# Number operator

References: Mark Srednicki, Quantum Field Theory, (Cambridge University Press, 2007) – Chapter 1, Problem 1.3.

The number operator is defined as

$\displaystyle N\equiv\int d^{3}x\;a^{\dagger}\left(\mathbf{x}\right)a\left(\mathbf{x}\right) \ \ \ \ \ (1)$

Applied to a quantum state, it counts the number of particles in that state:

$\displaystyle Na^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle =na^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (2)$

Another property of ${N}$ is that it commutes with any other operator that contains an equal number of creation and annihilation operators. To see this, look at the individual commutators as follows (where ${a_{i}\equiv a\left(\mathbf{x}_{i}\right)}$).

 $\displaystyle \left[N,a_{i}^{\dagger}\right]$ $\displaystyle =$ $\displaystyle \int d^{3}x\;\left(a^{\dagger}\left(\mathbf{x}\right)a\left(\mathbf{x}\right)a_{i}^{\dagger}-a_{i}^{\dagger}a^{\dagger}\left(\mathbf{x}\right)a\left(\mathbf{x}\right)\right)\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}x\;\left[a^{\dagger}\left(\mathbf{x}\right)\left(\delta\left(\mathbf{x}-\mathbf{x}_{i}\right)+a_{i}^{\dagger}a\left(\mathbf{x}\right)\right)-a_{i}^{\dagger}a^{\dagger}\left(\mathbf{x}\right)a\left(\mathbf{x}\right)\right]\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}x\;a^{\dagger}\left(\mathbf{x}\right)\delta\left(\mathbf{x}-\mathbf{x}_{i}\right)\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle a_{i}^{\dagger}\ \ \ \ \ (6)$ $\displaystyle \left[N,a_{i}\right]$ $\displaystyle =$ $\displaystyle \int d^{3}x\;\left(a^{\dagger}\left(\mathbf{x}\right)a\left(\mathbf{x}\right)a_{i}-a_{i}a^{\dagger}\left(\mathbf{x}\right)a\left(\mathbf{x}\right)\right)\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}x\;\left[a^{\dagger}\left(\mathbf{x}\right)a\left(\mathbf{x}\right)a_{i}-\left(\delta\left(\mathbf{x}-\mathbf{x}_{i}\right)+a^{\dagger}\left(\mathbf{x}\right)a_{i}\right)a\left(\mathbf{x}\right)\right]\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\int d^{3}x\;a\left(\mathbf{x}\right)\delta\left(\mathbf{x}-\mathbf{x}_{i}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -a_{i} \ \ \ \ \ (10)$

Here we’ve used the commutation relations

 $\displaystyle \left[a\left(\mathbf{x}\right),a\left(\mathbf{x}^{\prime}\right)\right]$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (11)$ $\displaystyle \left[a^{\dagger}\left(\mathbf{x}\right),a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right]$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (12)$ $\displaystyle \left[a\left(\mathbf{x}\right),a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right]$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (13)$

Now suppose we have an operator ${X}$ which contains ${n}$ creation operators ${a_{i}^{\dagger}}$, ${i=1,\ldots,n}$ and ${m}$ annihiliation operators ${a_{j}}$, ${j=1,\ldots,m}$:

$\displaystyle X=a_{i1}^{\dagger}\ldots a_{in}^{\dagger}a_{j1}\ldots a_{jm} \ \ \ \ \ (14)$

Then

 $\displaystyle \left[N,X\right]$ $\displaystyle =$ $\displaystyle Na_{i1}^{\dagger}\ldots a_{in}^{\dagger}a_{j1}\ldots a_{jm}-a_{i1}^{\dagger}\ldots a_{in}^{\dagger}a_{j1}\ldots a_{jm}N\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(a_{i1}^{\dagger}N+a_{i1}^{\dagger}\right)a_{i2}^{\dagger}\ldots a_{in}^{\dagger}a_{j1}\ldots a_{jm}-a_{i1}^{\dagger}\ldots a_{in}^{\dagger}a_{j1}\ldots a_{jm}N\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle X+a_{i1}^{\dagger}\left[N,a_{i2}^{\dagger}\ldots a_{in}^{\dagger}a_{j1}\ldots a_{jm}\right] \ \ \ \ \ (17)$

We can see that the commutator in the last line can be worked out recursively until we’ve processed all the creation operators up to ${a_{in}^{\dagger}}$, giving

$\displaystyle \left[N,X\right]=nX+a_{i1}^{\dagger}\ldots a_{in}^{\dagger}\left[N,a_{j1}\ldots a_{jm}\right] \ \ \ \ \ (18)$

The last commutator gives us

 $\displaystyle \left[N,a_{j1}\ldots a_{jm}\right]$ $\displaystyle =$ $\displaystyle Na_{j1}\ldots a_{jm}-a_{j1}\ldots a_{jm}N\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(a_{j1}N-a_{j1}\right)a_{j2}\ldots a_{jm}-a_{j1}\ldots a_{jm}N\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -a_{j1}\ldots a_{jm}+a_{j1}\left[N,a_{j2}\ldots a_{jm}\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -m\left(a_{j1}\ldots a_{jm}\right) \ \ \ \ \ (22)$

Therefore

 $\displaystyle a_{i1}^{\dagger}\ldots a_{in}^{\dagger}\left[N,a_{j1}\ldots a_{jm}\right]$ $\displaystyle =$ $\displaystyle -m\left(a_{i1}^{\dagger}\ldots a_{in}^{\dagger}a_{j1}\ldots a_{jm}\right)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -mX\ \ \ \ \ (24)$ $\displaystyle \left[N,X\right]$ $\displaystyle =$ $\displaystyle \left(n-m\right)X \ \ \ \ \ (25)$

So if ${n=m}$ (the numbers of creation and annihiliation operators are equal), the operator ${X}$ commutes with ${N}$. In particular, the hamiltonian we met last time satisfies this criterion, so ${\left[N,H\right]=0}$ and this hamiltonian conserves particle numbers.