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Calculate d<p> / dt

  1. May 18, 2013 #1
    1. The problem statement, all variables and given/known data
    Calculate ## \dfrac{d <p>}{dt} ##

    Answer: ## \left< -\dfrac{\partial V}{\partial x} \right> ##

    2. Relevant equations

    Schrodinger equation: ## i \hbar \dfrac{\partial \Psi}{\partial t} = -\dfrac{\hbar ^2}{2m} \frac{\partial ^2 \Psi}{\partial x^2} + V \Psi ##

    3. The attempt at a solution

    Here's what I did:

    ## \displaystyle \dfrac{\partial}{\partial t} \int^{\infty}_{- \infty} \Psi ^* \left( \dfrac{\hbar}{i} \dfrac{\partial}{\partial x} \right) \Psi dx ##

    ## \displaystyle \dfrac{\hbar}{i} \int^{\infty}_{- \infty} \dfrac{\partial}{\partial t} \Psi ^* \dfrac{\partial \Psi}{\partial x} dx ##

    ## \displaystyle \dfrac{\hbar}{i} \int^{\infty}_{- \infty} \left[ \dfrac{\partial \Psi ^*}{\partial t} \dfrac{\partial \Psi}{\partial x} + \Psi^* \dfrac{\partial}{\partial t} \dfrac{\partial \Psi}{\partial x} \right] dx ## (Differentiation by Product rule)


    From the Schrodinger equation we get that: ## \dfrac{\partial \Psi}{\partial t} = \dfrac{i \hbar}{2m} \dfrac{\partial ^2 \Psi}{\partial x^2} - \dfrac{i}{\hbar} V \Psi ##

    And it's conjugate: ## \dfrac{\partial \Psi ^*}{\partial t} = -\dfrac{i \hbar}{2m} \dfrac{\partial ^2 \Psi^*}{\partial x^2} + \dfrac{i}{\hbar} V \Psi^* ##

    Putting those into my integral I get:

    ## \displaystyle \dfrac{\hbar}{i} \int^{\infty}_{- \infty} \left[ \left( -\dfrac{i \hbar}{2m} \frac{\partial ^2 \Psi^*}{\partial x^2} + \dfrac{i}{\hbar} V \Psi^* \right) \dfrac{\partial \Psi}{\partial x} + \Psi^* \dfrac{\partial}{\partial x} \left( \dfrac{i \hbar}{2m} \frac{\partial ^2 \Psi}{\partial x^2} - \dfrac{i}{\hbar} V \Psi \right) \right] dx ##

    Expanding out everything:

    ## \displaystyle \dfrac{\hbar}{i} \int^{\infty}_{- \infty} \left[ -\dfrac{i \hbar}{2m} \frac{\partial ^2 \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x} + \dfrac{i}{\hbar} V \Psi^* \dfrac{\partial \Psi}{\partial x} + \dfrac{i \hbar}{2m} \frac{\partial ^3 \Psi}{\partial x^3} \Psi^* - \dfrac{i}{\hbar} \Psi^* \dfrac{\partial}{\partial x} (V \Psi) \right] dx ##

    ## \displaystyle \dfrac{\hbar}{i} \int^{\infty}_{- \infty} \left[ -\dfrac{i \hbar}{2m} \frac{\partial ^2 \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x} + \dfrac{i}{\hbar} V \Psi^* \dfrac{\partial \Psi}{\partial x} + \dfrac{i \hbar}{2m} \frac{\partial ^3 \Psi}{\partial x^3} \Psi^* - \dfrac{i}{\hbar} \dfrac{\partial V}{\partial x} \Psi \Psi * - \dfrac{i}{\hbar} \dfrac{\partial \Psi}{\partial x} V \Psi ^* \right] dx ##


    ## \displaystyle \dfrac{\hbar}{i} \int^{\infty}_{- \infty} \left[ \underbrace{-\dfrac{i \hbar}{2m} \frac{\partial ^2 \Psi^*}{\partial x^2} \dfrac{\partial \Psi}{\partial x}}_1 + \underbrace{\dfrac{i \hbar}{2m} \frac{\partial ^3 \Psi}{\partial x^3} \Psi^*}_2 - \underbrace{\dfrac{i}{\hbar} \dfrac{\partial V}{\partial x} \Psi \Psi *}_3 \right] dx ##

    I'm stuck at this point. I'm presuming there's a way to cancel out each of the integrals? I know the last integral is the one I want but I do not know how to cancel out the first 2.
     
  2. jcsd
  3. May 18, 2013 #2

    Simon Bridge

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    I'd have kept the operator notation for longer, vis...

    ##
    \renewcommand{H}{\hat{H}} \renewcommand{p}{\hat{p}}
    \renewcommand{\expn}[1]{\left \langle #1 \right \rangle}
    \renewcommand{dt}[1]{\frac{d #1}{dt}}
    \renewcommand{dx}[1]{\frac{d #1}{dx}}
    \renewcommand{intf}[1]{\int_{-\infty}^\infty #1 \; dx}
    \renewcommand{ddx}[1]{ \frac{d^2 #1}{dx^2} }## Need to show:$$\text{(1)... }\dt{}\expn{\hat{p}} = \expn{-\dx{V}}$$ (i.e. Newton's second law...)
    - expand: $$\begin{array}{rl}
    i\hbar \dt{}\expn{\p} & = i\hbar \dt{} \intf {\Psi^\star p\Psi} \\
    &= \intf { \left ( i\hbar \dt{} \Psi^\star \right ) \hat{p} \Psi + \Psi^\star \left ( i\hbar\dt{}(\hat{p}\Psi ) \right )} \; \text{ ...(2)}
    \end{array}$$ - from the Schrodinger equation: $$\text{(3)... }i\hbar\dt{} \Psi = \H\Psi\\
    \text{(4)... } \dt{}\expn{\p} = \frac{1}{i\hbar}\intf{ \H \Psi^\star \p\Psi + \Psi^\star \H\p\Psi} $$... hence, need to show that $$\text{(5)... } \H \Psi^\star \p\Psi + \Psi^\star \H\p\Psi = -i\hbar\Psi^\star \dx{V} \Psi$$ - which, I think, is pretty much where you are up to ;)
    (caveat: do not rely on me to get the math right - check!)

    - note that $$\text{(6)... }\p\H = - i\hbar \dx{} \left ( -\frac{\hbar^2}{2m}\ddx{}+V \right )$$ ... gives you third-order differentiation in x as well as the dV/dx you need.

    So how do you change the order of the operations?
     
    Last edited: May 18, 2013
  4. May 19, 2013 #3

    vela

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    If you want to continue on from where you got to, use integration by parts to move ##\frac{\partial}{\partial x}## between ##\Psi## and ##\Psi^*##.
     
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