Fock Space Exercises

These exercises practice Fock-space calculations in a fixed mode basis. They assume normalized number states. Bosonic creation and annihilation operators are denoted $a_i^\dagger,a_i$, fermionic operators are denoted $c_i^\dagger,c_i$, and $d_i^\dagger,d_i$ denotes either kind when the statistics do not affect the displayed formula.

Required background. Occupation-number states and number states fix normalization; creation and annihilation operators and number operators give their actions.

Occupation numbers and ordered mode operators

For fermions, use canonical mode order $1,2,3,\ldots$. The state convention is $\lvert n_1,n_2,\ldots\rangle=(c_1^\dagger)^{n_1}(c_2^\dagger)^{n_2}\cdots\lvert0\rangle$. The sign in $c_i$ or $c_i^\dagger$ is $(-1)^{S_i}$, where

$$S_i = \sum_{k<i} n_k$$

is the number of occupied modes before mode $i$ in the chosen order.

Calculations with occupation states

1. Bosonic transfer amplitudes

For two bosonic modes, compute

$$a_2^\dagger a_1\lvert3,1\rangle_B.$$

Solution

Operators act from right to left. First annihilate one boson in mode $1$:

$$a_1\lvert3,1\rangle_B = \sqrt3\,\lvert2,1\rangle_B.$$

Then create one boson in mode $2$:

$$a_2^\dagger\lvert2,1\rangle_B = \sqrt2\,\lvert2,2\rangle_B.$$

Therefore

$$a_2^\dagger a_1\lvert3,1\rangle_B = \sqrt6\,\lvert2,2\rangle_B.$$

The factor $\sqrt6$ is the product of the removal factor from mode $1$ and the creation factor into mode $2$.

2. Local and total number

Let

$$\lvert\Psi\rangle_B = \frac{1}{\sqrt2} \bigl( \lvert2,0\rangle_B+\lvert0,2\rangle_B \bigr).$$

Compute $N_1\lvert\Psi\rangle_B$, $N_{\mathrm{tot}}\lvert\Psi\rangle_B$, and $\langle N_1\rangle_\Psi$.

Solution

The mode number operator acts diagonally:

$$N_1\lvert2,0\rangle_B = 2\lvert2,0\rangle_B, \qquad N_1\lvert0,2\rangle_B = 0.$$

Thus

$$N_1\lvert\Psi\rangle_B = \frac{2}{\sqrt2}\lvert2,0\rangle_B = \sqrt2\,\lvert2,0\rangle_B.$$

This is not proportional to $\lvert\Psi\rangle_B$, so the state is not an eigenstate of $N_1$. The total number operator is

$$N_{\mathrm{tot}} = N_1+N_2,$$

and both components have two particles. Hence

$$N_{\mathrm{tot}}\lvert\Psi\rangle_B = 2\lvert\Psi\rangle_B.$$

The expectation value of $N_1$ is

$$\langle N_1\rangle_\Psi = \frac12(2)+\frac12(0) = 1.$$

This example separates “definite total particle number” from “definite occupation of each mode.”

3. Creation, removal, and exclusion

With fermionic mode order $1,2,3,4$, compute

$$c_2\lvert1,1,0,1\rangle_F, \qquad c_3^\dagger\lvert1,1,0,1\rangle_F, \qquad c_4^\dagger\lvert1,1,0,1\rangle_F.$$

Solution

For $c_2$, the mode is occupied, and the number of occupied modes before mode $2$ is

$$S_2=n_1=1.$$

Therefore

$$c_2\lvert1,1,0,1\rangle_F = -\lvert1,0,0,1\rangle_F.$$

For $c_3^\dagger$, mode $3$ is empty, and

$$S_3=n_1+n_2=2.$$

Thus the sign is positive:

$$c_3^\dagger\lvert1,1,0,1\rangle_F = \lvert1,1,1,1\rangle_F.$$

For $c_4^\dagger$, mode $4$ is already occupied. Pauli exclusion gives

$$c_4^\dagger\lvert1,1,0,1\rangle_F = 0.$$

4. The sign of a fermionic hop

With the same fermionic mode order, compute

$$c_1^\dagger c_3\lvert0,1,1,0\rangle_F.$$

Explain why the sign should be derived from the operator order rather than guessed from the final bitstring.

Solution

Act first with $c_3$. Since mode $3$ is occupied and one occupied mode lies before it,

$$S_3=n_1+n_2=1.$$

Therefore

$$c_3\lvert0,1,1,0\rangle_F = -\lvert0,1,0,0\rangle_F.$$

Now create in mode $1$. There are no earlier modes, so no additional sign appears:

$$c_1^\dagger \bigl( -\lvert0,1,0,0\rangle_F \bigr) = -\lvert1,1,0,0\rangle_F.$$

Thus

$$c_1^\dagger c_3\lvert0,1,1,0\rangle_F = -\lvert1,1,0,0\rangle_F.$$

Recompute the number of earlier occupied modes after each operator acts. The resulting sign is relative to the chosen mode convention; observable predictions are unchanged if states and operators are transformed together when that convention changes.

5. Checking a canonical anticommutator

Verify the fermionic identity

$$\{c_2,c_2^\dagger\} \lvert1,0,1\rangle_F = \lvert1,0,1\rangle_F$$

by acting on the state explicitly.

Solution

The anticommutator is

$$\{c_2,c_2^\dagger\} = c_2c_2^\dagger+c_2^\dagger c_2.$$

For the first term, mode $2$ is initially empty. Since $S_2=n_1=1$,

$$c_2^\dagger\lvert1,0,1\rangle_F = -\lvert1,1,1\rangle_F.$$

Acting again with $c_2$ on $\lvert1,1,1\rangle_F$ also contributes a minus sign:

$$c_2 \bigl( -\lvert1,1,1\rangle_F \bigr) = -\bigl( -\lvert1,0,1\rangle_F \bigr) = \lvert1,0,1\rangle_F.$$

For the second term,

$$c_2\lvert1,0,1\rangle_F = 0$$

because mode $2$ is empty. Hence

$$\{c_2,c_2^\dagger\} \lvert1,0,1\rangle_F = \lvert1,0,1\rangle_F.$$

6. Energies of number states

Let

$$\widehat H_0 = \sum_i \epsilon_i N_i.$$

Find the energy of $\lvert2,0,1\rangle_B$ and of $\lvert1,0,1\rangle_F$.

Solution

The number operators are diagonal on occupation-number states. For the bosonic state,

$$\widehat H_0\lvert2,0,1\rangle_B = (2\epsilon_1+\epsilon_3) \lvert2,0,1\rangle_B.$$

For the fermionic state,

$$\widehat H_0\lvert1,0,1\rangle_F = (\epsilon_1+\epsilon_3) \lvert1,0,1\rangle_F.$$

The formula is the same for bosons and fermions; the allowed occupations differ.

7. Bosonic enhancement and fermionic blocking

Consider two modes with

$$\widehat H = \epsilon_1N_1+\epsilon_2N_2 +t\,d_1^\dagger d_2 +t^*d_2^\dagger d_1.$$

Compute $\widehat H\lvert1,0\rangle$ and compare the action on $\lvert1,1\rangle_B$ and $\lvert1,1\rangle_F$.

Solution

On the one-particle state $\lvert1,0\rangle$, only mode $1$ is occupied. The diagonal terms give

$$(\epsilon_1N_1+\epsilon_2N_2) \lvert1,0\rangle = \epsilon_1\lvert1,0\rangle.$$

The hopping term $d_1^\dagger d_2$ gives zero because mode $2$ is empty. The term $d_2^\dagger d_1$ moves the particle from mode $1$ to mode $2$:

$$d_2^\dagger d_1\lvert1,0\rangle = \lvert0,1\rangle.$$

Therefore

$$\widehat H\lvert1,0\rangle = \epsilon_1\lvert1,0\rangle +t^*\lvert0,1\rangle.$$

For the two-boson state,

$$(\epsilon_1N_1+\epsilon_2N_2)\lvert1,1\rangle_B = (\epsilon_1+\epsilon_2)\lvert1,1\rangle_B,$$

and

$$d_1^\dagger d_2\lvert1,1\rangle_B = \sqrt2\,\lvert2,0\rangle_B, \qquad d_2^\dagger d_1\lvert1,1\rangle_B = \sqrt2\,\lvert0,2\rangle_B.$$

Thus

$$\widehat H\lvert1,1\rangle_B = (\epsilon_1+\epsilon_2)\lvert1,1\rangle_B +\sqrt2\,t\lvert2,0\rangle_B +\sqrt2\,t^*\lvert0,2\rangle_B.$$

For two fermionic modes, both modes are already occupied in $\lvert1,1\rangle_F$. A hopping term would try to create a fermion in an already occupied target mode, so both hopping terms vanish:

$$\widehat H\lvert1,1\rangle_F = (\epsilon_1+\epsilon_2)\lvert1,1\rangle_F.$$

This is Pauli blocking in a two-mode calculation.

8. Number conservation of a hopping term

Show that the bilinear $d_i^\dagger d_j$ conserves total particle number by proving

$$[N_{\mathrm{tot}},d_i^\dagger d_j]=0, \qquad N_{\mathrm{tot}}=\sum_k N_k.$$

Solution

For either bosons or fermions, the total number operator satisfies

$$[N_{\mathrm{tot}},d_i^\dagger] = d_i^\dagger, \qquad [N_{\mathrm{tot}},d_j] = -d_j.$$

Use the product rule for commutators:

$$[A,BC] = [A,B]C+B[A,C].$$

Then

$$\begin{aligned} [N_{\mathrm{tot}},d_i^\dagger d_j] &= [N_{\mathrm{tot}},d_i^\dagger]d_j +d_i^\dagger[N_{\mathrm{tot}},d_j] \\ &= d_i^\dagger d_j -d_i^\dagger d_j \\ &= 0. \end{aligned}$$

The physical content is simple: $d_i^\dagger d_j$ creates one particle and annihilates one particle, so the net particle number is unchanged.

9. One-body matrices and pair interactions

A one-particle Hamiltonian in a two-mode basis has matrix elements

$$h_{11}=\epsilon_1, \qquad h_{22}=\epsilon_2, \qquad h_{12}=t, \qquad h_{21}=t^*.$$

Write its second-quantized one-body operator. Then add a density interaction assigning energy $U$ to each pair with one particle in each mode. Explain why this equals a flat energy penalty $U$ for joint occupancy only when both occupations are restricted to $0$ or $1$.

Solution

The one-body rule is

$$\widehat H_1 = \sum_{ij}h_{ij}d_i^\dagger d_j.$$

For the given two-mode matrix,

$$\widehat H_1 = \epsilon_1d_1^\dagger d_1 +\epsilon_2d_2^\dagger d_2 +t\,d_1^\dagger d_2 +t^*d_2^\dagger d_1.$$

Using $N_i=d_i^\dagger d_i$, this is

$$\widehat H_1 = \epsilon_1N_1+\epsilon_2N_2 +t\,d_1^\dagger d_2 +t^*d_2^\dagger d_1.$$

A density interaction with energy $U$ per cross-mode pair is

$$\widehat H_{\mathrm{int}} = U N_1N_2.$$

The full model is therefore

$$\widehat H = \epsilon_1N_1+\epsilon_2N_2 +t\,d_1^\dagger d_2 +t^*d_2^\dagger d_1 +U N_1N_2.$$

For fermions, $N_1N_2$ is a projector onto the state with both modes occupied. For bosons, it counts cross-mode pairs, so the same expression gives an interaction energy $U n_1n_2$ on $\lvert n_1,n_2\rangle_B$.

For longer calculations, Operator Identities collects number commutators and bilinear algebra; Two-Body Operators explains pair counting.

References

• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• P. A. M. Dirac, The Principles of Quantum Mechanics, 4th ed., Oxford University Press, 1958.
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, McGraw-Hill, 1971.
• J. W. Negele and H. Orland, Quantum Many-Particle Systems, Addison-Wesley, 1988.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.