Two of these problems introduce the first real interaction in the chapter. Everything before them got its effects from symmetry alone, and the contrast is the point of the set.

Six problems, and they divide cleanly. 1 and 2 close the loops the chapter left open — that exchange symmetry survives time evolution, and that the two-particle oscillator really does separate. 3 is the important one: switch on a real interaction and the exchange integral appears, which is the quantity the whole chapter has been circling without naming. 4 and 5 are counting exercises in four statistical cases each. 6 turns the abstraction into a laboratory fact — hydrogen gas behaves as two different substances.

Closing the chapter’s open loops

1 Exchange symmetry is a constant of the motion derivation

Given that two-particle time evolution obeys

i\hbar\frac{\partial \Psi(\mathbf r_p,\mathbf r_q,t)}{\partial t} = \hat H(\mathbf r_p,\mathbf r_q)\,\Psi(\mathbf r_p,\mathbf r_q,t)

show that

\Psi(\mathbf r_p,\mathbf r_q,t+\mathrm dt) = \left[1 - \frac{i}{\hbar}\hat H(\mathbf r_p,\mathbf r_q)\,\mathrm dt\right]\Psi(\mathbf r_p,\mathbf r_q,t)

What general symmetry condition must $\hat H$ satisfy for the exchange symmetry to be the same at $t$ and $t+\mathrm dt$ ? Explain why that condition always holds for identical particles.

💡 Phillips' own hint

Exchange symmetry is a constant of motion if the Hamiltonian operator is unchanged when the particles are exchanged. Eq. (10.7) is an example of a two-particle Hamiltonian.

✓ Worked solution

The evolution step. For infinitesimal $\mathrm dt$ , expand to first order:

\Psi(t+\mathrm dt) = \Psi(t) + \frac{\partial\Psi}{\partial t}\mathrm dt = \Psi(t) + \frac{1}{i\hbar}\hat H\Psi(t)\,\mathrm dt = \left[1 - \frac{i}{\hbar}\hat H\,\mathrm dt\right]\Psi(t)

using $1/i = -i$ . That is the required result, and it is just Euler’s method on the Schrödinger equation.

The condition. Write $\hat P$ for exchange, so a state of definite symmetry has $\hat P\Psi = \pm\Psi$ . Apply $\hat P$ to the evolution step and push it through to the right:

\hat P\,\Psi(t+\mathrm dt) = \hat P\left[1 - \tfrac{i}{\hbar}\hat H\,\mathrm dt\right]\Psi(t) = \left[1 - \tfrac{i}{\hbar}\,\hat P\hat H\hat P\,\mathrm dt\right]\hat P\,\Psi(t)

using $\hat P^2 = \hat I$ to insert $\hat P\hat P$ after $\hat H$ . The bracket on the right is the evolution operator again only if $\hat P\hat H\hat P = \hat H$ , which is to say

\hat P\hat H = \hat H\hat P \qquad\text{equivalently}\qquad \hat H(\mathbf r_p,\mathbf r_q) = \hat H(\mathbf r_q,\mathbf r_p)

When that holds,

\hat P\,\Psi(t+\mathrm dt) = \left[1 - \tfrac{i}{\hbar}\hat H\,\mathrm dt\right]\hat P\,\Psi(t) = \pm\left[1 - \tfrac{i}{\hbar}\hat H\,\mathrm dt\right]\Psi(t) = \pm\,\Psi(t+\mathrm dt)

so the sign survives one step, and iterating carries it to all later times.

Why it always holds. For identical particles the Hamiltonian cannot distinguish $p$ from $q$ — the same mass, the same charge, the same potential. Any term involving $\mathbf r_p$ appears with a twin involving $\mathbf r_q$ . Eq. (10.7) shows the pattern: swapping $x_p \leftrightarrow x_q$ maps each term onto the other and leaves the sum unchanged. Being identical is exactly the statement that $\hat H$ is symmetric under exchange, so the condition is not an extra assumption — it is the definition, restated.

You have seen this proof twice already, and it is the same proof

The structure here is §7.5’s conservation argument, verbatim: an observable whose operator commutes with $\hat H$ does not change in time. There it produced conservation of energy and of angular momentum; here it produces conservation of exchange symmetry.

What is unusual is that $\hat P$ is not an observable anyone measures. You never build an exchange-symmetry meter. Yet the conservation law is what makes the whole chapter coherent: a state that starts antisymmetric stays antisymmetric, forever, so “electrons are fermions” is a statement that can be true once and remain true rather than something needing continual enforcement.

For a programmer the analogy is an invariant preserved by every state transition. You establish it at construction, prove every mutation preserves it, and thereafter you may assume it without checking. $[\hat H,\hat P]=0$ is that proof obligation, discharged once for all identical particles.

2 The two-particle oscillator really does separate derivation

The two-particle eigenvalue equation is

\left[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_p^2} -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_q^2} + \tfrac12 m\omega^2x_p^2 + \tfrac12 m\omega^2x_q^2\right]\psi = E\psi

Show by substitution that $\psi(x_p,x_q) = \psi_n(x_p)\psi_{n'}(x_q)$ satisfies it with $E = E_n + E_{n'}$ , where $\psi_n$ and $\psi_{n'}$ are one-particle oscillator eigenfunctions.

✓ Worked solution

Substitute the product. The first derivative term acts only on the first factor and the second only on the second, because each is a partial derivative:

-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x_p^2}\left[\psi_n(x_p)\psi_{n'}(x_q)\right] = \psi_{n'}(x_q)\left[-\frac{\hbar^2}{2m}\frac{\mathrm d^2\psi_n}{\mathrm dx_p^2}\right]

Group the four terms into two brackets, one per particle:

\psi_{n'}(x_q)\underbrace{\left[-\frac{\hbar^2}{2m}\frac{\mathrm d^2}{\mathrm dx_p^2} + \tfrac12 m\omega^2x_p^2\right]\psi_n(x_p)}_{= E_n\psi_n(x_p)} +\;\psi_n(x_p)\underbrace{\left[-\frac{\hbar^2}{2m}\frac{\mathrm d^2}{\mathrm dx_q^2} + \tfrac12 m\omega^2x_q^2\right]\psi_{n'}(x_q)}_{= E_{n'}\psi_{n'}(x_q)}

Each bracket is the one-particle oscillator Hamiltonian acting on its own eigenfunction, so each returns its eigenvalue times the function:

= E_n\psi_n(x_p)\psi_{n'}(x_q) + E_{n'}\psi_n(x_p)\psi_{n'}(x_q) = (E_n+E_{n'})\,\psi

So $\psi$ is an eigenfunction with $E = E_n + E_{n'}$ , as claimed.

Separability is the whole reason this chapter can use products

Every wave function in §10.2 was built from products $\psi_n(x_p)\psi_{n'}(x_q)$ , and problem 2 is the licence to do that. It works for one reason only: the Hamiltonian is a sum of terms each involving one particle. No $x_px_q$ anywhere.

The moment you add an interaction — problem 3 — the separation fails and the products stop being exact eigenfunctions. They remain a perfectly good basis, which is why problem 3 can still use them for a first-order energy shift, but the exact states are no longer products of anything.

This is the same principle as separation of variables in §4.4’s 3-D box, where the three coordinates decoupled for the same reason. A sum in the Hamiltonian becomes a product in the wave function, and a sum in the energy.

The one that matters

3 Normalization, and the exchange integral the book never names derivation

Take the three wave functions of §10.2 — $\Psi^{(D)}$ from Eq. (10.9), $\Psi^{(S)}$ from Eq. (10.12) and $\Psi^{(A)}$ from Eq. (10.13).

(a) Show that all three are normalized when $\psi_n$ and $\psi_{n'}$ are normalized and orthogonal.

(b) Suppose a weak repulsive interaction $V(|x_p-x_q|)$ shifts the energy by

\Delta E = \int_{-\infty}^{\infty}\mathrm dx_p\int_{-\infty}^{\infty}\mathrm dx_q\;\Psi^*(x_p,x_q,t)\,V(|x_p-x_q|)\,\Psi(x_p,x_q,t)

How does the shift for distinguishable particles compare with the shifts for identical ones?

💡 Phillips' own hint

(a) Show that $\int\mathrm dx_p\int\mathrm dx_q\,|\Psi|^2 = 1$ .

(b) Show that

\Delta E^{(S)} = \Delta E^{(D)} + K \qquad\text{and}\qquad \Delta E^{(A)} = \Delta E^{(D)} - K

where

K = \int_{-\infty}^{\infty}\mathrm dx_p\int_{-\infty}^{\infty}\mathrm dx_q\;\psi_n^*(x_q)\psi_{n'}^*(x_p)\,V(|x_p-x_q|)\,\psi_n(x_p)\psi_{n'}(x_q)

✓ Worked solution

(a) For $\Psi^{(D)}$ the double integral factorizes into $\int|\psi_n|^2\mathrm dx_p \int|\psi_{n'}|^2\mathrm dx_q = 1\cdot1$ .

For $\Psi^{(S)}$ and $\Psi^{(A)}$ , expand $|\Psi|^2$ . The two direct terms each integrate to 1, giving $\tfrac12(1+1)$ . The two cross terms are

\pm\tfrac12\int\psi_n^*(x_p)\psi_{n'}(x_p)\,\mathrm dx_p\int\psi_{n'}^*(x_q)\psi_n(x_q)\,\mathrm dx_q = 0

because $\psi_n$ and $\psi_{n'}$ are orthogonal. So the norm is 1 in both cases, and this is precisely why $1/\sqrt2$ is the right prefactor. The time factor $e^{-i(E_n+E_{n'})t/\hbar}$ has modulus 1 and never contributes.

(b) The same expansion, now with $V$ inserted. For the distinguishable case

\Delta E^{(D)} = \int\!\!\int |\psi_n(x_p)|^2\,V\,|\psi_{n'}(x_q)|^2 = \text{the \emph{direct} term}

which is just the average of $V$ over the classical-looking product density. For the identical cases the cross terms no longer vanish, because $V$ sits between them and destroys the orthogonality that killed them in part (a):

\Delta E^{(S)} = \Delta E^{(D)} + K \qquad \Delta E^{(A)} = \Delta E^{(D)} - K

$K$ is the exchange integral . Note its structure: the two wave functions on the left have their arguments swapped relative to those on the right. It has no classical counterpart — it is the energy of an interference term.

The comparison. For a repulsive $V$ , $K > 0$ , so

\Delta E^{(A)} < \Delta E^{(D)} < \Delta E^{(S)}

The antisymmetric state is cheapest, because those particles were already avoiding each other and so feel a short-range repulsion least. The symmetric state is dearest, for the mirror-image reason.

This is where a real force finally enters — and the contrast with §10.2 is the point

§10.2 was emphatic that huddling and avoidance are not forces, and the proof was that all three states had identical energies. Problem 3 adds $V$ , and now they do not. Both statements are true and the difference between them is worth being exact about:

Without an interaction, symmetry changes where the particles are and costs nothing. There is no exchange force.
With an interaction, symmetry changes where the particles are, and that changes how much of $V$ they feel. The energy difference $2K$ between the symmetric and antisymmetric cases is real — but it is $V$ ‘s energy, merely sorted by exchange symmetry.

So $K$ is not evidence of an exchange force after all. It is the ordinary interaction, re-weighted by an interference term. The distinction survives.

K is why iron is magnetic, and the book does not tell you

Follow the chain, because it ends somewhere surprising.

$K > 0$ for a repulsive interaction, so the antisymmetric spatial state lies lower. By Eq. (10.22), an antisymmetric spatial state pairs with a symmetric spin state — the triplet, $S = 1$ , spins parallel.

\text{electrons repel} \;\Rightarrow\; K > 0 \;\Rightarrow\; \text{parallel spins have lower energy}

That is Hund’s first rule: within a partly filled shell, electrons align their spins. And a solid whose atoms each carry a net aligned spin, with the alignment propagating between neighbours, is a ferromagnet.

So the reason a magnet sticks to your fridge traces back to a sign in an integral in a problem set — and to the fact that electrons avoiding each other costs less Coulomb energy than electrons huddling. Nothing magnetic causes ferromagnetism. Electrostatic repulsion does, filtered through exchange symmetry.

The book derives $K$ and moves on without naming it. It is called the exchange integral, and in the theory of magnetism it is the parameter — the $J$ in the Heisenberg model $\hat H = -J\sum \mathbf S_i\cdot\mathbf S_j$ .

Reproduce it in NumPy

Problem 3(b) is checked to machine precision with a concrete Gaussian repulsion — the identity ΔE(S,A) = ΔE(D) ± K is exact, not approximate, because it is pure algebra on the expansion of |Ψ|². The overlap prints as exactly zero: ψ₀ is even and ψ₁ odd, so the integrand is odd and cancels term by term on a symmetric grid.

import numpy as np

# Two oscillator states, n = 0 and n' = 1 (Table 6.1), in units where a = 1.
a = 1.0
psi0 = lambda z: (1/(a*np.sqrt(np.pi)))**0.5 * np.exp(-z*z/(2*a*a))
psi1 = lambda z: (1/(2*a*np.sqrt(np.pi)))**0.5 * (2*z/a) * np.exp(-z*z/(2*a*a))

N, L = 1400, 8.0
z = np.linspace(-L, L, N); dz = z[1]-z[0]
P, Q = np.meshgrid(z, z, indexing='ij'); dA = dz*dz

# a weak repulsive interaction depending only on the separation, as problem 3 asks
V0, d = 0.30, 0.8
V = V0*np.exp(-(P-Q)**2/(2*d*d))

f, g   = psi0(P), psi1(Q)        # psi_n(x_p) psi_n'(x_q)
fs, gs = psi0(Q), psi1(P)        # the exchanged product
D  = f*g
S  = (f*g + fs*gs)/np.sqrt(2)
A  = (f*g - fs*gs)/np.sqrt(2)

print("Problem 3(a): all three are normalized, given orthonormal single-particle states")
print("=" * 76)
for name, W in (("Psi(D)", D), ("Psi(S)", S), ("Psi(A)", A)):
    print(f"    {name} -> {(W*W).sum()*dA:.8f}")
print(f"    overlap <psi_0|psi_1> = {(psi0(z)*psi1(z)).sum()*dz:+.2e}  (orthogonal, which is what makes")
print("    the cross terms in |Psi(S)|^2 and |Psi(A)|^2 integrate to zero)")

dED = (D*D*V).sum()*dA
dES = (S*S*V).sum()*dA
dEA = (A*A*V).sum()*dA
K   = (fs*gs*V*f*g).sum()*dA                 # the hint's exchange integral

print("\nProblem 3(b): the energy shift splits into a direct part and an exchange part")
print("=" * 76)
print(f"    dE(D)  direct only            = {dED:.8f}")
print(f"    K      exchange integral      = {K:.8f}")
print(f"    dE(S)  = {dES:.8f}   dE(D) + K = {dED+K:.8f}   difference {abs(dES-(dED+K)):.1e}")
print(f"    dE(A)  = {dEA:.8f}   dE(D) - K = {dED-K:.8f}   difference {abs(dEA-(dED-K)):.1e}")
print(f"""
    K is POSITIVE for a repulsive V, so the ordering is

        dE(A) = {dEA:.4f}   <   dE(D) = {dED:.4f}   <   dE(S) = {dES:.4f}

    The antisymmetric spatial state costs least, because those particles were
    already avoiding each other and so feel the repulsion least. By Eq. (10.22)
    an antisymmetric SPATIAL state goes with a symmetric SPIN state -- the
    triplet, S = 1, parallel spins.

    So: parallel spins have the lower energy, and nothing magnetic caused it.
    That is Hund's first rule, and ultimately why iron is magnetic. The book
    derives K here and never names it.""")

printsProblem 3(a): all three are normalized, given orthonormal single-particle states ============================================================================ Psi(D) -> 1.00000000 Psi(S) -> 1.00000000 Psi(A) -> 1.00000000 overlap <psi_0|psi_1> = +0.00e+00 (orthogonal, which is what makes the cross terms in |Psi(S)|^2 and |Psi(A)|^2 integrate to zero) Problem 3(b): the energy shift splits into a direct part and an exchange part ============================================================================ dE(D) direct only = 0.13027177 K exchange integral = 0.05713674 dE(S) = 0.18740851 dE(D) + K = 0.18740851 difference 2.8e-17 dE(A) = 0.07313503 dE(D) - K = 0.07313503 difference 4.2e-17 K is POSITIVE for a repulsive V, so the ordering is dE(A) = 0.0731 < dE(D) = 0.1303 < dE(S) = 0.1874 The antisymmetric spatial state costs least, because those particles were already avoiding each other and so feel the repulsion least. By Eq. (10.22) an antisymmetric SPATIAL state goes with a symmetric SPIN state -- the triplet, S = 1, parallel spins. So: parallel spins have the lower energy, and nothing magnetic caused it. That is Hund's first rule, and ultimately why iron is magnetic. The book derives K here and never names it.

Counting states

4 Two particles in a cubical box, four ways derivation

Two non-interacting particles of mass $m$ sit in a cubical box of side $a$ with

E = \frac{3\hbar^2\pi^2}{2ma^2} + \frac{6\hbar^2\pi^2}{2ma^2}

Using the eigenfunctions of Eq. (4.43), write two-particle wave functions for the cases: (a) distinguishable spinless bosons; (b) identical spinless bosons; (c) identical spin-half fermions in a symmetric spin state; (d) identical spin-half fermions in the antisymmetric spin state.

💡 Phillips' own hint

The single-particle energies are $(n_x^2+n_y^2+n_z^2)\hbar^2\pi^2/2ma^2$ , so read off which two states are occupied. Then apply Eqs. (10.22)–(10.23): the spin symmetry fixes the spatial symmetry.

✓ Worked solution

Which states. $3 = 1^2+1^2+1^2$ and $6 = 2^2+1^2+1^2$ , so one particle is in $(1,1,1)$ and the other in a $(2,1,1)$ -type state. Write $\psi_1 \equiv \psi_{1,1,1}$ and $\psi_2 \equiv \psi_{2,1,1}$ , and abbreviate $\psi_1(p) \equiv \psi_{1,1,1}(x_p,y_p,z_p)$ .

(a) Distinguishable. No symmetry is required, so a plain product will do:

\Psi = \psi_1(p)\,\psi_2(q)

(b) Identical spinless bosons. Spinless means there is no spin factor, so the spatial function must itself be symmetric:

\Psi = \tfrac{1}{\sqrt2}\left[\psi_1(p)\psi_2(q) + \psi_1(q)\psi_2(p)\right]

(c) Identical spin-half fermions, symmetric (triplet) spin state. The total must be antisymmetric, so the spatial part is antisymmetric — Eq. (10.22). (In $\chi^{(S)}_{1,M_S}$ the parenthesised superscript means symmetric, while the subscript $1$ is the combined spin quantum number $S = 1$ — the same letter doing both jobs, as §10.3 warns.)

\Phi = \tfrac{1}{\sqrt2}\left[\psi_1(p)\psi_2(q) - \psi_1(q)\psi_2(p)\right]\chi^{(S)}_{1,M_S}(p,q)

(d) Identical spin-half fermions, antisymmetric (singlet) spin state. Now the spatial part must be symmetric — Eq. (10.23):

\Phi = \tfrac{1}{\sqrt2}\left[\psi_1(p)\psi_2(q) + \psi_1(q)\psi_2(p)\right]\chi^{(A)}_{0,0}(p,q)

5 Five particles in a three-dimensional oscillator numerical

Five non-interacting particles occupy the 3-D oscillator of §6.5, with $E_n = (n_x+n_y+n_z+\tfrac32)\hbar\omega$ . Find the lowest total energy when they are (a) distinguishable spinless bosons; (b) identical spinless bosons; (c) identical spin- $\tfrac12$ fermions; (d) identical spin- $\tfrac32$ fermions.

💡 Phillips' own hint

Shell $n = n_x+n_y+n_z$ has energy $(n+\tfrac32)\hbar\omega$ and orbital degeneracy $\tfrac12(n+1)(n+2)$ — count the ways to write $n$ as an ordered sum of three non-negative integers. A fermion shell holds $(2s+1)\times$ that many.

✓ Worked solution

The lowest shells are $n = 0$ with degeneracy 1, and $n = 1$ with degeneracy 3.

(a) and (b), both $\tfrac{15}{2}\hbar\omega$ . Nothing prevents all five from occupying $n=0$ at $\tfrac32\hbar\omega$ each: $5\times\tfrac32 = \tfrac{15}{2}\hbar\omega$ . Distinguishable particles and spinless bosons give the same answer, because being allowed to share and being obliged to symmetrize do not change the lowest energy available.

(c) Spin- $\tfrac12$ fermions, $\tfrac{21}{2}\hbar\omega$ . Here the Pauli exclusion principle finally bites: each orbital state holds 2. Shell $n=0$ takes $1\times2 = 2$ particles at $\tfrac32$ ; the remaining 3 go into shell $n=1$ (capacity $3\times2 = 6$ ) at $\tfrac52$ each:

2\times\tfrac32 + 3\times\tfrac52 = 3 + \tfrac{15}{2} = \tfrac{21}{2}\hbar\omega

(d) Spin- $\tfrac32$ fermions, $\tfrac{17}{2}\hbar\omega$ . Now each orbital state holds $2s+1 = 4$ . Shell $n=0$ takes 4 at $\tfrac32$ , and only 1 is left for shell $n=1$ at $\tfrac52$ :

4\times\tfrac32 + 1\times\tfrac52 = 6 + \tfrac52 = \tfrac{17}{2}\hbar\omega

Reproduce it in NumPy

The filling is shell by shell, so you can see WHERE the energy goes rather than only the total. It is the same greedy rule the diagram below runs, and the same one §10.4 used — capacity is (2s+1) times the shell's orbital degeneracy (n+1)(n+2)/2.

def ground(N, cap, label):
    """Fill 3-D oscillator shells from the bottom. Shell n holds
       (n+1)(n+2)/2 orbital states, each taking `cap` = 2s+1 fermions."""
    E, left, n, rows = 0.0, N, 0, []
    while left > 0:
        seats = ((n+1)*(n+2)//2)*cap
        take = min(seats, left)
        rows.append((n, seats, take, take*(n+1.5)))
        E += take*(n+1.5); left -= take; n += 1
    print(f"  {label}")
    for n, seats, take, e in rows:
        print(f"      shell n={n}: {take} of {seats if seats < 10**5 else 'unlimited':>9} seats at {n+1.5:.1f} hbar-omega -> {e:.1f}")
    print(f"      total = {E:.1f} = {int(round(E*2))}/2 hbar-omega\n")
    return E

print("Problem 5: the lowest energy of five particles, four kinds of statistics")
print("=" * 74)
ea = 5*1.5
print("  (a) distinguishable, spinless")
print(f"      nothing forbids sharing; all five sit in the n=0 shell at 1.5")
print(f"      total = {ea:.1f} = {int(round(ea*2))}/2 hbar-omega\n")
eb = ground(5, 10**6, "(b) identical spinless bosons")
ec = ground(5, 2,      "(c) identical spin-1/2 fermions  (2 per orbital state)")
ed = ground(5, 4,      "(d) identical spin-3/2 fermions  (4 per orbital state)")

print("=" * 74)
print(f"  (a) {int(ea*2)}/2      (b) {int(eb*2)}/2      (c) {int(ec*2)}/2      (d) {int(ed*2)}/2   hbar-omega")
print("""
  (a) and (b) agree. Distinguishable particles are not FORBIDDEN the ground
  level -- they merely need not share it, and the lowest state has them all
  there anyway. Symmetrization costs energy only when it forbids something.

  (c) and (d) differ from those and from each other, and the direction is the
  surprise: raising the spin LOWERS the energy, 21/2 down to 17/2. Spin carries
  no energy here. It only sets how many particles fit per spatial state, 2s+1 of
  them, so more spin means more room downstairs and fewer particles pushed up.""")

printsProblem 5: the lowest energy of five particles, four kinds of statistics ========================================================================== (a) distinguishable, spinless nothing forbids sharing; all five sit in the n=0 shell at 1.5 total = 7.5 = 15/2 hbar-omega (b) identical spinless bosons shell n=0: 5 of unlimited seats at 1.5 hbar-omega -> 7.5 total = 7.5 = 15/2 hbar-omega (c) identical spin-1/2 fermions (2 per orbital state) shell n=0: 2 of 2 seats at 1.5 hbar-omega -> 3.0 shell n=1: 3 of 6 seats at 2.5 hbar-omega -> 7.5 total = 10.5 = 21/2 hbar-omega (d) identical spin-3/2 fermions (4 per orbital state) shell n=0: 4 of 4 seats at 1.5 hbar-omega -> 6.0 shell n=1: 1 of 12 seats at 2.5 hbar-omega -> 2.5 total = 8.5 = 17/2 hbar-omega ========================================================================== (a) 15/2 (b) 15/2 (c) 21/2 (d) 17/2 hbar-omega (a) and (b) agree. Distinguishable particles are not FORBIDDEN the ground level -- they merely need not share it, and the lowest state has them all there anyway. Symmetrization costs energy only when it forbids something. (c) and (d) differ from those and from each other, and the direction is the surprise: raising the spin LOWERS the energy, 21/2 down to 17/2. Spin carries no energy here. It only sets how many particles fit per spatial state, 2s+1 of them, so more spin means more room downstairs and fewer particles pushed up.

Hydrogen is two different gases

6 Ortho- and para-hydrogen from the parity of a spherical harmonic derivation

A hydrogen molecule has two identical spin-half nuclei. In a rotational state with quantum numbers $l$ and $m_l$ , the separation $\mathbf r = \mathbf r_p - \mathbf r_q$ is governed by

\psi_{l,m_l}(\mathbf r) = R(r)\,Y_{l,m_l}(\theta,\phi)

Using the symmetry of the spin states in Eqs. (10.20)–(10.21) and the parity of the spherical harmonics from Table 8.1, explain why rotational states with odd $l$ have $S = 1$ and those with even $l$ have $S = 0$ .

✓ Worked solution

What exchanging the nuclei does to the spatial function. Swapping $p$ and $q$ sends $\mathbf r = \mathbf r_p - \mathbf r_q$ to $-\mathbf r$ . In spherical coordinates that is $\theta \to \pi-\theta$ and $\phi \to \phi+\pi$ , and $r$ is unchanged — so $R(r)$ is untouched and everything rests on the harmonic:

Y_{l,m_l}(\pi-\theta,\phi+\pi) = (-1)^l\,Y_{l,m_l}(\theta,\phi)

The spatial wave function is therefore symmetric for even $l$ and antisymmetric for odd $l$ .

What the nuclei require. Protons are spin-half fermions, so the total state must be antisymmetric. The spin factor must carry the opposite sign to the spatial one:

even $l$ → spatial symmetric → spin antisymmetric → the singlet, $S = 0$ . This is para-hydrogen.
odd $l$ → spatial antisymmetric → spin symmetric → the triplet, $S = 1$ . This is ortho-hydrogen.

which is what was to be shown.

Why it matters experimentally. Nuclear spins are barely affected by collisions, so a molecule cannot easily switch families. Hydrogen gas behaves as a mixture of two substances with different heat capacities, and it takes days for ortho and para to equilibrate. At room temperature the equilibrium mixture is about 3:1 ortho to para — the ratio of the degeneracies, 3 triplet states to 1 singlet.

A selection rule with industrial consequences

This is the most concrete result in the chapter: an abstract argument about exchange symmetry predicts that a bottle of hydrogen contains two gases.

It is also a genuine engineering problem. Liquid hydrogen is stored at 20 K, where equilibrium strongly favours para. Freshly liquefied hydrogen is still 75% ortho, and the slow ortho → para conversion releases more heat than the latent heat of vaporization — enough to boil away a large fraction of the tank over days. Industrial liquefiers therefore pass the gas over a catalyst to force the conversion before storage.

So a rocket-fuel depot has to care about $(-1)^l$ . The chain is short and each link is on this page: identical nuclei → antisymmetric total state → spatial parity locks to nuclear spin → two non-interconverting species → a thermal management problem.

⇅	rotational l⇅	spatial part⇅	nuclear spin⇅	statistical weight⇅
para-hydrogen	even — 0, 2, 4, …	symmetric	singlet, S = 0	1
ortho-hydrogen	odd — 1, 3, 5, …	antisymmetric	triplet, S = 1	3

Check yourself

0 / 6 answered

1.Problem 1: what must be true of $\hat{H}$ for a two-particle state's exchange symmetry to survive time evolution?
2.In problem 3(a), what makes the cross terms in $∣ Ψ^{(S)} ∣^{2}$ and $∣ Ψ^{(A)} ∣^{2}$ integrate to zero?
Problem 3(b) gives $Δ E^{(S)} = Δ E^{(D)} + K$ and $Δ E^{(A)} = Δ E^{(D)} - K$ , with $K > 0$ for a repulsive interaction.
3.What does that ordering imply about two electrons?
4.In problem 5, distinguishable particles and identical spinless bosons both give $\frac{15}{2} ℏ ω$ . Why do two different statistics agree?
5.Problem 5 finds spin- $\frac{3}{2}$ fermions ( $\frac{17}{2} ℏ ω$ ) have a **lower** ground-state energy than spin- $\frac{1}{2}$ fermions ( $\frac{21}{2} ℏ ω$ ). Why?
6.Problem 6: why must a hydrogen molecule with **odd** rotational $l$ have nuclear spin $S = 1$ ?