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Identical Particle Exercises

These exercises practice the exchange-symmetry layer added after forming a tensor-product space. The labels 1,2,…1,2,\ldots below are slot labels, not names of persistent particles. Physical states of identical bosons are symmetric under slot exchange; physical states of identical fermions are antisymmetric.

  1. Let ∣a⟩\lvert a\rangle and ∣b⟩\lvert b\rangle be distinct orthonormal one-particle states. Construct the normalized symmetric and antisymmetric two-particle states, and verify their exchange parity.
Solution

The normalized symmetric state is

∣a,b⟩S=12(∣a⟩1∣b⟩2+∣b⟩1∣a⟩2).\lvert a,b\rangle_S = \frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert b\rangle_2 +\lvert b\rangle_1\lvert a\rangle_2 \bigr).

The normalized antisymmetric state is

∣a,b⟩A=12(∣a⟩1∣b⟩2−∣b⟩1∣a⟩2).\lvert a,b\rangle_A = \frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert b\rangle_2 -\lvert b\rangle_1\lvert a\rangle_2 \bigr).

The exchange operator P12P_{12} swaps the two slots. Therefore

P12∣a,b⟩S=∣a,b⟩S,P12∣a,b⟩A=−∣a,b⟩A.P_{12}\lvert a,b\rangle_S = \lvert a,b\rangle_S, \qquad P_{12}\lvert a,b\rangle_A = -\lvert a,b\rangle_A.

The first state is allowed for identical bosons; the second is allowed for identical fermions.

  1. What happens if the two one-particle states in Exercise 1 are the same state ∣a⟩\lvert a\rangle?
Solution

For bosons, two particles can occupy the same one-particle state. The normalized two-slot state is simply

∣a⟩1∣a⟩2.\lvert a\rangle_1\lvert a\rangle_2.

Using the distinct-state formula with a plus sign would give 2 ∣a⟩1∣a⟩2\sqrt2\,\lvert a\rangle_1\lvert a\rangle_2, so it is not the correct normalized expression in the same-state limit.

For fermions, the antisymmetric projection vanishes:

12(∣a⟩1∣a⟩2−∣a⟩1∣a⟩2)=0.\frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert a\rangle_2 -\lvert a\rangle_1\lvert a\rangle_2 \bigr) = 0.

There is no normalized state with two identical fermions in the same complete one-particle state.

  1. Let ∣a⟩\lvert a\rangle and ∣b⟩\lvert b\rangle be normalized but not necessarily orthogonal, with
s=⟨a∣b⟩.s=\langle a\vert b\rangle.

Find the normalization factors for

∣a⟩1∣b⟩2±∣b⟩1∣a⟩2.\lvert a\rangle_1\lvert b\rangle_2 \pm \lvert b\rangle_1\lvert a\rangle_2.
Solution

Let

∣Φ±⟩=∣a⟩1∣b⟩2±∣b⟩1∣a⟩2.\lvert\Phi_\pm\rangle = \lvert a\rangle_1\lvert b\rangle_2 \pm \lvert b\rangle_1\lvert a\rangle_2.

Then

⟨Φ±∣Φ±⟩=1+1±⟨a∣b⟩⟨b∣a⟩±⟨b∣a⟩⟨a∣b⟩=2(1±∣s∣2).\begin{aligned} \langle\Phi_\pm\vert\Phi_\pm\rangle &= 1+1 \pm \langle a\vert b\rangle\langle b\vert a\rangle \pm \langle b\vert a\rangle\langle a\vert b\rangle\\ &= 2(1\pm\lvert s\rvert^2). \end{aligned}

Therefore the normalized states are

∣Φ±⟩norm=∣a⟩1∣b⟩2±∣b⟩1∣a⟩22(1±∣s∣2).\lvert\Phi_\pm\rangle_{\mathrm{norm}} = \frac{ \lvert a\rangle_1\lvert b\rangle_2 \pm \lvert b\rangle_1\lvert a\rangle_2 }{ \sqrt{2(1\pm\lvert s\rvert^2)} }.

If ∣s∣=1\lvert s\rvert=1, the antisymmetric denominator vanishes because the antisymmetric vector is zero.

  1. Write the two-fermion Slater determinant built from orthonormal spin-orbitals u(q)u(q) and v(q)v(q). Show that it changes sign under exchange of q1q_1 and q2q_2.
Solution

The determinant is

Ψ(q1,q2)=12∣u(q1)v(q1)u(q2)v(q2)∣\Psi(q_1,q_2) = \frac{1}{\sqrt2} \begin{vmatrix} u(q_1)&v(q_1)\\ u(q_2)&v(q_2) \end{vmatrix}

or

Ψ(q1,q2)=12[u(q1)v(q2)−v(q1)u(q2)].\Psi(q_1,q_2) = \frac{1}{\sqrt2} \bigl[ u(q_1)v(q_2) -v(q_1)u(q_2) \bigr].

Exchanging q1q_1 and q2q_2 swaps the two rows of the determinant, so

Ψ(q2,q1)=−Ψ(q1,q2).\Psi(q_2,q_1) = -\Psi(q_1,q_2).

If u=vu=v, the two columns are identical and the determinant is zero, which is Pauli exclusion in determinant form.

  1. Two electrons occupy the same spatial orbital ϕ(x)\phi(\mathbf x). Which spin state is allowed? Explain using exchange symmetry.
Solution

The spatial part is

ψspace(x1,x2)=ϕ(x1)ϕ(x2),\psi_{\mathrm{space}}(\mathbf x_1,\mathbf x_2) = \phi(\mathbf x_1)\phi(\mathbf x_2),

which is symmetric under exchange of the two electron slots. Electrons are fermions, so the total state must be antisymmetric. Therefore the spin part must be antisymmetric:

χsinglet=12(∣↑↓⟩−∣↓↑⟩).\chi_{\mathrm{singlet}} = \frac{1}{\sqrt2} \bigl( \lvert\uparrow\downarrow\rangle -\lvert\downarrow\uparrow\rangle \bigr).

The triplet spin states are symmetric and would make the total state symmetric, so they are not allowed for two electrons in the same spatial orbital.

  1. Suppose two electrons occupy distinct orthonormal spatial orbitals aa and bb. Define
ψ±(x1,x2)=12[a(x1)b(x2)±b(x1)a(x2)].\psi_\pm(\mathbf x_1,\mathbf x_2) = \frac{1}{\sqrt2} \bigl[ a(\mathbf x_1)b(\mathbf x_2) \pm b(\mathbf x_1)a(\mathbf x_2) \bigr].

Which spin symmetry must accompany ψ+\psi_+ and which must accompany ψ−\psi_-?

Solution

The state ψ+\psi_+ is spatially symmetric. To make the total two-electron state antisymmetric, it must be multiplied by the antisymmetric spin singlet.

The state ψ−\psi_- is spatially antisymmetric. It must be multiplied by a symmetric triplet spin state. Symbolically,

symmetric spatial×antisymmetric spin,antisymmetric spatial×symmetric spin.\begin{array}{ccl} \text{symmetric spatial} &\times& \text{antisymmetric spin},\\ \text{antisymmetric spatial} &\times& \text{symmetric spin}. \end{array}

The total spatial-spin product is antisymmetric in both cases.

  1. Translate the bosonic occupation state ∣2a,1c⟩B\lvert2_a,1_c\rangle_B into a normalized first-quantized slot wavefunction using orthonormal modes a,b,ca,b,c.
Solution

The occupation state has two bosons in mode aa and one boson in mode cc. The normalized symmetric slot state is

∣2a,1c⟩B=13(∣a⟩1∣a⟩2∣c⟩3+∣a⟩1∣c⟩2∣a⟩3+∣c⟩1∣a⟩2∣a⟩3).\lvert2_a,1_c\rangle_B = \frac{1}{\sqrt3} \bigl( \lvert a\rangle_1\lvert a\rangle_2\lvert c\rangle_3 +\lvert a\rangle_1\lvert c\rangle_2\lvert a\rangle_3 +\lvert c\rangle_1\lvert a\rangle_2\lvert a\rangle_3 \bigr).

There are three distinct slot arrangements, and they are orthonormal because aa and cc are orthonormal.

  1. Translate the fermionic occupation state ∣1a,1c⟩F\lvert1_a,1_c\rangle_F into a normalized two-slot state, assuming the mode order is a,b,ca,b,c.
Solution

With modes ordered as a,b,ca,b,c, the state with modes aa and cc occupied corresponds to

∣1a,1c⟩F=12(∣a⟩1∣c⟩2−∣c⟩1∣a⟩2).\lvert1_a,1_c\rangle_F = \frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert c\rangle_2 -\lvert c\rangle_1\lvert a\rangle_2 \bigr).

The minus sign is the two-particle form of fermionic antisymmetry. In occupation notation the same state is recorded compactly as the bitstring with occupations (1,0,1)(1,0,1).

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