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I know this question has likely been asked before, but I am horribly confused and need some help with this.

Let's say we have a system whose initial state at t = 0 is given in terms of a complete and orthonormal eigenvector of the Hamiltonian:

$$ | \Psi(0)\rangle = \frac{1}{\sqrt{3}} |\phi_1\rangle+ \frac{1}{\sqrt{2}} |\phi_2\rangle+ \frac{1}{\sqrt{6}} |\phi_3\rangle$$

How would you find the probability of finding the system, at a time t, in the state $|\phi_3\rangle$?

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You will need to evolve the state in time using the full Schrodinger equation.

However, if your Hamiltonian is time-independent then your system is "stationary" and all amplitudes are time-independent. Hence you can calculate the probability of the outcome of a measurement as $|c_n|^2$ at $t=0$ as usual and this is guaranteed to also hold at $t\neq 0$.

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  • $\begingroup$ So I got the state of the system equation at any time t (it was part B of the question) down to $$ \left(\frac{1}{\sqrt{3}} |\phi_1\rangle+ \frac{1}{\sqrt{2}} |\phi_2\rangle+ \frac{1}{\sqrt{6}} |\phi_3\rangle\right) e^{(-iEt/\hbar)},$$ though I'm not sure where to go from here for calculating probability $\endgroup$
    – TheAkashain
    Commented Oct 6, 2020 at 14:32
  • $\begingroup$ What do you get when you calculate $\bigl| \frac{1}{\sqrt 6}\cdot e^{-iEt/\hbar} \bigr|^2$? That is the coefficient of $|\phi_3\rangle$. $\endgroup$
    – Charlie
    Commented Oct 6, 2020 at 14:35
  • $\begingroup$ Okay it all just 'clicked' for me after seeing your comment. While my work may be wrong: $$|\frac{1}{\sqrt{6}} \cdot e^{-iEt/\hbar}|^2 = \frac{1}{6} \cdot |e^{-iE(0)/\hbar}|^2$$ Which should simplify to $\frac{1}{6}$ since it is a stationary state $\endgroup$
    – TheAkashain
    Commented Oct 6, 2020 at 14:42
  • $\begingroup$ Yes, note that even when $t\neq 0$ we have $|e^{-iEt/\hbar}|^2=1$ so it can be ignored for all $t$. $\endgroup$
    – Charlie
    Commented Oct 6, 2020 at 14:43
  • $\begingroup$ Perfect, thank you so much! I marked your answer as correct and gave it an upvote (though that won't appear until my score is higher)! $\endgroup$
    – TheAkashain
    Commented Oct 6, 2020 at 14:44

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