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.

One thought on “Number operator

  1. Pingback: Creation and annihilation operators: commutators and anticommutators | Physics pages

Leave a Reply

Your email address will not be published. Required fields are marked *