Particle Wave Equations · Quantum Mechanics (Phillips)

Being first order in time is the whole difference. It is why Ψ has to be complex, why one initial condition is enough, and why nothing here looks like a classical wave equation.

Everything is now in place. Chapter 1 gave us $p = h/\lambda$ ; §2.1 gave us packets, dispersion relations and group velocity. This section puts them together and gets the equation the rest of the book solves.

But before any of that, Phillips says something about method that is easy to read past and worth stopping on.

This equation is not derived. It is guessed.

In classical physics, fundamental laws of physics are used to derive the wave equations… In contrast, we shall view the wave equation governing the wave-like properties of a quantum particle as a fundamental equation which cannot be derived from underlying basic physical principles. We, like the inventors of quantum theory, can only guess the form of this wave equation and then test for consistency and agreement with experiment.

Maxwell’s equations give you the classical wave equation. Nothing gives you the Schrödinger equation. It sits at the same level as Newton’s second law: a postulate, justified entirely by the fact that its consequences keep turning out to be true.

What follows is therefore not a derivation but a construction — an argument that narrows the field until only one simple candidate is left. That is a weaker thing than a proof, and it is worth knowing which one you are reading.

The construction

Two requirements do all the work.

step 1 of 6

The only physics entering is de Broglie's relation and the demand that the packet move like the particle. Everything else is bookkeeping.

1Start from de Broglie, rewritten in terms of wave number rather than wavelength. This is the chapter's only physical input.
$p = ℏ k, ℏ = \frac{h}{2 π}$
Equation (2.9). λ = h/p and k = 2π/λ give this immediately.
2A packet spans a band of wave numbers, so it must span a band of momenta. Same relation, applied to the spread.
$Δ p \approx ℏ Δ k$
Equation (2.10).
3§2.1 showed a packet of bandwidth Δk has length 2π/Δk. Take that length as the uncertainty in position.
$Δ x \approx \frac{2 π}{Δ k}$
Equation (2.11). Keep the 2π — it is what makes the next line come out as h rather than ħ.
4Multiply the last two. The bandwidth cancels, and what is left is a statement about the particle alone.
$Δ x Δ p \approx \frac{2 π}{Δ k} \cdot ℏΔ k = 2 π ℏ = h$
This is the Heisenberg uncertainty principle, Eq. (1.14) — and nothing was assumed to get it beyond "a particle is described by a wave packet". A genuine consistency check, not a coincidence.
5Now the one physical demand: the packet must travel at the particle’s velocity. §2.1 established that a packet travels at dω/dk.
$\frac{d ω}{d k} = \frac{ℏ k}{m}$
Equation (2.12). The right-hand side is just p/m = v for a particle of momentum ħk.
6Integrate. The constant of integration is dropped because a constant shift in ω has no observable consequence in non-relativistic quantum mechanics.
$ω = \frac{ℏ k ^{2}}{2 m}$
Equation (2.13) — the dispersion relation for de Broglie waves. Notice it is ω ∝ k², not ω ∝ k: de Broglie waves are dispersive, and packets of them spread.

What just happened — and what did not

Read step 4 again. We assumed only that a particle is described by a wave packet, and out fell $\Delta x\,\Delta p \approx h$ . Chapter 1 obtained the same inequality from a completely different argument — a photon bouncing off a particle into a microscope — and got the same answer.

That is what it means for a theory to hang together. Two unrelated routes to the same bound is evidence that the bound is a property of the description and not of either story.

And notice what step 6 delivered: $\omega \propto k^2$ . Compare with light’s $\omega = ck$ . A de Broglie wave is dispersive, so a localized particle inevitably spreads out over time — which turns out to be a real physical prediction, not an artefact.

The equation

We need a wave equation whose sinusoidal solutions obey $\omega = \hbar k^2/2m$ . Since $\omega$ goes with $\partial/\partial t$ and $k^2$ goes with $\partial^2/\partial x^2$ , we want one time derivative on one side and two space derivatives on the other. The simplest such equation is the Schrödinger equation :

i\hbar\frac{\partial\Psi}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}\tag{2.14}

Check it by substitution. With $\Psi = A\,e^{i(kx-\omega t)}$ (Eq. 2.15):

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

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

The two agree precisely when $\hbar\omega = \hbar^2k^2/2m$ — which is Eq. (2.13), the dispersion relation we built the equation to have. So the sinusoid describes a free particle with sharply defined momentum $p = \hbar k$ and energy $E = p^2/2m = \hbar\omega$ .

The punchline: no real function solves it

Phillips draws the consequence out immediately, and problem 3 asks you to verify it. Try it here instead:

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.

Why the failure is structural, not bad luck

Look at what happens for the cosine. The left side takes one time derivative, turning $\cos$ into $\sin$ , and then multiplies by $i$ — landing on the imaginary axis. The right side takes two space derivatives, turning $\cos$ back into $\cos$ — staying on the real axis.

One side is real, the other imaginary. They can only be equal if both are zero. No value of $\omega$ , no amplitude, no combination of sine and cosine can fix this, because the problem is not the numbers — it is the parity of the derivative count.

The complex exponential escapes because it is an eigenfunction of differentiation: every derivative returns the same function times a constant. $\partial_t$ multiplies by $-i\omega$ , $\partial_x^2$ multiplies by $-k^2$ , and the shape is untouched. That is the whole reason quantum mechanics is written in complex exponentials rather than sines and cosines — not convenience, necessity.

Complex is not the same as unobservable-until-you-take-the-real-part

The trap, having seen the above, is to conclude “fine — I will work with the complex form and take the real part at the end, as I would for a classical wave.”

You cannot. Taking the real part of a solution does not give a solution; the widget above shows exactly that. $\mathrm{Re}\,\Psi$ and $\mathrm{Im}\,\Psi$ are coupled by Eq. (2.14) — write the equation out in terms of the two real functions $u$ and $v$ with $\Psi = u + iv$ and you get a pair of coupled equations, each driving the other:

\hbar\frac{\partial u}{\partial t} = -\frac{\hbar^2}{2m}\frac{\partial^2 v}{\partial x^2}, \qquad \hbar\frac{\partial v}{\partial t} = +\frac{\hbar^2}{2m}\frac{\partial^2 u}{\partial x^2}

Neither half is a wave on its own. You need both, and $i$ is the bookkeeping that keeps them locked together with the right relative sign.

So: what does a complex-valued description of a real particle mean? Phillips does not answer that here. Chapter 3 does, with the Born rule — and the answer is that $\Psi$ itself is never observed, only $|\Psi|^2$ .

It is first order in time — unlike every classical wave you know

Compare the two:

	classical wave equation	Schrödinger equation
time derivatives	$\partial^2/\partial t^2$	$\partial/\partial t$
solutions	real	necessarily complex
to predict the future you need	$\Psi(x,0)$ and $\dot\Psi(x,0)$	$\Psi(x,0)$ alone
dispersion	$\omega = ck$	$\omega = \hbar k^2/2m$

That third row is a real structural difference. Pluck a string and you must specify both its shape and its velocity to know what happens next. For a quantum particle the state at one instant fixes the entire future — no separate “velocity of the wave function” exists or is needed.

If that sounds like it makes quantum mechanics more deterministic than classical wave physics, that is because in one specific sense it is. The evolution of $\Psi$ is perfectly deterministic. The randomness enters only at measurement, which is not described by this equation at all.

Superposition and the general solution

Every term in Eq. (2.14) is linear in $\Psi$ , so if $\Psi_1$ and $\Psi_2$ are solutions, so is any combination of them. That is the superposition principle , and it is what allows a particle to be, in the two-slit sense, doing two things at once.

Superposing all wave numbers gives the general solution:

\Psi(x,t) = \int_{-\infty}^{\infty} A(k')\,e^{i(k'x-\omega' t)}\,dk' \qquad\text{with}\quad \hbar\omega' = \frac{\hbar^2 k'^2}{2m}\tag{2.16}

Engineer's bridge — this is an LTI system, and Eq. (2.16) is its transfer function

Read Eq. (2.16) as a signal-processing statement and it says something you have implemented:

$A(k)$ is the spectrum of the initial state — literally the Fourier transform of $\Psi(x,0)$ .
Each spectral component is multiplied by its own phase factor $e^{-i\omega(k)t}$ , with $\omega(k) = \hbar k^2/2m$ .
Inverse transform to get $\Psi(x,t)$ .

That is exactly the recipe for pushing a signal through a linear time-invariant system: transform, multiply by the transfer function, transform back. Here the transfer function is $H(k,t) = e^{-i\hbar k^2t/2m}$ — an all-pass filter with a quadratic phase response.

Two consequences drop out immediately:

Quadratic phase is exactly what a chirp filter does. Free-particle evolution is dispersive delay, which is why packets spread. If you have ever designed a dispersion compensator for an optical fibre, you have inverted this operator.
|H| = 1 for every $k$ . Nothing is attenuated, ever. Total probability is conserved automatically.

Where the analogy breaks. An LTI system is characterized by an impulse response that typically decays — energy leaves. This one is unitary: the norm is exactly preserved for all time, nothing dissipates, and the evolution is reversible. Run $t \to -t$ and you recover the initial state exactly. No real filter you have built behaves that way, and it is why “the wave function collapses on measurement” cannot be part of this equation — collapse is not unitary, and Eq. (2.14) has no room for it.

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 = ħk²/2m
length 2π/Δk: 8.4
ω/k: 3.00
dω/dk: 6.00(2.00×)

What this really says — one dispersion relation produces both the motion and the spreading

Press propagate. Two things happen at once, and they are both consequences of $\omega \propto k^2$ :

The packet moves at $d\omega/dk = \hbar k/m$ — the particle’s velocity, by construction. And it spreads, because the components at higher $k$ genuinely travel faster and outrun the rest. A particle localized now is less localized later; that is a physical prediction of the equation, and it is measured.

Watch the crest marker too. Crests appear at the back of the packet and sweep forward through it at twice the packet’s speed, because $v_{\text{phase}} = \hbar k/2m$ . There is nothing physical moving at that speed — it is a pattern in a superposition, exactly like the point of constructive interference in §2.1’s step-through.

Adding a potential

Free particles are a warm-up. Everything interesting happens when the particle sits in a potential energy field — an electron near a proton, say, with $V(r) = -e^2/4\pi\epsilon_0 r$ .

Physics you need first — what a potential energy field is

A conservative force is one where the work done moving between two points does not depend on the route. Gravity and the Coulomb force are both conservative; friction is not. For such a force you can define a potential energy $V$ — a single number attached to each point in space — with the force being its downhill slope:

F = -\frac{dV}{dx}

The minus sign says the force points downhill, towards lower potential energy. A ball on a hillside rolls down; an electron near a proton is pulled in.

Two consequences worth having:

$V$ has an arbitrary zero. Only differences matter, because only the slope is physical. Choosing $V \to 0$ far from a nucleus is a convention, and it is why bound states come out with negative energies (as in ch01 §1.3).
$V(x)$ replaces the whole force law. Instead of tracking vectors you carry one scalar function, and everything about the interaction is in its shape. For a hydrogen atom, $V(r) = -e^2/4\pi\epsilon_0 r$ — a well that deepens without limit as $r \to 0$ .

If you have used a potential field in graphics or path planning, it is the same object: a scalar landscape whose gradient is the force.

In 1926 Schrödinger generalized Eq. (2.14) in the obvious way: add the potential energy inside the bracket.

i\hbar\frac{\partial\Psi}{\partial t} = \left[-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{r})\right]\Psi\tag{2.17}

and in one dimension

i\hbar\frac{\partial\Psi}{\partial t} = \left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\right]\Psi\tag{2.18}

What this really says — and why §0.3 already showed you this

The bracket is total energy: kinetic (curvature) plus potential (position). It has a name — the Hamiltonian $\hat H$ — and chapter 7 will make that formal.

But you have already met this object. In §0.3 the same bracket, discretized on a grid, became a tridiagonal matrix: the kinetic term the off-diagonals, $V(x)$ the diagonal. Eq. (2.18) says nothing more than

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

and once $\hat H$ is a matrix that is an ordinary linear ODE system. Every chapter from 4 to 11 is this one equation with a different $V$ — which is why one numerical recipe solves all of them.

Put a number on it — a constant potential

The simplest non-free case: a particle on a step of constant potential $V_0$ . Try the same plane wave $\Psi = A\,e^{i(kx-\omega t)}$ in Eq. (2.18). The potential term just multiplies, so the substitution gives

\hbar\omega = \frac{\hbar^2k^2}{2m} + V_0

Reading $E = \hbar\omega$ and $p = \hbar k$ , that is

E = \frac{p^2}{2m} + V_0

which is conservation of energy — kinetic plus potential — recovered from a wave equation. That is the first real check that the guess was a good one.

E has now meant three different things

Phillips reuses the letter without comment, and the meanings are genuinely different. Keep them straight:

Where	Symbol	Means
ch01 §1.1–1.2	$\epsilon$	total relativistic energy, including rest mass — and also a photon’s energy
ch01 §1.2	$E$	kinetic energy alone, $p^2/2m$ , in the non-relativistic reduction
here, ch02 §2.2	$E$	total energy, $p^2/2m + V_0$ — kinetic plus potential, no rest mass
ch04 onward	$E$	the energy eigenvalue — an allowed total energy of a confined system

The rest-mass energy quietly disappears the moment the book goes non-relativistic, which is why $E$ can switch from “kinetic” to “total” without anyone announcing it: with no rest mass in play and no potential yet, the two coincided.

When an equation surprises you, check which energy it means before checking your algebra. The notation reference records all four.

There is something to notice, though. Raising $V_0$ at fixed total $E$ lowers $k$ , so the wavelength gets longer: a particle slows down and its de Broglie wave stretches out as it climbs a potential hill. And if $V_0$ exceeds $E$ then $k^2$ turns negative, so $k$ becomes imaginary and $e^{ikx}$ becomes a real exponential $e^{-|k|x}$ — a decaying, non-oscillating wave in a region a classical particle could never enter.

That is tunnelling, and it is chapter 5.

Reproduce it in NumPy

Substituting the trial functions numerically. The complex exponential matches to machine precision at exactly one omega; the cosine never matches at any omega.

import numpy as np

hbar = m = 1.0
k = 2.0
x = np.linspace(-np.pi, np.pi, 400)

def sides(psi, dpsi_dt, d2psi_dx2):
    lhs = 1j * hbar * dpsi_dt
    rhs = -hbar**2 / (2*m) * d2psi_dx2
    return np.max(np.abs(lhs - rhs))

print("checking that A exp(i(kx - wt)) solves Eq. (2.14), with hbar = m = 1, k = 2")
print(f"  required dispersion: w = hbar k^2 / 2m = {hbar*k**2/(2*m):.1f}")
for w in (2.0, 1.4):
    psi = np.exp(1j*k*x)
    err = sides(psi, -1j*w*psi, -k**2*psi)
    tag = "<- a solution" if err < 1e-12 else "<- not a solution"
    print(f"  max |LHS - RHS| at w = {w:.1f}      : {err:.1e}   {tag}")

print("\nand the same test for cos(kx - wt), scanning every w:")
best, best_w = np.inf, None
for w in np.linspace(0, 6, 601):
    psi = np.cos(k*x).astype(complex)
    err = sides(psi, w*np.sin(k*x), -k**2*np.cos(k*x))
    if err < best:
        best, best_w = err, w
print(f"  best w found        : {best_w:.2f}")
print(f"  smallest mismatch   : {best:.1e}   <- no w works; LHS is imaginary, RHS is real")

printschecking that A exp(i(kx - wt)) solves Eq. (2.14), with hbar = m = 1, k = 2 required dispersion: w = hbar k^2 / 2m = 2.0 max |LHS - RHS| at w = 2.0 : 0.0e+00 <- a solution max |LHS - RHS| at w = 1.4 : 6.0e-01 <- not a solution and the same test for cos(kx - wt), scanning every w: best w found : 0.00 smallest mismatch : 2.0e+00 <- no w works; LHS is imaginary, RHS is real

Where this leaves us

We have an equation. What we do not have is any idea what $\Psi$ means.

Phillips is explicit about the gap: the sinusoidal solution “represents a free particle”, the packet “represents a quantum particle which moves with velocity $\hbar k/m$ ” — but represents is doing a lot of unexamined work, and the wave function $\Psi$ is a complex number at every point in space. It is not a displacement, not a pressure, not a field strength. Nothing is waving.

There is also the unresolved business from §2.2’s own text: a measurement transforms the packet — measure position precisely and it must instantly become a short packet made of many wavelengths. Phillips’ comment on how that happens is one line long and entirely honest:

No one knows how this happens.

The next chapter supplies the missing interpretation. It is due to Max Born, it is one line long, and it is the bridge between this equation and anything you can measure: $|\Psi|^2$ is a probability density .

Check yourself

0 / 6 answered

Try every real option in the trial-solution widget.
1.Why can no real function solve the free-particle Schrödinger equation?
2.The construction gets $Δ x Δ p \approx h$ from the packet in step 4. Why does that matter?
3.The Schrödinger equation is first order in time; the classical wave equation is second order. What follows?
4.Reading Eq. (2.16) as a signal-processing operation, what is the transfer function?
5.Substituting a plane wave into Eq. (2.18) with constant $V_{0}$ gives $ℏ ω = ℏ^{2} k^{2} /2 m + V_{0}$ . What is that?
6.Phillips says the equation "cannot be derived from underlying basic physical principles". What is the status of the argument on this page, then?