDE, so a superposition of solutions is a solution. For a particle in a one-dimensional box:

\Psi(x,t) = \sum_{n=1,2,3\dots}c_n\Psi_n(x,t)\tag{4.47}

with $\Psi_n = N\sin k_nx\,e^{-iE_nt/\hbar}$ from Eq. (4.39), and the $c_n$ arbitrary complex constants.

For more complicated potentials the sines are replaced by whatever $\hat H$ ‘s eigenfunctions happen to be, and the series becomes a generalized Fourier series. That works because the eigenfunctions of a Hamiltonian form a complete orthonormal set of basis functions — which means three things:

They can be normalized:

\int|\psi_n(\mathbf r)|^2\,\mathrm d^3\mathbf r = 1\tag{4.48}

They are orthogonal — eigenfunctions belonging to different eigenvalues satisfy

\int\psi_m^*(\mathbf r)\psi_n(\mathbf r)\,\mathrm d^3\mathbf r = 0 \quad\text{if}\quad E_m \ne E_n\tag{4.49}

as problem 2 proves. For a degenerate eigenvalue the eigenfunctions are not uniquely determined, and Phillips notes you can use that freedom to make them orthogonal — so Eq. (4.49) holds generally.

They are complete — any wave function can be written as

\Psi(\mathbf r,t) = \sum_{n=1,2,3\dots}c_n\psi_n(\mathbf r)\,e^{-iE_nt/\hbar}\tag{4.50}

Engineer's bridge — this is an orthonormal basis, and you extract coefficients by projection

Every word here is linear algebra you already use:

§4.5	linear algebra
eigenfunctions $\psi_n$	orthonormal basis vectors $\mathbf e_n$
Eq. (4.48), normalized	$\mathbf e_n\cdot\mathbf e_n = 1$
Eq. (4.49), orthogonal	$\mathbf e_m\cdot\mathbf e_n = 0$
complete	the basis spans the space
$c_n = \int\psi_n^*\Psi\,\mathrm d^3\mathbf r$	$c_n = \mathbf e_n\cdot\mathbf v$ — projection
$\sum\lvert c_n\rvert^2 = 1$	Parseval / $\lVert\mathbf v\rVert = 1$

The coefficient formula is derived exactly as you would derive it for vectors: write $\Psi$ as the sum, dot both sides with $\psi_3^*$ , and orthogonality kills every term but one.

The degeneracy remark is Gram–Schmidt. When an eigenvalue is repeated, any basis of that eigenspace will do, and you are free to orthogonalize it — the same freedom, and the same fix, as in a matrix eigenproblem with repeated eigenvalues.

Where it breaks. These are infinite-dimensional spaces, and completeness is a genuine theorem rather than a counting argument — for some potentials the spectrum has a continuum part where the “basis vectors” cannot be normalized at all. Phillips flags this in a footnote and defers it to chapter 7, where the normalization involves a Dirac delta function and the sum becomes an integral. Finite-dimensional intuition gets you a long way here and then quietly stops.

Extracting the coefficients

step 1 of 4

Exactly how you would find a component of a vector — and the same reason it works.

1Write the wave function at t = 0 as its expansion. Every time factor is 1 at t = 0, so this is just the spatial superposition.
$Ψ (r, 0) = c_{1} ψ_{1} (r) + c_{2} ψ_{2} (r) + c_{3} ψ_{3} (r) + \dots$
2To pick out c₃, multiply both sides by ψ₃* and integrate over all space.
$\int ψ_{3}^{*} Ψ (r, 0) d^{3} r = c_{1} \int ψ_{3}^{*} ψ_{1} + c_{2} \int ψ_{3}^{*} ψ_{2} + c_{3} \int ψ_{3}^{*} ψ_{3} + \dots$
3Orthogonality (4.49) kills every cross term, and normalization (4.48) makes the survivor equal to 1.
$\int ψ_{3}^{*} (r) Ψ (r, 0) d^{3} r = c_{3}$
4Nothing was special about 3.
$c_{n} = \int ψ_{n}^{*} (r) Ψ (r, 0) d^{3} r$
Equation (4.51). Armed with these, Eq. (4.50) tracks the wave function for all later times — the initial shape is enough.

What the coefficients mean

Take the superposition

\Psi(\mathbf r,t) = \sum_{n=1,2,\dots}c_n\psi_n(\mathbf r)\,e^{-iE_nt/\hbar}\tag{4.52}

and ask when it is normalized. Multiplying out $\int\Psi^*\Psi\,\mathrm d^3\mathbf r$ gives terms like $c_2^*c_1e^{i(E_2-E_1)t/\hbar}\int\psi_2^*\psi_1\,\mathrm d^3\mathbf r$ , which vanish by orthogonality, and terms like $c_1^*c_1\int\psi_1^*\psi_1\,\mathrm d^3\mathbf r$ , which give $|c_1|^2$ . So

\sum_{n=1,2,\dots}|c_n|^2 = 1\tag{4.53}

The same collapse applied to Eqs. (4.26) and (4.27), using $\hat H\psi_n = E_n\psi_n$ and $\hat H^2\psi_n = E_n^2\psi_n$ , gives

\langle E\rangle = \sum_{n=1,2,\dots}|c_n|^2E_n \quad\text{and}\quad \langle E^2\rangle = \sum_{n=1,2,\dots}|c_n|^2E_n^2\tag{4.54}

Now compare with §3.1. A set of non-negative numbers that sums to 1, and which weights each outcome in an expectation value, is a probability distribution:

p_n = |c_n|^2 \quad\text{with}\quad n = 1, 2, 3\dots\tag{4.55}

so $|c_n|^2$ is the probability that a measurement of the energy returns $E_n$ , and the $c_n$ are called energy probability amplitudes .

4.6 Time Dependence

A stationary state

Take a single eigenstate,

\Psi_n(\mathbf r,t) = \psi_n(\mathbf r)\,e^{-iE_nt/\hbar}\tag{4.56}

Its wave function oscillates at angular frequency $E_n/\hbar$ — and yet nothing observable changes:

\Psi_n^*\Psi_n = \psi_n^*e^{+iE_nt/\hbar}\,\psi_ne^{-iE_nt/\hbar} = \psi_n^*\psi_n

The exponentials cancel identically. Not the probabilities, not the expectation value of any observable, nothing changes with time. Such a state is called a stationary state .

A non-stationary state

Now mix two energies equally:

\Psi(\mathbf r,t) = \sqrt{\tfrac12}\,\psi_1(\mathbf r)e^{-iE_1t/\hbar} + \sqrt{\tfrac12}\,\psi_2(\mathbf r)e^{-iE_2t/\hbar}\tag{4.57}

Two outcomes are possible, $E_1$ and $E_2$ , each with probability $\tfrac12$ . So $\langle E\rangle = \tfrac12E_1 + \tfrac12E_2$ , $\langle E^2\rangle = \tfrac12E_1^2 + \tfrac12E_2^2$ , and

\Delta E = \sqrt{\langle E^2\rangle - \langle E\rangle^2} = \tfrac12|E_1 - E_2|

The probability density is now

\Psi^*\Psi = \tfrac12\left[|\psi_1|^2 + |\psi_2|^2 + \psi_1^*\psi_2e^{+i(E_1-E_2)t/\hbar} + \psi_1\psi_2^*e^{-i(E_1-E_2)t/\hbar}\right]

Why the cross terms survive when the diagonal ones cancelled

In the stationary case each term paired with its own conjugate, so the phases cancelled exactly. Here the cross terms pair $\psi_1$ with $\psi_2^*$ — two different energies — and what survives is the difference of the two rotations, $e^{\pm i(E_1-E_2)t/\hbar}$ .

That is the whole mechanism. An absolute phase is unobservable; a phase difference is not. §4.3 ended by noting that an electron’s wave function rotates $10^{14}$ times a second with nothing detectable happening — this is where that rotation finally becomes visible, by being compared against another one.

Exactly the same structure as §3.2’s two-slit interference: two amplitudes, a cross term, and the cross term is the entire physics.

So $|\Psi|^2$ oscillates with angular frequency $|E_1-E_2|/\hbar$ — period $2\pi\hbar/|E_1-E_2|$ , which in terms of the energy uncertainty is $\pi\hbar/\Delta E$ . Such states are called non-stationary states , and in general they change more rapidly when the energy is more uncertain:

\delta t\,\Delta E \approx \hbar\tag{4.58}

⟨x⟩ over one revival period T = 4ma²/πħ

c₁1.00c₂1.00c₃0.00c₄0.00

⟨E⟩ 12.337

ΔE 7.402

period πħ/ΔE 0.424

t 0.000

2 energies — a non-stationary state. The cross terms carry e^{±i(Eₘ−Eₙ)t/ħ} and do not cancel, so |Ψ|² sloshes and ⟨x⟩ oscillates. More energy uncertainty means faster change — that is Eq. (4.58), δt·ΔE ≈ ħ, and you can watch the period shorten as you push ΔE up.

δt is not an uncertainty, and this is not a fourth uncertainty principle

Equation (4.58) is often called the Heisenberg uncertainty relation for time and energy. Phillips says outright that a better name is the time and uncertainty in energy relation, and the distinction matters:

Time, unlike position, momentum or energy, is not an observable in quantum mechanics. It is a parameter which is used to label a changing system.

So $\delta t$ in Eq. (4.58) is not the spread in the outcome of a measurement. There is no operator $\hat t$ , no eigenfunctions of it, and no sandwich integral $\int\Psi^*\hat t\Psi$ . $\delta t$ is the time scale on which observable properties change — a property of the state’s evolution, not of any measurement.

Compare honestly:

	$\Delta x\,\Delta p \ge \hbar/2$	$\delta t\,\Delta E \approx \hbar$
both factors are measurement spreads	yes	no — only $\Delta E$ is
derived from operators	yes (ch. 7)	no
an exact inequality	yes	an order-of-magnitude estimate
$x$ , $p$ , $E$ are observables	yes	$t$ is not

Read correctly it says something clean: a short-lived state cannot have a sharp energy. Infinite time scale ⇒ stationary state ⇒ certain energy; short time scale ⇒ rapid change ⇒ large $\Delta E$ .

The book returns to this at the very end of the chapter, noting that treating time and position so differently is against the spirit of special relativity — and that in relativistic quantum physics neither is an observable; both become labels on quantum field operators.

Why an excited atom is not quite stationary

Phillips closes with a case where the distinction does real work, and it is subtle.

An atom’s ground state is a state of definite energy, hence stationary, and accordingly the electrons — despite having kinetic energy — have no time-dependent properties. That is the answer to §1.3’s puzzle about why an orbiting electron does not radiate away.

An atom in an excited state looks like the same thing: its wave function is an energy eigenfunction of the Hamiltonian describing the interactions inside the atom. It ought to be stationary and timeless.

But excited atoms decay. They emit radiation and drop to a lower state, so they are at best almost stationary, and by Eq. (4.58) their energy must have a small uncertainty.

The resolution is that the eigenfunction belongs to the wrong Hamiltonian. The true one describes not only the particles inside the atom but their interaction with fluctuating electromagnetic fields that are present even in empty space. Those interactions give

\Delta E = \frac{\hbar}{\tau}\tag{4.59}

with $\tau$ the mean lifetime of the excited state. The emitted wavelength is therefore uncertain, and the spectral line has a natural line width — though in most situations that is smaller than the broadening from atoms moving and colliding.

Put a number on it — three line widths, over 25 orders of magnitude

A typical atomic transition, $\tau \approx 10^{-8}$ s:

\Delta E = \frac{\hbar}{\tau} \approx 6.6\times10^{-8}\ \mathrm{eV}

Against a visible photon of ~2 eV that is a fractional width of $3\times10^{-8}$ — extremely sharp, which is why atomic spectra are useful as fingerprints at all. Doppler broadening at room temperature is around $10^{-6}$ relative, so the book is right that thermal motion usually dominates.

The 2s state of hydrogen, $\tau \approx 0.12$ s because the decay is strongly forbidden:

\Delta E \approx 5\times10^{-15}\ \mathrm{eV}

An extraordinarily well-defined energy — and the reason metastable states are so useful in precision spectroscopy and atomic clocks.

The Z boson, at the other extreme, has $\Delta E = 2.5$ GeV — a “particle” whose mass is uncertain by 2.7%, because it lives about $2.6\times10^{-25}$ s. Problem 11 asks for exactly this. In particle physics the relation is used backwards: measure the width of a resonance peak and read off the lifetime, because no clock can time $10^{-25}$ s directly.

Reproduce it in NumPy

Every claim of §4.6, evaluated for a box in natural units ħ = m = a = 1 — that a stationary state does not move, that a superposition does, that ΔE for the two-state mix is |E₂−E₁|/2, and that the revival time of problem 9 reconstructs the state to machine precision.

import numpy as np

hbar = m = a = 1.0
E = lambda n: n**2*np.pi**2/2
x = np.linspace(0, a, 4001)
psi = lambda n: np.sqrt(2/a)*np.sin(n*np.pi*x/a)

def state(c, t):
    out = np.zeros_like(x, dtype=complex)
    for j, cn in enumerate(c, start=1):
        out += cn*psi(j)*np.exp(-1j*E(j)*t/hbar)
    return out

def stats(c, t):
    P = state(c, t)
    nrm = np.trapezoid(np.abs(P)**2, x)
    return nrm, np.trapezoid(x*np.abs(P)**2, x)/nrm

print("a stationary state does not move; a superposition does")
print("=" * 66)
for lab, c in [("n=1 alone", [1, 0]), ("equal 1+2", [1/np.sqrt(2)]*2)]:
    print(f"  {lab}")
    for t in (0.0, 0.2, 0.5, 0.9):
        nrm, xm = stats(c, t)
        print(f"    t = {t:.1f}:  integral |Psi|^2 = {nrm:.6f}   <x> = {xm:.6f}")

print("\n<E> and Delta E from the coefficients, Eq. (4.54)")
print(f"  {'state':14} {'<E>':>10} {'Delta E':>10} {'period pi*hbar/DeltaE':>22}")
for lab, c in [("n=1 alone", [1, 0, 0, 0]), ("equal 1+2", [1, 1, 0, 0]),
               ("equal 1+3", [1, 0, 1, 0]), ("1:0.7:0.5:0.3", [1, .7, .5, .3])]:
    cn = np.array(c)/np.linalg.norm(c)
    mE = sum(v**2*E(j) for j, v in enumerate(cn, 1))
    m2 = sum(v**2*E(j)**2 for j, v in enumerate(cn, 1))
    dE = np.sqrt(max(0, m2-mE**2))
    per = np.pi*hbar/dE if dE > 1e-12 else np.inf
    print(f"  {lab:14} {mE:10.4f} {dE:10.4f} {per:22.4f}")

print("\nEq. (4.57): equal mix of 1 and 2 -- Delta E should be |E2-E1|/2")
c = np.array([1, 1])/np.sqrt(2)
print(f"  E1 = {E(1):.4f}   E2 = {E(2):.4f}   |E2-E1|/2 = {abs(E(2)-E(1))/2:.4f}")
mE = sum(v**2*E(j) for j, v in enumerate(c, 1))
m2 = sum(v**2*E(j)**2 for j, v in enumerate(c, 1))
print(f"  computed Delta E = {np.sqrt(m2-mE**2):.4f}   period = {np.pi/np.sqrt(m2-mE**2):.4f}")
print(f"  2*pi*hbar/|E2-E1| = {2*np.pi/abs(E(2)-E(1)):.4f}   (same thing)")

print("\nproblem 9's revival: T = 4ma^2/(pi*hbar) rebuilds the state exactly")
T = 4*m*a**2/(np.pi*hbar)
print(f"  T = {T:.6f}")
for n in (1, 2, 3, 4):
    print(f"    E_{n}*T/hbar = {E(n)*T/hbar:9.5f} = 2*pi*{n}^2 = {2*np.pi*n**2:9.5f}")
c = np.array([1, .7, .5, .3]); c = c/np.linalg.norm(c)
d = np.abs(state(c, T) - state(c, 0)).max()
print(f"  max |Psi(T) - Psi(0)| over the whole box = {d:.2e}")

printsa stationary state does not move; a superposition does ================================================================== n=1 alone t = 0.0: integral |Psi|^2 = 1.000000 <x> = 0.500000 t = 0.2: integral |Psi|^2 = 1.000000 <x> = 0.500000 t = 0.5: integral |Psi|^2 = 1.000000 <x> = 0.500000 t = 0.9: integral |Psi|^2 = 1.000000 <x> = 0.500000 equal 1+2 t = 0.0: integral |Psi|^2 = 1.000000 <x> = 0.319873 t = 0.2: integral |Psi|^2 = 1.000000 <x> = 0.677193 t = 0.5: integral |Psi|^2 = 1.000000 <x> = 0.421363 t = 0.9: integral |Psi|^2 = 1.000000 <x> = 0.369140 <E> and Delta E from the coefficients, Eq. (4.54) state <E> Delta E period pi*hbar/DeltaE n=1 alone 4.9348 0.0000 inf equal 1+2 12.3370 7.4022 0.4244 equal 1+3 24.6740 19.7392 0.1592 1:0.7:0.5:0.3 17.9325 19.2908 0.1629 Eq. (4.57): equal mix of 1 and 2 -- Delta E should be |E2-E1|/2 E1 = 4.9348 E2 = 19.7392 |E2-E1|/2 = 7.4022 computed Delta E = 7.4022 period = 0.4244 2*pi*hbar/|E2-E1| = 0.4244 (same thing) problem 9's revival: T = 4ma^2/(pi*hbar) rebuilds the state exactly T = 1.273240 E_1*T/hbar = 6.28319 = 2*pi*1^2 = 6.28319 E_2*T/hbar = 25.13274 = 2*pi*2^2 = 25.13274 E_3*T/hbar = 56.54867 = 2*pi*3^2 = 56.54867 E_4*T/hbar = 100.53096 = 2*pi*4^2 = 100.53096 max |Psi(T) - Psi(0)| over the whole box = 4.16e-15

Chapter 4 in one line

$\hat H$ measures energy and generates time evolution, so a state of definite energy cannot change and a state of uncertain energy must.

The problems prove the two facts §4.5 leaned on — that energy eigenvalues are real and that eigenfunctions of different eigenvalues are orthogonal — and work the sudden-expansion and revival problems that this machinery makes possible.

After that, chapter 5 applies all of it to the harmonic oscillator: a new potential, the same eigenvalue problem, and a spectrum that is evenly spaced instead of going as $n^2$ .

Check yourself

0 / 7 answered

1.How is the coefficient $c_{n}$ extracted from a wave function?
2.Why is $∣ c_{n} ∣^{2}$ entitled to be called a probability?
3.In a stationary state the phases cancel; in a superposition the cross terms survive. What is the difference?
4.Why is $δ t Δ E \approx ℏ$ *not* a fourth uncertainty principle?
5.An atom in an excited state has an energy eigenfunction, so it ought to be stationary. Why does it decay?
6.The $⟨ x ⟩$ trace in the widget closes exactly after $T = 4 m a^{2} / π ℏ$ . Why does the state reconstruct itself?
7.In §4.5, eigenfunctions belonging to a *degenerate* eigenvalue are said not to be uniquely determined, and that this freedom can be used to make them orthogonal. What is the linear-algebra name for that?