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.

Helpful background. Number States explains factorial normalization; Creation and Annihilation Operators supplies the operator action.

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

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

In the ordered basis $(\lvert2,0\rangle,\lvert1,1\rangle,\lvert0,2\rangle)$ the complete matrix is

$$H=\begin{pmatrix}U&-\sqrt2J&0\\-\sqrt2J&0&-\sqrt2J\\0&-\sqrt2J&U\end{pmatrix}.$$

The antisymmetric combination $(\lvert2,0\rangle-\lvert0,2\rangle)/\sqrt2$ has energy $U$; the remaining two energies are $(U\pm\sqrt{U^2+16J^2})/2$. These are parity combinations under exchanging the modes, not antisymmetric particle states: every vector here is bosonic.

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 for an ideal lossless balanced splitter when the photons match in their unobserved internal, spectral, and temporal states. The two alternatives in which the photons leave separately destructively interfere. This is the Hong–Ou–Mandel effect (Hong, Ou, and Mandel, 1987). For pure normalized internal wavepackets $f,g$, an ideal splitter with detectors insensitive to those labels gives coincidence probability $P_{\rm coin}=(1-|\langle f|g\rangle|^2)/2$. Orthogonal wavepackets recover $1/2$; identical wavepackets give zero.

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.

For these free modes one often chooses a vacuum-subtracted Hamiltonian:

$$H_{\rm sub}=H-\langle0|H|0\rangle I = \sum_i \hbar\omega_i N_i.$$

This changes the reference energy and preserves occupation algebra. It agrees with normal ordering the symmetric quadratic oscillator expression; Normal Ordering explains why reordering explicit operator monomials and subtracting vacuum expectations are distinct operations in general.

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.

Exercises

1. 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

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

2. 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.

3. **Opposite hopping signs.** In the stated four-mode order, compute $H_t\lvert0,1,1,0\rangle_F$ and compare it with the result for $\lvert1,0,0,1\rangle_F$. What happens to the spin triplet combination?

Solution

Acting with the ordered operators gives

$$H_t\lvert0,1,1,0\rangle_F =+t(\lvert1,1,0,0\rangle_F+\lvert0,0,1,1\rangle_F).$$

The amplitudes cancel for the plus combination of the two singly occupied states, which is the $S_z=0$ triplet. They add for their minus combination, the singlet. With $D_+=(\lvert1,1,0,0\rangle_F+\lvert0,0,1,1\rangle_F)/\sqrt2$, one finds $H_t\lvert S\rangle=-2t\lvert D_+\rangle$. This is a useful check on fermionic signs before solving the Hubbard dimer.

References

• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.
• A. Auerbach, Interacting Electrons and Quantum Magnetism, Springer, 1994.
• 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.
• C. K. Hong, Z. Y. Ou, and L. Mandel, “Measurement of Subpicosecond Time Intervals between Two Photons by Interference”, Physical Review Letters 59, 2044–2046 (1987).
• M. O. Scully and M. S. Zubairy, Quantum Optics, Cambridge University Press, 1997.
• R. Shankar, Principles of Quantum Mechanics, 2nd ed., Springer, 1994.