Number Operators

A number operator measures the occupation of a mode. For a bosonic mode $i$,

$$N_i = a_i^\dagger a_i.$$

For a fermionic mode $i$,

$$N_i = c_i^\dagger c_i.$$

In both cases, number states are eigenstates:

$$N_i \lvert n_1,n_2,\ldots\rangle = n_i \lvert n_1,n_2,\ldots\rangle.$$

The same symbol $N_i$ is often used for bosons and fermions because the interpretation is the same: it counts how many particles or excitations occupy mode $i$. The allowed eigenvalues depend on the statistics.

Definition

Creation and annihilation operators raise and lower occupation. The number operator combines them so that the net effect is to return to the same occupation state with a numerical eigenvalue.

For bosons,

$$a_i \lvert\ldots,n_i,\ldots\rangle_B = \sqrt{n_i}\, \lvert\ldots,n_i-1,\ldots\rangle_B,$$

and

$$a_i^\dagger \lvert\ldots,n_i-1,\ldots\rangle_B = \sqrt{n_i}\, \lvert\ldots,n_i,\ldots\rangle_B.$$

Therefore

$$a_i^\dagger a_i \lvert\ldots,n_i,\ldots\rangle_B = n_i \lvert\ldots,n_i,\ldots\rangle_B.$$

The fermionic formula has the same eigenvalue statement, with $n_i=0$ or $1$.

Bosonic Eigenvalues

For a bosonic mode,

$$n_i=0,1,2,\ldots .$$

The spectrum of $N_i=a_i^\dagger a_i$ is the set of nonnegative integers. On one mode,

$$N\lvert n\rangle_B = n\lvert n\rangle_B.$$

For many modes,

$$N_i \lvert n_1,\ldots,n_i,\ldots\rangle_B = n_i \lvert n_1,\ldots,n_i,\ldots\rangle_B.$$

Since bosons can repeatedly occupy the same mode, $N_i$ is not a projector. For example,

$$N_i^2 \lvert\ldots,n_i,\ldots\rangle_B = n_i^2 \lvert\ldots,n_i,\ldots\rangle_B,$$

which is generally not the same as $N_i$ unless $n_i=0$ or $1$.

Fermionic Eigenvalues

For a fermionic mode,

$$n_i\in\{0,1\}.$$

Thus

$$N_i\lvert\ldots,0_i,\ldots\rangle_F = 0,$$

and

$$N_i\lvert\ldots,1_i,\ldots\rangle_F = \lvert\ldots,1_i,\ldots\rangle_F.$$

The fermionic number operator is a projector:

$$N_i^2=N_i.$$

Using the anticommutation relation $\{c_i,c_i^\dagger\}=I$,

$$\begin{aligned} N_i^2 &= c_i^\dagger c_i c_i^\dagger c_i\\ &= c_i^\dagger(I-c_i^\dagger c_i)c_i\\ &= c_i^\dagger c_i - c_i^\dagger c_i^\dagger c_i c_i\\ &= N_i. \end{aligned}$$

The last term vanishes because $(c_i^\dagger)^2=0$ and $c_i^2=0$.

Total Number Operator

The total number operator is the sum over modes:

$$N_{\mathrm{tot}} = \sum_i N_i.$$

On a number state,

$$N_{\mathrm{tot}} \lvert n_1,n_2,\ldots\rangle = \left(\sum_i n_i\right) \lvert n_1,n_2,\ldots\rangle.$$

For fixed-particle-number sectors, this eigenvalue is constant. For a general Fock-space state

$$\lvert\Psi\rangle = \sum_{\mathbf n} C_{\mathbf n}\lvert\mathbf n\rangle,$$

the state has definite total number only if all basis states with nonzero $C_{\mathbf n}$ have the same value of $\sum_i n_i$.

Commutators with Creation and Annihilation

For bosonic modes,

$$[N_i,a_j^\dagger] = \delta_{ij}a_j^\dagger, \qquad [N_i,a_j] = -\delta_{ij}a_j.$$

For fermionic modes, the same commutators with the number operator hold:

$$[N_i,c_j^\dagger] = \delta_{ij}c_j^\dagger, \qquad [N_i,c_j] = -\delta_{ij}c_j.$$

The algebra used to prove the formulas differs for bosons and fermions, but the meaning is identical:

• creating in mode $j$ raises $N_j$ by one;
• annihilating in mode $j$ lowers $N_j$ by one;
• occupations of other modes are unchanged by that single-mode operation.

For the total number operator,

$$[N_{\mathrm{tot}},a_j^\dagger] = a_j^\dagger, \qquad [N_{\mathrm{tot}},a_j] = -a_j,$$

and similarly for fermionic $c_j^\dagger,c_j$.

Number-Conserving Bilinears

Let $d_i$ denote either bosonic or fermionic annihilation operators when only number-counting identities are being discussed. A bilinear operator

$$d_i^\dagger d_j$$

moves one quantum from mode $j$ to mode $i$. It changes individual mode occupations but preserves total number:

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

For a particular mode $k$,

$$[N_k,d_i^\dagger d_j] = (\delta_{ki}-\delta_{kj})d_i^\dagger d_j.$$

This says exactly what the bilinear does: it raises the occupation of mode $i$ and lowers the occupation of mode $j$.

The one-body Hamiltonian

$$H_0 = \sum_i \epsilon_i N_i$$

is diagonal in the occupation basis. More general number-conserving one-body operators have the form

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

They can mix modes while preserving total particle number.

Particle-Number Conservation

If a Hamiltonian has no explicit time dependence, the Heisenberg equation gives

$$\frac{d}{dt}N_{\mathrm{tot}}(t) = \frac{i}{\hbar} [H,N_{\mathrm{tot}}](t).$$

Thus total particle number is conserved when

$$[H,N_{\mathrm{tot}}]=0.$$

Number-conserving many-particle Hamiltonians do not connect sectors with different total occupation. If a state begins in the $N=3$ sector, it stays in that sector.

Terms such as

$$d_i^\dagger d_j^\dagger \quad \text{or} \quad d_i d_j$$

change total number by two and do not commute with $N_{\mathrm{tot}}$. They appear in effective descriptions such as pairing Hamiltonians or driven bosonic systems, but then particle number is not conserved in the simple mode-counting sense.

Mode Basis Dependence

The operator $N_i$ counts occupation of mode $i$. If the one-particle mode basis is changed, the meaning of $N_i$ changes too. A state with definite occupation in one basis can be a superposition of number states in another basis.

The total number operator is basis-independent under unitary changes of the one-particle mode basis, but individual mode occupations are not. This is why one must state the modes before assigning physical meaning to a number operator.

In a position-space basis, the same counting idea becomes the local density $\psi^\dagger(\mathbf x)\psi(\mathbf x)$ introduced in Field Operators; the basis expansion connecting $N_i$ and field notation is treated in Mode Expansions.

Common Mistakes

• Treating $N_i$ as a particle label rather than a mode-occupation operator.
• Forgetting that bosonic $N_i$ can have eigenvalues $2,3,\ldots$.
• Forgetting that fermionic $N_i$ is a projector with eigenvalues $0$ and $1$.
• Assuming mode occupation is conserved just because total number is conserved.
• Writing a Hamiltonian with pair-creation terms and still claiming particle number is conserved.
• Forgetting that individual number operators depend on the chosen mode basis.

Cross-Links

• Number States
• Creation and Annihilation Operators
• Bosonic Commutation Relations
• Fermionic Anticommutation Relations
• Occupation-Number Basis
• Bosonic Fock Space
• Fermionic Fock Space
• Particle-Number Superselection Preview
• Fock Space Examples
• Mode Expansions
• Field Operators
• One-Body Operators
• Many-Particle Hamiltonians
• Normal Ordering
• Second Quantization: Bridge to QFT
• Reference Bridge: Second Quantization
• Formula Sheet
• 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.
• 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. Compute $N_{\mathrm{tot}}\lvert2,0,1\rangle_B$.

Solution

The total occupation is $2+0+1=3$, so

$$N_{\mathrm{tot}}\lvert2,0,1\rangle_B = 3\lvert2,0,1\rangle_B.$$

2. Use the bosonic action on number states to show that $N\lvert n\rangle_B=n\lvert n\rangle_B$.

Solution

For one bosonic mode,

$$a\lvert n\rangle_B = \sqrt n\,\lvert n-1\rangle_B.$$

Then

$$a^\dagger a\lvert n\rangle_B = a^\dagger\sqrt n\,\lvert n-1\rangle_B = \sqrt n\sqrt n\,\lvert n\rangle_B = n\lvert n\rangle_B.$$

3. Prove that a fermionic number operator is a projector.

Solution

Let $N_i=c_i^\dagger c_i$. Then

$$\begin{aligned} N_i^2 &= c_i^\dagger c_i c_i^\dagger c_i\\ &= c_i^\dagger(I-c_i^\dagger c_i)c_i\\ &= c_i^\dagger c_i - c_i^\dagger c_i^\dagger c_i c_i\\ &= N_i. \end{aligned}$$

The last term vanishes because $(c_i^\dagger)^2=0$.

4. Show that $d_i^\dagger d_j$ preserves total number.

Solution

Use

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

Then

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

5. Does the term $d_i^\dagger d_j^\dagger$ conserve total number?

Solution

No. Using the same commutator rule,

$$\begin{aligned} [N_{\mathrm{tot}},d_i^\dagger d_j^\dagger] &= [N_{\mathrm{tot}},d_i^\dagger]d_j^\dagger + d_i^\dagger[N_{\mathrm{tot}},d_j^\dagger]\\ &= d_i^\dagger d_j^\dagger + d_i^\dagger d_j^\dagger\\ &= 2d_i^\dagger d_j^\dagger. \end{aligned}$$

The operator creates two quanta, so it changes total number by two.