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!


1 Answer 1


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 $|\psi\rangle$:


But, before finding this projection, you should make sure that your states are normalized, otherwise statistical interpretation of wave function won't make any sense.

  • $\begingroup$ Thank you for your answer, I was just wondering is there any intuitive way of seeing why this gives us the probability? I'm aware that the coefficients in the wave function of each eign state give us the probability of being in that state $\endgroup$
    – Wooster
    Commented May 23, 2014 at 7:43
  • $\begingroup$ Well, it is one of the principles of quantum mechanics. See this wiki page for details. $\endgroup$
    – Ruslan
    Commented May 23, 2014 at 8:19
  • $\begingroup$ I asked this question a while ago but have just come back to it. In this case, how can I normalise the states given above? Because am I right in thinking there are no arbitrary constants to fix anymore? $\endgroup$
    – Wooster
    Commented Jun 16, 2014 at 14:24
  • $\begingroup$ 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 $e^{i\alpha}$. $\endgroup$
    – Ruslan
    Commented Jun 16, 2014 at 19:56

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