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.

Helpful background. Creation and Annihilation Operators supplies the action, and the bosonic and fermionic relations justify the identities.

Counting mode occupation

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

For a number operator with no explicit Schrödinger-picture time dependence, the Heisenberg equation gives, where the operator products and time derivative are defined,

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

For a Hamiltonian whose self-adjoint realization preserves the number sectors, the algebraic conservation test is

$$[H(t),N_{\mathrm{tot}}]=0 \quad\text{at every time}.$$

For unbounded operators, a formal commutator on an arbitrary small domain is insufficient. The precise conservation statement is that the well-defined propagator $U(t,s)$ commutes with every total-number spectral projector $P_N$: $U(t,s)P_N=P_NU(t,s)$. Thus all number probabilities are preserved, even for a state without a finite mean number. Conserved Quantities explains the domain and spectral conditions.

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.

Nonzero 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 $N_{\mathrm{tot}}$. The same-mode fermionic pair is zero and is not a number-changing term. 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.

Number fluctuations and phase rotations

For a fermionic mode, $N_i^2=N_i$ implies $\operatorname{Var}N_i=\langle N_i\rangle(1-\langle N_i\rangle)$. A fractional mean lies between zero and one and describes binary measurement statistics; it does not mean a fractional outcome. Bosonic variances are not fixed by the mean.

Total number generates a phase rotation. With $R(\theta)=e^{i\theta N_{\rm tot}}$,

$$R(\theta)d_iR(\theta)^\dagger=e^{-i\theta}d_i, \qquad R(\theta)d_i^\dagger R(\theta)^\dagger=e^{i\theta}d_i^\dagger.$$

A product with equal numbers of creation and annihilation operators is invariant. More generally, a monomial with $p$ creation and $q$ annihilation factors has $[N_{\rm tot},O]=(p-q)O$, on the common domain in the bosonic case. This is a quick way to recognize a global U(1) symmetry. Conservation of this generator is distinct from a superselection restriction.

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.

Exercises

1. 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 $(c_i^\dagger)^2=0$.

2. 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

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

3. A time-dependent hopping amplitude. Let $H(t)=-J(t)(d_1^\dagger d_2+d_2^\dagger d_1)$ with real, piecewise-continuous $J(t)$. Use the standard two-mode Fock realization. Is total number conserved? Is $N_1$ conserved?

Solution

Each bilinear preserves the finite-dimensional sectors of $N_1+N_2$. Writing $B=d_1^\dagger d_2+d_2^\dagger d_1$, the propagator is $U(t,s)=\exp[(i/\hbar)B\int_s^t J(t')\,dt']$, which commutes with every $P_N$. Total number is therefore conserved for this time-dependent drive. But $[H(t),N_1]=J(t)(d_1^\dagger d_2-d_2^\dagger d_1)$ is generally nonzero: particles can move between modes while remaining in the same total-number sector.

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.