2.1.2 Symmetrization#

Worked solutions for the homework problems in the 2.1.2 Symmetrization lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Three-particle statistics. For three identical particles, consider the transpositions \(\hat{\mathcal{P}}_{12}, \hat{\mathcal{P}}_{23}, \hat{\mathcal{P}}_{13}\).

(a) Verify the braid-like relation \(\hat{\mathcal{P}}_{12}\hat{\mathcal{P}}_{23}\hat{\mathcal{P}}_{12} = \hat{\mathcal{P}}_{13}\) by acting on a generic tensor basis state \(\vert\alpha\rangle\otimes\vert\beta\rangle\otimes\vert\gamma\rangle\).

(b) Suppose a three-particle state \(\vert\Psi\rangle\) is a simultaneous eigenstate of every transposition, \(\hat{\mathcal{P}}_{ij}\vert\Psi\rangle = \lambda_{ij}\vert\Psi\rangle\) with \(\lambda_{ij} \in \{\pm 1\}\). Use the identity in (a) to show that \(\lambda_{13} = \lambda_{23}\), and analogous relations to conclude \(\lambda_{12} = \lambda_{23} = \lambda_{13}\).

(c) Conclude that the only consistent exchange statistics for three or more identical particles in three dimensions are bosonic (\(\lambda = +1\)) or fermionic (\(\lambda = -1\))—there is no third option.

Solution.

(a) The transposition \(\hat{\mathcal{P}}_{ij}\) exchanges the contents of tensor slots \(i\) and \(j\). Apply the three factors of \(\hat{\mathcal{P}}_{12}\hat{\mathcal{P}}_{23}\hat{\mathcal{P}}_{12}\) to \(\vert\alpha\beta\gamma\rangle\), working right to left:

The end result \(\vert\gamma\beta\alpha\rangle\) has slots \(1\) and \(3\) interchanged relative to the start, which is exactly \(\hat{\mathcal{P}}_{13}\vert\alpha\beta\gamma\rangle = \vert\gamma\beta\alpha\rangle\). Since the basis states \(\vert\alpha\rangle\otimes\vert\beta\rangle\otimes\vert\gamma\rangle\) span the three-particle space and both sides are linear operators, agreement on every basis state is agreement as operators:

(b) Let \(\vert\Psi\rangle\neq 0\) be a simultaneous eigenstate, \(\hat{\mathcal{P}}_{ij}\vert\Psi\rangle = \lambda_{ij}\vert\Psi\rangle\). Act with the operator identity from (a) and peel off one factor at a time:

where the last step uses \(\lambda_{12}^2 = 1\) (each eigenvalue is \(\pm1\) because \(\hat{\mathcal{P}}_{ij}^2 = \hat{I}\)). But the left side is also \(\hat{\mathcal{P}}_{13}\vert\Psi\rangle = \lambda_{13}\vert\Psi\rangle\). Equating and cancelling the nonzero \(\vert\Psi\rangle\),

The same braid identity with relabelled indices, \(\hat{\mathcal{P}}_{23}\hat{\mathcal{P}}_{12}\hat{\mathcal{P}}_{23} = \hat{\mathcal{P}}_{13}\) (check on \(\vert\alpha\beta\gamma\rangle\): \(\vert\alpha\beta\gamma\rangle\to\vert\alpha\gamma\beta\rangle\to\vert\gamma\alpha\beta\rangle\to\vert\gamma\beta\alpha\rangle\)), gives by the identical manipulation \(\lambda_{23}^2\,\lambda_{12} = \lambda_{12} = \lambda_{13}\). Combining the two results,

All three transpositions must share one common eigenvalue. (This is no accident: in \(S_3\) every transposition is conjugate to every other — e.g. \(\hat{\mathcal{P}}_{13} = \hat{\mathcal{P}}_{12}\hat{\mathcal{P}}_{23}\hat{\mathcal{P}}_{12}^{-1}\) — and conjugate operators have equal eigenvalues on a common eigenstate.)

(c) Each \(\hat{\mathcal{P}}_{ij}\) squares to the identity, so \(\lambda^2 = 1\) and \(\lambda \in \{+1, -1\}\) — there is no continuous or intermediate value. Part (b) forces this single \(\lambda\) to be the same for every pair. Hence a physical state of three (or, by the same conjugacy argument, \(N \ge 3\)) identical particles is either

• totally symmetric, \(\lambda = +1\) for all transpositions — bosons, or

• totally antisymmetric, \(\lambda = -1\) for all transpositions — fermions.

No “mixed” assignment (some \(+1\), some \(-1\)) and no third species survives: those are exactly the only two one-dimensional representations of the permutation group \(S_N\). The clause “in three dimensions” is essential. The decisive input was \(\hat{\mathcal{P}}_{ij}^2 = \hat{I}\) — exchanging two particles twice is the identity. In two dimensions, a double exchange is a closed loop that can wind, the exchange operators generate the braid group rather than \(S_N\), and \(\hat{\mathcal{P}}^2 \neq \hat{I}\) is allowed. That loophole admits anyons with \(\lambda = \mathrm{e}^{\mathrm{i}\theta}\) for arbitrary \(\theta\) (see §2.3.1); it is closed in 3D.

2. Exchange projectors. For two particles, define \(\hat{\Sigma}_\pm = \tfrac{1}{2}(\hat{I} \pm \hat{\mathcal{P}}_{12})\). Show that \(\hat{\Sigma}_\pm^2 = \hat{\Sigma}_\pm\), \(\hat{\Sigma}_+ \hat{\Sigma}_- = 0\), and \(\hat{\Sigma}_+ + \hat{\Sigma}_- = \hat{I}\), and verify that \(\hat{\Sigma}_+\vert\Psi\rangle\) is symmetric and \(\hat{\Sigma}_-\vert\Psi\rangle\) is antisymmetric under \(\hat{\mathcal{P}}_{12}\). Conclude that every two-particle state decomposes uniquely as \(\vert\Psi\rangle = \hat{\Sigma}_+\vert\Psi\rangle + \hat{\Sigma}_-\vert\Psi\rangle\).

Solution.

The single fact driving everything is \(\hat{\mathcal{P}}_{12}^2 = \hat{I}\).

Idempotence. Expand the square, using \(\hat{\mathcal{P}}_{12}^2 = \hat{I}\):

Orthogonality. The cross terms cancel and \(\hat{\mathcal{P}}_{12}^2 = \hat{I}\):

(The same computation with the factors reversed gives \(\hat{\Sigma}_-\hat{\Sigma}_+ = 0\).)

Completeness. The \(\hat{\mathcal{P}}_{12}\) terms cancel on addition:

Idempotent, mutually orthogonal, and summing to \(\hat{I}\): \(\hat{\Sigma}_\pm\) are complementary projectors.

Symmetry of the projected states. Act with \(\hat{\mathcal{P}}_{12}\) and again use \(\hat{\mathcal{P}}_{12}^2 = \hat{I}\):

So \(\hat{\Sigma}_+\vert\Psi\rangle\) is an eigenstate with eigenvalue \(+1\) (symmetric) and \(\hat{\Sigma}_-\vert\Psi\rangle\) with eigenvalue \(-1\) (antisymmetric).

Unique decomposition. Completeness gives the decomposition immediately:

a symmetric piece plus an antisymmetric piece. Uniqueness: suppose \(\vert\Psi\rangle = \vert\psi_+\rangle + \vert\psi_-\rangle\) with \(\hat{\mathcal{P}}_{12}\vert\psi_\pm\rangle = \pm\vert\psi_\pm\rangle\). Then \(\hat{\Sigma}_+\vert\psi_+\rangle = \tfrac12(\hat I + \hat{\mathcal{P}}_{12})\vert\psi_+\rangle = \vert\psi_+\rangle\) and \(\hat{\Sigma}_+\vert\psi_-\rangle = \tfrac12(\hat I + \hat{\mathcal{P}}_{12})\vert\psi_-\rangle = \tfrac12(\vert\psi_-\rangle - \vert\psi_-\rangle) = 0\), so \(\hat{\Sigma}_+\vert\Psi\rangle = \vert\psi_+\rangle\); likewise \(\hat{\Sigma}_-\vert\Psi\rangle = \vert\psi_-\rangle\). The two components are forced to be \(\hat{\Sigma}_\pm\vert\Psi\rangle\), so the decomposition is unique. This is the explicit statement that the two-particle space splits cleanly, \(\mathcal{H}^{\otimes 2} = \mathcal{H}_+ \oplus \mathcal{H}_-\), into bosonic and fermionic sectors. (For \(N \ge 3\) the analogous projectors still exist, but \(\mathcal{H}^{\otimes N}\) also contains mixed-symmetry sectors that are neither; only the totally symmetric and totally antisymmetric pieces are physical — cf. Problem 1.)

3. Generalized Pauli exclusion. The lecture shows the two-fermion state vanishes when \(\vert\alpha_1\rangle = \vert\alpha_2\rangle\). Here we extend this vanishing to linear dependence.

(a) Show that \(\vert\alpha_1, \alpha_2\rangle_- = 0\) whenever \(\vert\alpha_2\rangle = c\,\vert\alpha_1\rangle\) for any \(c \in \mathbb{C}\).

(b) For three fermions, show that \(\vert\alpha_1, \alpha_2, \alpha_3\rangle_-\) vanishes whenever \(\vert\alpha_3\rangle \in \mathrm{span}\{\vert\alpha_1\rangle, \vert\alpha_2\rangle\}\). (Hint: use linearity in the third slot and the two-fermion result.)

(c) Argue from multilinearity and antisymmetry that \(\vert\alpha_1, \ldots, \alpha_N\rangle_-\) depends only on the \(N\)-dimensional subspace spanned by \(\{\vert\alpha_i\rangle\}\), up to an overall scalar, and vanishes whenever these states are linearly dependent. This is the geometric form of Pauli exclusion: the physical fermionic state is determined by the filled subspace, not by a choice of basis inside it.

(d) Contrast with bosons. Show by explicit counterexample that \(\vert\alpha, \alpha\rangle_+ \ne 0\) for any nonzero \(\vert\alpha\rangle\), and that \(\vert\alpha_1, c\,\alpha_1\rangle_+ \ne 0\) for any \(c \ne 0\). Bosons admit no analogous linear-dependence vanishing.

Solution.

(a) The two-fermion antisymmetric state is \(\vert\alpha_1,\alpha_2\rangle_- = \tfrac{1}{\sqrt2}\big(\vert\alpha_1\rangle\otimes\vert\alpha_2\rangle - \vert\alpha_2\rangle\otimes\vert\alpha_1\rangle\big)\). It is linear in each slot. Substitute \(\vert\alpha_2\rangle = c\,\vert\alpha_1\rangle\) and pull the scalar out:

The two terms are identical and cancel — antisymmetry kills any state with proportional arguments.

(b) The three-fermion state \(\vert\alpha_1,\alpha_2,\alpha_3\rangle_-\) is multilinear and totally antisymmetric. First note that a three-fermion state with two equal arguments vanishes: if slots \(i\) and \(j\) carry the same single-particle state, then exchanging those slots leaves the state unchanged, \(\hat{\mathcal{P}}_{ij}\vert\cdots\rangle = +\vert\cdots\rangle\), while antisymmetry demands \(\hat{\mathcal{P}}_{ij}\vert\cdots\rangle = -\vert\cdots\rangle\); a vector equal to minus itself is zero. (Equivalently, the Slater-determinant sum has two identical columns.)

Now let \(\vert\alpha_3\rangle = a\,\vert\alpha_1\rangle + b\,\vert\alpha_2\rangle\) lie in \(\mathrm{span}\{\vert\alpha_1\rangle,\vert\alpha_2\rangle\}\). Linearity in the third slot gives

In the first term slots \(1\) and \(3\) are equal; in the second, slots \(2\) and \(3\) are equal. By the observation just made, both terms vanish, so \(\vert\alpha_1,\alpha_2,\alpha_3\rangle_- = 0\).

(c) The antisymmetrized state \(\vert\alpha_1,\ldots,\alpha_N\rangle_-\) is multilinear in each slot and totally antisymmetric under any pairwise exchange. Two structural facts follow:

• Adding a multiple of another argument changes nothing. Replacing \(\vert\alpha_i\rangle \to \vert\alpha_i\rangle + c\,\vert\alpha_j\rangle\) (\(j\neq i\)) produces an extra term with \(\vert\alpha_j\rangle\) in both slots \(i\) and \(j\); that term has a repeated argument and vanishes (part b’s mechanism). So this “shear” leaves the state invariant.

• Rescaling an argument rescales the state. Replacing \(\vert\alpha_i\rangle \to c\,\vert\alpha_i\rangle\) multiplies the whole state by \(c\).

These are exactly the elementary column operations on a determinant. Any change of basis inside \(V \equiv \mathrm{span}\{\vert\alpha_i\rangle\}\) is a composition of such operations, i.e. a matrix \(T\); it multiplies the state by the single scalar \(\det T\). Therefore the state depends only on the subspace \(V\), not on which basis of \(V\) was chosen — up to an overall scalar (and after normalization, up to a phase). If the \(\{\vert\alpha_i\rangle\}\) are linearly dependent, some \(\vert\alpha_k\rangle\) is a combination of the others; expanding that slot by linearity produces only terms with a repeated argument, each zero, so the whole state vanishes. Physically: a fermionic many-body state is labelled by the filled subspace \(V\) (which \(N\)-dimensional subspace of single-particle states is occupied), and \(N\) fermions cannot be packed into fewer than \(N\) independent states.

(d) Bosons use the plus sign, so the two terms reinforce instead of cancelling. First, \(\vert\alpha,\alpha\rangle_+ = \vert\alpha\rangle\otimes\vert\alpha\rangle\) (the lecture’s normalized form for a doubly-occupied mode); if \(\vert\alpha\rangle\neq 0\) then \(\big\|\,\vert\alpha\rangle\otimes\vert\alpha\rangle\,\big\| = \big\|\,\vert\alpha\rangle\,\big\|^2 \neq 0\), so \(\vert\alpha,\alpha\rangle_+ \neq 0\). Second, for proportional arguments the unnormalized symmetric combination is

which is nonzero for every \(c\neq 0\). (Its norm \(\sqrt2\,\vert c\vert\) exceeds \(1\), signalling the repeated-mode normalization correction of Problem 4 — but it certainly does not vanish.)

There is no bosonic linear-dependence vanishing: the symmetrizer never cancels terms. Many bosons piling into one state is allowed — indeed it is the basis of Bose–Einstein condensation — whereas the antisymmetrizer forbids the fermionic analog outright.

4. Bosonic normalization factor. Work at the state level, with single-particle modes drawn from an orthonormal basis.

(a) For two bosons in distinct modes \(\vert\alpha\rangle, \vert\beta\rangle\) with \(\alpha \ne \beta\), verify that the lecture’s prefactor \(1/\sqrt{2!}\) gives a unit-norm state \(\vert\alpha, \beta\rangle_+\). For two bosons in the same mode, verify that the normalized state is \(\vert\alpha, \alpha\rangle_+ = \vert\alpha\rangle \otimes \vert\alpha\rangle\), whose prefactor differs from the naive \(1/\sqrt{2!}\) by an extra factor of \(\sqrt{2!}\).

(b) Three bosons with occupations \(n_\alpha = 2,\ n_\beta = 1\). Enumerate the six terms in the unnormalized sum \(\sum_{\pi \in S_3} \vert\alpha_{\pi(1)} \alpha_{\pi(2)} \alpha_{\pi(3)}\rangle\) (with \(\alpha_1 = \alpha_2 = \alpha\), \(\alpha_3 = \beta\)), observe that only three distinct orthonormal orderings appear (each with multiplicity \(2!\)), and compute the norm squared of the full sum: \((2!)^2 \cdot 3 = 3!\cdot 2!\).

(c) General formula. For \(N\) bosons with occupation numbers \(\{n_\alpha\}\) on an orthonormal basis, count the number of distinct orderings (\(N!/\prod_\alpha n_\alpha!\)) and the multiplicity of each (\(\prod_\alpha n_\alpha!\)). Deduce that the unnormalized sum has norm squared \(N!\,\prod_\alpha n_\alpha!\), hence

(d) Explain in one sentence why fermions require no analogous \(n_\alpha!\) correction.

Solution.

(a) Distinct modes. With the prefactor \(1/\sqrt{2!}\), \(\vert\alpha,\beta\rangle_+ = \tfrac{1}{\sqrt2}\big(\vert\alpha\beta\rangle + \vert\beta\alpha\rangle\big)\). Its norm squared, with tensor inner products factorizing as \(\langle\mu\nu\vert\mu'\nu'\rangle = \langle\mu\vert\mu'\rangle\langle\nu\vert\nu'\rangle\), is

For orthonormal modes with \(\alpha\neq\beta\): the “diagonal” terms are \(\langle\alpha\beta\vert\alpha\beta\rangle = \langle\beta\alpha\vert\beta\alpha\rangle = 1\), and the “cross” terms are \(\langle\alpha\beta\vert\beta\alpha\rangle = \langle\alpha\vert\beta\rangle\langle\beta\vert\alpha\rangle = 0\). Hence \(\langle\alpha,\beta\vert\alpha,\beta\rangle_+ = \tfrac12(1+0+0+1) = 1\) — unit norm. The naive \(1/\sqrt{2!}\) is correct here.

Same mode. Now both bosons occupy \(\vert\alpha\rangle\), so \(\alpha_1 = \alpha_2 = \alpha\) and the permanent sum has two identical terms:

The naive prefactor \(1/\sqrt{2!}\) would give \(\tfrac{2}{\sqrt2}\vert\alpha\alpha\rangle = \sqrt2\,\vert\alpha\alpha\rangle\), which has norm \(\sqrt2 \neq 1\). The correctly normalized state is just \(\vert\alpha,\alpha\rangle_+ = \vert\alpha\rangle\otimes\vert\alpha\rangle\), obtained from the unnormalized sum \(2\vert\alpha\alpha\rangle\) by dividing by \(2 = \sqrt{2!}\cdot\sqrt{2!}\). Compared with the naive \(1/\sqrt{2!}\), the true prefactor carries one extra \(1/\sqrt{2!} = 1/\sqrt{n_\alpha!}\) — the repeated-occupation correction. Equivalently \(\mathcal{N}_+ = 1/\sqrt{N!\,\prod n_\alpha!} = 1/\sqrt{2!\cdot 2!} = 1/2\), and \(\tfrac12\cdot 2\vert\alpha\alpha\rangle = \vert\alpha\alpha\rangle\). ✓

(b) With \(\alpha_1 = \alpha_2 = \alpha\) and \(\alpha_3 = \beta\), the term for permutation \(\pi\) is \(\vert\alpha_{\pi(1)}\rangle\otimes\vert\alpha_{\pi(2)}\rangle\otimes\vert\alpha_{\pi(3)}\rangle\); the mode \(\beta\) sits in whichever slot \(k\) has \(\pi(k)=3\). Running over \(S_3\):

Only three distinct orderings appear — \(\vert\alpha\alpha\beta\rangle\), \(\vert\alpha\beta\alpha\rangle\), \(\vert\beta\alpha\alpha\rangle\) — each occurring with multiplicity \(2! = 2\) (the \(2!\) ways of permuting the two identical \(\alpha\) labels among themselves). So

The three distinct orderings are orthonormal (distinct arrangements of orthonormal modes), so the norm squared is

(c) For \(N\) bosons with occupations \(\{n_\alpha\}\), \(\sum_\alpha n_\alpha = N\), on orthonormal modes: the \(N!\) permutations in the sum produce orderings of the multiset of mode labels. The number of distinct orderings is the multinomial coefficient

and each distinct ordering is reached by exactly \(\prod_\alpha n_\alpha!\) permutations — the ways of shuffling identical labels within their occupied slots. Hence the unnormalized sum equals \(\big(\prod_\alpha n_\alpha!\big)\) times the sum over distinct orderings. Those distinct orderings are mutually orthonormal, so

To make the state unit-norm we divide by the square root of this, giving

(Checks: \(n_\alpha=2,n_\beta=1\) gives \(3!\cdot2! = 12\), matching part b; the all-distinct case \(n_\alpha\le1\) gives \(\prod n_\alpha! = 1\) and recovers \(1/\sqrt{N!}\).)

(d) Fermions obey the Pauli exclusion principle, so every occupation number is \(n_\alpha \in \{0,1\}\), making \(n_\alpha! = 1\) and the correction factor \(\prod_\alpha n_\alpha! = 1\) trivially — there is never a repeated mode to over-count (a repeated mode would make the antisymmetric state vanish outright, not merely rescale it).

5. Insertion and deletion as inverses. Work in a single-mode Fock space throughout.

(a) Fermionic case. The fermionic Fock space for one mode contains only the vacuum \(\mathbb{1}\) and the occupied state \(\vert\alpha\rangle\). Apply the insertion and deletion rules to compute all four of \(\vert\alpha\rangle \rhd_- \mathbb{1}\), \(\vert\alpha\rangle \rhd_- \vert\alpha\rangle\), \(\vert\alpha\rangle \lhd_- \mathbb{1}\), and \(\vert\alpha\rangle \lhd_- \vert\alpha\rangle\). Verify that \(\vert\alpha\rangle \rhd_- \vert\alpha\rangle = 0\) (Pauli) and \(\vert\alpha\rangle \lhd_- \mathbb{1} = 0\).

(b) Verify that deletion undoes insertion on the vacuum, and insertion undoes deletion on the occupied state:

Combining both, conclude that \(\vert\alpha\rangle \lhd_- \bigl(\vert\alpha\rangle \rhd_- \vert\Psi\rangle\bigr) + \vert\alpha\rangle \rhd_- \bigl(\vert\alpha\rangle \lhd_- \vert\Psi\rangle\bigr) = \vert\Psi\rangle\) for every \(\vert\Psi\rangle\) in the single-mode fermionic Fock space.

(c) Bosonic case. Let \(\vert\alpha^n\rangle\) denote the \(n\)-boson state in a single mode (with \(\vert\alpha^0\rangle = \mathbb{1}\) and \(\vert\alpha^1\rangle = \vert\alpha\rangle\)). Using the recursive definition, prove by induction on \(n\):

(d) Compute \(\vert\alpha\rangle \lhd_+ \bigl(\vert\alpha\rangle \rhd_+ \vert\alpha^n\rangle\bigr) - \vert\alpha\rangle \rhd_+ \bigl(\vert\alpha\rangle \lhd_+ \vert\alpha^n\rangle\bigr)\) and show the result is \((2n+1)\,\vert\alpha^n\rangle\), not \(\vert\alpha^n\rangle\). Conclude that the bare bosonic insertion and deletion rules are mutual inverses only up to a particle-number-dependent multiplicity, whereas the fermionic rules are already exact inverses. The bosonic normalization is addressed in §2.1.3.

Solution.

Recall the rules. Insertion: \(\vert\alpha\rangle\rhd_\pm\mathbb{1} = \vert\alpha\rangle\) and \(\vert\alpha\rangle\rhd_\pm(\vert\beta\rangle\otimes\vert\Psi\rangle) = \vert\alpha\rangle\otimes\vert\beta\rangle\otimes\vert\Psi\rangle \pm \vert\beta\rangle\otimes(\vert\alpha\rangle\rhd_\pm\vert\Psi\rangle)\). Deletion: \(\vert\alpha\rangle\lhd_\pm\mathbb{1} = 0\) and \(\vert\alpha\rangle\lhd_\pm(\vert\beta\rangle\otimes\vert\Psi\rangle) = \delta_{\alpha\beta}\vert\Psi\rangle \pm \vert\beta\rangle\otimes(\vert\alpha\rangle\lhd_\pm\vert\Psi\rangle)\).

(a) Evaluate the four fermionic (\(-\)) operations in turn. The two base cases are immediate: \(\vert\alpha\rangle\rhd_-\mathbb{1} = \vert\alpha\rangle\) (insertion into the vacuum) and \(\vert\alpha\rangle\lhd_-\mathbb{1} = 0\) (nothing can be deleted from the vacuum).

For \(\vert\alpha\rangle\rhd_-\vert\alpha\rangle\), write the target as \(\vert\alpha\rangle = \vert\alpha\rangle\otimes\mathbb{1}\) (so \(\beta=\alpha,\ \Psi=\mathbb{1}\)) and apply the recursive insertion rule:

The minus sign of the fermionic rule cancels the two copies — this is the Pauli exclusion principle. For \(\vert\alpha\rangle\lhd_-\vert\alpha\rangle\), again with \(\beta=\alpha,\ \Psi=\mathbb{1}\), the recursive deletion rule gives

Summary: \(\vert\alpha\rangle\rhd_-\mathbb{1} = \vert\alpha\rangle\), \(\ \vert\alpha\rangle\rhd_-\vert\alpha\rangle = 0\), \(\ \vert\alpha\rangle\lhd_-\mathbb{1} = 0\), \(\ \vert\alpha\rangle\lhd_-\vert\alpha\rangle = \mathbb{1}\).

(b) Compose the results of (a):

Deletion undoes insertion on the vacuum; insertion undoes deletion on the occupied state. Now define, for any \(\vert\Psi\rangle\) in the two-dimensional single-mode fermionic Fock space,

Evaluate on the two basis states:

• \(\vert\Psi\rangle = \mathbb{1}\): first term \(= \vert\alpha\rangle\lhd_-\vert\alpha\rangle = \mathbb{1}\); second term \(= \vert\alpha\rangle\rhd_-(\vert\alpha\rangle\lhd_-\mathbb{1}) = \vert\alpha\rangle\rhd_-0 = 0\). Sum \(= \mathbb{1}\).

• \(\vert\Psi\rangle = \vert\alpha\rangle\): first term \(= \vert\alpha\rangle\lhd_-(\vert\alpha\rangle\rhd_-\vert\alpha\rangle) = \vert\alpha\rangle\lhd_-0 = 0\); second term \(= \vert\alpha\rangle\rhd_-(\vert\alpha\rangle\lhd_-\vert\alpha\rangle) = \vert\alpha\rangle\rhd_-\mathbb{1} = \vert\alpha\rangle\). Sum \(= \vert\alpha\rangle\).

So \(\mathcal{D}[\Psi] = \vert\Psi\rangle\) on both basis states; since insertion and deletion are linear, \(\mathcal{D}[\Psi] = \vert\Psi\rangle\) for every \(\vert\Psi\rangle\) in the space. This is the first-quantized precursor of the fermionic anticommutator \(\{\hat{a}_\alpha, \hat{a}^\dagger_\alpha\} = \hat{I}\) (§2.1.3): the sum of “insert-then-delete” and “delete-then-insert” is the identity.

(c) Write \(\vert\alpha^n\rangle = \vert\alpha\rangle^{\otimes n}\), the \(n\)-fold tensor product (already symmetric), with \(\vert\alpha^0\rangle = \mathbb{1}\).

Insertion, \(\vert\alpha\rangle\rhd_+\vert\alpha^n\rangle = (n+1)\vert\alpha^{n+1}\rangle\). Base case \(n=0\): \(\vert\alpha\rangle\rhd_+\mathbb{1} = \vert\alpha\rangle = 1\cdot\vert\alpha^1\rangle\). ✓ Inductive step: assume the claim for \(n\). Write \(\vert\alpha^{n+1}\rangle = \vert\alpha\rangle\otimes\vert\alpha^n\rangle\) (so \(\beta=\alpha,\ \Psi=\alpha^n\)) and apply the bosonic (\(+\)) rule:

This is the claim with \(n\to n+1\), completing the induction.

Deletion, \(\vert\alpha\rangle\lhd_+\vert\alpha^n\rangle = n\,\vert\alpha^{n-1}\rangle\). Base case \(n=0\): \(\vert\alpha\rangle\lhd_+\mathbb{1} = 0 = 0\cdot\vert\alpha^{-1}\rangle\). ✓ Inductive step: assume the claim for \(n\). With \(\vert\alpha^{n+1}\rangle = \vert\alpha\rangle\otimes\vert\alpha^n\rangle\) and \(\delta_{\alpha\alpha}=1\),

the claim with \(n\to n+1\). Both formulas hold for all \(n\ge 0\). (Intuitively: insertion drops \(\vert\alpha\rangle\) into each of \(n+1\) gaps of \(\vert\alpha^{n+1}\rangle\), all identical, giving the multiplicity \(n+1\); deletion removes \(\vert\alpha\rangle\) from each of the \(n\) occupied slots, giving \(n\).)

(d) Substitute the formulas from (c), being careful that the operations are linear so scalars pull through:

Their difference is

The result is \((2n+1)\vert\alpha^n\rangle\), not \(\vert\alpha^n\rangle\): the bare bosonic insertion/deletion rules fail to be exact mutual inverses, missing by an \(n\)-dependent multiplicity. Contrast the fermionic case (part b), where the analogous combination — there a sum, the anticommutator pattern — returned \(\vert\Psi\rangle\) exactly. The cure is to rescale: defining \(\hat{a}^\dagger_\alpha,\hat{a}_\alpha\) to carry \(\sqrt{n}\)-type normalization factors converts the bare rules into operators obeying \([\hat{a}_\alpha,\hat{a}^\dagger_\alpha] = \hat{I}\) exactly. That construction is the subject of §2.1.3.

6. Slater determinant overlap. Derive the central identity relating many-fermion inner products to determinants of single-particle overlaps.

(a) For arbitrary (possibly non-orthogonal) single-particle states \(\vert\alpha_1\rangle, \vert\alpha_2\rangle, \vert\beta_1\rangle, \vert\beta_2\rangle\), compute \(\langle\alpha_1, \alpha_2 \vert \beta_1, \beta_2\rangle_-\) directly from the two-fermion antisymmetric definition, and show it equals \(\det M\) with \(M_{ij} = \langle\alpha_i \vert \beta_j\rangle\).

(b) Specialize to orthonormal \(\{\vert\alpha_i\rangle\}\) and verify \(\langle\alpha_1, \alpha_2 \vert \alpha_1, \alpha_2\rangle_- = 1\).

(c) Hydrogen molecule. Let \(\vert L\rangle, \vert R\rangle\) be unit-norm but non-orthogonal left and right atomic orbitals with real overlap \(\langle L \vert R\rangle = s \in [0, 1]\). Compute \(\langle L, R \vert L, R\rangle_-\) and interpret the limits \(s \to 0\) (atoms far apart) and \(s \to 1\) (atoms merge; linear dependence forces the antisymmetric state to vanish, a generalized form of Pauli exclusion).

(d) State without proof the \(N\)-fermion generalization \(\langle\alpha_1, \ldots, \alpha_N \vert \beta_1, \ldots, \beta_N\rangle_- = \det\langle\alpha_i \vert \beta_j\rangle\). Briefly note the bosonic analog, which replaces the determinant with the permanent of the same matrix.

Solution.

(a) The two-fermion antisymmetric states are \(\vert\beta_1,\beta_2\rangle_- = \tfrac{1}{\sqrt2}\big(\vert\beta_1\beta_2\rangle - \vert\beta_2\beta_1\rangle\big)\) and, dualizing, \(\langle\alpha_1,\alpha_2\vert = \tfrac{1}{\sqrt2}\big(\langle\alpha_1\alpha_2\vert - \langle\alpha_2\alpha_1\vert\big)\). Their overlap, expanding the product of two binomials, is

Each two-particle inner product factorizes over the tensor slots, \(\langle\mu\nu\vert\mu'\nu'\rangle = \langle\mu\vert\mu'\rangle\langle\nu\vert\nu'\rangle\):

The first and fourth products are identical; so are the second and third. The \(\tfrac12\) cancels the resulting factor of \(2\):

With \(M_{ij} \equiv \langle\alpha_i\vert\beta_j\rangle\) this is \(M_{11}M_{22} - M_{12}M_{21} = \det M\), exactly the \(2\times2\) determinant. No orthogonality was assumed.

(b) For orthonormal \(\{\vert\alpha_i\rangle\}\) taken as the bra states and with \(\beta_j = \alpha_j\), the matrix is \(M_{ij} = \langle\alpha_i\vert\alpha_j\rangle = \delta_{ij}\), the identity. Hence

so the prefactor \(1/\sqrt{2!}\) correctly normalizes the antisymmetric state when the modes are orthonormal.

(c) Here \(\vert\alpha_1\rangle = \vert\beta_1\rangle = \vert L\rangle\) and \(\vert\alpha_2\rangle = \vert\beta_2\rangle = \vert R\rangle\), with \(\langle L\vertL\rangle = \langle R\vertR\rangle = 1\) and \(\langle L\vertR\rangle = \langle R\vertL\rangle = s\) real. The overlap matrix and its determinant are

Limit \(s\to 0\) (atoms far apart, orbitals effectively orthogonal): \(\langle L,R\vert L,R\rangle_- \to 1\). The antisymmetric two-electron state is properly normalized — two well-separated, independent fermions, one on each atom.

Limit \(s\to 1\) (atoms merged, \(\vert R\rangle\to\vert L\rangle\)): \(\langle L,R\vert L,R\rangle_- \to 0\). The orbitals become linearly dependent, and the antisymmetric state collapses to the zero vector — one cannot place two fermions in what has become a single orbital. (A unit physical state would require dividing by \(\sqrt{1-s^2}\), which diverges as \(s\to1\), signalling that the state ceases to exist.) This is the generalized Pauli exclusion of Problem 3 made quantitative: the norm \(1-s^2\) interpolates smoothly between “two distinct states available” (\(s=0\)) and “exclusion” (\(s=1\)).

(d) The \(N\)-fermion generalization is the Slater-determinant overlap identity

an \(N\times N\) determinant of single-particle overlaps. (It follows from the antisymmetric definition: the \(N!\) permutations in bra and ket combine, by orthogonality of distinct slot assignments, into the Leibniz sum \(\sum_\pi \mathrm{sgn}(\pi)\prod_i \langle\alpha_i\vert\beta_{\pi(i)}\rangle\).) The vanishing of a determinant with dependent columns reproduces Pauli exclusion automatically. The bosonic analog uses the same overlap matrix but with the all-plus signs, replacing the determinant by the permanent:

which has no sign cancellation and therefore no linear-dependence vanishing — bosons may share states freely.

7. Fermion counting. How many independent antisymmetric states can be formed by placing \(N\) fermions into \(D\) single-particle modes (with \(D \geq N\))? Give the combinatorial formula as a binomial coefficient. What happens when \(N > D\), and how does this connect to the Pauli exclusion principle?

Solution.

By the geometric form of Pauli exclusion (Problem 3c), an \(N\)-fermion antisymmetric state is determined entirely by which \(N\) of the \(D\) modes are occupied — each mode holds \(0\) or \(1\) fermion, and the occupied set fixes the Slater determinant up to a phase. Counting independent states is therefore counting size-\(N\) subsets of a \(D\)-element set:

For example, \(D = 5\) modes and \(N = 3\) fermions give \(\binom{5}{3} = 10\) states.

When \(N > D\), the formula gives \(\binom{D}{N} = 0\) — no antisymmetric states exist. The reason is the pigeonhole principle: placing more than \(D\) fermions into \(D\) modes forces at least two fermions into the same mode, so the Slater determinant has two identical columns and vanishes. This is exactly the Pauli exclusion principle in counting form: a system of \(D\) single-particle modes can hold at most \(D\) fermions. (It is the microscopic origin of atomic shell filling and of the degeneracy pressure that supports white dwarfs and neutron stars.)

8. Boson counting. How many independent symmetric states can be formed by placing \(N\) bosons into \(D\) single-particle modes? Derive the combinatorial formula (stars and bars) as a binomial coefficient, and contrast with the fermionic count from the previous problem. Explain why there is no upper bound on \(N\) at fixed \(D\) for bosons.

Solution.

A symmetric \(N\)-boson state is determined by its occupation numbers \((n_1, n_2, \ldots, n_D)\), the number of bosons in each of the \(D\) modes, subject to \(n_i \ge 0\) and \(\sum_{i=1}^D n_i = N\). (Bosons carry no labels and have no exclusion constraint, so only the occupation tuple matters.) Counting independent states means counting such tuples.

Stars and bars. Represent the configuration as a row of \(N\) stars (the bosons) and \(D-1\) bars (dividers separating the \(D\) modes):

Every arrangement of these \(N + (D-1)\) symbols corresponds to exactly one occupation tuple (the stars before the first bar fill mode \(1\), those between bars \(1\) and \(2\) fill mode \(2\), and so on; an empty gap means \(n_i = 0\)). The number of arrangements is the number of ways to choose which of the \(N + D - 1\) positions hold the stars:

For example, \(N = 3\) bosons in \(D = 2\) modes give \(\binom{4}{3} = 4\) states: occupations \((3,0), (2,1), (1,2), (0,3)\).

Contrast with fermions. The fermion count \(\binom{D}{N}\) requires \(D \ge N\) and vanishes for \(N > D\). The boson count \(\binom{N+D-1}{N}\) is positive for every \(N \ge 0\) at any fixed \(D \ge 1\), and grows without bound — polynomially in \(N\), like \(N^{D-1}/(D-1)!\) for large \(N\). The difference is the Pauli exclusion principle: fermion occupations are capped at \(n_i \le 1\) (choosing a subset of modes), whereas boson occupations \(n_i \in \{0,1,2,\ldots\}\) are unrestricted (choosing a multiset). Because any number of bosons can pile into a single mode, there is no ceiling on \(N\) at fixed \(D\) — the physical content of Bose–Einstein statistics and, in the extreme, of Bose–Einstein condensation.