How can I find the probability and expectation value in quantum mechanics?

[Martin Osborne]

Homework Statement

[/B]

Hi, I have a problem I have been trying to do for a few days and I am not getting it. Any hints would be greatly appreciated. The question is from "The physics of Quantum Mechanics" by Binney and Skinner.

The question states:
Let $\psi$(x) be a properly normalised wavefunction and Q an operator on wavefunctions. Let {q_r} be the spectrum of Q and let {U_r(x)} be the corresponding correctly normalised eigenfunctions. Write down an expression for the probability that a measure of Q will yeild the value q_r.

Show that $\Sigma_r P(q_r |\psi) = 1$.

Show further that the expectation of Q is $\langle Q \rangle = \int _{-\infty} ^\infty \psi^* Q \psi dx$ .

Homework Equations

and attempt[/B]

So for the first part, the probability amplitude of measuring q_r given the system is in the state $|\psi\rangle$ is given by $\langle q _ r | \psi \rangle = \int _{-\infty} ^\infty u_r^*(x) \psi(x) dx$ .

and taking the mod squared of this gives the probability the question is asking for.

The next part says that summing these probabilities over all r = 1? I understand what this means (probability of finding a value of q within the spectrum given = 1), but don't know how to show this.

As for the last part, the expectation value is the sum of the probabilities of getting each value of q multiplied by the value q_r, so $$ \langle Q \rangle = \Sigma _ r q_r | \int _ {-\infty}^\infty u_r^*(x) \psi(x) dx |^2 $$Cant get any further...

 

[Martin Osborne]

sorry, latex not working let me try again...

fixed it...

 

[vela]

Try expanding $\lvert \psi \rangle$ in terms of the eigenstates.

 

[Martin Osborne]

Thanks Vela,

I am thinking $|\psi\rangle = \int_{-\infty} ^\infty \psi(x) |x\rangle$ But is it also the case that $\psi(x) = \sum a_r u_r(x)$ where the $a_r$s are probability amplitudes in Q space.

Can I say that $|\psi\rangle = \int_{-\infty} ^\infty (\sum a_r u_r(x)) |x\rangle$

 

[vela]

Yes, and that would be equivalent to saying $\lvert \psi \rangle = \sum_r a_r \lvert q_r \rangle$.