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

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.

## 1 Answer

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

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

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

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.

• Why can one just set h to 1 though? – DJA Oct 28 '20 at 8:23
• when I do it also get (2|<phin | H | phi n>| )^2 so something not quite right – DJA Oct 28 '20 at 9:03
• @DJA Sorry, I forgot a few $\frac{1}{2}$ factors. I edited the answer, now you are going to obtain your result. Regarding the fact $\hbar$ is set to one, that is just due to the formalism of natural units, some other textbooks would include the $\hbar$ terms in the same expression. – Milarepa Oct 28 '20 at 12:03
• Thanks so much for that! – DJA Oct 28 '20 at 12:40