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.

Helpful background. Occupation-Number Basis supplies mode labels; Direct Sums versus Tensor Products explains the sector construction.

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 norm zero 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- and two-particle sectors

The one-particle sector is $\operatorname{Sym}^1\mathcal h=\mathcal h$. In an orthonormal two-mode basis $a,b$, its basis is $\lvert1,0\rangle,\lvert0,1\rangle$.

The two-particle sector has basis $\lvert2,0\rangle,\lvert1,1\rangle,\lvert0,2\rangle$. In particle-slot notation, the middle vector is $(\lvert a,b\rangle+\lvert b,a\rangle)/\sqrt2$, while $\lvert2,0\rangle$ is $\lvert a,a\rangle$. Repeated occupation is allowed, and no extra particle label is needed.

Fixed-number symmetric sectors

For general $N$, the vectors in $\operatorname{Sym}^N\mathcal h$ satisfy $U(\pi)\lvert\Psi\rangle=\lvert\Psi\rangle$ for every permutation. The symmetrizer constructs this subspace from the unrestricted tensor product.

For $M$ modes, its dimension is $\binom{M+N-1}{N}$. This counts nonnegative solutions of $\sum_i n_i=N$: with two modes the dimensions are $1,2,3,4,\ldots$ as $N=0,1,2,3,\ldots$. Even a single bosonic mode has an infinite-dimensional full Fock space because $N$ is unbounded.

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

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

The Hilbert direct sum contains exactly the square-summable sequences, with normalization $\sum_N\|\psi_N\|^2=1$ for a pure state. Finite norm does not guarantee finite mean number: $\langle N\rangle=\sum_N N\|\psi_N\|^2$ needs a separate convergence condition. 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 fixed $N$, these vectors form an orthonormal basis of $\operatorname{Sym}^N\mathcal h$; their closed linear span is the whole sector. Allowing all finite $N$ gives an orthonormal basis of the full Fock space. Even when there are only finitely many modes, a general Fock vector may be an infinite square-summable superposition of these basis vectors. No basis vector with an actual infinite occupation is added by this completion.

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.

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. 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 $\operatorname{Sym}^2\mathbb C^M$.

3. Finite norm, divergent mean. In a single mode let $\lvert\Psi\rangle=C\sum_{n\ge0}(n+1)^{-1}\lvert n\rangle$. Does it belong to Fock space? Does it have finite mean number?

Solution

Yes: $\sum_{n\ge0}(n+1)^{-2}=\pi^2/6$, so $C=\sqrt6/\pi$ normalizes it. But $\langle N\rangle=C^2\sum_{n\ge0}n/(n+1)^2$ diverges as a harmonic series. A normalized state need not lie in the domain needed for every unbounded observable.

References

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