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.

Milarepa · Accepted Answer

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

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

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