# Expectation value for p

1. Sep 3, 2008

### kde2520

1. The problem statement, all variables and given/known data
First off, this is my first time posting here so please excuse any editing mistakes or guidelines I may have overlooked.

This is problem 1.17(c) from Griffiths, Introduction to Quantum Mechanics 2nd edition. It reads: $$\Psi$$(x, 0) = A(a^2 - x^2), -a$$\leq$$x$$\leq$$a. $$\Psi$$(x, 0) = 0, otherwise. What is the expectation value for p? (Note that you cannot get it from p = md<x>/dt. Why not?)

2. Relevant equations
So far we've derived the expression <p>=$$\int$$$$\Psi$$*($$\frac{h}{i}$$$$\frac{d}{dx}$$)$$\Psi$$dx

3. The attempt at a solution
I found the expectation value for position to be <x>=0. Also, t=0. These seem to explain why I can't get <p> from md<x>/dt. But since the function is not complex I can't see how to interpret the above expression for <p>. The operator acts on the real part, but there is no imaginary part to deal with. Any clues on how to interpret this?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 4, 2008

### Domnu

Are you talking about the $$\hbar / i$$ in the integration? You can pull this out of the integral, as this is just a constant. It's okay that there's no imaginary part... what would $$\psi^*$$ be in this case?

3. Sep 4, 2008

### kde2520

Thanks for the reply. It wasn't the [tex]\hbar / i[tex] giving me trouble, but the idea of [tex]\psi^*[tex] being the conjugate of something with only real parts. Might sound silly I guess... Then I was having trouble seeing how we actually arived at the partial derivative with respect to x; but I see now that it was borrowed directly from Shrodinger's equation.

Thanks again.

4. Sep 4, 2008

### Domnu

Any time =] But yes, the complex conjugate of a real function is just the function itself.

5. Sep 22, 2008

### lycogen

eem
what about <p^2>?

i need to write d^2/dx^2? do I?

6. Sep 22, 2008

### Domnu

You could do that, or you could also consider the Hamiltonian... what is the Hamiltonian in terms of <p^2>?

7. Sep 22, 2008

### lycogen

i am doing the <p^2> for harmonic oscillator (ground state)
when I use second derivative I end up with part which contains x after integration :(

8. Sep 22, 2008

### Domnu

okay if you're doing it for the harmonic oscillator, a *very* useful thing to do is to express all observables in terms of the annihilation/creation operators: a+ and a- in the book. go do that and see what happens.

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