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.


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

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

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

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.

  • $\begingroup$ 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

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