Nothing is added to the Hamiltonian and the particles still avoid each other. The whole effect comes from a requirement on the wave function’s symmetry, and from nothing physical at all.

Two identical billiard balls are still two different billiard balls: you can watch them, and say which is which. Quantum particles cannot be watched — ch02 removed the trajectory — so identical quantum particles are not merely hard to tell apart. They are indistinguishable in principle, and that single sentence, taken seriously, produces the Pauli principle, the rigidity of matter and the laser.

§10.1 turns the sentence into an equation in about half a page. §10.2 spends four pages on what it costs, and the answer is startling: identical particles cluster together or avoid each other with no force acting between them.

The argument: three lines, and no third option

Take two particles $p$ and $q$ , described by a two-particle wave function $\Psi(\mathbf r_p, \mathbf r_q, t)$ . The joint probability of finding $p$ in $\mathrm d^3a$ at $\mathbf a$ and $q$ in $\mathrm d^3b$ at $\mathbf b$ is

|\Psi(\mathbf a, \mathbf b, t)|^2\,\mathrm d^3a\,\mathrm d^3b\tag{10.1}

If the particles are identical, no measurement can tell ” $p$ here, $q$ there” from ” $q$ here, $p$ there”. The two must be equally probable:

|\Psi(\mathbf a, \mathbf b, t)|^2 = |\Psi(\mathbf b, \mathbf a, t)|^2\tag{10.2}

Two complex numbers with the same modulus differ by a phase, so

\Psi(\mathbf a, \mathbf b, t) = e^{i\delta}\,\Psi(\mathbf b, \mathbf a, t)\tag{10.3}

and now the argument closes on itself.

step 1 of 4

This is the entire content of section 10.1. It uses nothing but Eq. (10.3) and the fact that swapping twice gets you back where you started.

1Start from Eq. (10.3): exchanging the two arguments multiplies the wave function by some phase. We do not yet know what the phase is.
$Ψ (a, b, t) = e^{i δ} Ψ (b, a, t)$
The particles are identical, so the SAME phase must work whichever way round we write it — there is nothing to distinguish the two cases.
2Apply the same relation again, with a and b interchanged. This is legitimate precisely because the phase does not depend on which label is which.
$Ψ (b, a, t) = e^{i δ} Ψ (a, b, t)$
3Substitute the second into the first. Exchanging twice has returned the function to itself, so whatever multiplies it must be 1.
$Ψ (a, b, t) = e^{i δ} e^{i δ} Ψ (a, b, t) ⟹ e^{2 i δ} = 1$
4A phase whose square is one is plus or minus one. There is no third solution, and no continuous range in between.
$e^{i δ} = + 1 or e^{i δ} = - 1 (10.4)$
Everything in this chapter — and the whole periodic table — descends from this line.

So every acceptable wave function for two identical particles is either symmetric

\Psi(\mathbf a, \mathbf b, t) = \Psi(\mathbf b, \mathbf a, t)\tag{10.5}

or antisymmetric

\Psi(\mathbf a, \mathbf b, t) = -\Psi(\mathbf b, \mathbf a, t)\tag{10.6}

These are said to have definite exchange symmetry . Nature will turn out to assign one or the other by species, and never to mix them — but that is §10.4’s news, and it is empirical. What §10.1 establishes is only that the choice is binary.

You already know this theorem: it is the spectrum of an involution

Define the exchange operator $\hat P$ by $\hat P\Psi(\mathbf a,\mathbf b) = \Psi(\mathbf b,\mathbf a)$ . Swapping twice is doing nothing, so

\hat P^2 = \hat I

An operator equal to its own inverse — an involution — satisfies the polynomial $\lambda^2 - 1 = 0$ , so its eigenvalues are $\pm1$ and nothing else. Eq. (10.4) is the minimal polynomial of $\hat P$ , and Eqs. (10.5)–(10.6) are its two eigenspaces.

That reframing pays immediately. $\hat P$ is Hermitian as well as unitary, so its eigenspaces are orthogonal and together they span: any two-particle function decomposes uniquely into a symmetric and an antisymmetric part,

\Psi = \underbrace{\tfrac12(\Psi + \hat P\Psi)}_{\text{symmetric}} + \underbrace{\tfrac12(\Psi - \hat P\Psi)}_{\text{antisymmetric}}

which is exactly the projector decomposition

\hat I = \tfrac12(\hat I + \hat P) + \tfrac12(\hat I - \hat P)

If you have ever split a signal into its even and odd parts, you have done this. And because the Hamiltonian of identical particles must commute with $\hat P$ — the particles being identical is precisely the statement that the physics is unchanged by relabelling — exchange symmetry is a constant of the motion by §7.5’s argument. A state that starts antisymmetric stays antisymmetric forever. Problem 1 is this in the book’s own notation.

The book admits its own logic is contrived — and it is right to

Footnote 1 on p. 214 is unusually candid:

Some readers may consider our treatment of identical quantum particles to be contrived, because we label the identical particles and then we only accept wave functions for which the labelling has no physical consequences.

That is exactly what has just happened. We wrote $\mathbf r_p$ and $\mathbf r_q$ , giving the particles names, and then spent the section forbidding those names from mattering. It works, but it is a workaround: the labels are scaffolding erected only to be discarded.

The book names the alternative — quantum field theory, where the fields themselves carry the symmetry and no particle is ever labelled, so the question cannot arise. Nothing in this book needs it. But when you meet creation and annihilation operators later and find that antisymmetry is built into the algebra rather than imposed on the answer, this footnote is why.

What it costs: two particles in one oscillator

§10.2 takes the simplest system that can show the effect — two identical particles of mass $m$ in the same one-dimensional oscillator, with no interaction between them at all:

\hat H(x_p, x_q) = -\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\tag{10.7}

Read that Hamiltonian carefully, because everything below depends on what is absent from it. There is no $x_p - x_q$ anywhere. The two particles do not push, pull, scatter or notice each other. Each independently occupies an oscillator eigenfunction $\psi_n$ with $E_n = (n + \tfrac12)\hbar\omega$ (problem 2 verifies this by substitution).

Both particles in the same state — and Pauli arrives four pages early

If both sit in state $n$ , the total energy is $E = 2E_n = (2n+1)\hbar\omega$ and the wave function is

\Psi^{(S)}(x_p, x_q, t) = \psi_n(x_p)\,\psi_n(x_q)\,e^{-i(E_n+E_n)t/\hbar}\tag{10.8}

Swap $x_p$ and $x_q$ : nothing changes. The state is symmetric, and it is the only one available — there is no way to build an antisymmetric function out of one single-particle state, because the antisymmetric combination of $\psi_n$ with itself is $\psi_n\psi_n - \psi_n\psi_n = 0$ .

So: particles requiring antisymmetric wave functions can never share a single-particle state. That is the Pauli exclusion principle , and it has arrived here as a triviality — not as a postulate, not as a rule imposed on the theory, but as the observation that the state you were trying to build is the zero vector. It has no wave function because it is not a state.

Two different states — and the particles become entangled whether you like it or not

With one particle in $n$ and one in $n'$ , the energy is $E = (n + n' + 1)\hbar\omega$ . For distinguishable particles you could write

\Psi_1^{(D)} = \psi_n(x_p)\psi_{n'}(x_q)e^{-i(E_n+E_{n'})t/\hbar}\tag{10.9}

\Psi_2^{(D)} = \psi_n(x_q)\psi_{n'}(x_p)e^{-i(E_n+E_{n'})t/\hbar}\tag{10.10}

or any combination of them,

\Psi^{(D)} = c_1\Psi_1^{(D)} + c_2\Psi_2^{(D)}\tag{10.11}

with $|c_1|^2$ the probability that $p$ is in state $n$ , and $|c_2|^2$ that $q$ is. Eq. (10.11) is an entangled state — both particles associated with both states — and the book coins untangled for the plain products (10.9) and (10.10). For distinguishable particles, which one you get depends on how the state was prepared.

For identical particles there is no choice. Neither (10.9) nor (10.10) has definite exchange symmetry — swapping the arguments turns one into the other — so neither is acceptable. Only the two symmetric combinations survive:

\Psi^{(S)} = \tfrac{1}{\sqrt2}\left[\psi_n(x_p)\psi_{n'}(x_q) + \psi_n(x_q)\psi_{n'}(x_p)\right]e^{-i(E_n+E_{n'})t/\hbar}\tag{10.12}

\Psi^{(A)} = \tfrac{1}{\sqrt2}\left[\psi_n(x_p)\psi_{n'}(x_q) - \psi_n(x_q)\psi_{n'}(x_p)\right]e^{-i(E_n+E_{n'})t/\hbar}\tag{10.13}

The punchline: put both particles at the same place

Set $x_p = x_q = x_0$ in each. The untangled functions give $\psi_n(x_0)\psi_{n'}(x_0)$ , and the two terms of (10.12) and (10.13) become identical, so

\Psi^{(S)}(x_0,x_0,t) = \sqrt2\,\psi_n(x_0)\psi_{n'}(x_0)\,e^{-i(E_n+E_{n'})t/\hbar} \qquad\text{and}\qquad \Psi^{(A)}(x_0,x_0,t) = 0

Square them. Identical particles with a symmetric wave function are twice as likely to be found at the same point as distinguishable ones; identical particles with an antisymmetric wave function are never found at the same point. The mechanism is interference — constructive in one case, destructive in the other — and the destructive case is total, because at $x_p = x_q$ the two terms are not merely similar but exactly equal.

Two views make this concrete. The plane shows $|\Psi(x_1,x_2)|^2$ over both positions at once, with the dashed line marking $x_1 = x_2$ ; exchange is reflection across that line, so “symmetric” and “antisymmetric” become properties you can simply look at. The separation view integrates along that line to reproduce Fig. 10.1. Set both quantum numbers equal with antisymmetry selected and the whole map goes blank — that is the exclusion principle, drawn.

Two identical particles: exchange is reflection across the diagonal

systemstate n0state n′1

max |Ψ| on the diagonal x₁ = x₂: 0.00000

state: n = 0 and n′ = 1

Look along the dashed diagonal. Ψ is exactly zero there, because a function equal to minus its own mirror image must vanish on the mirror. That is the Fermi hole: two identical fermions are never found at the same point — and nothing in the Hamiltonian pushes them apart.

The gap that opens along the diagonal in the antisymmetric case has a standard name the book does not use: the Fermi hole . Its cause is worth being precise about, because the obvious explanation is wrong.

This is not a force. The Hamiltonian contains no interaction at all.

The book calls it “a surprising tendency for identical particles to either huddle together or to avoid each other”, and the language invites a false conclusion. There is no exchange force. Look again at Eq. (10.7): it is a sum of two independent one-particle Hamiltonians, with not one term coupling $x_p$ to $x_q$ . Nothing attracts and nothing repels.

What produces the clustering is that the wave function has two terms, and at $x_p = x_q$ those two terms are equal — so they add to $\sqrt2$ times either one, or cancel exactly. It is interference in a two-dimensional configuration space, the same phenomenon as ch02’s two-slit pattern, which also produced bright and dark fringes with no force acting.

The test is decisive and the calculation below performs it: all three states have exactly the same energy. A force between the particles would raise the energy of the huddling state and lower the avoiding one, or the reverse. Instead $\langle\hat H\rangle$ comes out at $2\hbar\omega$ for the distinguishable, the symmetric and the antisymmetric state alike. There is nothing there to be a force.

(Add a real interaction and something genuinely new happens: the huddling and the avoidance now cost different amounts of energy, and the difference is the exchange integral $K$ of problem 3 — which is why parallel spins have lower energy in a partly filled shell, and ultimately why iron is magnetic. That is a real energy from a real interaction, made visible by exchange symmetry. It is not what is happening on this page.)

How far apart do they sit?

Take the book’s own example, $n = 0$ and $n' = 1$ , and change variables to the separation and the centre of mass,

x = x_p - x_q \qquad\text{and}\qquad X = \frac{x_p + x_q}{2}

Table 6.1’s eigenfunctions then collapse into a strikingly simple pair:

\Psi^{(S)}(x,X,t) = \frac{2}{a^2\sqrt\pi}\,e^{-x^2/4a^2}\,X\,e^{-X^2/a^2}\,e^{-i(E_0+E_1)t/\hbar}

\Psi^{(A)}(x,X,t) = \frac{-1}{a^2\sqrt\pi}\,x\,e^{-x^2/4a^2}\,e^{-X^2/a^2}\,e^{-i(E_0+E_1)t/\hbar}

The difference between them is one factor: $\Psi^{(S)}$ carries $X$ , and $\Psi^{(A)}$ carries $x$ . That is the whole story in one symbol — the antisymmetric state has a factor of the separation multiplying it, so it must vanish when the separation does.

Integrating $|\Psi|^2$ over all $X$ leaves the probability of a separation of magnitude between $|x|$ and $|x + \mathrm dx|$ :

P^{(S)}(x)\,\mathrm dx = \frac{2}{a\sqrt{2\pi}}\,e^{-x^2/2a^2}\,\mathrm dx\tag{10.14}

P^{(A)}(x)\,\mathrm dx = \frac{2}{a\sqrt{2\pi}}\,\frac{x^2}{a^2}\,e^{-x^2/2a^2}\,\mathrm dx\tag{10.15}

P^{(D)}(x)\,\mathrm dx = \frac{1}{a\sqrt{2\pi}}\left(1 + \frac{x^2}{a^2}\right)e^{-x^2/2a^2}\,\mathrm dx\tag{10.16}

These are the three curves of Fig. 10.1, and the separation view of the widget above draws them from the wave function rather than from these formulas — the two agree to $4\times10^{-11}$ .

Reproduce it in NumPy

The energies are obtained by applying Eq. (10.7) to each state on a grid with a fourth-order stencil, not by quoting E = (n + n' + 1)ħω — the point being tested is precisely that the three states share an energy, so it had to be measured rather than assumed.

import numpy as np

a = 1.0                                          # oscillator length; hbar = m = omega = 1
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 = 900, 8.0
z = np.linspace(-L, L, n); dz = z[1] - z[0]
P, Q = np.meshgrid(z, z, indexing='ij'); dA = dz*dz

def d2(F, ax):                                   # 4th-order second derivative
    r = lambda k: np.roll(F, k, axis=ax)
    return (-r(-2) + 16*r(-1) - 30*F + 16*r(1) - r(2)) / (12*dz*dz)

def H(F):                                        # Eq. (10.7), in natural units
    return -0.5*(d2(F, 0) + d2(F, 1)) + 0.5*(P*P + Q*Q)*F

D = psi0(P)*psi1(Q)                                       # Eq. (10.9), untangled
S = (psi0(P)*psi1(Q) + psi0(Q)*psi1(P))/np.sqrt(2)        # Eq. (10.12)
A = (psi0(P)*psi1(Q) - psi0(Q)*psi1(P))/np.sqrt(2)        # Eq. (10.13)

print("Three states of the SAME Hamiltonian -- Eq. (10.7) has no interaction term")
print("=" * 78)
print(f"  {'state':<18}{'norm':>10}{'<H> / hbar-omega':>20}{'max |Psi| at x_p = x_q':>26}")
diag = {}
for name, W in (("distinguishable", D), ("symmetric", S), ("antisymmetric", A)):
    d = np.abs(np.diag(W)).max(); diag[name] = d
    print(f"  {name:<18}{(W*W).sum()*dA:>10.6f}{(W*H(W)).sum()*dA:>20.6f}{d:>26.6f}")
print(f"""
  <H> is computed by applying the Hamiltonian operator on the grid, not assumed.
  All three come out at exactly 2 hbar-omega, because Eq. (10.7) is two
  independent oscillators and every one of the three states puts one particle in
  n = 0 and one in n = 1.  Exchange symmetry rearranges WHERE the particles are
  without changing the energy -- so whatever makes them huddle or avoid, it is
  not a force.  A force would have to show up right here, as an energy gap.

  On the diagonal x_p = x_q:
    symmetric / distinguishable = {diag['symmetric']/diag['distinguishable']:.6f}   (the book's sqrt(2) = {np.sqrt(2):.6f})
    squared, that is {(diag['symmetric']/diag['distinguishable'])**2:.4f} -- the book's 'twice as likely'
    antisymmetric               = {diag['antisymmetric']:.2e}   (identically zero)""")

print("\nThe book's (x, X) forms of Eqs. (10.12)-(10.13), prefactors included")
print("=" * 78)
x_, X_ = P - Q, (P + Q)/2
Sb = (2/(a*a*np.sqrt(np.pi))) * np.exp(-x_**2/(4*a*a)) * X_ * np.exp(-X_**2/(a*a))
Ab = (-1/(a*a*np.sqrt(np.pi))) * x_ * np.exp(-x_**2/(4*a*a)) * np.exp(-X_**2/(a*a))
print(f"  Psi(S): max |printed form - built from Table 6.1| = {np.abs(Sb-S).max():.2e}")
print(f"  Psi(A): max |printed form - built from Table 6.1| = {np.abs(Ab-A).max():.2e}")
print("  Both exact -- the change of variables is only a change of variables.")

print("\nEqs. (10.14)-(10.16): the separation distributions plotted in Fig. 10.1")
print("=" * 78)
s = np.linspace(0, 14, 200001)
PS = 2/(a*np.sqrt(2*np.pi)) * np.exp(-s**2/(2*a*a))
PA = 2/(a*np.sqrt(2*np.pi)) * (s*s/(a*a)) * np.exp(-s**2/(2*a*a))
PD = 1/(a*np.sqrt(2*np.pi)) * (1 + s*s/(a*a)) * np.exp(-s**2/(2*a*a))
print(f"  {'':<18}{'integral over |x|':>18}{'<|x|> / a':>12}{'value at x = 0':>16}")
for name, f in (("P(S) symmetric", PS), ("P(D) distinguish.", PD), ("P(A) antisym.", PA)):
    print(f"  {name:<18}{np.trapezoid(f,s):>18.6f}{np.trapezoid(f*s,s):>12.4f}{f[0]:>16.4f}")
print(f"""
  Each normalizes over the MAGNITUDE |x| from 0 to infinity, not over the line.
  The book says 'magnitude' once, in passing, and never repeats it.

  P(D) is exactly the average of P(S) and P(A): max difference {np.abs(PD-(PS+PA)/2).max():.1e}
  A distinguishable pair therefore always sits BETWEEN the two identical cases.
  That is what 'untangled' costs you: no interference, so no preference either.

  Mean separation, antisymmetric / symmetric = {np.trapezoid(PA*s,s)/np.trapezoid(PS*s,s):.4f}
  Two identical fermions sit exactly twice as far apart as two identical bosons
  in the same two states of the same potential, with nothing pushing either.""")

printsThree states of the SAME Hamiltonian -- Eq. (10.7) has no interaction term ============================================================================== state norm <H> / hbar-omega max |Psi| at x_p = x_q distinguishable 1.000000 2.000000 0.342187 symmetric 1.000000 2.000000 0.483925 antisymmetric 1.000000 2.000000 0.000000 <H> is computed by applying the Hamiltonian operator on the grid, not assumed. All three come out at exactly 2 hbar-omega, because Eq. (10.7) is two independent oscillators and every one of the three states puts one particle in n = 0 and one in n = 1. Exchange symmetry rearranges WHERE the particles are without changing the energy -- so whatever makes them huddle or avoid, it is not a force. A force would have to show up right here, as an energy gap. On the diagonal x_p = x_q: symmetric / distinguishable = 1.414214 (the book's sqrt(2) = 1.414214) squared, that is 2.0000 -- the book's 'twice as likely' antisymmetric = 0.00e+00 (identically zero) The book's (x, X) forms of Eqs. (10.12)-(10.13), prefactors included ============================================================================== Psi(S): max |printed form - built from Table 6.1| = 2.78e-16 Psi(A): max |printed form - built from Table 6.1| = 2.22e-16 Both exact -- the change of variables is only a change of variables. Eqs. (10.14)-(10.16): the separation distributions plotted in Fig. 10.1 ============================================================================== integral over |x| <|x|> / a value at x = 0 P(S) symmetric 1.000000 0.7979 0.7979 P(D) distinguish. 1.000000 1.1968 0.3989 P(A) antisym. 1.000000 1.5958 0.0000 Each normalizes over the MAGNITUDE |x| from 0 to infinity, not over the line. The book says 'magnitude' once, in passing, and never repeats it. P(D) is exactly the average of P(S) and P(A): max difference 1.7e-16 A distinguishable pair therefore always sits BETWEEN the two identical cases. That is what 'untangled' costs you: no interference, so no preference either. Mean separation, antisymmetric / symmetric = 2.0000 Two identical fermions sit exactly twice as far apart as two identical bosons in the same two states of the same potential, with nothing pushing either.

Two numbers worth carrying out of this section

The distinguishable case is the exact average of the other two. $P^{(D)} = \tfrac12\left(P^{(S)} + P^{(A)}\right)$ , to $2\times10^{-16}$ — you can read it straight off Eqs. (10.14)–(10.16), since $\tfrac12(1 + x^2/a^2)$ is half the sum of $1$ and $x^2/a^2$ . A distinguishable pair sits exactly between the huddlers and the avoiders, because it does not interfere at all, and so has no preference either way.

The fermion pair sits exactly twice as far apart as the boson pair. $\langle|x|\rangle = 0.798a$ symmetric, $1.197a$ distinguishable, $1.596a$ antisymmetric. Same potential, same energy, same two single-particle states — and a factor of two in mean separation, produced entirely by a sign.

Capital S now means three different things

From here to the end of the chapter, watch the symbol:

$\Psi^{(S)}$ — the superscript means symmetric (and $(A)$ antisymmetric, $(D)$ distinguishable). This page.
$S$ — the combined spin quantum number of two particles, $S = 0$ or $1$ . Arrives in §10.3.
$S$ — the magnitude of a spin angular momentum, $\sqrt{s(s+1)}\hbar$ , from §8.1.

The book’s own footnote 3 concedes the first two collide. The rescue is that the superscript is always parenthesised and always on a $\Psi$ or a $\chi$ ; a bare $S$ is a quantum number. When §10.3 writes $\chi^{(S)}_{1,M_S}$ , both meanings appear in one symbol — the superscript says the spin state is symmetric, the subscript $1$ says $S = 1$ .

Where this is going

The chapter’s own introduction says these concepts “need not be studied in detail in order to understand atoms which is the topic covered in Chapter 11”. Treat that disclaimer with suspicion. Chapter 11 is the periodic table, and the periodic table is the exclusion principle established above; Hund’s rules are the exchange integral of problem 3. This section is load-bearing for everything that follows it.

Still missing is the physical input. Nothing so far says which particles get Eq. (10.5) and which get Eq. (10.6) — the mathematics permits both and chooses neither. §10.3 brings in spin and finds that the spatial symmetry and the spin symmetry are locked together, and §10.4 gives the empirical rule that ties the choice to spin itself.

Check yourself

0 / 6 answered

1.Eq. (10.4) allows the exchange phase only the values $+ 1$ and $- 1$ . What rules out, say, $e^{iπ /3}$ ?
2.Two identical particles occupy two *different* single-particle states. What does exchange symmetry force?
The Hamiltonian of Eq. (10.7) is a sum of two independent oscillator Hamiltonians, with no term coupling $x_{p}$ to $x_{q}$ .
3.So what makes identical bosons cluster and identical fermions avoid each other?
4.Two identical particles requiring antisymmetric wave functions are put into the *same* single-particle state $ψ_{n}$ . What is the resulting wave function?
5.Over what range must Eqs. (10.14)–(10.16) be integrated to give 1?
6.The chapter opens by saying its concepts "need not be studied in detail in order to understand atoms which is the topic covered in Chapter 11". How should you take that?