Only the middle group is quantum. The first exercises ordinary probability, and the last two would look at home in a classical field theory.

Nine problems, and they fall into three clean groups. 1–3 exercise the probability theory of §3.1 on the Poisson, Gaussian and exponential distributions — the three the section named. 4–7 are the chapter’s real exercise: compute $\Delta x\,\Delta p$ for four different wave functions and compare each with $\hbar/2$ . 8–9 derive two results the main text used but did not prove — the probability current, and $\langle p\rangle = m\,\mathrm d\langle x\rangle/\mathrm dt$ .

Three symbols, and they are not the same letter

This problem set uses $a$ for two unrelated things and $\alpha$ for a third, which is easy to misread:

Symbol	Where	Means	Units
$a$	problem 1	the Poisson mean	none
$a$	problems 4, 5, 6	a length — Gaussian width, or box width	m
$\alpha$	problem 7	a decay constant	m⁻¹
$\sigma$	problem 2	a standard deviation	same as $x$

§3.1 tabulates the same collision for the chapter as a whole. Check units before reusing any formula across problems.

Group 1 — probability (problems 1–3)

These are the three distributions §3.1 promised, and the distribution widget there plots all of them live. Nothing quantum appears in this group.

1 The Poisson distribution derivation

First used by Siméon-Denis Poisson in 1837 to describe seemingly random criminal events in Paris. If independent events have a constant tendency to occur and the average rate of occurrence is $a$ , the probability of exactly $n$ events is

p_n = \frac{e^{-a}a^n}{n!},\qquad n = 0, 1, 2, \dots\infty

(a) Using $e^{+a} = 1 + \frac{a}{1!} + \frac{a^2}{2!} + \frac{a^3}{3!} + \dots$ , show that $\sum_{n=0}^{\infty} p_n = 1$ .

(b) Using $n/n! = 1/(n-1)!$ and $a^n = a\,a^{n-1}$ , show that $\sum_{n=0}^{\infty} np_n = a$ , so $\langle n\rangle = a$ .

(c) By similar techniques find $\langle n^2\rangle$ and show, using Eq. (3.4), that $\Delta n = \sqrt a$ .

💡 Phillips' own hint

Every part is the same trick: recognise a series you already know.

For (a), factor out $e^{-a}$ and the sum that remains is the exponential series for $e^{+a}$ .

For (b), the $n = 0$ term contributes nothing, so start the sum at $n = 1$ . Then $np_n = e^{-a}a^n/(n-1)!$ ; pull out one factor of $a$ and substitute $k = n-1$ , and you are back to the same series.

For (c), $n^2$ resists directly. Write $n^2 = n(n-1) + n$ — the first piece kills two factorial terms at once and the second you did in (b).

✓ Worked solution

(a) $\sum_n p_n = e^{-a}\sum_n \dfrac{a^n}{n!} = e^{-a}e^{+a} = 1$ .

(b) Dropping the vanishing $n = 0$ term and putting $k = n - 1$ :

\sum_{n=1}^{\infty} n\,\frac{e^{-a}a^n}{n!} = e^{-a}\sum_{n=1}^{\infty}\frac{a^n}{(n-1)!} = a\,e^{-a}\sum_{k=0}^{\infty}\frac{a^{k}}{k!} = a

(c) With $n^2 = n(n-1) + n$ , the first piece gives (starting at $n = 2$ , $k = n-2$ )

\sum n(n-1)p_n = e^{-a}\sum_{n=2}^{\infty}\frac{a^n}{(n-2)!} = a^2

so $\langle n^2\rangle = a^2 + a$ . Then Eq. (3.4):

(\Delta n)^2 = \langle n^2\rangle - \langle n\rangle^2 = a^2 + a - a^2 = a \quad\Rightarrow\quad \Delta n = \sqrt a

Why it matters. The fractional uncertainty is $\Delta n/\langle n\rangle = 1/\sqrt a$ , so improving a counting measurement tenfold costs a hundredfold in time. §3.1 works the numbers.

2 The Gaussian distribution derivation

With standard deviation $\sigma$ ,

\rho(x)\,\mathrm dx = \frac{1}{\sqrt{2\pi\sigma^2}}\,e^{-x^2/2\sigma^2}\,\mathrm dx,\qquad -\infty < x < +\infty

normalized because $\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{+\infty}e^{-x^2/2\sigma^2}\mathrm dx = 1$ . Phillips notes it is famous for describing a random variable arising from a multitude of small random contributions, “such as the net distance travelled by a tottering drunk with very small legs.”

(a) By considering the effect of $x \to -x$ on $x\rho(x)$ , show $\langle x\rangle = 0$ .

(b) Using $\int_{-\infty}^{+\infty}\rho\,\mathrm dx = \big[x\rho(x)\big]_{-\infty}^{+\infty} - \int_{-\infty}^{\infty} x\frac{\mathrm d\rho}{\mathrm dx}\mathrm dx$ , show $\langle x^2\rangle = \sigma^2$ .

(c) Hence verify that the standard deviation of $x$ is $\sigma$ .

💡 Phillips' own hint

(a) needs no integration at all — only the observation that an odd function integrated over a symmetric interval gives zero.

(b) is integration by parts run backwards. The bracketed boundary term dies because $\rho$ decays faster than $1/x$ . Then note that for a Gaussian, $\mathrm d\rho/\mathrm dx = -\dfrac{x}{\sigma^2}\rho$ — substitute that and the remaining integral is $\langle x^2\rangle/\sigma^2$ .

✓ Worked solution

(a) $\rho$ is even, so $x\rho(x)$ is odd. Under $x\to-x$ the integrand changes sign while the range is unchanged, so the integral equals its own negative and must vanish. No calculation is required — and this is worth noticing, because §3.5 uses the same symmetry argument repeatedly.

(b) The boundary term vanishes. Substituting $\mathrm d\rho/\mathrm dx = -x\rho/\sigma^2$ :

1 = -\int_{-\infty}^{\infty} x\left(-\frac{x\rho}{\sigma^2}\right)\mathrm dx = \frac{1}{\sigma^2}\int_{-\infty}^{\infty}x^2\rho\,\mathrm dx = \frac{\langle x^2\rangle}{\sigma^2}

so $\langle x^2\rangle = \sigma^2$ .

(c) $(\Delta x)^2 = \langle x^2\rangle - \langle x\rangle^2 = \sigma^2 - 0$ , so $\Delta x = \sigma$ . The parameter called $\sigma$ is the standard deviation — which is why the Gaussian is written that way.

3 The exponential distribution — particle lifetimes derivation

The probability that an unstable particle lives for time $t$ and then decays in $t$ to $t + \mathrm dt$ is

p(t)\,\mathrm dt = e^{-\lambda t}\lambda\,\mathrm dt

with $\lambda$ a positive decay constant.

(a) Show that the probability the particle eventually decays is 1.

(b) Find the mean lifetime.

(c) Find the probability that the particle lives for at least a time $T$ .

💡 Phillips' own hint

All three are the same integral, $\int e^{-\lambda t}\,\mathrm dt$ , over different ranges — plus one integration by parts for (b).

For (c), “lives at least $T$ ” means it decays at some time after $T$ , so integrate from $T$ to $\infty$ rather than from 0.

✓ Worked solution

(a) $\displaystyle\int_0^{\infty}\lambda e^{-\lambda t}\mathrm dt = \big[-e^{-\lambda t}\big]_0^{\infty} = 1$ . The particle decays with certainty — it just does not say when.

(b) $\displaystyle\langle t\rangle = \int_0^{\infty}t\lambda e^{-\lambda t}\mathrm dt = \frac{1}{\lambda}$ .

(c) $\displaystyle P(t > T) = \int_T^{\infty}\lambda e^{-\lambda t}\mathrm dt = e^{-\lambda T}$ .

The memoryless property. Given survival to time $T$ , the probability of surviving a further $s$ is $e^{-\lambda(T+s)}/e^{-\lambda T} = e^{-\lambda s}$ — independent of $T$ . A nucleus that has waited a billion years is exactly as likely to decay in the next second as a freshly made one. Nothing ages; there is no internal clock. It is also why $\Delta t = 1/\lambda = \langle t\rangle$ , as §3.1’s widget shows: mean and spread coincide.

Group 2 — the uncertainty product (problems 4–7)

Four wave functions, four values of $\Delta x\,\Delta p$ . Worked one at a time they are four disconnected integrals; the point only appears when you line them up.

N e^(−x²/2a²) — The only shape that touches the bound. Infinitely smooth, with no edges and no kinks anywhere — and its Fourier transform is another Gaussian, so squeezing one side widens the other by exactly the reciprocal factor.

4 The Gaussian wave function — the minimum numerical

A particle of mass $m$ with

\Psi(x) = N e^{-x^2/2a^2}

where $a$ is a constant length. Use the properties of the Gaussian distribution from problem 2 rather than integrating from scratch.

(a) Confirm $\langle x\rangle = 0$ and $\langle x^2\rangle = a^2/2$ .

(b) Show, without lengthy calculation, that $\langle p\rangle = 0$ and $\langle p^2\rangle = \hbar^2/2a^2$ . Hint from the book: use integration by parts to show $\int_{-\infty}^{+\infty}\Psi^*\frac{\mathrm d^2\Psi}{\mathrm dx^2}\mathrm dx = -\int_{-\infty}^{+\infty}\frac{\mathrm d\Psi^*}{\mathrm dx}\frac{\mathrm d\Psi}{\mathrm dx}\mathrm dx$ , then reuse the integrals of part (a).

(c) Hence show $\Delta x\,\Delta p = \hbar/2$ .

💡 Phillips' own hint

For (a), note $|\Psi|^2 \propto e^{-x^2/a^2}$ . Match that against problem 2’s $e^{-x^2/2\sigma^2}$ : you need $2\sigma^2 = a^2$ , so $\sigma = a/\sqrt2$ and $\langle x^2\rangle = \sigma^2 = a^2/2$ immediately.

For (b), $\langle p\rangle = 0$ by symmetry — $\Psi$ is real and even, so $\mathrm d\Psi/\mathrm dx$ is odd and the integrand is odd. For $\langle p^2\rangle$ , the book’s hint converts the second derivative into $|\mathrm d\Psi/\mathrm dx|^2$ , and $\mathrm d\Psi/\mathrm dx = -(x/a^2)\Psi$ turns the result into $\langle x^2\rangle/a^4$ .

✓ Worked solution

(a) $|\Psi|^2 \propto e^{-x^2/a^2}$ is a Gaussian with $2\sigma^2 = a^2$ , i.e. $\sigma = a/\sqrt2$ . Problem 2 then gives $\langle x\rangle = 0$ and $\langle x^2\rangle = \sigma^2 = a^2/2$ with no integration.

(b) $\langle p\rangle = 0$ by parity. For the square, the hint gives

\langle p^2\rangle = -\hbar^2\int\Psi^*\frac{\mathrm d^2\Psi}{\mathrm dx^2}\mathrm dx = \hbar^2\int\left|\frac{\mathrm d\Psi}{\mathrm dx}\right|^2\mathrm dx

With $\mathrm d\Psi/\mathrm dx = -(x/a^2)\Psi$ ,

\langle p^2\rangle = \frac{\hbar^2}{a^4}\int x^2|\Psi|^2\mathrm dx = \frac{\hbar^2}{a^4}\langle x^2\rangle = \frac{\hbar^2}{a^4}\cdot\frac{a^2}{2} = \frac{\hbar^2}{2a^2}

(c) $\Delta x = a/\sqrt2$ and $\Delta p = \hbar/a\sqrt2$ , so

\Delta x\,\Delta p = \frac{a}{\sqrt2}\cdot\frac{\hbar}{a\sqrt2} = \frac{\hbar}{2}

Exactly the floor — every $a$ cancels, so it holds for a Gaussian of any width. This is the only wave function in the book that achieves it.

Reproduce it in NumPy

The one wave function in the book that touches the floor — every factor of $a$ cancels, so it holds for a Gaussian of any width.

import numpy as np
hbar = 1.0

def stats(psi, x):
    psi = psi/np.sqrt(np.trapezoid(np.abs(psi)**2, x))
    r = np.abs(psi)**2
    mx, mx2 = np.trapezoid(x*r, x), np.trapezoid(x*x*r, x)
    d = np.gradient(psi, x, edge_order=2)
    mp  = np.trapezoid(np.conj(psi)*(-1j*hbar*d), x).real
    mp2 = np.trapezoid(np.abs(hbar*d)**2, x)
    return mx, np.sqrt(mx2-mx*mx), mp, np.sqrt(mp2-mp*mp)

a = 1.0
x = np.linspace(-40*a, 40*a, 800001)
mx, dx, mp, dp = stats(np.exp(-x**2/(2*a**2)) + 0j, x)
print(f"problem 4, the Gaussian   (a = {a:.0f})")
print(f"  <x> = {abs(mx):.6f}      <p> = {abs(mp):.6f}")
print(f"  Dx  = {dx:.6f}   book: a/sqrt(2)      = {a/np.sqrt(2):.6f}")
print(f"  Dp  = {dp:.6f}   book: hbar/a sqrt(2) = {hbar/(a*np.sqrt(2)):.6f}")
print(f"  Dx*Dp = {dx*dp:.6f} hbar  <- exactly hbar/2, the floor")

printsproblem 4, the Gaussian (a = 1) <x> = 0.000000 <p> = 0.000000 Dx = 0.707107 book: a/sqrt(2) = 0.707107 Dp = 0.707107 book: hbar/a sqrt(2) = 0.707107 Dx*Dp = 0.500000 hbar <- exactly hbar/2, the floor

5 A parabola in a box numerical

A particle confined to $0 \le x \le a$ with

\psi(x) = Nx(a - x)

(a) Normalize the wave function and find the average position.

(b) Show that

\Delta x = \sqrt{\frac{1}{28}}\,a \quad\text{and}\quad \Delta p = \sqrt{10}\,\frac{\hbar}{a}

💡 Phillips' own hint

Everything reduces to $\int_0^a x^n(a-x)^m\mathrm dx$ , which you can expand and integrate term by term.

$\langle x\rangle = a/2$ follows from symmetry about the midpoint — the parabola is symmetric, so no integration is needed.

For $\Delta p$ , use the same trick as problem 4: $\langle p^2\rangle = \hbar^2\int|\psi'|^2\mathrm dx$ , and $\psi' = N(a - 2x)$ is simple. Note $\langle p\rangle = 0$ because $\psi$ is real and bound.

✓ Worked solution

(a) $N^2\int_0^a x^2(a-x)^2\mathrm dx = N^2a^5/30 = 1$ , so $N = \sqrt{30/a^5}$ . By symmetry $\langle x\rangle = a/2$ .

(b) $\langle x^2\rangle = N^2\int_0^a x^4(a-x)^2\mathrm dx = \dfrac{30}{a^5}\cdot\dfrac{a^7}{105} = \dfrac{2a^2}{7}$ , so

(\Delta x)^2 = \frac{2a^2}{7} - \frac{a^2}{4} = \frac{8a^2 - 7a^2}{28} = \frac{a^2}{28}

For momentum, $\psi' = N(a-2x)$ and

\langle p^2\rangle = \hbar^2N^2\int_0^a(a-2x)^2\mathrm dx = \hbar^2\cdot\frac{30}{a^5}\cdot\frac{a^3}{3} = \frac{10\hbar^2}{a^2}

so $\Delta p = \sqrt{10}\,\hbar/a$ and

\Delta x\,\Delta p = \frac{a}{\sqrt{28}}\cdot\frac{\sqrt{10}\,\hbar}{a} = \sqrt{\frac{10}{28}}\,\hbar = 0.5976\,\hbar

Comfortably above $\hbar/2$ , and — see the ladder above — slightly worse than the true ground state it approximates.

Reproduce it in NumPy

Both stated answers, plus the normalization constant the problem asks you to find.

import numpy as np
hbar = 1.0

def stats(psi, x):
    psi = psi/np.sqrt(np.trapezoid(np.abs(psi)**2, x))
    r = np.abs(psi)**2
    mx, mx2 = np.trapezoid(x*r, x), np.trapezoid(x*x*r, x)
    d = np.gradient(psi, x, edge_order=2)
    mp  = np.trapezoid(np.conj(psi)*(-1j*hbar*d), x).real
    mp2 = np.trapezoid(np.abs(hbar*d)**2, x)
    return mx, np.sqrt(mx2-mx*mx), mp, np.sqrt(mp2-mp*mp)

a = 1.0
x = np.linspace(0, a, 800001)
psi = x*(a-x) + 0j
N = 1/np.sqrt(np.trapezoid(np.abs(psi)**2, x))
mx, dx, mp, dp = stats(psi, x)
print(f"problem 5, the parabola   (a = {a:.0f})")
print(f"  N   = {N:.6f}   analytic sqrt(30/a^5) = {np.sqrt(30/a**5):.6f}")
print(f"  <x> = {mx:.6f}   a/2 = {a/2:.6f}")
print(f"  Dx  = {dx:.6f}   book: a/sqrt(28)  = {np.sqrt(1/28)*a:.6f}")
print(f"  Dp  = {dp:.6f}   book: sqrt(10)hbar/a = {np.sqrt(10)*hbar/a:.6f}")
print(f"  Dx*Dp = {dx*dp:.6f} hbar   = sqrt(10/28) = {np.sqrt(10/28):.6f}")

printsproblem 5, the parabola (a = 1) N = 5.477226 analytic sqrt(30/a^5) = 5.477226 <x> = 0.500000 a/2 = 0.500000 Dx = 0.188982 book: a/sqrt(28) = 0.188982 Dp = 3.162278 book: sqrt(10)hbar/a = 3.162278 Dx*Dp = 0.597614 hbar = sqrt(10/28) = 0.597614

6 A cosine in a box — the ground state relocated numerical

A particle of mass $m$ confined to $-a/2 < x < +a/2$ , zero outside, and inside

\Psi = \sqrt{\frac{2}{a}}\cos\frac{\pi x}{a}

(a) Explain the physical significance of $\int_{-\infty}^{+\infty}\Psi^*x\Psi\,\mathrm dx$ and $\int_{-\infty}^{+\infty}\Psi^*\left(-i\hbar\frac{\mathrm d}{\mathrm dx}\right)\Psi\,\mathrm dx$ , and show that both are zero.

(b) Show that

\Delta x\,\Delta p = \sqrt{\frac{\pi^2 - 6}{12}}\,\hbar

using $\int_{-a/2}^{+a/2}x^2\cos^2\frac{\pi x}{a}\mathrm dx = \dfrac{a^3}{\pi^3}\left(\dfrac{\pi^3}{24} - \dfrac{\pi}{4}\right)$ .

💡 Phillips' own hint

For (a), both integrals are $\langle x\rangle$ and $\langle p\rangle$ — Eqs. (3.25) and (3.27). Both vanish by parity: $\Psi$ is real and even, so $x|\Psi|^2$ is odd, and $\Psi\,\mathrm d\Psi/\mathrm dx$ is odd too.

For (b), $\langle p^2\rangle$ is easiest from $\hbar^2\int|\Psi'|^2$ , with $\Psi' \propto \sin(\pi x/a)$ , giving $\langle p^2\rangle = (\pi\hbar/a)^2$ directly.

✓ Worked solution

(a) They are $\langle x\rangle$ and $\langle p\rangle$ : the average position and average momentum. Both vanish by symmetry, since $\Psi$ is even about the origin. Physically, the particle sits symmetrically in its box and goes nowhere on average — though neither $\Delta x$ nor $\Delta p$ is zero.

(b) From the supplied integral,

\langle x^2\rangle = \frac{2}{a}\cdot\frac{a^3}{\pi^3}\left(\frac{\pi^3}{24}-\frac{\pi}{4}\right) = a^2\left(\frac{1}{12} - \frac{1}{2\pi^2}\right)

and $\langle p^2\rangle = \pi^2\hbar^2/a^2$ . Hence

\Delta x\,\Delta p = a\sqrt{\frac{1}{12}-\frac{1}{2\pi^2}}\cdot\frac{\pi\hbar}{a} = \sqrt{\frac{\pi^2-6}{12}}\,\hbar = 0.5679\,\hbar

This is the box ground state of §3.4, shifted so the box is centred on the origin. Sure enough it reproduces the $0.5679\,\hbar$ computed there — a useful check that shifting the origin changes $\langle x\rangle$ but not $\Delta x$ .

Reproduce it in NumPy

Part (a)’s two zeros and part (b)’s product, and the cross-check against §3.4.

import numpy as np
hbar = 1.0

def stats(psi, x):
    psi = psi/np.sqrt(np.trapezoid(np.abs(psi)**2, x))
    r = np.abs(psi)**2
    mx, mx2 = np.trapezoid(x*r, x), np.trapezoid(x*x*r, x)
    d = np.gradient(psi, x, edge_order=2)
    mp  = np.trapezoid(np.conj(psi)*(-1j*hbar*d), x).real
    mp2 = np.trapezoid(np.abs(hbar*d)**2, x)
    return mx, np.sqrt(mx2-mx*mx), mp, np.sqrt(mp2-mp*mp)

a = 1.0
x = np.linspace(-a/2, a/2, 800001)
mx, dx, mp, dp = stats(np.sqrt(2/a)*np.cos(np.pi*x/a) + 0j, x)
print(f"problem 6, the cosine   (a = {a:.0f})")
print(f"  <x> = {abs(mx):.6f}   <- part (a): zero by symmetry")
print(f"  <p> = {abs(mp):.6f}   <- part (a): zero by symmetry")
print(f"  Dx  = {dx:.6f}   a*sqrt(1/12 - 1/2pi^2) = {a*np.sqrt(1/12-1/(2*np.pi**2)):.6f}")
print(f"  Dp  = {dp:.6f}   pi*hbar/a = {np.pi*hbar/a:.6f}")
print(f"  Dx*Dp = {dx*dp:.6f} hbar   book: sqrt((pi^2-6)/12) = {np.sqrt((np.pi**2-6)/12):.6f}")
print(f"  and section 3.4 got {0.5679:.4f} for the same state in the box 0..a")

printsproblem 6, the cosine (a = 1) <x> = 0.000000 <- part (a): zero by symmetry <p> = 0.000000 <- part (a): zero by symmetry Dx = 0.180756 a*sqrt(1/12 - 1/2pi^2) = 0.180756 Dp = 3.141593 pi*hbar/a = 3.141593 Dx*Dp = 0.567862 hbar book: sqrt((pi^2-6)/12) = 0.567862 and section 3.4 got 0.5679 for the same state in the box 0..a

7 A one-sided exponential numerical

A particle with normalized wave function

\Psi(x) = \begin{cases} Nxe^{-\alpha x/2} & \text{if } 0 \le x < \infty\\ 0 & \text{elsewhere}\end{cases}

where $\alpha$ is a positive real constant and $N = \sqrt{\alpha^3/2}$ .

(a) Write down the probability of finding the particle between $x$ and $x + \mathrm dx$ , describe how it depends on $x$ , and find the most probable value of $x$ .

(b) Find $\langle x\rangle$ and $\langle x^2\rangle$ .

(c) Find $\langle p\rangle$ and $\langle p^2\rangle$ .

(d) Show these give uncertainties consistent with the Heisenberg relation. The identity $\int_0^{\infty}e^{-\alpha x}x^n\mathrm dx = \dfrac{n!}{\alpha^{n+1}}$ for $n > -1$ is useful.

💡 Phillips' own hint

Every integral here is the supplied identity with a different $n$ — set them up and read off the answer.

For (a), the most probable $x$ maximises $|\Psi|^2 = N^2x^2e^{-\alpha x}$ ; differentiate and set to zero.

For (c), $\langle p\rangle = 0$ because $\Psi$ is real and normalizable (a bound state carries no net momentum), and $\langle p^2\rangle = \hbar^2\int|\Psi'|^2$ with $\Psi' = N e^{-\alpha x/2}(1 - \alpha x/2)$ .

✓ Worked solution

(a) $|\Psi|^2\mathrm dx = \dfrac{\alpha^3}{2}x^2e^{-\alpha x}\mathrm dx$ . It vanishes at the origin (the $x^2$ ), rises to a peak, then decays exponentially. Setting $\mathrm d(x^2e^{-\alpha x})/\mathrm dx = 0$ gives $2x = \alpha x^2$ , so the most probable position is $x = 2/\alpha$ .

(b) With the identity,

\langle x\rangle = \frac{\alpha^3}{2}\cdot\frac{3!}{\alpha^4} = \frac{3}{\alpha},\qquad \langle x^2\rangle = \frac{\alpha^3}{2}\cdot\frac{4!}{\alpha^5} = \frac{12}{\alpha^2}

so $(\Delta x)^2 = 12/\alpha^2 - 9/\alpha^2 = 3/\alpha^2$ and $\Delta x = \sqrt3/\alpha$ .

Note the mean $3/\alpha$ exceeds the mode $2/\alpha$ : the long right-hand tail drags the average past the peak. That is what skew does, and it is why this state ends up far from the uncertainty floor.

(c) $\langle p\rangle = 0$ . For the square, with $\Psi' = Ne^{-\alpha x/2}(1 - \alpha x/2)$ ,

\langle p^2\rangle = \hbar^2N^2\int_0^{\infty}e^{-\alpha x}\left(1 - \alpha x + \frac{\alpha^2x^2}{4}\right)\mathrm dx = \hbar^2N^2\left(\frac{1}{\alpha} - \frac{1}{\alpha} + \frac{1}{2\alpha}\right) = \frac{\hbar^2\alpha^2}{4}

so $\Delta p = \hbar\alpha/2$ . Note how the first two terms cancel exactly.

(d)

\Delta x\,\Delta p = \frac{\sqrt3}{\alpha}\cdot\frac{\hbar\alpha}{2} = \frac{\sqrt3}{2}\hbar \approx 0.866\,\hbar

which is $> \hbar/2$ , consistent with Heisenberg — and the furthest from the floor of the four, which is the ladder’s point about skewness.

Reproduce it in NumPy

The book asks for these but states no answer, so every line here is derived rather than checked.

import numpy as np
hbar = 1.0

def stats(psi, x):
    psi = psi/np.sqrt(np.trapezoid(np.abs(psi)**2, x))
    r = np.abs(psi)**2
    mx, mx2 = np.trapezoid(x*r, x), np.trapezoid(x*x*r, x)
    d = np.gradient(psi, x, edge_order=2)
    mp  = np.trapezoid(np.conj(psi)*(-1j*hbar*d), x).real
    mp2 = np.trapezoid(np.abs(hbar*d)**2, x)
    return mx, np.sqrt(mx2-mx*mx), mp, np.sqrt(mp2-mp*mp)

al = 1.0
x = np.linspace(0, 120/al, 1200001)
psi = x*np.exp(-al*x/2) + 0j
N = 1/np.sqrt(np.trapezoid(np.abs(psi)**2, x))
mx, dx, mp, dp = stats(psi, x)
rho = np.abs(psi/np.sqrt(np.trapezoid(np.abs(psi)**2,x)))**2
print(f"problem 7, the one-sided exponential   (alpha = {al:.0f})")
print(f"  N    = {N:.6f}   book: sqrt(alpha^3/2) = {np.sqrt(al**3/2):.6f}")
print(f"  most probable x = {x[int(np.argmax(rho))]:.6f}   analytic 2/alpha = {2/al:.6f}")
print(f"  <x>  = {mx:.6f}   analytic 3/alpha  = {3/al:.6f}")
print(f"  Dx   = {dx:.6f}   analytic sqrt(3)/alpha = {np.sqrt(3)/al:.6f}")
print(f"  Dp   = {dp:.6f}   analytic hbar*alpha/2  = {hbar*al/2:.6f}")
print(f"  Dx*Dp = {dx*dp:.6f} hbar = sqrt(3)/2 = {np.sqrt(3)/2:.6f}  <- furthest from the floor")

printsproblem 7, the one-sided exponential (alpha = 1) N = 0.707107 book: sqrt(alpha^3/2) = 0.707107 most probable x = 2.000000 analytic 2/alpha = 2.000000 <x> = 3.000000 analytic 3/alpha = 3.000000 Dx = 1.732051 analytic sqrt(3)/alpha = 1.732051 Dp = 0.500000 analytic hbar*alpha/2 = 0.500000 Dx*Dp = 0.866025 hbar = sqrt(3)/2 = 0.866025 <- furthest from the floor

What the four problems add up to

The book asks for four numbers and never says what they have in common. Ranked:

Wave function	$\Delta x\,\Delta p / \hbar$	excess over the floor
Gaussian	0.5000	0%
cosine in a box	0.5679	14%
parabola in a box	0.5976	20%
one-sided exponential	0.8660	73%

Smoothness is what buys you closeness to the bound. The Gaussian is infinitely differentiable with no edges anywhere. The cosine is continuous and vanishes at both walls, but its derivative jumps there, and that kink costs 14%. The parabola has the same kink and a slightly worse shape. The exponential is badly skewed, with a long tail that inflates $\langle x^2\rangle$ far more than it moves $\langle x\rangle$ .

The ordering is not an accident of these four examples: sharp features in $x$ mean broad content in $p$ , which is the entire content of §3.3’s Fourier argument. The problems are a numerical demonstration of it, one wave function at a time.

Group 3 — two derivations (problems 8–9)

8 The probability current derivation

$\rho(x,t)$ is the position probability density. Its value in a region changes with time, and that change should be attributable to a flow of probability in and out — so we expect a probability current density $j(x,t)$ obeying the continuity equation

\frac{\partial\rho}{\partial t} = -\frac{\partial j}{\partial x}

(a) Using the Schrödinger equation $i\hbar\frac{\partial\Psi}{\partial t} = \left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x)\right]\Psi$ , derive an expression for the time derivative of $\rho(x,t) = \Psi^*(x,t)\Psi(x,t)$ .

(b) Hence show that the probability current is given by

j(x,t) = \frac{i\hbar}{2m}\left[\Psi\frac{\partial\Psi^*}{\partial x} - \Psi^*\frac{\partial\Psi}{\partial x}\right]

💡 Phillips' own hint

For (a), differentiate the product: $\partial_t\rho = \Psi^*\partial_t\Psi + \Psi\,\partial_t\Psi^*$ . Get $\partial_t\Psi$ from the Schrödinger equation and $\partial_t\Psi^*$ by conjugating it — the conjugate flips the sign of every $i$ , which is what makes the potential terms cancel.

Then recognise that $\Psi^*\Psi'' - \Psi\Psi^{*\prime\prime}$ is a perfect derivative: it equals $\partial_x(\Psi^*\Psi' - \Psi\Psi^{*\prime})$ .

For (b), compare what you have with $-\partial_x j$ and read off $j$ . Watch the sign — and see the Caution below.

✓ Worked solution

(a) From the equation, $\partial_t\Psi = \dfrac{i\hbar}{2m}\Psi'' - \dfrac{i}{\hbar}V\Psi$ , and conjugating, $\partial_t\Psi^* = -\dfrac{i\hbar}{2m}\Psi^{*\prime\prime} + \dfrac{i}{\hbar}V\Psi^*$ . Hence

\frac{\partial\rho}{\partial t} = \Psi^*\partial_t\Psi + \Psi\partial_t\Psi^* = \frac{i\hbar}{2m}\left[\Psi^*\Psi'' - \Psi\Psi^{*\prime\prime}\right]

The potential terms cancel exactly — $V$ is real, so it contributes $-\frac{i}{\hbar}V|\Psi|^2 + \frac{i}{\hbar}V|\Psi|^2 = 0$ . Probability is conserved whatever the potential.

(b) The bracket is a perfect derivative, so

\frac{\partial\rho}{\partial t} = \frac{\partial}{\partial x}\left\{\frac{i\hbar}{2m}\left[\Psi^*\frac{\partial\Psi}{\partial x} - \Psi\frac{\partial\Psi^*}{\partial x}\right]\right\}

Matching against $\partial_t\rho = -\partial_x j$ gives

j(x,t) = \frac{i\hbar}{2m}\left[\Psi\frac{\partial\Psi^*}{\partial x} - \Psi^*\frac{\partial\Psi}{\partial x}\right] = \frac{\hbar}{m}\,\mathrm{Im}\left(\Psi^*\frac{\partial\Psi}{\partial x}\right)

Check it on a plane wave. For $\Psi = e^{ikx}$ , $\mathrm{Im}(\Psi^*\Psi') = \mathrm{Im}(ik) = k$ , so $j = \hbar k/m = v$ — the density (1) times the velocity. Exactly what a current should be, and positive for a right-moving wave.

Reproduce it in NumPy

Two independent tests of the sign. The packet is evolved with an exact split-step propagator, so the left-hand side of the continuity equation is measured rather than assumed.

import numpy as np
hbar = m = 1.0

print("problem 8: is the printed current the right sign?")
print("  book's 8(b):  j = (i*hbar/2m)[Psi* dPsi/dx - Psi dPsi*/dx]")
print("  preamble   :  drho/dt = -dj/dx")

# test 1 -- a right-moving plane wave must carry a rightward current
k = 2.0
x = np.linspace(0, 10, 200001)
Psi = np.exp(1j*k*x)
d = lambda f: np.gradient(f, x, edge_order=2)
j_book = ((1j*hbar/(2*m))*(np.conj(Psi)*d(Psi) - Psi*d(np.conj(Psi)))).real
j_std  = (hbar/m)*np.imag(np.conj(Psi)*d(Psi))
print(f"\n  plane wave e^(ikx), k = +{k:.0f} (moving RIGHT), |Psi|^2 = 1")
print(f"    expected  rho * v = hbar*k/m = {hbar*k/m:+.4f}")
print(f"    book's j          = {j_book.mean():+.4f}   <- wrong sign")
print(f"    (hbar/m) Im(Psi* dPsi/dx) = {j_std.mean():+.4f}   <- correct")

# test 2 -- continuity, with drho/dt measured from an exact propagator
L, N = 40.0, 4000
xx = np.linspace(-L/2, L/2, N, endpoint=False); dxg = xx[1]-xx[0]
p0 = np.exp(-xx**2/2)*np.exp(3j*xx); p0 /= np.sqrt(np.sum(abs(p0)**2)*dxg)
kk = 2*np.pi*np.fft.fftfreq(N, d=dxg)
evolve = lambda p, dt: np.fft.ifft(np.exp(-1j*hbar*kk**2/(2*m)*dt)*np.fft.fft(p))
dt = 1e-4
rho_t = (np.abs(evolve(p0, dt))**2 - np.abs(evolve(p0, -dt))**2)/(2*dt)
D = lambda f: np.gradient(f, dxg, edge_order=2)
jb = ((1j*hbar/(2*m))*(np.conj(p0)*D(p0) - p0*D(np.conj(p0)))).real
js = (hbar/m)*np.imag(np.conj(p0)*D(p0))
c = slice(200, -200)
print(f"\n  a Gaussian packet evolved by the Schrodinger equation:")
print(f"    max |drho/dt|                    = {np.abs(rho_t[c]).max():.4f}")
print(f"    max |drho/dt + dj/dx|, book's j  = {np.abs(rho_t[c]+D(jb)[c]).max():.4f}   <- violates continuity")
print(f"    max |drho/dt + dj/dx|, correct j = {np.abs(rho_t[c]+D(js)[c]).max():.4f}   <- satisfies it")
print("    the residual is exactly twice the scale: the signature of a flipped sign.")

printsproblem 8: is the printed current the right sign? book's 8(b): j = (i*hbar/2m)[Psi* dPsi/dx - Psi dPsi*/dx] preamble : drho/dt = -dj/dx plane wave e^(ikx), k = +2 (moving RIGHT), |Psi|^2 = 1 expected rho * v = hbar*k/m = +2.0000 book's j = -2.0000 <- wrong sign (hbar/m) Im(Psi* dPsi/dx) = +2.0000 <- correct a Gaussian packet evolved by the Schrodinger equation: max |drho/dt| = 1.4518 max |drho/dt + dj/dx|, book's j = 2.9032 <- violates continuity max |drho/dt + dj/dx|, correct j = 0.0004 <- satisfies it the residual is exactly twice the scale: the signature of a flipped sign.

The book prints problem 8(b) with the two terms the wrong way round

Page 57 gives

j(x,t) = \frac{i\hbar}{2m}\left[\Psi^*\frac{\partial\Psi}{\partial x} - \Psi\frac{\partial\Psi^*}{\partial x}\right]

which is the negative of the correct result. Two independent checks, both in the Compute block below:

The plane-wave test. For $\Psi = e^{ikx}$ with $k > 0$ — a wave moving to the right — the printed formula gives $j = -\hbar k/m$ . A right-moving wave with a leftward current is not a bookkeeping convention; it is a sign error.

The continuity test. Evolve a Gaussian packet with the actual Schrödinger equation and measure $\partial\rho/\partial t$ numerically. The correct $j$ satisfies $\partial\rho/\partial t + \partial j/\partial x = 0$ to $4\times10^{-4}$ (numerical noise); the printed $j$ leaves a residual of 2.90 against a $|\partial\rho/\partial t|$ of 1.45 — exactly twice the scale, the signature of a flipped sign.

So the correct expression is the one in the solution above, equivalently $j = \frac{\hbar}{m}\mathrm{Im}(\Psi^*\partial_x\Psi)$ , which is the standard form. The derivation in part (a) is unaffected; only the final line’s ordering is wrong. This is the fourth misprint in chapter 3 — see §3.2, §3.5–3.6 and problem 4 above for the others.

Engineer's bridge — this is a conservation law, and you have written one

$\partial\rho/\partial t + \partial j/\partial x = 0$ is the one-dimensional continuity equation, and in 3-D the book’s footnote writes it $\partial\rho/\partial t = -\nabla\cdot\mathbf j$ . You have met it as charge conservation in electromagnetism, as mass conservation in fluid dynamics, and as the governing equation of every finite-volume solver ever written.

The footnote’s own argument is the finite-volume argument exactly: integrate over a cell $[x_1, x_2]$ and

\frac{\mathrm d}{\mathrm dt}\int_{x_1}^{x_2}\rho\,\mathrm dx = j(x_1,t) - j(x_2,t)

what is inside changes only by what crosses the boundary. That is a flux balance, and it is why conservative schemes are built around fluxes at cell faces rather than derivatives at cell centres.

Where it breaks. In fluid dynamics $\rho$ is a density of stuff you could in principle scoop out and weigh. Here it is a density of probability for a single particle — there is no fluid, and $j$ is not a flow of anything material. Phillips is careful about this: he notes that with many particles you get an actual current by multiplying $j$ by the actual particle density. For one particle it is bookkeeping for where the particle is likely to turn up, and the “flow” is a flow of expectation, not of substance.

The mathematics is nonetheless load-bearing: it is what guarantees that normalizing $\Psi$ once, at $t = 0$ , keeps it normalized for ever — the promise §3.2 made and deferred to here.

9 Ehrenfest's relation derivation

Use Eqs. (3.25) and (3.27) to show that the expectation values of position and momentum for a particle of mass $m$ are related by

m\frac{\mathrm d\langle x\rangle}{\mathrm dt} = \langle p\rangle

The method is similar to problem 8.

(a) Show that $\dfrac{\mathrm d(\Psi^*x\Psi)}{\mathrm dt} = \dfrac{i\hbar}{2m}\left[x\Psi^*\dfrac{\partial^2\Psi}{\partial x^2} - x\Psi\dfrac{\partial^2\Psi^*}{\partial x^2}\right]$ , and rewrite it as $\dfrac{i\hbar}{2m}\dfrac{\partial}{\partial x}\left[x\Psi^*\dfrac{\partial\Psi}{\partial x} - x\Psi\dfrac{\partial\Psi^*}{\partial x}\right] - \dfrac{i\hbar}{2m}\left[\Psi^*\dfrac{\partial\Psi}{\partial x} - \Psi\dfrac{\partial\Psi^*}{\partial x}\right]$ .

(b) Assuming $\Psi\to0$ sufficiently rapidly at $x = \pm\infty$ , show that $\dfrac{\mathrm d}{\mathrm dt}\int\Psi^*x\Psi\,\mathrm dx = -\dfrac{i\hbar}{2m}\int\left[\Psi^*\dfrac{\partial\Psi}{\partial x} - \Psi\dfrac{\partial\Psi^*}{\partial x}\right]\mathrm dx$ .

(c) Integrate by parts to show $m\dfrac{\mathrm d}{\mathrm dt}\int\Psi^*x\Psi\,\mathrm dx = \int\Psi^*\left(-i\hbar\dfrac{\partial}{\partial x}\right)\Psi\,\mathrm dx$ .

💡 Phillips' own hint

Part (a) is problem 8(a) with an extra factor of $x$ carried along; the potential terms cancel for the same reason. The rewriting is just the product rule run backwards — differentiating the first bracket regenerates the second-derivative terms plus the leftover you are subtracting.

In (b), the first term of (a) is a perfect $x$ -derivative, so it integrates to a boundary term that dies.

In (c), integrating $\int\Psi\,\partial_x\Psi^*\mathrm dx$ by parts turns it into $-\int\Psi^*\partial_x\Psi\,\mathrm dx$ , so the two terms in the bracket combine rather than cancel.

✓ Worked solution

(a) As in problem 8, $V$ cancels, leaving the stated form with $x$ carried through. The rewriting is the product rule:

\frac{\partial}{\partial x}\left[x\Psi^*\Psi' - x\Psi\Psi^{*\prime}\right] = \left[x\Psi^*\Psi'' - x\Psi\Psi^{*\prime\prime}\right] + \left[\Psi^*\Psi' - \Psi\Psi^{*\prime}\right]

so subtracting the second bracket recovers (a).

(b) Integrating, the perfect-derivative term contributes $\big[\,\cdot\,\big]_{-\infty}^{+\infty} = 0$ since $\Psi\to0$ . What survives is

\frac{\mathrm d\langle x\rangle}{\mathrm dt} = -\frac{i\hbar}{2m}\int\left[\Psi^*\frac{\partial\Psi}{\partial x} - \Psi\frac{\partial\Psi^*}{\partial x}\right]\mathrm dx

(c) Integrating the second term by parts (boundary term again zero):

\int\Psi\frac{\partial\Psi^*}{\partial x}\mathrm dx = -\int\Psi^*\frac{\partial\Psi}{\partial x}\mathrm dx

so the bracket becomes $2\int\Psi^*\partial_x\Psi\,\mathrm dx$ and

m\frac{\mathrm d\langle x\rangle}{\mathrm dt} = -i\hbar\int\Psi^*\frac{\partial\Psi}{\partial x}\mathrm dx = \int\Psi^*\hat p\,\Psi\,\mathrm dx = \langle p\rangle

which is Eq. (3.28).

Why problem 9 matters more than it looks

The result $m\,\mathrm d\langle x\rangle/\mathrm dt = \langle p\rangle$ is the classical definition of momentum — for averages.

This is the first piece of Ehrenfest’s theorem, and it is how the correspondence principle gets its teeth. Expectation values obey the classical equations of motion even though nothing else in the theory does. A baseball is described by a wave function too; its $\langle x\rangle$ moves exactly as Newton says, and its $\Delta x$ is around $10^{-34}$ m, so the quantum description is indistinguishable from the classical trajectory.

Note carefully what it does not say. It does not say the particle has a position moving with velocity $\langle p\rangle/m$ — the $n = 2$ box state of §3.5 has $\langle x\rangle$ sitting on a node where the particle is never found. The average moves classically; no individual outcome has to.

The second half of Ehrenfest’s theorem, $\mathrm d\langle p\rangle/\mathrm dt = -\langle\mathrm dV/\mathrm dx\rangle$ — Newton’s second law for averages — is not in this book’s problems, but follows by the same method with $\hat p$ in the sandwich instead of $\hat x$ .

Chapter 3 is complete

Position and momentum both come from one wave function; expectation values and uncertainties both come from sandwich integrals; observables are operators; and probability is conserved, so normalizing once is enough.

Chapter 4 applies the same machinery to energy, and gets something new out of it: for a confined particle the possible outcomes are discrete. That is where quantization finally comes from — and where the $E_n = \hbar^2k_n^2/2m$ that §3.4 borrowed on trust gets derived.

Check yourself

0 / 8 answered

1.For the Poisson distribution, $⟨ n ⟩ = a$ and $Δ n = a$ . What follows for a counting experiment?
2.Problem 2(a) asks you to show $⟨ x ⟩ = 0$ for a Gaussian. What is the intended method?
3.An exponential lifetime distribution has $⟨ t ⟩ = Δ t = 1/ λ$ . What does the equality signify?
4.Ranking problems 4–7 by $Δ x Δ p$ gives 0.500, 0.568, 0.598, 0.866 in units of $ℏ$ . What explains the ordering?
5.Problem 6's cosine gives $Δ x Δ p = 0.5679 ℏ$ — the same number §3.4 found for the box ground state. Why?
6.The book prints problem 8(b)'s probability current as $j = \frac{i ℏ}{2 m} [Ψ^{*} \partial_{x} Ψ - Ψ \partial_{x} Ψ^{*}]$ . What is wrong with it?
7.Problem 8(a) shows the potential terms cancel in $\partial ρ / \partial t$ . What does that establish?
8.Problem 9 proves $m d ⟨ x ⟩ / d t = ⟨ p ⟩$ . What does this NOT say?