Finding the probability of measuring a particular eigenvalue of an operator for a system after time evolution

Question

Consider a quantum system with Hamiltonian H and consider the measurement of an observable $a_n$ associated with a different operator A.

Initially the system is an eigenstate $|\phi_n \rangle$ with eigenvalue $a_n$ and we begin to take measurements of the observable A.

We can approximate the probability of measuring an eigenvalue of $a_n$ at time t as:

$$1-t^2( \langle \phi_n| H^2|\phi_n \rangle - \langle \phi_n| H|\phi_n \rangle^2))$$

I am very confused as to where this equation has come from and any guidance to deduce it would be appreciated.

Answer 1

$|\psi(t)\rangle= U(t)|\phi_{n}\rangle = e^{-\frac{i}{\hbar}Ht}|\phi_{n}\rangle \approx (1-\frac{i}{\hbar}Ht -\frac{1}{2\hbar^{2}}H^2t^2) |\phi_{n}\rangle$

$\langle \phi_{n} | \psi(t)\rangle = 1 -\frac{i}{\hbar}t\langle \phi_{n}|H|\phi_{n}\rangle -\frac{1}{2\hbar^2}t^2 \langle \phi_{n}|H^2|\phi_{n}\rangle$

$p_{n}(t)=|\langle \phi_{n} | \psi(t)\rangle|^2= (1 -\frac{i}{\hbar}t\langle \phi_{n}|H|\phi_{n}\rangle -\frac{1}{2\hbar^2}t^2 \langle \phi_{n}|H^2|\phi_{n}\rangle)(1 +\frac{i}{\hbar}t\langle \phi_{n}|H|\phi_{n}\rangle -\frac{1}{2\hbar^2}t^2 \langle \phi_{n}|H^2|\phi_{n}\rangle)$

If you now do the algebra, neglect all $t^3$ and $t^4$ terms and set $\hbar=1$ you should get to the expression you are looking for.