# Finding an approximate expectation value $\langle E_0|\hat{O}|E_0\rangle$ when i dont know the ground state?

I'm assuming that i know the hamiltonian although i don't know it's ground state $|E_0\rangle$ and that i have a way to find $|\psi(s)\rangle\equiv e^{-\hat{H}s}|{\psi}\rangle$, $\forall s\in\mathbb{R}$ with units one over energy. I want to find an approximate expectation value $\langle E_0|\hat{O}|E_0\rangle$ for some hermitian operator $\hat{O}$.

I have, so far, taylor expanded the exponential operator and truncated the taylor expansion with an error that goes like $\mathcal{O}(s^2)$, that is \begin{align*} e^{-\hat{H}s} \approx \mathbb{I} - \hat{H}s + \mathcal{O}(s^2) \end{align*}

Also, if we expand $|\psi\rangle$ in terms of the energy eigenstates we can find try to find the expectation value of some hermitian operator $\hat{O}$, or try to find the norm of $|\psi(s)\rangle$, but to be honest, I'm a bit stuck. We haven't gone through the Variational principle yet, but is that what we're asked to work with?

• If your hamiltonian can be split in two parts: $H = H_0 + \lambda H_{\text{interaction}}$, $\lambda$ is small and you know the ground state of $H_0$, then you can use perturbation theory. Otherwise your expansion is probably useless. – Prof. Legolasov Oct 12 '16 at 11:14
• Thanks, but we haven't gotten to perturbation theory either. There's nothing stopping me from learning it and apply it, but it seems to me that i should be able to solve this without the use of perturbation due to the simple fact that we haven't learned it yet. – QuantumMechanic Oct 12 '16 at 12:27

Can we assume that all eigenvalues of $\hat H$ are bounded from below and/or positive and there is a gap $\Delta$ between the lowest and second lowest eigenvalue and every $| \psi \rangle$ can be expanded into sum of eigenvectors? If yes then we could explore the limit $$\lim_{s \rightarrow \infty} \frac{\langle \psi(s)| \hat O | \psi(s)}{\langle \psi(s)|\psi(s) \rangle}.$$
The underlying observation is that $|\psi(s)\rangle \approx e^{-sE_0} \left(\alpha_0|E_0\rangle + e^{-s\Delta}\alpha_1|E_1\rangle + \dots \right)$. Knowing the value of $\Delta$ also allows to quantify how big $s$ has to be to provide sensible approximation.
• @QuantumMechanic I added the coefficients of the linear combination that were missing in the expression for $\psi(s)$. It should be ok now. – Korf Oct 14 '16 at 12:51