How to proceed with expectation values of $x^n$ operators in the quantum harmonic oscillator? Harmonic oscillator is in the state $\Phi=\frac{1}{\sqrt{2}}(\vert 0\rangle+\vert 1\rangle)$. How can I calculate all the non vanishing values of $\langle \Phi \vert x^n \vert \Phi \rangle$?
I tried to express $x^n$ as ladder operators and distinguish the even and odd situations. Then I try to express the terms as a function of previous known terms $(n-1$ terms $)$. But it didn't help.
Any ideas?
 A: There is a powerful theorem for this situations. And is very simple. The Wick theorem.  The idea is do all kinds of commutations to find only normal orderings plus c-numbers. This c-numbers are called contractions.
A hint: try to work with vacuum expectation value only, expressing your state in terms of creation and annihilation acting on the vacuum state:
$$
|\psi\rangle=\frac{1}{\sqrt{2}}(1+a^{\dagger})|0\rangle
$$
A: I recently come across a similar question, which involves in calculating the vacuum expectation value as well. For simplicity, we denote
$$\langle x^n \rangle =\langle{0}\vert x^n\vert0\rangle.$$
Your question can be converted to calculating the vacuum expectation as well.
Because
$$\vert{1}\rangle=a^\dagger \vert{0}\rangle =(a+a^\dagger)\vert0\rangle=\sqrt{\frac{2\hbar}{m\omega}}x\vert0\rangle,$$
we have
$$\langle{\Phi}\vert x^n\vert\Phi\rangle=\frac{1}{2}\left(\langle{x^n}\rangle+2\sqrt{\frac{2\hbar}{m\omega}} \langle{x^{n+1}\rangle} +\frac{2\hbar}{m\omega}\langle{x^{n+2}\rangle}  \right).$$
Now how do you calculate $\langle x^n \rangle$, or equivalently $\langle(a+a^\dagger)^{n}\rangle$? The result is zero if $n$ is odd, which is easy to imagine. For even numbers,
$$\langle(a+a^\dagger)^{2k}\rangle =(2k-1)!!,$$
where $(2k-1)!!=1\times 3\times\cdots\times(2k-1)$.
I derive this by  expanding $(a+a^\dagger)^{2k}$. Obviously, the only term left are those with $k$ raising and $k$ lowering operators. For those terms, we can apply Wick's theorem to convert them to normal ordering. But all normal ordered sequence has vanishing expectation value. Therefore the only terms left are those full contracted. The number of possible nonzero contractions gives the expectation value of each term. Since the only nonzero contraction is of the form $\dot a \dot a^\dagger=1$, we need to count the number of ways to connect an earlier $a$ with a later $a^\dagger$.
For example, there are two possible contractions for $ aaa^\dagger a^\dagger$, so the expectation value is 2:
$$
\langle aaa^\dagger a^\dagger \rangle = \langle \dot a \ddot a 
\dot a^\dagger \ddot a^\dagger +  \dot a \ddot a 
\ddot a^\dagger \dot a^\dagger \rangle =2.$$
How about the total number of non-zeros contractions? This is equivalent to the problem where we put $k$ pairs of objects labeled by their order of appearance   in a queue and ask the total number of combinations. Let's just represent them as numbers. For example, when $k=3$, two such queues are
$$1,2,2,3,1,3; \quad \text{and}\quad 1,1,2,3,3,2.$$
We can count the total number of combinations by "filling the slot".
There are $2k$ slots in total, but the first $1$ is in the first slot by definition. For the second $1$ there are $(2k-1)$ possibilities. Afterwards the first $2$ is immediately filled to the first non-empty slot, which left $(2k-3) $ possible slots to write the second $2$. Going on like this, until finally there is only one way to fill the second $k$. We can see there are $(2k-1)!!$ different combinations.
