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About the expectation value of position of a particle

  1. Jan 19, 2016 #1
    I am following Griffiths' intro to quantum mechanics and struggling(already) on page 16. When a particle is in state ##\Psi##,
    $$\frac{d<x>}{dt} = \frac{i\hbar}{2m}\int_{-\infty}^{\infty} x\frac{\partial}{\partial t}\bigg (\Psi^*\frac{\partial \Psi}{\partial x}-\frac{\partial \Psi^*}{\partial x}\Psi\bigg )dx$$
    By integration-by-parts, this becomes
    $$\frac{i\hbar}{2m}\bigg [x\bigg (\Psi^*\frac{\partial \Psi}{\partial x}-\frac{\partial \Psi^*}{\partial x}\Psi \bigg)-\int_{-\infty}^{\infty} \bigg (\Psi^*\frac{\partial \Psi}{\partial x}-\frac{\partial \Psi^*}{\partial x}\Psi\bigg )dx\bigg ]$$

    But the author throws the first term away by reasoning "##\Psi## goes to zero at (##\pm\infty##)"
    but as at infinity, x also goes to infinity, so the first term is infinity x 0, which is not clear to me if the whole term could be thrown away. Can someone please explain this?
     
  2. jcsd
  3. Jan 19, 2016 #2

    blue_leaf77

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    The units between the first and second equations seem to be off.
     
  4. Jan 19, 2016 #3

    PeroK

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    Yes, there is more to it than Griffiths says. If you want you can check what kind of functions this step is not valid for. But, Griffiths point would be that all functions he'll be dealing with meet the criterion that that term in the integration by parts is 0.
     
    Last edited: Jan 19, 2016
  5. Jan 19, 2016 #4

    PeroK

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    I can give you a better answer. Some square integrable functions will fail that step. E.g

    ##\Psi(x) = \frac{1}{x} + i\frac{sin(x^3)}{x}##

    You better check that, but I think it does the job. Somewhere else Griffiths says that "any good mathematician will be able to provide pathological counterexamples" or words to that effect.

    Note that's for large ##x##. You'd need to patch the function up with something finite around ##x = 0##.
     
    Last edited: Jan 19, 2016
  6. Jan 19, 2016 #5
    Yeah I tried that function and it did the job. Thank you. I wonder what are the functions Griffiths will be dealing with and how it makes sense to throw the term away for those functions...
     
  7. Jan 19, 2016 #6

    A. Neumaier

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    add ''sufficiently fast'' to make the statement precise. You can easily work out how fast (in terms of powers of ##x##) the decay must be in order that the argument works. In practice, decay is exponential so there is no problem.
     
  8. Jan 20, 2016 #7

    PeroK

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    It's the derivative that is the problem. The Schrodinger Equation involves a second derivative. If the derivative and the second derivative go to 0 then finding a counterexample would be harder!

    Or, if the derivative is square integrable that might do.

    But, perhaps, just keep an eye out for these things and move on. It's QM you want to learn, not functional analysis!
     
    Last edited: Jan 20, 2016
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