# What is the expectation value of number operator squared $\hat{N}^2$? I am confused with normal ordering

## Definition of terms

\begin{align} \hat{N} & = a^\dagger a, \\ a^\dagger|n\rangle & = \sqrt{n+1}|n+1\rangle, \\ a|n\rangle & = \sqrt{n}|n-1\rangle, \end{align}

where $$|n\rangle$$ is the wavefunction of Fock state with $$n$$ photons.

$$a^\dagger a - a a^\dagger=1.$$

## The expectation value of $$\hat{N}$$

\begin{align} \langle n|\hat{N}|n\rangle & =\langle n|a^\dagger a|n\rangle \\ & =\langle n|a^\dagger \sqrt{n}|n-1\rangle \\ & =\langle n|\sqrt{n-1+1} \sqrt{n}|n-1+1\rangle \\ & =n\langle n|n\rangle \\ & =n \end{align}

## The expectation value of $$\hat{N}^2$$

Result 1 without normal ordering:

\begin{align} \langle n|\hat{N}^2|n\rangle & =\langle n|a^\dagger a a^\dagger a|n\rangle \\ & =\langle n|a^\dagger a a^\dagger \sqrt{n}|n-1\rangle \\ & =\langle n|a^\dagger a\sqrt{n} \sqrt{n}|n\rangle \\ & =n^2 \end{align}

Result 2a with normal ordering (for $$n\geq2$$): \begin{align} \langle n|\hat{N}^2|n\rangle & =\langle n|a^\dagger a a^\dagger a|n\rangle \\ & =\langle n|a^\dagger(1+a a^\dagger)a|n-1\rangle \\ & =\langle n|a^\dagger a +a^\dagger a^\dagger a a|n\rangle \\ & =n+\sqrt{n+1}\sqrt{n}\sqrt{n-1}\sqrt{n} \\ & =n+n\sqrt{n^2-1} \end{align}

Result 2b using $$\langle n|a^\dagger a^\dagger a a|n\rangle=|aa|n\rangle|^2=n(n-1)$$: \begin{align} \langle n|\hat{N}^2|n\rangle & =\langle n|a^\dagger a a^\dagger a|n\rangle \\ & =\langle n|a^\dagger(1+a a^\dagger)a|n-1\rangle \\ & =\langle n|a^\dagger a +a^\dagger a^\dagger a a|n\rangle \\ & =n+n(n-1) \\ & =n^2 \end{align}

## The question

Result 1 and 2b are the same I am assuming correct result. What am I doing wrong in 2a?

• check the $\sqrt{n}$ factors when you apply creation and annihilation operators
– fqq
Aug 26, 2020 at 15:51
• oh I see $a^\dagger |n-2>=\sqrt{n-1}|n-1>$ I went a bit too fast there thought it was acting on $|n-1>$ Aug 26, 2020 at 15:58
• Please use \langle and \rangle instead of < and > to typeset brakets. Aug 26, 2020 at 18:27

\begin{align} \langle n|a^\dagger a +a^\dagger a^\dagger a a|n\rangle & =n+\langle n|a^\dagger a^\dagger \sqrt{n-1}\sqrt{n}|n-2\rangle \\ & =\sqrt{n}\sqrt{n-1}\sqrt{n-1}\sqrt{n} \\ & =n+n(n-1) \end{align}