Problems 2 · Quantum Mechanics (Phillips)

Four of these six problems attack the same equation from different sides, and the interesting answers are the ones where a plausible wave turns out not to solve it.

Six problems, each with Phillips’ own hint and a full worked solution, both collapsed so the page still works as practice.

1 Capillary waves — the ratio flips derivation

Waves on the surface of water are dispersive. If the wavelength is short enough that surface tension rather than gravity provides the restoring force, the dispersion relation is

\omega = \sqrt{\frac{Tk^3}{\rho}}

where $T$ is the surface tension and $\rho$ the density of water.

Find the phase velocity of a sinusoidal wave with wave number $k$ , and the group velocity of a packet with wave numbers near $k$ .

💡 Phillips' own hint

Use $v_{\text{phase}} = \omega/k$ and $v_{\text{group}} = d\omega/dk$ . Verify that $v_{\text{group}} = \tfrac32 v_{\text{phase}}$ .

✓ Worked solution

Write the dispersion relation as $\omega = (T/\rho)^{1/2}k^{3/2}$ so the $k$ -dependence is explicit. Then

v_{\text{phase}} = \frac{\omega}{k} = \sqrt{\frac{Tk}{\rho}}

v_{\text{group}} = \frac{d\omega}{dk} = \frac{3}{2}\left(\frac{T}{\rho}\right)^{1/2}k^{1/2} = \frac{3}{2}\sqrt{\frac{Tk}{\rho}} = \frac{3}{2}\,v_{\text{phase}}

The general rule is worth extracting: for any power law $\omega \propto k^n$ , $v_{\text{group}} = n\,v_{\text{phase}}$ . Gravity waves have $n = \tfrac12$ (ratio $\tfrac12$ ), capillary waves $n = \tfrac32$ , light $n = 1$ , and a free quantum particle $n = 2$ .

What this really says — the sign flips for capillary waves, so crests run backwards through the packet

The sign of $v_{\text{group}} - v_{\text{phase}}$ decides which way crests move through the packet, and here it flips relative to §2.1’s gravity waves.

Gravity waves ( $\tfrac12$ ): crests are faster than the group, so they appear at the back, sweep forward and die at the front.
Capillary ripples ( $\tfrac32$ ): crests are slower, so they appear at the front and drift backwards through the packet.

You can see both on the same pond. Drop a stone and the long gravity waves at the outside of the ring have crests running outward through the group; the fine ripples ahead of them have crests running inward. Same water — only the restoring force changed.

Select ripples (surface tension) in the widget on §2.1 and watch the markers cross the other way.

Reproduce it in NumPy

import numpy as np
T, rho, g = 0.072, 1000.0, 9.81

print("capillary waves, T = 0.072 N/m, rho = 1000 kg/m^3")
for lam in (0.002, 0.005, 0.010):
    k = 2*np.pi/lam
    vp = np.sqrt(T*k/rho)                       # omega/k
    h = 1e-6
    om = lambda kk: np.sqrt(T*kk**3/rho)
    vg = (om(k+h) - om(k-h)) / (2*h)            # d(omega)/dk, numerically
    print(f" lambda = {lam*1000:4.1f} mm   v_phase = {vp:.3f} m/s   "
          f"v_group = {vg:.3f} m/s   ratio = {vg/vp:.2f}")

k_c = np.sqrt(rho*g/T)                          # where the two effects balance
print(f"\ngravity and surface tension contribute equally at "
      f"lambda = {2*np.pi/k_c*1000:.1f} mm")
print(" -> use the capillary formula well below this, the gravity one well above")

printscapillary waves, T = 0.072 N/m, rho = 1000 kg/m^3 lambda = 2.0 mm v_phase = 0.476 m/s v_group = 0.713 m/s ratio = 1.50 lambda = 5.0 mm v_phase = 0.301 m/s v_group = 0.451 m/s ratio = 1.50 lambda = 10.0 mm v_phase = 0.213 m/s v_group = 0.319 m/s ratio = 1.50 gravity and surface tension contribute equally at lambda = 17.0 mm -> use the capillary formula well below this, the gravity one well above

2 A packet that keeps its shape derivation

Consider the wave packet

\Psi(x,t) = \int_{k-\Delta k}^{k+\Delta k} A\cos(k'x - \omega' t)\,dk'

with the non-dispersive relation $\omega' = ck'$ . By integrating over $k'$ , show that

\Psi(x,t) = S(x-ct)\cos k(x-ct), \qquad S(x-ct) = 2A\Delta k\,\frac{\sin[\Delta k(x-ct)]}{[\Delta k(x-ct)]}

and describe how this packet propagates.

💡 Phillips' own hint

Use

\int\cos k'(x-ct)\,dk' = \frac{\sin k'(x-ct)}{(x-ct)}

and $\sin A - \sin B = 2\cos\!\big(\tfrac{A+B}{2}\big)\sin\!\big(\tfrac{A-B}{2}\big)$ .

✓ Worked solution

The whole trick is that $\omega' = ck'$ lets you write the integrand’s phase as $k'x - ck't = k'(x-ct)$ — a single variable $u = x - ct$ . Substituting:

\Psi = \int_{k-\Delta k}^{k+\Delta k} A\cos(k'u)\,dk' = \frac{A}{u}\Big[\sin(k'u)\Big]_{k-\Delta k}^{k+\Delta k} = \frac{A}{u}\big[\sin((k{+}\Delta k)u) - \sin((k{-}\Delta k)u)\big]

Apply the sine difference identity with $A' = (k+\Delta k)u$ and $B' = (k-\Delta k)u$ , so $\tfrac{A'+B'}{2} = ku$ and $\tfrac{A'-B'}{2} = \Delta k\,u$ :

\Psi = \frac{A}{u}\cdot 2\cos(ku)\sin(\Delta k\,u) = 2A\Delta k\,\frac{\sin(\Delta k\,u)}{(\Delta k\,u)}\,\cos(ku)

which is the required form with $u = x - ct$ .

How it propagates. Both the carrier and the envelope depend on $x$ and $t$ only through the single combination $x - ct$ . So the entire packet — shape and all — translates rigidly to the right at speed $c$ , without any change of form, for ever.

Ψ(x,t)
envelope S(x), Eq. (2.5)
a crest — moves at ω/k
envelope peak — moves at dω/dk

band Δkk/8components41

ω(k): ω = ck
length 2π/Δk: 8.4
ω/k: 1.00
dω/dk: 1.00(1.00×)

3 Real functions are not solutions verification

Verify by direct substitution that the real functions

\Psi = A\cos(kx-\omega t) \qquad\text{and}\qquad \Psi = A\sin(kx-\omega t)

are not solutions of the Schrödinger equation for a free particle.

💡 Phillips' own hint

Use relations like

\frac{\partial\cos(kx-\omega t)}{\partial t} = \omega\sin(kx-\omega t), \qquad \frac{\partial^2\cos(kx-\omega t)}{\partial x^2} = -k^2\cos(kx-\omega t)

✓ Worked solution

Take the cosine and put each side of Eq. (2.14) together separately.

Left side.

i\hbar\frac{\partial\Psi}{\partial t} = i\hbar A\omega\sin(kx-\omega t)

Right side.

-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = -\frac{\hbar^2}{2m}\big(-k^2A\cos(kx-\omega t)\big) = \frac{\hbar^2k^2}{2m}A\cos(kx-\omega t)

Now look at what they are. The left side is $i$ times a real quantity — purely imaginary. The right side is purely real. Two complex numbers are equal only if their real and imaginary parts match separately, so we would need

\hbar\omega A\sin(kx-\omega t) = 0 \quad\text{and}\quad \frac{\hbar^2k^2}{2m}A\cos(kx-\omega t) = 0

for all $x$ and $t$ — which forces $A = 0$ . The only “solution” is no wave at all.

The sine works out identically: the time derivative produces a cosine and the second space derivative reproduces the sine, so the two sides again land on perpendicular axes.

Confirm it for yourself — try every real option and drag $\omega$ across its whole range:

left side: iħ ∂Ψ/∂t

right side: −(ħ²/2m) ∂²Ψ/∂x²

real part
imaginary part

ω1.40(k = 2, ħ = m = 1)

✗ not a solutionmismatch 100.0%

The left side is purely IMAGINARY and the right side purely REAL. No choice of ω can reconcile them — the failure is structural.

4 A complex standing wave verification

Verify that

\Psi(x,t) = A\,e^{i(kx-\omega t)} - A\,e^{-i(kx+\omega t)}

with $A$ an arbitrary complex constant, is a solution of the free-particle Schrödinger equation if $\hbar\omega = \hbar^2k^2/2m$ . Show it can be rewritten

\Psi(x,t) = 2iA\sin kx\;e^{-i\omega t}

What sort of wave is this?

💡 Phillips' own hint

The rewriting uses $e^{i\theta} - e^{-i\theta} = 2i\sin\theta$ . It is a complex standing wave with wave number $k$ and angular frequency $\omega$ .

✓ Worked solution

It is a solution. The first term is a plane wave with wave number $+k$ ; the second is a plane wave with wave number $-k$ , since $e^{-i(kx+\omega t)} = e^{i((-k)x - \omega t)}$ . Both satisfy the equation when $\hbar\omega = \hbar^2k^2/2m$ — and crucially the same $\omega$ works for both, because the dispersion relation depends on $k^2$ and is blind to the sign of $k$ . The equation is linear, so their difference is also a solution.

The rewriting. Factor out the common time dependence:

\Psi = A\,e^{-i\omega t}\big(e^{ikx} - e^{-ikx}\big) = A\,e^{-i\omega t}\cdot 2i\sin kx = 2iA\sin kx\;e^{-i\omega t}

What sort of wave. A standing wave — $x$ and $t$ have separated into different factors, exactly as in §2.1’s classical standing wave. The nodes sit at $x = n\pi/k$ and stay there for ever.

Why this problem is placed here

Compute the probability density:

|\Psi|^2 = |2iA\sin kx|^2\,|e^{-i\omega t}|^2 = 4|A|^2\sin^2 kx

The time factor has modulus one, so it drops out entirely: $|\Psi|^2$ does not depend on $t$ at all. The wave function keeps changing — the whole thing rotates in the complex plane at rate $\omega$ — but nothing observable about it ever changes.

That is the definition of a stationary state , and this problem is a preview of the single most important structure in the book. Chapter 4 shows that every state of definite energy has exactly this form, $\psi(x)e^{-iEt/\hbar}$ : a fixed spatial shape multiplied by a phase that turns but never shows.

It is also worth noticing what the two counter-propagating waves have become. A particle with momentum $+\hbar k$ superposed with one of momentum $-\hbar k$ gives a state with no definite momentum at all but a perfectly definite energy. That trade — you may have one or the other — is the subject of chapter 7.

5 Physicists on another planet conceptual

In quantum mechanics it is the convention to represent a free particle of momentum $p$ and energy $E$ by

\Psi(x,t) = A\,e^{i(px-Et)/\hbar}

Physicists on another planet may have chosen instead

\Psi(x,t) = A\,e^{-i(px-Et)/\hbar}

What is the form of the Schrödinger equation on that planet?

💡 Phillips' own hint

Verify that their wave function solves

-i\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}

✓ Worked solution

Substitute their convention and see what equation it satisfies. With $\Psi = A e^{-i(px-Et)/\hbar}$ :

\frac{\partial\Psi}{\partial t} = \frac{iE}{\hbar}\Psi, \qquad \frac{\partial^2\Psi}{\partial x^2} = \left(\frac{-ip}{\hbar}\right)^2\Psi = -\frac{p^2}{\hbar^2}\Psi

We want an equation that reproduces $E = p^2/2m$ . Try multiplying the time derivative by $-i\hbar$ :

-i\hbar\frac{\partial\Psi}{\partial t} = -i\hbar\cdot\frac{iE}{\hbar}\Psi = E\Psi

and the right side is unchanged:

-\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = \frac{p^2}{2m}\Psi

So their Schrödinger equation is

\boxed{\;-i\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}\;}

— the complex conjugate of ours. Their wave functions are the complex conjugates of ours, and every physical prediction is identical, because $|\Psi|^2 = |\Psi^*|^2$ .

What is conventional here, and what is not

This is a better question than it looks. Three things are worth separating.

Conventional: the sign of $i$ in the exponent. Nothing measurable distinguishes the two planets. Swap $i \to -i$ everywhere — in the wave functions and in the equation — and every observable comes out the same.

Not conventional: that the equation must contain $i$ at all, and that the time derivative is first order while the space derivative is second. Those are forced (problem 3), and neither planet can escape them.

Also not conventional: the relative sign between the two sides. If you flipped the sign on only one side you would get $E = -p^2/2m$ — negative kinetic energy for a free particle, which is not a convention but a different and wrong physics.

Engineer's bridge — this is an endianness problem

A global sign convention is a free choice that must be applied consistently everywhere, and mixing two conventions in one calculation produces silent nonsense. You have met this exact failure mode:

Endianness. Big-endian and little-endian are equally valid; a file written in one and read in the other is garbage, and nothing in the bytes tells you which happened.
FFT sign conventions. numpy.fft uses $e^{-2\pi i kn/N}$ for the forward transform; plenty of textbooks and DSP libraries use $e^{+2\pi i kn/N}$ . Mix them and your phases come out conjugated — a bug that typically survives every magnitude-only test you write.
Row-major vs column-major, quaternion $w$ -first vs $w$ -last, and every other convention that is invisible until two codebases meet.

The physics analogue is exact. Textbooks that write $e^{+i(\omega t - kx)}$ are on the other planet, and every formula they give will differ from Phillips’ by complex conjugation. Both are right; a calculation that uses one for the wave function and the other for the equation is right about nothing.

Where the analogy breaks. Byte order is pure representation — there is no sense in which one is more natural. Here there is a genuine physical asymmetry underneath: the relative sign between the two sides of the equation is fixed by requiring positive kinetic energy, and no convention can move it. What is free is only the overall direction of rotation you call “positive”, not the structure it describes.

Select e^{−i(kx − ωt)} in the widget in problem 3 above: the two sides have the right shape but the wrong sign, which is exactly the statement that it needs the other planet’s equation.

6 Trying it relativistically — the Klein–Gordon equation derivation

According to relativity, the momentum $p$ and energy $E$ of a particle of mass $m$ satisfy

E^2 - p^2c^2 = m^2c^4

and the velocity of the particle is $u = pc^2/E$ .

(a) Assume the motion is described by a wave packet with $E = \hbar\omega$ and $p = \hbar k$ . Derive the group velocity of the packet and show it equals the particle velocity.

(b) Show that the wave equation

\frac{\partial^2\Psi}{\partial t^2} - c^2\frac{\partial^2\Psi}{\partial x^2} + \frac{m^2c^4}{\hbar^2}\Psi = 0

has solutions $\Psi = A\,e^{-i(\omega t - kx)}$ which could describe a relativistic particle of energy $E = \hbar\omega$ and momentum $p = \hbar k$ .

💡 Phillips' own hint

(a) Use $v_{\text{group}} = d\omega/dk$ . (b) Show that $\Psi = A e^{-i(\omega t - kx)}$ is a solution if $E^2 - p^2c^2 = m^2c^4$ .

✓ Worked solution

(a) Write the energy–momentum relation in wave variables using $E = \hbar\omega$ , $p = \hbar k$ :

\hbar^2\omega^2 - \hbar^2k^2c^2 = m^2c^4

Differentiate both sides with respect to $k$ , treating $\omega$ as $\omega(k)$ :

2\hbar^2\omega\frac{d\omega}{dk} - 2\hbar^2kc^2 = 0 \quad\Longrightarrow\quad \frac{d\omega}{dk} = \frac{kc^2}{\omega} = \frac{\hbar k c^2}{\hbar\omega} = \frac{pc^2}{E} = u

The group velocity is the particle velocity — exactly as in the non-relativistic case, and by the same reasoning. The requirement that fixed §2.2’s equation is not special to non-relativistic physics.

(b) With $\Psi = A e^{-i(\omega t - kx)}$ ,

\frac{\partial^2\Psi}{\partial t^2} = -\omega^2\Psi, \qquad \frac{\partial^2\Psi}{\partial x^2} = -k^2\Psi

Substituting into the equation:

-\omega^2\Psi + c^2k^2\Psi + \frac{m^2c^4}{\hbar^2}\Psi = 0

Divide by $\Psi$ and multiply by $\hbar^2$ :

\hbar^2\omega^2 - \hbar^2c^2k^2 = m^2c^4 \quad\Longleftrightarrow\quad E^2 - p^2c^2 = m^2c^4

which is exactly the relativistic energy–momentum relation. So the plane wave is a solution precisely when it describes a relativistic particle.

And yet it does not work — the book's own warning

Everything above went through cleanly, which makes what Phillips says next the most interesting sentence in the chapter:

At first sight this wave equation, which is called the Klein–Gordon equation , provides a basis for a theory for relativistic particles which is similar to that provided by the Schrödinger equation for non-relativistic particles. However, a sensible interpretation of the Klein–Gordon equation cannot be obtained if its solutions are treated as wave functions. A sensible interpretation only emerges if the solutions are treated as quantum field operators.

Why it fails is worth knowing even though the book does not say. Two reasons, both traceable to that second time derivative:

Negative energies. $\hbar^2\omega^2 = \hbar^2c^2k^2 + m^2c^4$ is quadratic in $\omega$ , so for every solution with $E > 0$ there is one with $E < 0$ , and nothing forbids a particle cascading down through them without limit.
Negative probabilities. The conserved density that goes with a second-order equation is not $|\Psi|^2$ but something involving $\partial\Psi/\partial t$ — and it can come out negative. A negative probability means nothing.

Both problems are absent from the Schrödinger equation for the same structural reason its solutions must be complex: it is first order in time. That apparently minor difference is the whole reason a wave-function interpretation works non-relativistically and collapses relativistically.

The resolution — quantum field theory, where $\Psi$ becomes an operator that creates and destroys particles — is exactly what §1.4 said this book would not cover, and why it restricts itself to non-relativistic particles.

Reproduce it in NumPy

import numpy as np
hbar = c = 1.0
m = 2.0
omega = lambda k: np.sqrt(k**2*c**2 + m**2*c**4/hbar**2)

print("Klein-Gordon: check d(omega)/dk = p c^2 / E   (hbar = c = 1, m = 2)")
ks = (0.5, 2.0, 10.0)
for k in ks:
    h = 1e-7
    vg = (omega(k+h) - omega(k-h)) / (2*h)
    E, p = hbar*omega(k), hbar*k
    print(f"   k = {k:5.1f}   d(omega)/dk = {vg:.6f}   p c^2 / E = {p*c**2/E:.6f}   "
          f"{'match' if abs(vg - p*c**2/E) < 1e-6 else 'MISMATCH'}")
print("   note every group velocity is < c = 1, as relativity requires")

print("\nbut the phase velocities exceed c:")
for k in ks:
    print(f"   k = {k:5.1f}   v_phase = {omega(k)/k:.6f}")

printsKlein-Gordon: check d(omega)/dk = p c^2 / E (hbar = c = 1, m = 2) k = 0.5 d(omega)/dk = 0.242536 p c^2 / E = 0.242536 match k = 2.0 d(omega)/dk = 0.707107 p c^2 / E = 0.707107 match k = 10.0 d(omega)/dk = 0.980581 p c^2 / E = 0.980581 match note every group velocity is < c = 1, as relativity requires but the phase velocities exceed c: k = 0.5 v_phase = 4.123106 k = 2.0 v_phase = 1.414214 k = 10.0 v_phase = 1.019804

The phase velocity exceeds c, and that is fine

The numbers above show every group velocity below $c$ and every phase velocity above it. This alarms people, and it should not.

A single plane wave is infinite in extent and has existed for ever. It carries no information — there is no beginning to it, no edge, nothing you could modulate to send a message. Relativity forbids signals faster than $c$ , not patterns.

Anything that could carry information has to be a packet with a beginning and an end, and packets move at the group velocity — which comes out below $c$ here, automatically. §2.1’s caution made the same point for the non-relativistic case, where $v_{\text{phase}} = v_{\text{particle}}/2$ : the phase velocity is a property of a mathematical idealization, not of anything that moves.

Check yourself

0 / 6 answered

1.For a dispersion relation $ω \propto k^{n}$ , what is the ratio $v_{group} / v_{phase}$ ?
2.In problem 2, why does the packet keep its shape exactly?
3.Problem 3 shows $cos (k x - ω t)$ fails. What exactly goes wrong?
4.Problem 4's wave is $Ψ = 2 i A sin k x e^{- iω t}$ . Why is it worth more than the algebra suggests?
5.The other planet writes $Ψ = A e^{- i (p x - E t) /ℏ}$ . Which of these is genuinely a convention?
6.The Klein–Gordon equation reproduces $E^{2} - p^{2} c^{2} = m^{2} c^{4}$ correctly, yet Phillips says its solutions cannot be treated as wave functions. Why?