# Probability of being in the same initial state

If the Hamiltonian has basis of eigenvectors $\phi _1, \phi_2,..$ with corresponding eigenvalues $E_1,E_2,...$.

I then define an observable $A$ by:

$$A\phi_1 = \cos(\beta)\phi_1 + \sin(\beta)\phi_2$$ $$A\phi_2 = \sin(\beta)\phi_1 -\cos(\beta)\phi_2$$ And $A\phi_n=0$ for all $n=3,4...$

Now I can calculate that this observable has eigenvalues $\lambda = -1,1$ and corresponding eigenvectors $(1, \frac{-1-\cos(\beta)}{\sin(\beta)})$ and $(1, \frac{1-\cos(\beta)}{\sin(\beta)})$. If initially the system is in the state corresponding to $\lambda = -1$, so the wave function looks like:

$$\psi(t) = \exp \left(\frac{-iE_1t}{h} \right)\phi_1 + \frac{-1-\cos(\beta)}{\sin(\beta)}\exp \left(\frac{-iE_2t}{h} \right)\phi_2$$.

How would I go about finding the probability that at a later time, the particle is in the state corresponding to $\lambda = 1$? And in general, how do you find the probability that a wave function is in a certain state at a later time?

Any help is much appreciated

P.s Apologies if I have used any strange notation, I am studying mathematics and am not entirely sure on the correct way to write things in physics!

Probability of finding a system with state $|\psi\rangle$ in state $|\phi\rangle$ is defined to be a square of norm of projection of $|\phi\rangle$ on $|\psi\rangle$:
$$P_\phi=|\langle\phi|\psi\rangle|^2.$$
• Just divide the state by square root of its norm. There will always remain a constant to fix: global phase. A normalized state vector is always defined up to unimportant phase factor $e^{i\alpha}$. – Ruslan Jun 16 '14 at 19:56