N-Representation of an Operator

Question

Calculating $\langle{n} \; {| \; \hat{X}^2 \; |} \; m\rangle$ in the N-representation, where $| \; m\rangle$ and $| \; n\rangle$ are harmonic oscillator states and $\hat{X} = \sqrt{\frac{\hbar}{2mw}}( \hat{a} + \hat{a}^\dagger )$, I find that :

$ \; \hat{X}^2 = \frac{\hbar}{2mw} [\hat{a}^2 + \hat{a}^{\dagger2} + 2\hat{a}^{\dagger}\hat{a} + 1] \quad $ (1)

Next, omitting the work, I find :

$\langle{n} \; {| \; \hat{X}^2 \; |} \; m\rangle = \frac{\hbar}{2mw} [ \sqrt{m(m-1)}\delta_{n,m-2} + \sqrt{(m+1)(m+2)}\delta_{n,m+2} + (2n+1)\delta_{n,m}] \quad $ (2)

The problem I'm facing is figuring out how to approach the N-representation of $\langle{m} \; {| \; \hat{X}^4 \; |} \; n\rangle$ and would appreciate some guidance.

If I attempt to first expand $\hat{X}^4$ I have problems trying to reduce it, which leads me to wonder if I could simply just multiply the RHS of (2) by itself?

UPDATE

Upon reviewing the suggestions below I realise that a combination of misunderstanding and intimidation led me astray ...

With interest in the coefficients, as well as the form, I find that :

\begin{align} \langle{n} \; {| \; \hat{X}^2 \; |} \; m\rangle & = \sqrt{n(n-1)(n-2)(n-3)}\delta_{m,n-4} \; + \\ & \quad [ 2\sqrt{n(n-1)} + 2n\sqrt{n(n-1)} + 2(n-2)\sqrt{n(n-1)} ]\delta_{m,n-2} \; + \\ & \quad [4n + 1 + \sqrt{(n+1)(n+2)(n+1)(n+2)} + \sqrt{n(n-1)(n-1)n} + 4n^2]\delta_{m,n} \; + \\ & \quad [ 2\sqrt{(n+1)(n+2)} + 2n\sqrt{(n+1)(n+2)} + 2(n+2)\sqrt{(n+1)(n+2)} ]\delta_{m,n+2} \; + \\ & \quad \sqrt{(n+1)(n+2)(n+3)(n+4)}\delta_{m,n+4} \\ \end{align}

Which is then further simplified.

Answer 1

Try this for (maybe) less painful algebra $$ \langle m|X^4|n\rangle=\sum_{l_1, l_2, l_3} \langle m|X|l_1\rangle \langle l_1|X|l_2\rangle \langle l_2|X|l_3\rangle \langle l_3|X|n\rangle $$ And use the fact that $$ \langle p|X|q\rangle = \sqrt{\dfrac{\hbar}{2m\omega}}[\sqrt{q}\delta_{p,q-1}+\sqrt{q+1}\delta_{p, q+1}] $$ You will get a lot of Kronecker deltas. The end result is of the form $$A\delta_{m,n-4}+ B\delta_{m,n-2}+C \delta_{m,n}+D \delta_{m,n+2}+E\delta_{m,n+4}$$ Actually if the form interests you and not the coefficients it is quite clear why the form is as above: Every time you have a pairing $a^\dagger a$ or $aa^\dagger$ in $X^4$ you can ignore it since it will give just a number which goes into the coefficients.

Answer 2

Let $\phi = ( \hat{a} + \hat{a}^\dagger )$ and $[\hat{a}, \hat{a}^\dagger]=1$ for which \begin{align} \phi^2 &= \hat{a}^2 + \hat{a} \hat{a}^\dagger + \hat{a}^\dagger \hat{a} + (\hat{a}^\dagger)^2 \\ &= \hat{a}^2 + 2 \, \hat{a} \hat{a}^\dagger + (\hat{a}^\dagger)^2 - 1 \end{align} Now, \begin{align} \phi^3 &= \hat{a} \, \phi^2 + \hat{a}^\dagger \, \phi^2 \\ &= \hat{a}^3 + 2 \, \hat{a}^2 \hat{a}^\dagger + \hat{a} \, (\hat{a}^\dagger)^2 + \hat{a}^\dagger (\hat{a})^2 + 2 \, \hat{a}^\dagger \hat{a} \hat{a}^\dagger + (\hat{a}^\dagger)^3 - \hat{a} - \hat{a}^\dagger \\ &= (\hat{a})^3 + 3 \, (\hat{a})^2 \hat{a}^\dagger + 3 \, \hat{a} (\hat{a}^\dagger)^2 + (\hat{a})^3 - 3 \, (\hat{a} + \hat{a}^\dagger) \end{align} Following the pattern again one finds the general form $$\phi^{2m} = \sum_{s=0}^{m} (-1)^s b_{s}^{2m} \, \left(\sum_{k=0}^{2m-2s} \binom{2m-2s}{k} (\hat{a})^{2m-2s-k} (\hat{a}^\dagger)^{k} \right) $$ which is valid for the even case.

An example is $$\phi^4 = \left(\sum_{k=0}^{4} \binom{4}{k} (\hat{a})^{4-k} (\hat{a}^\dagger)^{k} \right) - 6 \, \left(\sum_{k=0}^{2} \binom{2}{k} (\hat{a})^{2-k} (\hat{a}^\dagger)^{k} \right) + 3$$