1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expectation value for p

  1. Sep 3, 2008 #1
    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: [tex]\Psi[/tex](x, 0) = A(a^2 - x^2), -a[tex]\leq[/tex]x[tex]\leq[/tex]a. [tex]\Psi[/tex](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>=[tex]\int[/tex][tex]\Psi[/tex]*([tex]\frac{h}{i}[/tex][tex]\frac{d}{dx}[/tex])[tex]\Psi[/tex]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. jcsd
  3. Sep 4, 2008 #2
    Are you talking about the [tex]\hbar / i[/tex] 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 [tex]\psi^*[/tex] be in this case?
  4. Sep 4, 2008 #3
    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.
  5. Sep 4, 2008 #4
    Any time =] But yes, the complex conjugate of a real function is just the function itself.
  6. Sep 22, 2008 #5
    what about <p^2>?

    i need to write d^2/dx^2? do I?
  7. Sep 22, 2008 #6
    You could do that, or you could also consider the Hamiltonian... what is the Hamiltonian in terms of <p^2>?
  8. Sep 22, 2008 #7
    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 :(
  9. Sep 22, 2008 #8
    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.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?