# Probability of Finding a Quantum System in a Specific State

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 $$|\phi_3\rangle$$?

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 $$t\neq 0$$.
• 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 Oct 6, 2020 at 14:32
• What do you get when you calculate $\bigl| \frac{1}{\sqrt 6}\cdot e^{-iEt/\hbar} \bigr|^2$? That is the coefficient of $|\phi_3\rangle$. Oct 6, 2020 at 14:35
• 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 Oct 6, 2020 at 14:42
• Yes, note that even when $t\neq 0$ we have $|e^{-iEt/\hbar}|^2=1$ so it can be ignored for all $t$. Oct 6, 2020 at 14:43