Bosonic Fock Space

Bosonic Fock space is the Hilbert space that collects all possible particle-number sectors for identical bosons. If $\mathcal h$ is the one-particle Hilbert space, the bosonic Fock space is

$$\mathcal F_B(\mathcal h) = \bigoplus_{N=0}^{\infty} \operatorname{Sym}^N\mathcal h.$$

Here $\operatorname{Sym}^N\mathcal h$ is the symmetric $N$-particle subspace of $\mathcal h^{\otimes N}$. The direct sum lets a state have a definite particle number, or a superposition of different particle-number sectors when the physics allows it.

Vacuum Sector

The $N=0$ sector is the vacuum sector:

$$\operatorname{Sym}^0\mathcal h \cong \mathbb C.$$

A normalized basis vector for this one-dimensional sector is written

$$\lvert 0\rangle.$$

The vacuum is not the zero vector. It is a physical no-particle state in Fock space. The zero vector has no norm and represents no state; the vacuum is normalized and can be acted on by creation operators in later notation.

The vacuum also does not necessarily have zero energy. A Hamiltonian may assign it zero energy by convention, or it may include zero-point energies depending on the model and normal-ordering convention.

One-Particle Sector

The $N=1$ sector is just the one-particle Hilbert space:

$$\operatorname{Sym}^1\mathcal h = \mathcal h.$$

If $\{\lvert\varphi_i\rangle\}$ is an orthonormal mode basis of $\mathcal h$, then the one-boson occupation states are

$$\lvert 1_i\rangle_B,$$

meaning one boson in mode $i$ and no bosons in the other modes.

For one particle, there is no distinction between symmetric and antisymmetric exchange behavior because there is no second particle to exchange. The distinction begins in the two-particle sector.

Two-Particle Symmetric Sector

The two-boson sector is the symmetric subspace

$$\operatorname{Sym}^2\mathcal h \subset \mathcal h\otimes\mathcal h.$$

For two distinct orthonormal modes $\lvert a\rangle$ and $\lvert b\rangle$, the state with one boson in each mode is

$$\lvert 1_a,1_b\rangle_B \longleftrightarrow \frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert b\rangle_2 + \lvert b\rangle_1\lvert a\rangle_2 \bigr).$$

For two bosons in the same mode,

$$\lvert 2_a\rangle_B \longleftrightarrow \lvert a\rangle_1\lvert a\rangle_2.$$

Bosons allow repeated occupation of the same mode. This is the structural reason that photon number states, phonon occupation states, Bose-Einstein condensates, and oscillator excitations use nonnegative integer occupation numbers.

The two-particle symmetrizer is

$$\Pi_S = \frac12(I+P_{12}).$$

It projects a two-slot vector onto the symmetric subspace.

N-Particle Sector

For $N$ identical bosons, the fixed-particle-number Hilbert space is

$$\operatorname{Sym}^N\mathcal h.$$

Equivalently, these are the vectors in $\mathcal h^{\otimes N}$ that satisfy

$$U(\pi)\lvert\Psi\rangle = \lvert\Psi\rangle \qquad \text{for every }\pi\in S_N.$$

The full symmetrizer is

$$\Pi_S^{(N)} = \frac{1}{N!} \sum_{\pi\in S_N}U(\pi).$$

The explicit slot form becomes large quickly because the sum contains $N!$ permutations. Fock-space occupation notation is designed to hide that bookkeeping while preserving the same physics.

Direct Sum Over Particle Number

A vector in bosonic Fock space is a sequence of fixed-number components:

$$\lvert\Psi\rangle = \psi_0\oplus\psi_1\oplus\psi_2\oplus\cdots, \qquad \psi_N\in\operatorname{Sym}^N\mathcal h.$$

The norm is

$$\lVert\Psi\rVert^2 = \sum_{N=0}^{\infty} \lVert\psi_N\rVert^2.$$

Physical Fock-space vectors have finite norm. A state of definite particle number has only one nonzero component. A superposition of different particle numbers has more than one nonzero component.

The direct sum is different from a tensor product over $N$. It is not a space with every particle-number sector simultaneously occupied as separate subsystems. It is a Hilbert space whose alternatives are different total particle numbers.

Occupation-Number Basis

Choose an orthonormal mode basis $\{\lvert\varphi_i\rangle\}$ for $\mathcal h$. A bosonic occupation-number basis vector is

$$\lvert n_1,n_2,n_3,\ldots\rangle_B,$$

with

$$n_i=0,1,2,\ldots, \qquad N=\sum_i n_i.$$

For a fixed total $N$, these vectors span $\operatorname{Sym}^N\mathcal h$. Allowing all finite $N$ gives a basis for the finite-particle subspace of $\mathcal F_B(\mathcal h)$, with Hilbert-space completion for infinite-dimensional cases.

Once creation operators are introduced, the normalized occupation vector is written

$$\lvert n_1,n_2,\ldots\rangle_B = \prod_i \frac{(a_i^\dagger)^{n_i}}{\sqrt{n_i!}} \lvert0\rangle,$$

where only finitely many $n_i$ are nonzero in the finite-particle sector. The factorials are the normalization factors associated with repeated bosonic occupation of the same mode.

The definite-occupation basis vectors themselves are treated in Number States.

Examples

For two bosonic modes $a$ and $b$, the $N=0$ sector has

$$\lvert0_a,0_b\rangle_B.$$

The $N=1$ sector has

$$\lvert1_a,0_b\rangle_B, \qquad \lvert0_a,1_b\rangle_B.$$

The $N=2$ sector has

$$\lvert2_a,0_b\rangle_B, \qquad \lvert1_a,1_b\rangle_B, \qquad \lvert0_a,2_b\rangle_B.$$

This three-dimensional $N=2$ sector is already easier to read in occupation notation than in explicit symmetrized slot notation.

In quantum optics, $\lvert n\rangle$ often denotes $n$ photons in a single mode. In lattice boson models, $\lvert n_1,n_2,\ldots,n_L\rangle$ records how many bosons occupy each lattice site. In phonon language, occupation numbers count excitations of normal modes rather than atoms as individual particles.

Common Mistakes

• Confusing the vacuum vector with the zero vector.
• Treating the direct sum over particle number as a tensor product over particle numbers.
• Forgetting that $\operatorname{Sym}^N\mathcal h$ is the fixed-$N$ bosonic sector, not the full Fock space.
• Thinking a bosonic mode can only hold one particle.
• Forgetting that occupation numbers depend on the chosen mode basis.
• Using bosonic factorial normalization for fermionic states.

Cross-Links

• Occupation-Number Basis
• Direct Sums versus Tensor Products
• Bosons
• Permanents
• Vacuum State
• Number States
• Mode Occupations
• Particle-Number Superselection Preview
• Fock Space Examples
• Fermionic Fock Space
• Creation and Annihilation Operators
• Bosonic Commutation Relations
• Fock Space Exercises
• Symmetrization Postulate
• Indistinguishability
• Why Composite Systems Matter
• Reference Bridge: Second Quantization
• Harmonic Oscillator to Fields

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.
• L. E. Ballentine, Quantum Mechanics: A Modern Development, 2nd ed., World Scientific, 2014.
• A. Altland and B. Simons, Condensed Matter Field Theory, 2nd ed., Cambridge University Press, 2010.

Exercises

1. For two bosonic modes $a$ and $b$, list the occupation basis states in the $N=3$ sector.

Solution

The nonnegative occupations must satisfy $n_a+n_b=3$. The basis states are

$$\lvert3_a,0_b\rangle_B, \qquad \lvert2_a,1_b\rangle_B, \qquad \lvert1_a,2_b\rangle_B, \qquad \lvert0_a,3_b\rangle_B.$$

2. Why is the vacuum vector not the same as the zero vector?

Solution

The vacuum is a normalized state in the $N=0$ sector. It represents no particles. The zero vector has norm zero and is not a physical state. Operators can act nontrivially on the vacuum, while the zero vector remains zero under every linear operator.

3. Write the slot-language state corresponding to $\lvert1_a,1_b\rangle_B$ for orthonormal modes $a$ and $b$.

Solution

The two-boson state is symmetric:

$$\frac{1}{\sqrt2} \bigl( \lvert a\rangle_1\lvert b\rangle_2 + \lvert b\rangle_1\lvert a\rangle_2 \bigr).$$

4. If there are $M$ bosonic modes and total particle number $N=2$, how many occupation basis states are there?

Solution

One can put both bosons in the same mode, giving $M$ states, or put them in two distinct modes, giving $M(M-1)/2$ states. The total is

$$M+\frac{M(M-1)}{2} = \frac{M(M+1)}{2}.$$

This is also the dimension of $\operatorname{Sym}^2\mathbb C^M$.