1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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

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

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

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.

$\endgroup$
4
  • $\begingroup$ Why can one just set h to 1 though? $\endgroup$ – DJA Oct 28 '20 at 8:23
  • $\begingroup$ when I do it also get (2|<phin | H | phi n>| )^2 so something not quite right $\endgroup$ – DJA Oct 28 '20 at 9:03
  • 1
    $\begingroup$ @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. $\endgroup$ – Milarepa Oct 28 '20 at 12:03
  • $\begingroup$ Thanks so much for that! $\endgroup$ – DJA Oct 28 '20 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.