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QM, show a relation (velocity of a free particle related)

  1. Apr 5, 2013 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Hi guys, I'm stuck at some step in a QM exercise. Here it is: Consider a free particle of mass m that moves along the x-axis (1 dimensional). Show that ##\frac{dA}{dt}=\frac{2 \hbar ^2}{m^2}\int \frac{\partial \psi}{\partial x} \frac{\partial \psi ^*}{\partial x}dx=\text{constant}##.
    Where ##A(t)=\frac{i\hbar}{m} \int x \left ( \psi \frac{\partial \psi ^*}{\partial x} - \psi ^* \frac{\partial \psi}{\partial x} \right ) dx##.
    In fact ##A(t)=\frac{d \langle x^2 \rangle}{dt}## but I already showed that it's worth the integral I just wrote (the result matches the answer).
    2. Relevant equations
    (1)Schrödinger's equation: ##i\hbar \frac{\partial \psi}{\partial t}=\frac{(p^2 \psi)}{2m}##, where p is the momentum operator and worth ##-ih \frac{\partial \psi}{\partial x }##.
    Taking the complex conjugate on both sides, I get that ##\psi ^*## satisfies:
    (2) ##-i \hbar \frac{\partial \psi ^*}{\partial t}=\frac{p^2 \psi ^*}{2m}##
    From (1) I get that
    (3) ##\frac{\partial \psi}{\partial t }=-\frac{i(p^2\psi)}{2m\hbar}##
    and
    (4)##\frac{\partial \psi ^*}{\partial t }=\frac{i(p^2\psi ^*)}{2m\hbar}##

    3. The attempt at a solution
    ##\frac{dA(t)}{dt}= \frac{d}{dt} \left [ \frac{i\hbar}{m} \int x \left ( \psi \frac{\partial \psi ^*}{\partial x} - \psi ^* \frac{\partial \psi }{\partial x} \right ) dx \right ]= \frac{ih}{m} \int x \underbrace{ \frac{d}{dt} \left ( \psi \frac{\partial \psi ^*}{\partial x} - \psi ^* \frac{\partial \psi }{\partial x} \right ) dx } _{K}## (line 5)
    I try to calculate K first.
    ##K=\frac{\partial \psi}{\partial t} \frac{\partial \psi ^*}{\partial x}+\psi \frac{\partial ^2 \psi ^*}{\partial x^2}- \left ( \frac{\partial \psi^*}{\partial t} \frac{\partial \psi }{\partial x} + \psi ^* \frac{\partial ^2 \psi }{\partial x ^2}\right )##. (line 6)
    Now I use (3) and (4) to get ##K=-\frac{i}{\hbar} \frac{(p^2\psi)}{2m} \frac{\partial \psi ^*}{\partial x}+\psi \frac{\partial ^2 \psi ^*}{\partial t \partial x }- \left ( \frac{i}{\hbar} \frac{(p^2 \psi ^* )}{2m} \frac{\partial \psi }{\partial x }+ \psi ^* \frac{\partial ^2 \psi}{ \partial t \partial x } \right )## (line 7)
    Here I assume that ##\frac{\partial ^2 \psi }{\partial t \partial x }=\frac{\partial ^2 \psi}{\partial x \partial t}##.
    If so, I get ##K=-\frac{i}{\hbar} \frac{(p^2 \psi)}{2m} \frac{\partial \psi ^*}{\partial x}+\psi \frac{\partial ^2 \psi ^*}{\partial x \partial t}- \left ( \frac{i}{\hbar} \frac{(p^2 \psi ^*)}{2m} \frac{\partial \psi}{\partial x} + \psi ^* \frac{\partial ^2 \psi }{\partial x \partial t} \right )## (line 8)
    Now I use (3) and (4) to get ##K=-\frac{i}{\hbar} \frac{(p^2 \psi)}{2m} \frac{\partial \psi ^*}{\partial x} + \psi \frac{\partial}{\partial x} \left [ \frac{i(p^2 \psi ^*)}{2m \hbar} \right ] - \{ \frac{i}{\hbar} \frac{(p^2 \psi ^*)}{2m\hbar} \frac{\partial \psi}{\partial x} + \psi ^* \frac{\partial }{\partial x } \left [ -\frac{i (p^2 \psi)}{2m\hbar} \right ] \} ## (line 9)
    I factor out i/(2m hbar) to get ##K=\frac{i}{2m\hbar} \{ \psi \frac{\partial }{\partial x} (p^2 \psi ^*) - \frac{\partial \psi ^*}{\partial x} (p^2 \psi ) - \frac{\partial \psi }{\partial x} (p^2 \psi ^*) -\psi ^* \frac{\partial }{\partial x} (p^2 \psi) \} ## (line 10)
    Now I replace ##p^2## by ##-\hbar ^2 \frac{\partial ^2 }{\partial x^2}## to get ##K=-\frac{i\hbar}{2m} \left ( \psi \frac{\partial ^3 \psi ^*}{\partial x^3} - \frac{\partial \psi ^*}{\partial x} \frac{\partial ^2 \psi }{\partial x^2} - \frac{\partial \psi}{\partial x} \frac{\partial ^2 \psi ^*}{\partial x^2} - \psi ^* \frac{\partial ^3 \psi}{\partial x^3} \right ) ## (line 11)
    Now I replace the value of K into where I took it (that is, line 11 into line 5), to get ##\frac{dA(t)}{dt} = \frac{\hbar^2}{2m^2} \int x \left ( \psi \frac{\partial ^3 \psi ^*}{\partial x^3} - \frac{\partial \psi ^*}{\partial x} \frac{\partial ^2 \psi }{\partial x^2} - \frac{\partial \psi}{\partial x} \frac{\partial ^2 \psi ^*}{\partial x^2} - \psi ^* \frac{\partial ^3 \psi}{\partial x^3} \right ) dx##. (line 12).
    This is where I'm stuck. I thought about integration by parts, but I'm not seeing anything nice.
    I've rewritten ##\frac{dA(t)}{dt}## as ##\frac{\hbar^2}{2m^2} \int x \left [ \frac{\partial }{\partial x} \left ( \psi \frac{\partial ^2 \psi ^* }{\partial x^2} \right ) -2 \frac{\partial \psi}{\partial x} \frac{\partial ^2 \psi ^*}{\partial x^2 } - \frac{\partial }{\partial x} \left ( \psi ^* \frac{\partial ^2 \psi}{\partial x^2} \right ) \right ] dx##. I'm still totally stuck.
    I've been extremely careful and what I wrote in latex contains no typo error. All is exactly as in my draft so any mistake/error you see, it's not a typo but an algebra one that I made.

    Any help is greatly appreciated, I really want to solve that exercise. Thanks :)
     
  2. jcsd
  3. Apr 5, 2013 #2

    TSny

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    Homework Helper
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    Check the sign of the last term on the right. Otherwise, things look good to me up to this point. After substituting for p2 I think you have the right idea to continue with some integration by parts.
     
  4. Apr 5, 2013 #3

    fluidistic

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    Oh thank you very much TSny, I didn't see this sign mistake. I fixed it and in consequences this changes all the following lines.
    I've reached the result! I had to integrate by parts twice for an integral and once for another. I also had to assume that ##x\frac{\partial \psi ^*}{\partial x} \frac{\partial \psi}{\partial \psi} \big | _{-\infty}^\infty## is worth 0 (and another 2 similar relations), which isn't that obvious to me. I mean I have a big feeling that it's worth 0 but I'm not really sure why. I know that ##\frac{\partial \psi}{\partial x}## and ##\frac{\partial \psi ^*}{\partial x}## must tend to 0 at infinities but if I multiply their product by x, then I'm not really sure.

    I also have to show that ##\frac{dA(t)}{dt}## is constant. I guess I just have to derivate it with respect to time and see that it's worth 0... I'll do it, I don't think this will be hard nor that long.

    All in all the problem was extremely lengthy in terms of algebra!
     
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