Quantum mechanics: Expectation values

1. Aug 8, 2008

Niles

1. The problem statement, all variables and given/known data
Hi all.

Let's say that i have a wave function
$$\Psi (x,t) = A \cdot \exp ( - \lambda \cdot \left| x \right|) \cdot \exp ( - i\omega t)$$

I want to find the expectation value for x. For this I use
$$\left\langle x \right\rangle = \int_{ - \infty }^\infty x \left| \Psi \right|^2 dx$$.

Why must I set t = 0 when doing this?

2. Aug 8, 2008

nrqed

You don't.

Just leave t arbitrary and you will see that it will drop out by itself.

3. Aug 9, 2008

Niles

Yes, you are right. But that is when I have normalized the wave function first.

In my book, when he normalizes this function he uses t=0, so A = sqrt(lambda). This way he uses t=0 when finding <x>.

But I can't do both (both use t=t and t=0). So what is more correct?

4. Aug 9, 2008

Dick

??? The normalization is independent of t. exp(i*w*t) times it's complex conjugate is 1. For all t.

5. Aug 9, 2008

Niles

My mistake. I just squared the whole expression, when I should have multiplied with the conjugate (we are dealing with complex numbers, of course).

Thanks to both of you.

6. Aug 10, 2008

nrqed

You are welcome.

By the way, it is a general property of wavefunctions: if you normalize them at any time, they will remain normalized at all times (this assumes that the wavefunction is square integrable, that the potential is real, etc). So one can in principle fix the time to be some value (any value!) before normalizaing but this is never necessary because the time dependence *always* drops out when we normalize.