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Derivation of Schrodinger's equation in momentum space

  1. Dec 4, 2015 #1
    1. The problem statement, all variables and given/known data
    Screen Shot 2015-12-04 at 4.39.10 pm.png
    How do you get from (3.171) to (3.172)? In particular, why is

    ##\int e^{-ip.r/{\hbar}}\frac{p_{op}^2}{2m}\Psi(r,t)\,dr=\int\frac{p_{op}^2}{2m}[e^{-ip.r/{\hbar}}\Psi(r,t)]\,dr##? ##\,\,\,\,\,##-- (1)

    2. Relevant equations
    Screen Shot 2015-12-04 at 4.39.54 pm.png
    Screen Shot 2015-12-04 at 4.40.19 pm.png

    3. The attempt at a solution
    For (1) to be true, it must be true that
    ##\int\frac{\partial ^2}{\partial x^2}[e^{-ip.x/{\hbar}}\Psi(r,t)]\,dr=\int e^{-ip.x/{\hbar}}\frac{\partial ^2}{\partial x^2}\Psi(r,t)\,dr##. ##\,\,\,\,\,##-- (2)

    Using integration by parts and the fact that ##\Psi(r,t)=0## at infinity, I am able to show that (2) is true provided

    ##\int\frac{-ip}{\hbar}e^{-ipx/\hbar}\frac{\partial}{\partial x}\Psi(r,t)\,dr=0## ##\,\,\,\,\,##-- (3) or

    ##\int\Psi(r,t)\,(\frac{-ip}{\hbar})^2e^{-ipx/\hbar}\,dr=0##.##\,\,\,\,\,##-- (4)

    But I don't know how (3) or (4) is true?
     
  2. jcsd
  3. Dec 4, 2015 #2

    DrClaude

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    Staff: Mentor

    The author has left out steps leading to (3.171). The way it is written there, p is already an operator in momentum space, meaning that it is a simple multiplication.

    Edit: Maybe it is not missing steps, but the author has taken the problem backward. I would start with the Schrödinger equation for ##\Psi(\mathbf{r},t)##, write ##\Psi(\mathbf{r},t)## in terms of ##\Phi(\mathbf{p},t)##, and get the Schrödinger equation for ##\Phi(\mathbf{p},t)##.
     
  4. Dec 4, 2015 #3
    Since the integration in (3.171) is done with respect to r, shouldn't p still be an operator in configuration space?

    How I see the steps leading to (3.171) is as follows:

    Since ##\Psi(r,t)## satisfies the Schrodinger equation, any linear combination of ##\Psi##'s satisfies the Schrodinger equation too by the principle of superposition (or by the linearity of the operators in the Schrodinger equation).
     
  5. Dec 4, 2015 #4
    Oh I got it now. The author used

    ##H\Psi(r,t)=E\Psi(r,t)=[\frac{p^2}{2m}+V(r,t)]\Psi(r,t)##.

    So the p##^2## in (3.171) is a constant and not an operator.
     
  6. Dec 4, 2015 #5

    DrClaude

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    Staff: Mentor

    In the equation you wrote, p2 has got to be an operator. It is in momentum space that it is a number.

    I'm still trying to figure how to best approach the way it is presented in your book.
     
  7. Dec 4, 2015 #6
    I think I got it.

    ##(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}\frac{p_{op}^2}{2m}\Psi(r,t)\,dr\\##

    first substitute ##\Psi(r,t)## in terms of ##\Phi(p',t)##
    ##=(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}\frac{p_{op}^2}{2m}(2\pi\hbar)^{-3/2}\int e^{ip'.r/\hbar}\Phi(p',t)\,dp'\,dr\\##

    then group the exponential terms together
    ##=\int\frac{p_{op}^2}{2m}(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\Phi(p',t)\,dp'\,dr\\##

    since ##p_{op}## and the r integration commute, we have
    ##=\frac{p_{op}^2}{2m}\int(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\Phi(p',t)\,dp'\,dr\\##

    since the r integration and the p' integration commute, we have
    ##=\frac{p_{op}^2}{2m}\int(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\,dr\,\Phi(p',t)\,dp'\\##

    next perform the r integration
    ##=\frac{p_{op}^2}{2m}\int\delta(p'-p)\Phi(p',t)\,dp'\\
    =\frac{p_{op}^2}{2m}\Phi(p,t)\\
    ##
     
    Last edited: Dec 4, 2015
  8. Dec 4, 2015 #7
    I still have one unresolved question:

    How does ##p_{op}^2## change from being ##-\hbar^2\nabla^2## to being ##p^2##, where ##p## is a number? At which step does it occur and why?
     
  9. Dec 4, 2015 #8

    nrqed

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    You cannot pull ##p_{or}^2## outside of the integral over r, since it is acting on that variable! You must apply it to the exponential and therefore it will generate a factor ##p^2## where now p is the variable, not an operator.
     
  10. Dec 5, 2015 #9
    I see the mistake.

    ##(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}\frac{p_{op}^2}{2m}\Psi(r,t)\,dr\\##

    first substitute ##\Psi(r,t)## in terms of ##\Phi(p',t)##
    ##=(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}\frac{p_{op}^2}{2m}(2\pi\hbar)^{-3/2}\int e^{ip'.r/\hbar}\Phi(p',t)\,dp'\,dr\\##

    then group the exponential terms together (This step is wrong because ##p_{op}^2## is acting on ##r## but the exponential term has ##r## so they don't commute.)
    ##=\int\frac{p_{op}^2}{2m}(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\Phi(p',t)\,dp'\,dr\\##

    instead it should be
    ##=(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}(2\pi\hbar)^{-3/2}\int\frac{p_{op}^2}{2m}e^{ip'.r/\hbar}\Phi(p',t)\,dp'\,dr\\##
    ##=(2\pi\hbar)^{-3/2}\int e^{-ip.r/\hbar}(2\pi\hbar)^{-3/2}\int\frac{p'^2}{2m}e^{ip'.r/\hbar}\Phi(p',t)\,dp'\,dr\\##

    since the r integration and the p' integration commute, we have
    ##=\int\frac{p'^2}{2m}(2\pi\hbar)^{-3}\int e^{i(p'-p).r/\hbar}\,dr\,\Phi(p',t)\,dp'\\##

    next perform the r integration
    ##=\int\frac{p'^2}{2m}\delta(p'-p)\Phi(p',t)\,dp'\\
    =\frac{p^2}{2m}\Phi(p,t)\\
    ##
     
  11. Dec 5, 2015 #10

    nrqed

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    Perfect!
     
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