Fock Space Examples

Fock-space notation becomes useful when it is used in calculations. This page collects worked examples in which the same idea appears in different clothes: occupation strings, creation operators, photon modes, oscillator quanta, and lattice-site occupations.

The examples assume normalized number states, a fixed mode order for fermions, and the operator conventions introduced on the Creation and Annihilation Operators page.

Two Bosons in Two Modes

Let $a_L^\dagger,a_R^\dagger$ create bosons in two orthonormal modes $L$ and $R$. The fixed-$N=2$ bosonic basis is

$$\lvert2,0\rangle, \qquad \lvert1,1\rangle, \qquad \lvert0,2\rangle.$$

These states are normalized as

$$\lvert2,0\rangle = \frac{(a_L^\dagger)^2}{\sqrt2}\lvert0\rangle, \qquad \lvert1,1\rangle = a_L^\dagger a_R^\dagger\lvert0\rangle,$$

and

$$\lvert0,2\rangle = \frac{(a_R^\dagger)^2}{\sqrt2}\lvert0\rangle.$$

The factor $\sqrt2$ in the repeated-occupation states is not optional. It is the bosonic factorial normalization.

Consider the tunneling operator

$$T = -J \left( a_L^\dagger a_R + a_R^\dagger a_L \right).$$

It preserves total particle number but changes mode occupation. Acting on the basis states gives

$$\begin{aligned} T\lvert2,0\rangle &= -J\sqrt2\,\lvert1,1\rangle,\\ T\lvert1,1\rangle &= -J\sqrt2 \left( \lvert2,0\rangle+\lvert0,2\rangle \right),\\ T\lvert0,2\rangle &= -J\sqrt2\,\lvert1,1\rangle. \end{aligned}$$

The square-root factors are occupation factors. For example,

$$a_R^\dagger a_L\lvert2,0\rangle = \sqrt2\,\lvert1,1\rangle,$$

because two bosons are available to remove from mode $L$, and the empty mode $R$ receives one boson.

If an on-site interaction is added,

$$U_{\mathrm{int}} = \frac U2 \left[ N_L(N_L-1)+N_R(N_R-1) \right],$$

then

$$U_{\mathrm{int}}\lvert2,0\rangle = U\lvert2,0\rangle, \qquad U_{\mathrm{int}}\lvert1,1\rangle = 0,$$

and

$$U_{\mathrm{int}}\lvert0,2\rangle = U\lvert0,2\rangle.$$

This example shows the basic many-body pattern: hopping mixes occupation configurations while interactions often assign energies according to occupation.

Two Fermions in Four Spin-Orbitals

For fermions, a mode is usually a complete spin-orbital, not merely a spatial orbital. Take four modes in the fixed order

$$1=L\uparrow, \qquad 2=L\downarrow, \qquad 3=R\uparrow, \qquad 4=R\downarrow.$$

A two-fermion occupation state such as

$$\lvert1,0,0,1\rangle_F$$

means modes $L\uparrow$ and $R\downarrow$ are occupied. In operator notation,

$$\lvert1,0,0,1\rangle_F = c_1^\dagger c_4^\dagger\lvert0\rangle.$$

The state

$$\lvert1,1,0,0\rangle_F = c_1^\dagger c_2^\dagger\lvert0\rangle$$

has two electrons on the left spatial site, but it does not violate Pauli exclusion because the occupied spin-orbitals differ: $L\uparrow$ and $L\downarrow$ are distinct modes.

By contrast,

$$(c_1^\dagger)^2\lvert0\rangle=0.$$

One complete fermionic mode cannot be doubly occupied.

A spin-singlet state with one fermion in each spatial region is

$$\lvert S\rangle = \frac{ c_1^\dagger c_4^\dagger - c_2^\dagger c_3^\dagger }{\sqrt2} \lvert0\rangle.$$

In occupation strings this is

$$\lvert S\rangle = \frac{ \lvert1,0,0,1\rangle_F - \lvert0,1,1,0\rangle_F }{\sqrt2}.$$

The minus sign is the spin singlet sign. The fermionic anticommutation relations are already encoded in the ordered creation operators. One should not add a second antisymmetrization rule on top of the Fock-space expression.

Photons in Two Modes

Photons are bosons. In quantum optics, the modes might be spatial paths, polarizations, frequencies, or cavity modes. Let $a^\dagger$ and $b^\dagger$ create one photon in two input modes, and let $c^\dagger,d^\dagger$ create photons in two output modes of a balanced beam splitter.

Choose the phase convention

$$a^\dagger = \frac{c^\dagger+d^\dagger}{\sqrt2}, \qquad b^\dagger = \frac{c^\dagger-d^\dagger}{\sqrt2}.$$

The input state with one photon in each input mode is

$$\lvert1_a,1_b\rangle = a^\dagger b^\dagger\lvert0\rangle.$$

After substituting the beam-splitter transformation,

$$\begin{aligned} a^\dagger b^\dagger\lvert0\rangle &= \frac12 \left( c^\dagger+d^\dagger \right) \left( c^\dagger-d^\dagger \right) \lvert0\rangle\\ &= \frac12 \left[ (c^\dagger)^2-(d^\dagger)^2 \right] \lvert0\rangle. \end{aligned}$$

Using

$$(c^\dagger)^2\lvert0\rangle = \sqrt2\,\lvert2_c,0_d\rangle,$$

and the analogous expression for $d$, the output is

$$\lvert1_a,1_b\rangle \longmapsto \frac{ \lvert2_c,0_d\rangle - \lvert0_c,2_d\rangle }{\sqrt2}.$$

There is no $\lvert1_c,1_d\rangle$ term in this ideal convention. The two alternatives in which the photons leave separately destructively interfere. This is the occupation-number form of two-photon bunching.

Harmonic-Oscillator Mode Occupations

A single quantum harmonic oscillator has number states $\lvert n\rangle$ with Hamiltonian

$$H = \hbar\omega \left( a^\dagger a+\frac12 \right).$$

In the one-particle oscillator problem, $\lvert n\rangle$ is the $n$th excitation state of one particle in a quadratic potential. In a field-mode or phonon setting, the same algebra is used to describe $n$ quanta occupying one mode.

For two independent oscillator modes with frequencies $\omega_1$ and $\omega_2$,

$$H = \hbar\omega_1 \left( N_1+\frac12 \right) + \hbar\omega_2 \left( N_2+\frac12 \right).$$

The two-mode occupation state $\lvert n_1,n_2\rangle$ has energy

$$E_{n_1,n_2} = \hbar\omega_1 \left( n_1+\frac12 \right) + \hbar\omega_2 \left( n_2+\frac12 \right).$$

For example,

$$\lvert3,1\rangle$$

means three quanta in mode $1$ and one quantum in mode $2$. It does not mean four distinguishable particles have been given private labels.

In many field-theory and many-body conventions one normal-orders the Hamiltonian and drops the zero-point term:

$$:H: = \sum_i \hbar\omega_i N_i.$$

Normal ordering changes the reference energy, not the occupation-number algebra.

Simple Hubbard-Site Occupations

The two-site Hubbard model is a compact example where site, spin, fermionic signs, and interactions meet. Use the same mode order

$$1=L\uparrow, \qquad 2=L\downarrow, \qquad 3=R\uparrow, \qquad 4=R\downarrow.$$

The Hamiltonian is

$$H = -t \sum_{\sigma=\uparrow,\downarrow} \left( c_{L\sigma}^\dagger c_{R\sigma} + c_{R\sigma}^\dagger c_{L\sigma} \right) + U \left( N_{L\uparrow}N_{L\downarrow} + N_{R\uparrow}N_{R\downarrow} \right).$$

The hopping part $H_t$ preserves total particle number but moves fermions between sites. The interaction term assigns energy $U$ when both spin-orbitals on the same site are occupied.

For example,

$$\lvert1,0,0,1\rangle_F = c_{L\uparrow}^\dagger c_{R\downarrow}^\dagger\lvert0\rangle$$

has one fermion on each site, so the interaction term gives zero. Hopping can produce doubly occupied configurations:

$$H_t\lvert1,0,0,1\rangle_F = -t \left( \lvert1,1,0,0\rangle_F + \lvert0,0,1,1\rangle_F \right),$$

with the displayed sign following from the chosen mode order. The two resulting configurations have double occupancy on $L$ or $R$ and therefore acquire interaction energy $U$ from the Hubbard term.

This example is small, but it contains the main bookkeeping problems of fermionic many-body theory:

• the mode order fixes signs;
• site occupations are not particle labels;
• hopping mixes configurations;
• interaction energies depend on occupation patterns;
• total number can be conserved even when site occupations fluctuate.

What the Examples Have in Common

Across these examples, the same questions recur:

• What are the modes?
• What statistics do the quanta obey?
• What is the fixed mode order, if the particles are fermions?
• Which number operators are diagonal in the chosen basis?
• Which Hamiltonian terms preserve total number?
• Which terms mix occupation configurations?

Fock-space notation is not a new physical postulate. It is a compact way of doing the same many-particle quantum mechanics after the relevant modes, statistics, and operator conventions have been chosen.

Common Mistakes

• Forgetting the $\sqrt{n}$ and $\sqrt{n+1}$ factors for bosonic operators.
• Treating fermionic bitstrings as particle labels rather than mode occupations.
• Applying Pauli exclusion to spatial orbitals without including spin.
• Guessing fermionic signs from the final occupation string instead of applying operators in order.
• Confusing harmonic-oscillator excitation number with the number of particles in a one-particle potential.
• Treating a beam-splitter mode transformation as a classical probability split rather than an amplitude transformation.
• Assuming total particle-number conservation implies every mode occupation is conserved.

Cross-Links

• Occupation-Number Basis
• Bosonic Fock Space
• Fermionic Fock Space
• Vacuum State
• Number States
• Mode Occupations
• Particle-Number Superselection Preview
• Creation and Annihilation Operators
• Number Operators
• Mode Expansions
• Field Operators
• Bosonic Commutation Relations
• Fermionic Anticommutation Relations
• One-Body Operators
• Two-Body Operators
• Many-Particle Hamiltonians
• Ladder-Operator Solution
• Fock Space Exercises

References

• 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.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.
• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• A. Auerbach, Interacting Electrons and Quantum Magnetism, Springer, 1994.

Exercises

1. Bosonic tunneling factor. Verify that

$$a_R^\dagger a_L\lvert2,0\rangle = \sqrt2\,\lvert1,1\rangle.$$

Solution

Act from right to left:

$$a_L\lvert2,0\rangle = \sqrt2\,\lvert1,0\rangle,$$

and then

$$a_R^\dagger\lvert1,0\rangle = \lvert1,1\rangle.$$

Therefore

$$a_R^\dagger a_L\lvert2,0\rangle = \sqrt2\,\lvert1,1\rangle.$$

2. Fermionic double occupation. In the four-mode order $L\uparrow,L\downarrow,R\uparrow,R\downarrow$, explain why $\lvert1,1,0,0\rangle_F$ is allowed but $\lvert2,0,0,0\rangle_F$ is not.

Solution

$\lvert1,1,0,0\rangle_F$ occupies two distinct spin-orbitals, $L\uparrow$ and $L\downarrow$. Pauli exclusion forbids double occupation of a complete fermionic mode, not double occupation of a spatial site when spin is included. The string $\lvert2,0,0,0\rangle_F$ would require two fermions in the same complete mode $L\uparrow$, so it is not a fermionic basis state.

3. Beam splitter output. Using

$$a^\dagger = \frac{c^\dagger+d^\dagger}{\sqrt2}, \qquad b^\dagger = \frac{c^\dagger-d^\dagger}{\sqrt2},$$

derive the output of $\lvert1_a,1_b\rangle$.

Solution

Substitute into $a^\dagger b^\dagger\lvert0\rangle$:

$$\begin{aligned} a^\dagger b^\dagger\lvert0\rangle &= \frac12 \left[ (c^\dagger)^2-(d^\dagger)^2 \right] \lvert0\rangle\\ &= \frac{ \lvert2_c,0_d\rangle - \lvert0_c,2_d\rangle }{\sqrt2}. \end{aligned}$$

The mixed term cancels because the two indistinguishable alternatives for one photon in each output have opposite amplitudes.

4. Oscillator energy. For two independent oscillator modes, compute the energy of $\lvert3,1\rangle$ under

$$H = \hbar\omega_1 \left( N_1+\frac12 \right) + \hbar\omega_2 \left( N_2+\frac12 \right).$$

Solution

Since $N_1=3$ and $N_2=1$ on this state,

$$E = \hbar\omega_1 \left( 3+\frac12 \right) + \hbar\omega_2 \left( 1+\frac12 \right).$$

Thus

$$E = \frac72\hbar\omega_1 + \frac32\hbar\omega_2.$$

5. Hubbard interaction energy. Which two-particle basis states in the four-mode two-site Hubbard example have interaction energy $U$?

Solution

The interaction is

$$U \left( N_{L\uparrow}N_{L\downarrow} + N_{R\uparrow}N_{R\downarrow} \right).$$

It gives energy $U$ to states with both spin-orbitals occupied on one site:

$$\lvert1,1,0,0\rangle_F, \qquad \lvert0,0,1,1\rangle_F.$$

States with one fermion on each site have zero interaction energy from this term.