Spin turns one requirement into two options. The total wave function must be antisymmetric, and the spatial and spin parts can each supply the sign in either order.

§10.1 proved the choice is binary — symmetric or antisymmetric, nothing else — and then stopped, because nothing in that argument says which. Two things are still missing, and this page supplies both.

§10.3 brings in spin, and finds that the spatial and spin symmetries are not independent: fix one and the other is forced. §10.4 supplies the empirical input that the mathematics cannot: integer spin takes the plus sign, half-integer spin the minus sign. From those two facts follow the periodic table, the rigidity of the chair you are sitting on, and the laser.

A state needs a spin as well as a place

So far a single-particle state has been a wave function $\psi(\mathbf r_p)$ . A real particle also has spin, and when the two are independent the state is just their product:

\Phi(p) = \psi(\mathbf r_p)\,\chi(p)

with $\chi_{s,m_s}(p)$ one of the $2s+1$ spin eigenvectors — the exact analogue of the $2l+1$ spherical harmonics $Y_{l,m_l}$ from §8.3. Labelling both halves,

\Phi_{n,l,m_l,m_s}(p) = \psi_{n,l,m_l}(\mathbf r_p)\,\chi_{s,m_s}(p)

Everything from §10.1 now runs again with $\Phi$ in place of $\psi$ . Both particles in the same state gives only

\Phi^{(S)}(p,q) = \Phi_{n,l,m_l,m_s}(p)\,\Phi_{n,l,m_l,m_s}(q)\tag{10.17}

with no antisymmetric partner — the exclusion principle again, now counting spin among the labels, which is what makes it two electrons per orbital rather than one. Two different states give the familiar pair:

\Phi^{(S)}(p,q) = \tfrac{1}{\sqrt2}\left[\Phi_{n,l,m_l,m_s}(p)\Phi_{n',l',m_l',m_s'}(q) + \Phi_{n,l,m_l,m_s}(q)\Phi_{n',l',m_l',m_s'}(p)\right]\tag{10.18}

\Phi^{(A)}(p,q) = \tfrac{1}{\sqrt2}\left[\Phi_{n,l,m_l,m_s}(p)\Phi_{n',l',m_l',m_s'}(q) - \Phi_{n,l,m_l,m_s}(q)\Phi_{n',l',m_l',m_s'}(p)\right]\tag{10.19}

Spin eigenvectors are two-component columns, and the book won't say so twice

Footnote 2 gives the whole representation and then declines to use it:

\chi_{\frac12,+\frac12}(p) = \begin{pmatrix}1\\0\end{pmatrix}_p \qquad \chi_{\frac12,-\frac12}(p) = \begin{pmatrix}0\\1\end{pmatrix}_p

“The mathematics of the representation of spin quantum states is covered in more advanced books, but this mathematics will not be needed here.” For a reader who knows linear algebra that is a gift left unopened — because with it, everything below stops being a list of formulas to accept.

Two spin-halves live in the tensor product $\mathbb C^2 \otimes \mathbb C^2 = \mathbb C^4$ , spanned by $|{\uparrow\uparrow}\rangle$ , $|{\uparrow\downarrow}\rangle$ , $|{\downarrow\uparrow}\rangle$ , $|{\downarrow\downarrow}\rangle$ . The exchange operator $\hat P$ of §10.1 acts on this space by swapping the two factors — a $4\times4$ permutation matrix. The total spin operator is $\hat{\mathbf S} = \hat{\mathbf s}_p \otimes \hat I + \hat I \otimes \hat{\mathbf s}_q$ .

Then the entire content of §10.3 is one linear-algebra fact: $\hat S^2$ , $\hat S_z$ and $\hat P$ commute, so they share an eigenbasis, and that basis splits $\mathbb C^4$ into a 3-dimensional and a 1-dimensional piece. The calculation below builds those matrices from Pauli matrices and diagonalizes them. If you have met $\mathbf 2 \otimes \mathbf 2 = \mathbf 3 \oplus \mathbf 1$ in any other context, this is that.

Two spin-halves: three symmetric states and one antisymmetric

Eq. (8.6) applied to $s = \tfrac12$ twice gives

S = \tfrac12 + \tfrac12 = 1 \qquad\text{or}\qquad S = \tfrac12 - \tfrac12 = 0

Read the next four equations twice: S appears in two different roles at once

§10.1–§10.2 warned that capital $S$ carries three meanings. This is the page where two of them land in the same symbol, so it is worth being explicit before the equations rather than after:

$S$ on its own — the combined spin quantum number, $0$ or $1$ here. A quantity, and the thing Eq. (8.6) computes.
$(S)$ in a parenthesised superscript — “symmetric under exchange”. A label, and it is equally at home on $\chi$ , $\psi$ or $\Psi$ .
and from §8.1, $S = \sqrt{s(s+1)}\,\hbar$ was the magnitude of a spin angular momentum — not used on this page, but it is the same letter.

So in $\chi^{(S)}_{1,M_S}$ the superscript says the state is symmetric, the subscript $1$ says $S = 1$ , and both are true of the same object. The book’s footnote 3 admits the collision and keeps the notation anyway; the rule that rescues it is that the symmetry label is always parenthesised and always a superscript, while a bare $S$ is always a quantum number.

Writing $\chi_+$ for $m_s = +\tfrac12$ and $\chi_-$ for $-\tfrac12$ , the four combinations are three symmetric ones — the triplet —

\begin{aligned} \chi^{(S)}_{1,1}(p,q) &= \chi_+(p)\chi_+(q)\\ \chi^{(S)}_{1,0}(p,q) &= \tfrac{1}{\sqrt2}\left[\chi_+(p)\chi_-(q) + \chi_+(q)\chi_-(p)\right]\\ \chi^{(S)}_{1,-1}(p,q) &= \chi_-(p)\chi_-(q) \end{aligned}\tag{10.20}

and one antisymmetric — the singlet:

\chi^{(A)}_{0,0}(p,q) = \tfrac{1}{\sqrt2}\left[\chi_+(p)\chi_-(q) - \chi_+(q)\chi_-(p)\right]\tag{10.21}

Only $\chi^{(S)}_{1,0}$ and $\chi^{(A)}_{0,0}$ needed building; the other two were symmetric already, having nothing to swap. Note that the count works out: four basis products in, four combined states out, split three-and-one.

Reproduce it in NumPy

The spin operators are built from 2×2 Pauli matrices by Kronecker product — no quantum mechanics is assumed beyond Eq. (8.6), and the singlet/triplet split falls out of diagonalizing a 4×4 matrix. The truth table at the end is generated by multiplying two signs, which is all Eqs. (10.22)–(10.23) actually say.

import numpy as np

# One spin-half, in units of hbar. Basis: |+> = (1,0), |-> = (0,1).
I2 = np.eye(2)
sx = np.array([[0, 1], [1, 0]], dtype=float)/2
sy = np.array([[0, -1j], [1j, 0]])/2
sz = np.array([[1, 0], [0, -1]], dtype=float)/2
K = np.kron

# Two spin-halves live in a 4-dimensional product space.
# Basis order: |++>, |+->, |-+>, |-->
Sx, Sy, Sz = K(sx,I2)+K(I2,sx), K(sy,I2)+K(I2,sy), K(sz,I2)+K(I2,sz)
S2 = (Sx@Sx + Sy@Sy + Sz@Sz).real

# The exchange operator, built by permuting basis labels -- nothing more.
pairs = [(0,0),(0,1),(1,0),(1,1)]
P = np.zeros((4,4))
for i,(a,b) in enumerate(pairs): P[pairs.index((b,a)), i] = 1

print("The exchange operator on two spin-halves")
print("=" * 74)
print(f"  P is its own inverse, P^2 = I          : {np.allclose(P@P, np.eye(4))}")
print(f"  P commutes with S^2                    : {np.allclose(S2@P, P@S2)}")
print(f"  P commutes with S_z                    : {np.allclose(Sz@P, P@Sz)}")
print("  So S^2, S_z and P are simultaneously diagonalizable: every state can be")
print("  labelled by S, M_S AND its exchange symmetry at once. That is why the")
print("  book can call the S = 1 states symmetric and the S = 0 state antisymmetric.")

r = 1/np.sqrt(2)
states = [("chi(S)_1,+1  Eq.(10.20)", np.array([1,0,0,0.])),
          ("chi(S)_1, 0  Eq.(10.20)", np.array([0,r,r,0.])),
          ("chi(S)_1,-1  Eq.(10.20)", np.array([0,0,0,1.])),
          ("chi(A)_0, 0  Eq.(10.21)", np.array([0,r,-r,0.]))]
print(f"\n  {'state':<24}{'norm':>7}{'S^2/hbar^2':>12}{'=> S':>6}{'M_S':>6}{'P':>5}")
for name, v in states:
    s2 = float(v@S2@v); S = (-1+np.sqrt(1+4*s2))/2
    print(f"  {name:<24}{v@v:>7.3f}{s2:>12.3f}{S:>6.0f}{float(v@Sz@v):>6.1f}{float(v@P@v):>5.0f}")

M = np.array([v for _, v in states])
print(f"\n  the four are orthonormal              : {np.allclose(M@M.T, np.eye(4))}")
print(f"  spectrum of S^2 on the 4-dim space    : {np.round(np.linalg.eigvalsh(S2),6)}")
print(f"  spectrum of P                         : {np.round(np.linalg.eigvalsh(P),6)}")
print("""
  Read the two spectra together. S^2 has the value 2 hbar^2 three times and 0
  once: the triplet and the singlet, S(S+1) with S = 1 and S = 0. P has +1 three
  times and -1 once, on exactly the same states. Four dimensions in, 3 + 1 out.

  2 x 2 = 4 = 3 + 1 is the whole of Eq. (8.6) applied to two spin-halves, and
  the symmetric subspace is the LARGER one -- which is why an electron pair has
  three ways to be spin-symmetric and only one way to be antisymmetric.""")

print("\nThe consequence for electrons: choosing one symmetry forces the other")
print("=" * 74)
print(f"  {'total state':<16}{'spin part':<22}{'S':>3}{'spatial part':>17}{'allowed?':>10}")
for sname, sS, sp in (("triplet", 1, +1), ("singlet", 0, -1)):
    for wname, wp in (("symmetric", +1), ("antisymmetric", -1)):
        tot = sp*wp                                  # symmetry of the product
        ok = "YES" if tot == -1 else "no"
        print(f"  {'antisymmetric':<16}{sname+' ('+('sym' if sp>0 else 'antisym')+')':<22}{sS:>3}{wname:>17}{ok:>10}")
print("""
  Only two rows survive, and they are Eqs. (10.22) and (10.23). The product of
  the two symmetries must be -1, so fixing the spin fixes the space:

    S = 1 (spin symmetric)     ->  wave function ANTIsymmetric  ->  they avoid
    S = 0 (spin antisymmetric) ->  wave function symmetric      ->  they huddle

  Nothing here is a force, and nothing here is a new postulate. It is the sign
  of a product, plus the empirical fact that electrons take the minus sign.""")

printsThe exchange operator on two spin-halves ========================================================================== P is its own inverse, P^2 = I : True P commutes with S^2 : True P commutes with S_z : True So S^2, S_z and P are simultaneously diagonalizable: every state can be labelled by S, M_S AND its exchange symmetry at once. That is why the book can call the S = 1 states symmetric and the S = 0 state antisymmetric. state norm S^2/hbar^2 => S M_S P chi(S)_1,+1 Eq.(10.20) 1.000 2.000 1 1.0 1 chi(S)_1, 0 Eq.(10.20) 1.000 2.000 1 0.0 1 chi(S)_1,-1 Eq.(10.20) 1.000 2.000 1 -1.0 1 chi(A)_0, 0 Eq.(10.21) 1.000 0.000 0 0.0 -1 the four are orthonormal : True spectrum of S^2 on the 4-dim space : [0. 2. 2. 2.] spectrum of P : [-1. 1. 1. 1.] Read the two spectra together. S^2 has the value 2 hbar^2 three times and 0 once: the triplet and the singlet, S(S+1) with S = 1 and S = 0. P has +1 three times and -1 once, on exactly the same states. Four dimensions in, 3 + 1 out. 2 x 2 = 4 = 3 + 1 is the whole of Eq. (8.6) applied to two spin-halves, and the symmetric subspace is the LARGER one -- which is why an electron pair has three ways to be spin-symmetric and only one way to be antisymmetric. The consequence for electrons: choosing one symmetry forces the other ========================================================================== total state spin part S spatial part allowed? antisymmetric triplet (sym) 1 symmetric no antisymmetric triplet (sym) 1 antisymmetric YES antisymmetric singlet (antisym) 0 symmetric YES antisymmetric singlet (antisym) 0 antisymmetric no Only two rows survive, and they are Eqs. (10.22) and (10.23). The product of the two symmetries must be -1, so fixing the spin fixes the space: S = 1 (spin symmetric) -> wave function ANTIsymmetric -> they avoid S = 0 (spin antisymmetric) -> wave function symmetric -> they huddle Nothing here is a force, and nothing here is a new postulate. It is the sign of a product, plus the empirical fact that electrons take the minus sign.

The lock: choose the spin, and the space is decided

A total state is a spatial part times a spin part, and its exchange symmetry is the product of their two symmetries. Electrons must be antisymmetric overall (§10.4 will say why), so the product must be $-1$ — which leaves exactly two possibilities:

\Phi^{(A)}(p,q) = \psi^{(A)}(\mathbf r_p,\mathbf r_q)\,\chi^{(S)}_{S,M_S}(p,q)\tag{10.22}

\Phi^{(A)}(p,q) = \psi^{(S)}(\mathbf r_p,\mathbf r_q)\,\chi^{(A)}_{S,M_S}(p,q)\tag{10.23}

The spin decides the geometry. Two electrons with parallel spins ( $S = 1$ ) are forced into an antisymmetric wave function and therefore avoid each other; two with opposed spins ( $S = 0$ ) are permitted a symmetric one and may huddle. Neither is a force. Both are the sign of a product.

Switch the widget to its spin view and pick a row: the spatial symmetry — and with it the Fermi hole in every other view — is set for you.

Two identical particles: exchange is reflection across the diagonal

Two spin-halves span a four-dimensional space. Combining them (Eq. 8.6) splits it into three states with S = 1 and one with S = 0 — and the split is exactly the split by exchange symmetry. Pick one and the spatial symmetry of an electron pair is forced, because the total state must be antisymmetric.

spin state	χ	S	M_S	under exchange
\|↑↑⟩	χ⁽ˢ⁾₁,₊₁	1	+1	symmetric (+1)
(\|↑↓⟩ + \|↓↑⟩)/√2	χ⁽ˢ⁾₁,₀	1	0	symmetric (+1)
\|↓↓⟩	χ⁽ˢ⁾₁,₋₁	1	-1	symmetric (+1)
(\|↑↓⟩ − \|↓↑⟩)/√2	χ⁽ᴬ⁾₀,₀	0	0	antisymmetric (−1)

Spin triplet, S = 1 — spin part symmetric. The total state of two electrons must be antisymmetric, so the spatial part is forced antisymmetric: Eq. (10.22). The other views now show the Fermi hole. This is why solids resist compression.

Three symmetric states and one antisymmetric: 2 × 2 = 4 = 3 + 1. The symmetric subspace is the larger one, so an electron pair has three ways to be spin-symmetric and only one way not to be.

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.

This is the hinge the whole of chemistry turns on

Read Eqs. (10.22)–(10.23) backwards and they become a rule about bonding.

Two hydrogen atoms approach. If their electrons have opposed spins, the singlet permits a symmetric spatial state, the electrons huddle between the two nuclei, and that negative charge sitting between two positive ones binds the molecule. If the spins are parallel, the triplet forces an antisymmetric spatial state, the electrons are excluded from the midpoint, and there is no bond.

That is why the covalent bond is a pair bond, and why electrons in molecules are drawn as paired arrows pointing opposite ways. Not because opposite spins attract — they do not, magnetically they barely interact at all — but because opposite spins are the only ones allowed to share the space between the nuclei.

Which particles get which sign?

Nothing derived so far chooses. §10.4 supplies the answer as an experimental fact:

type⇅	spin⇅	exchange symmetry⇅	examples⇅
boson	integer — 0, 1, 2, …	symmetric	photon (1), ⁴He nucleus (0), deuteron (1)
fermion	half-integer — ½, 3/2, …	antisymmetric	electron, proton, neutron, quark, ³He nucleus

This is the spin–statistics theorem . It is worth being clear about its status: within this book it is an empirical rule, added from outside because the mathematics permits both options and observation reports only one pairing. It was later derived, but only from relativistic quantum field theory — the same place §10.1’s footnote pointed when apologising for particle labels.

What follows for electrons

Electrons are spin-half, so every multi-electron state is antisymmetric. Three consequences, in the book’s order:

One — the Pauli exclusion principle . At most one electron per single-particle state, since two would make the state symmetric. This is the periodic table.

Two — the rigidity of matter. Symmetric spin forces antisymmetric space, so electrons avoid each other. A solid resists compression because its electrons cannot be pushed into the same states.

Three — covalent bonding. Antisymmetric spin permits symmetric space, so electrons huddle between two nuclei and bind them.

The second one is worth making quantitative, because “electrons avoid each other” sounds like a force and is not. Drop $N$ identical particles into the same ladder of levels and let statistics do the only work:

Reproduce it in NumPy

The filling is greedy from the bottom, respecting each level's capacity for fermions and ignoring it for bosons — the same rule the two diagrams above run, so the picture and the numbers cannot drift apart. Problem 5's four answers come out as a side effect.

def ground(N, stat, cap=2, dim=1):
    """Greedy fill of an oscillator ladder. cap = particles per orbital level."""
    if dim == 1:
        deg = lambda n: 1
        E_of = lambda n: n + 0.5
    else:
        deg = lambda n: (n + 1) * (n + 2) // 2          # 3-D isotropic shell
        E_of = lambda n: n + 1.5
    if stat == 'boson':
        return N * E_of(0)
    E, left, n = 0.0, N, 0
    while left > 0:
        take = min(deg(n) * cap, left)
        E += take * E_of(n); left -= take; n += 1
    return E

print("Ground-state energy of N identical particles in one oscillator")
print("=" * 70)
print(f"  {'N':>4}{'bosons':>12}{'spin-1/2 fermions':>20}{'(N/2)^2':>10}{'ratio':>8}")
for N in (2, 4, 6, 8, 10, 20, 100):
    b, f = ground(N, 'boson'), ground(N, 'fermion', 2)
    print(f"  {N:>4}{b:>12.1f}{f:>20.1f}{(N/2)**2:>10.1f}{f/b:>8.1f}")
print("""
  Bosons cost N/2 hbar-omega -- every particle in the ground level, no limit.
  Fermions cost exactly (N/2)^2, because filling k levels two at a time gives
  2 * sum(n + 1/2, n = 0..k-1) = k^2 with k = N/2.  LINEAR versus QUADRATIC.

  The ratio is N/2, and it grows without bound: at N = 100 the fermions already
  sit fifty times higher, and for the electrons in a real solid the factor is
  astronomical.  Nothing repels them -- the low levels simply run out, and each
  new pair must go somewhere more expensive.

  This is the rigidity of ordinary matter. Compressing a solid squeezes the
  levels apart, and because the fermion total grows as N^2 rather than as N,
  that costs enormously more than it would if electrons could share a state.
  The same energy holds up a white dwarf with no fuel burning at all.""")

print("\nProblem 5: five particles in a 3-D oscillator, four kinds of statistics")
print("=" * 70)
cases = [("distinguishable", 5 * 1.5),
         ("spinless bosons",  ground(5, 'boson', dim=3)),
         ("spin-1/2 fermions (cap 2)", ground(5, 'fermion', 2, dim=3)),
         ("spin-3/2 fermions (cap 4)", ground(5, 'fermion', 4, dim=3))]
for name, E in cases:
    print(f"  {name:<28}{E:>6.1f} = {int(round(E*2)):>2}/2 hbar-omega")
print("""
  Note the last two. Raising the spin from 1/2 to 3/2 LOWERS the ground-state
  energy, from 21/2 to 17/2, because each spatial state now holds four particles
  instead of two and fewer of them are forced upstairs. More spin, less crowding.""")

printsGround-state energy of N identical particles in one oscillator ====================================================================== N bosons spin-1/2 fermions (N/2)^2 ratio 2 1.0 1.0 1.0 1.0 4 2.0 4.0 4.0 2.0 6 3.0 9.0 9.0 3.0 8 4.0 16.0 16.0 4.0 10 5.0 25.0 25.0 5.0 20 10.0 100.0 100.0 10.0 100 50.0 2500.0 2500.0 50.0 Bosons cost N/2 hbar-omega -- every particle in the ground level, no limit. Fermions cost exactly (N/2)^2, because filling k levels two at a time gives 2 * sum(n + 1/2, n = 0..k-1) = k^2 with k = N/2. LINEAR versus QUADRATIC. The ratio is N/2, and it grows without bound: at N = 100 the fermions already sit fifty times higher, and for the electrons in a real solid the factor is astronomical. Nothing repels them -- the low levels simply run out, and each new pair must go somewhere more expensive. This is the rigidity of ordinary matter. Compressing a solid squeezes the levels apart, and because the fermion total grows as N^2 rather than as N, that costs enormously more than it would if electrons could share a state. The same energy holds up a white dwarf with no fuel burning at all. Problem 5: five particles in a 3-D oscillator, four kinds of statistics ====================================================================== distinguishable 7.5 = 15/2 hbar-omega spinless bosons 7.5 = 15/2 hbar-omega spin-1/2 fermions (cap 2) 10.5 = 21/2 hbar-omega spin-3/2 fermions (cap 4) 8.5 = 17/2 hbar-omega Note the last two. Raising the spin from 1/2 to 3/2 LOWERS the ground-state energy, from 21/2 to 17/2, because each spatial state now holds four particles instead of two and fewer of them are forced upstairs. More spin, less crowding.

Why a white dwarf does not collapse

Six particles, one oscillator, no interaction: bosons cost $3\hbar\omega$ and spin-half fermions cost $9\hbar\omega$ . The ratio is $N/2$ , and it grows without limit — the boson total is linear in $N$ , the fermion total quadratic.

For a white dwarf the same arithmetic runs with $N \sim 10^{57}$ electrons. The star has exhausted its fuel and is radiating away what heat it has, so there is no thermal pressure left; what holds it up is that compressing it would force electrons into ever higher levels, and that costs energy quadratically. It is called degeneracy pressure, and it is not a pressure in the kinetic sense at all — no collisions, no repulsion, just levels running out.

Above about 1.4 solar masses gravity wins anyway, because the electrons become relativistic and the exponent changes. That number is the Chandrasekhar limit, and it is a consequence of the minus sign in Eq. (10.6).

And for bosons

Many bosons may share one state, and when a macroscopic number do, quantum mechanics becomes visible at human scale:

Laser light. Photons are spin-1 bosons, so they have an enhanced probability of taking the same energy and momentum as photons already present — the same enhancement as §10.2’s two particles at one point, in momentum space instead of position space. Coherence is boson huddling.
Superfluid ⁴He below 2.2 K. A ⁴He atom is a spin-0 nucleus plus two electrons with combined spin 0, hence a boson. A large fraction condense into the lowest state and flow without friction.
Bose–Einstein condensates. Almost pure condensates were first produced in 1995 by cooling trapped atoms; the 2001 Nobel Prize went to Cornell, Ketterle and Wieman for it.

The twist: fermions can borrow bosonic behaviour, in pairs

Two fermions bound together make a composite boson — half-integer plus half-integer is an integer. The consequences are not small:

Superconductivity. Electrons pair up (Cooper pairs), and the pairs behave as bosons that condense into a single coherent state carrying current without resistance.
Superfluid ³He. A ³He atom is a fermion, unlike ⁴He, so it cannot condense — until pairs of atoms act as bosons, at temperatures about a thousand times lower than ⁴He needs.

So “boson” and “fermion” describe how a thing behaves under exchange at the scale you are looking at, not an immutable label. Bind two fermions tightly enough that nothing probes their internal structure, and the pair exchanges like a boson.

Where this is going

The chapter ends by noting the obvious generalization: for more than two particles, the state must be symmetric or antisymmetric under exchange of any pair. That is all chapter 11 needs, and it uses it immediately — the exclusion principle fills shells, and the space/spin lock of Eqs. (10.22)–(10.23) becomes Hund’s rules.

One thread is still loose. On this page the two electrons never interacted, so huddling and avoiding cost nothing. Turn on a real repulsion between them and the two arrangements have genuinely different energies — the difference is the exchange integral $K$ , which problem 3 derives and which the book never names. It is why parallel spins are favoured in a partly filled shell, and ultimately why iron is magnetic.

Check yourself

0 / 6 answered

1.Why can the four two-particle spin states be labelled by $S$ , $M_{S}$ **and** an exchange symmetry all at once?
Two electrons are in a spin triplet state, $S = 1$ .
2.What does that force about their spatial wave function, and why?
3.What is the logical status of the spin–statistics theorem within this book?
Six identical particles are placed in the same one-dimensional oscillator, with no interaction between them. As bosons the ground-state energy is 3ħω; as spin-½ fermions it is 9ħω.
4.Where does the extra 6ħω come from?
5.Why is a covalent bond formed by a *pair* of electrons with opposed spins?
6.⁴He becomes a superfluid below 2.2 K, but ³He only at temperatures about a thousand times lower. Why the difference?