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I'm examining the Hamiltonian

$$\hat{H} = \frac{\hat{p}^2}{2 m} + \frac{1}{2} m \omega^2 \hat{x}^2 + \lambda \hat{x}^4 \text{,}$$

where $\lambda$ is small and $\hat{x} = \sqrt{\frac{\hbar}{2 m \omega}} (\hat{a} + \hat{a}^{\dagger})$ is the position operator in terms of the creation and annihilation operators. If this oscillator is perturbed by a small amount, I get

$$\lambda \hat{x}^4 = \frac{\lambda \hbar^2}{4 m^2 \omega^2} (\hat{a} + \hat{a}^{\dagger})^4$$

for the last term in the Hamiltonian. Now, a first-order shift in the ground state energy (or vacuum expectation value, VEV) is $\langle 0 \text{|} \lambda \hat{x}^4 \text{|} 0 \rangle​$, which means that I will need to expand $(\hat{a} + \hat{a}^{\dagger})^4​$, but keep only those terms which have equal numbers of $\hat{a}^{\dagger}​$ (up) and $\hat{a}​$ (down). My physical interpretation is that since the ground state is our reference, then only those terms will contribute to the VEV. Classically, I think of myself being on a stepladder, and having the choice of moving up two rungs and then down two rungs to get back to the bottom (but this analogy falls apart quickly as you will see).

There are a lot of relevant terms; for example, $\hat{a} \hat{a} \hat{a}^{\dagger} \hat{a}^{\dagger}$. I will need to go work evaluating what their contribution will be. It looks like the result $\hat{a} \hat{a}^{\dagger} - \hat{a}^{\dagger} \hat{a} = 1$ (from the commutator) will be useful. The text I'm looking at states that the normalized operators $\hat{a} \text{|} n \rangle = \sqrt{n} \text{|} n - 1 \rangle$ and $\hat{a}^{\dagger} \text{|} n \rangle = \sqrt{n + 1} \text{|} n + 1 \rangle$ will also be essential. And that's my problem. For example, $\langle n \text{|} \hat{a} \hat{a} \hat{a}^{\dagger} \hat{a}^{\dagger} \text{|} n \rangle = (n + 1)(n + 2)$ somehow. Shankar isn't much help. I don't understand how this and the other terms are evaluated.

I'm also struggling with the following. Two sample VEVs for differing terms give, for example, $\langle 0 \text{|} \hat{a} \hat{a} \hat{a}^{\dagger} \hat{a}^{\dagger} \text{|} 0 \rangle = 2$ (from above) and $\langle 0 \text{|} \hat{a} \hat{a}^{\dagger} \hat{a} \hat{a}^{\dagger} \text{|} 0 \rangle = 1$. I know these are operators, and order matters, but I get two different vacuum expectation values for, conceptually, the same number of steps up and down the ladder picture I have in my head. That's very counterintuitive.

Does anyone have any insight into (1) computing the expectation values of the terms in the perturbation above, and (2) how I can conceptually resolve two different VEVs for the same number of ups and downs? Thank you in advance.

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\begin{align} \langle n \vert \hat{a} \hat{a} \hat{a}^{\dagger} \hat{a}^{\dagger} \vert n \rangle &= \sqrt{n+1}\langle n \vert \hat{a} \hat{a} \hat{a}^{\dagger} \vert n+1 \rangle\, ,\\ &= \sqrt{n+1} \sqrt{n+2}\langle n \vert \hat{a} \hat{a} \vert n+2 \rangle\, \\ &=\sqrt{n+1} \sqrt{n+2}\sqrt{n+2}\langle n \vert \hat{a} \vert n+1 \rangle\, \\ &=\sqrt{n+1} \sqrt{n+2}\sqrt{n+2}\sqrt{n+1}\langle n \vert n \rangle\, \\ &= (n+1)(n+2) \end{align} should help.

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Compute $(a+a^\dagger)^2| 0\rangle$, then square.

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