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I wish to estimate the expectation value of the number operator : $$\langle \hat{N} \rangle =\langle \hat{a}^{\dagger} \hat{a} \rangle$$

I know the Hamiltonian operator of my system, and thus want the expectation value of this operator when taken between eigenstates of the Hamiltonian.

The Hamiltonian reads : $$\hat{H}=E_0 \hat{I_d} + i E_1 (\hat{a}^{\dagger}+\hat{a}) + E_2 (\hat{a}^{\dagger}+\hat{a})^2 $$ Or equivalently : $$\hat{H}=E_0 \hat{I_d} + i E_1 (\hat{a}^{\dagger}+\hat{a}) + E_2 ({\hat{a}^{\dagger}}^2+\hat{a}^{\dagger}\hat{a}+\hat{a}\hat{a}^{\dagger}+{\hat{a}}^2)$$

Where $\hat{I_d}$ is the identity operator, $\hat{a}^{\dagger}$ and $\hat{a}$ are respectively creation and anihilation operators, and where $E_0$,$E_1$ and $E_2$ are real numbers.

How would I go about finding the eigenstates $\vert \Psi_i \rangle$ of $\hat{H}$ ?

In order to then compute the expectation value of $\hat{N}$ when the system is in an eigenstate of $\hat{H}$: $$\frac{\langle \Psi_i \vert \hat{N} \vert \Psi_i \rangle}{\langle \Psi_i \vert\Psi_i \rangle}$$

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    $\begingroup$ Your question title compared to the body of the question is very confusing. You want to get the eigenstates of $H$ - as far as I can see, this is the question. What you do with the eigenstate then is up to you... $\endgroup$
    – Tobias Fünke
    Mar 14 at 17:36
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    $\begingroup$ Your Hamiltonian is not Hermitian (unless $E_1$ is purely imaginary). Are you sure about that factor of $i$? -- or perhaps you mean $\hat{a}-\hat{a}^{\dagger}$ in that term. In any case, to diagonalize $\hat{H}$, you'll need a Bogoliubov transformation. $\endgroup$
    – march
    Mar 14 at 17:46
  • $\begingroup$ Does this answer your question? Ladder operator identity for $\langle n | (a+a^\dagger)^k | m \rangle$ $\endgroup$
    – ZeroTheHero
    Mar 14 at 19:03

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The Hamiltonian reads : $$\hat{H}=E_0 \hat{I_d} + i E_1 (\hat{a}^{\dagger}+\hat{a}) + E_2 (\hat{a}^{\dagger}+\hat{a})^2 $$ ...

How would I go about finding the eigenstates $\vert \Psi_i \rangle$ of $\hat{H}$ ?

In order to then compute the expectation value of $\hat{N}$ when the system is in an eigenstate of $\hat{H}$: $$\frac{\langle \Psi_i \vert \hat{N} \vert \Psi_i \rangle}{\langle \Psi_i \vert\Psi_i \rangle}$$

First, it may be helpful to recognize that your Hamiltonian can be written in terms of a position operator $\hat X$ as: $$ \hat H = \alpha + \beta\hat X + \gamma \hat X^2\;, $$ since $$ \hat X \propto \hat a + \hat a^\dagger\;. $$

Therefore, your Hamiltonian is like a shifted simple harmonic oscillator, but with no kinetic term.

You can shift the creation operators like: $$ \hat b = \hat a + \delta\;, $$ where $\delta$ is a number (not an operator), and where the shift can be chosen to put the Hamiltonian into the form: $$ \hat H = \epsilon + \zeta (\hat b^\dagger + b)^2 $$ $$ =\epsilon + \eta {\hat Y}^2\;, $$ where $$ \hat Y \propto (\hat b + \hat b^\dagger) $$ where $\epsilon$ and $\zeta$ are numbers.

Therefore, eigenstates of the Hamiltonian are eigenstates of the $\hat Y$ operator, which can be written in the form: $$ \left|y\right>\propto e^{\sqrt{2}~yb^\dagger-(b^\dagger)^2/2}\left|0\right>\;. $$ and where: $$ \hat H |y\rangle = |y\rangle(\epsilon + \eta y^2)\;. $$

The expectation value of the number operator is thus: $$ \langle N\rangle =\langle a^\dagger a\rangle = \frac{\langle 0 |e^{\sqrt{2}~yb-(b)^2/2} a^\dagger a e^{\sqrt{2}~yb^\dagger-(b^\dagger)^2/2}\left|0\right>}{\langle 0 |e^{\sqrt{2}~yb-(b)^2/2}e^{\sqrt{2}~yb^\dagger-(b^\dagger)^2/2}\left|0\right>} $$

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