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Momentum Operator Integration by Parts

  1. May 15, 2006 #1
    Hello

    I am teaching myself Quantum Mechanics from Griffiths. I have run into a mathematical problem which I need help with. As I have found no convincing answer, I am posting all the details here.

    Ref :Section 1.5 (Momentum) in "Introduction to Quantum Mechanics (2nd Edition)" by David J Griffiths

    [tex]\frac{d<x>}{dt} = \int_{-\infty}^{+\infty}
    x\frac{\partial}{\partial t}|\psi|^2 dx=
    \frac{i\hbar}{2m}\int_{-\infty}^{+\infty} x\frac{\partial}{\partial
    x}(\psi^{*}\frac{\partial \psi}{\partial x}-\frac{\partial
    \psi^{*}}{\partial x}\psi)dx[/tex] (1)

    If we denote the integral by [itex]I[/itex] then

    [tex]\frac{d<x>}{dt} = \frac{i\hbar}{2m}I[/tex] (2)

    Integrating by parts,

    [tex]I = x(\psi^{*}\frac{\partial \psi}{\partial
    x}-\frac{\partial \psi^{*}}{\partial x}\psi)|_{-\infty}^{+\infty} -
    \int_{-\infty}^{+\infty} \frac{\partial}{\partial
    x}(\psi^{*}\frac{\partial \psi}{\partial x}-\frac{\partial
    \psi^{*}}{\partial x}\psi)dx[/tex] (3)

    The second term in (3) is easily handled based on the steps described on Page 28 (a second integration by parts). However, the first term in this equation is the troublesome term.

    The expression in parentheses of the first term goes to zero as [itex]x\rightarrow \pm \infty[/itex]. As [itex]x \rightarrow \pm \infty[/itex] the first term in (3) is of the form [itex]\infty * 0[/itex]. This term would tend to zero as [itex]x \rightarrow \pm \infty[/itex] only if the terms in the parentheses were to go to zero faster than [itex]x[/itex] goes to [itex]\pm \infty[/itex].

    The first term in (3) has been stated in the book to be equal to zero at [itex]x = \pm \infty[/itex] in its entirety. Is this because [itex]\psi(x,t)[/itex] is square integrable and must go to zero faster than [itex]1/\sqrt{|x|}[/itex] as [itex]|x| \rightarrow \infty[/itex] (as given in the footnote to section 1.4 on page 25)?

    Or is this because all physically meaningful wavefunctions behave this way? Am I integrating correctly?

    Mathematical Counterexample

    If
    [tex]f(x) = x^{-a}[/tex]

    where [itex]1/2 < a < 1[/itex]

    then

    [tex]xf(x) = x^{1-a}[/tex]

    and [itex]0<1-a<1/2[/itex] so as [itex]x\rightarrow \pm \infty[/itex], [itex]xf(x)[/itex] can tend to [itex]\pm \infty[/itex]. Of course I understand that [itex]f(x)[/itex] cannot be a wavefunction because it is discontinuous at [itex]x = 0[/itex]. This is just a counterexample.

    Note to the Moderator: Please shift this post to the correct forum if this is not the right location.
     
    Last edited: May 15, 2006
  2. jcsd
  3. May 15, 2006 #2
    Your counterexample is quite pointless for the reason you stated yourself. You're integrating correctly and both of the reasons for psi to tend to zero at infinity are ok. The physical argument is pretty intuitive and saves one the trouble of a more detailed mathematical analysis.
     
  4. May 15, 2006 #3

    George Jones

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    What does

    [tex]\left[x \left(\psi^{*}\frac{\partial \psi}{\partial x}-\frac{\partial \psi^{*}}{\partial x}\psi \right) \right]^{+ \infty}_{- \infty}[/tex]

    equal if

    [tex]\psi \left( x \right) = \frac{i}{\sqrt{\left| x \right|}}?[/tex]

    Regards,
    George
     
    Last edited: May 15, 2006
  5. May 15, 2006 #4

    George:

    That equals zero. But what is your point? The wave function you have proposed is discontinuous at x = 0. Its not a meaningful wavefunction because the wavefunction should be well-defined and continuous everywhere.

    inha:

    Yes I know its useless. The real question is whether [itex]xf(x)[/itex] tends to zero when [itex]x\rightarrow\pm\infty[/itex] in this case (in general it need not).
     
    Last edited: May 15, 2006
  6. May 15, 2006 #5

    George Jones

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    We don't care what happens to the wavefunction at x = 0, since the offending expression is being evaluated in the limit as x approaches +- infinity. As you say, In order for psi to be square-intregrable, psi must die faster that 1/sqrt(|x}) in this limit.

    To make things nice, define psi as I have for |x| > 1, and use anything that makes psi nice for |x| < 1. The part of psi for |x| < 1 has no effect on the offending expression.

    Regards,
    George
     
  7. May 15, 2006 #6
    Okay I understand what you're saying mathematically. But physically how can I define [itex]\psi[/itex] to suit my needs??? [itex]\psi[/itex] is supposed to be the solution to an equation. I can't control it. Why is it that the function multiplying x in the "offending" expression--as you say--is dropping to zero so fast that multiplying x which is tending to [itex]\pm\infty[/itex] does not affect it?? What is the physical explanation?
     
  8. May 15, 2006 #7

    George Jones

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    If [itex]\psi[/itex] doesn't die off at least this fast, then psi is not square-integrable, and the probablity of finding the particle between [itex]x = -\infty[/itex] and [itex]x = \infty[/itex] cannot be normalized to 1.

    Regards,
    George
     
  9. May 15, 2006 #8
    Well, we assume that [itex]\psi[/itex] is square-integrable but its not necessary that it is already normalized so we can find some constant A (to within an undetermined phase) so that [itex]A\psi(x,t)[/itex] is normalized. Thats why we can say that we want

    [tex]\int_{-\infty}^{+\infty}|\psi|^2 dx < \infty[/tex]

    and the integral to be nonzero ([itex]\psi(x,t) \neq 0[/itex]).

    The problem is how does one say that in general the "offending term" is zero without using a particular kind of wavefunction.
     
  10. May 15, 2006 #9

    George Jones

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    I think you're asking how to show

    [tex]\int_{-\infty}^{+\infty}|\psi|^2 dx < \infty \Rightarrow \lim_{x \rightarrow \pm \infty} \frac{\left| \psi \left( x \right) \right|}{\frac{1}{\sqrt{|x|}}} = 0 \Rightarrow \left[ x \left(\psi^{*}\frac{\partial \psi}{\partial x}-\frac{\partial \psi^{*}}{\partial x}\psi \right) \right]^{+ \infty}_{- \infty} = 0.[/tex]

    Is this what you're actually asking?

    Regards,
    George
     
    Last edited: May 15, 2006
  11. May 16, 2006 #10
    Is that true? (Esp from the integral to the right hand side, not the limit...the limit seems to be a property of most physically meaningful wavefunctions.)
     
  12. May 16, 2006 #11

    George Jones

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    If the integral implies the limit, and the limit implies the RHS, then the integral implies the the RHS. This is just a logical syllogism, i.e., if A => B and B => C, then A => C.

    If you want, I can supply some of the steps that show either implication.

    Regards,
    George
     
  13. May 16, 2006 #12
    Yes please do so.
     
  14. May 18, 2006 #13
    Okay I get it now...in addition to being square integrable, we also want the expectation value of x to be defined, that is the integral

    [tex]<x> = \int_{-\infty}^{+\infty} \psi^{*}(x)[x]\psi(x) dx[/tex]

    to be defined.
     
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