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Homework Help: Very Basic QM problem: Commuter of position and momentum operators

  1. Apr 18, 2009 #1
    I'm not exactly sure if this belongs in introductory or advanced physics help.
    1. The problem statement, all variables and given/known data
    In my book, the author was explaining the proof of the Uncertainty relation between po
    position and momentum.

    It simply stated that [x,p]= ih(h is reduced)
    But when I tried to verify it I got -ih. I now it would give the same result, but it still won't be good for me to mess up a fundamental concept so early.

    2. Relevant equations
    [tex]\hat{p}[/tex]=-ih d/dx
    3. The attempt at a solution

    It would become this:
    -ihx d[tex]\psi[/tex]/dx -ihx-(-ihx d[tex]\psi[/tex]/dx)=

    Which is not the answer my book gave me.
  2. jcsd
  3. Apr 18, 2009 #2


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    Science Advisor
    Homework Helper

    Your answer isn't right. The px part means you have to differentiate x*psi. Use the product rule.
  4. Apr 20, 2009 #3
    Hello, Pinu7 ! I wish I could do some help.

    Firstly, I'd like to tell you that, the given attempt solution is incorrect, for the sign of $i\hbar x$ is plus rather than minus. It is a slight carelessness in extending the derivative of product. Now, I will re-perform the calculating in details. And please translate the tex codes yourself.

    For one-dimensional simplified case, as put forward in the question (or for the x-component of 3-dimensional analyses):
    \hat{p}= -i\hbar \frac{d}{dx}
    Hence, with the quantum Poisson bracket operator:

    [\hat{x}, \hat{p}] \phi = \hat{x} \hat{p} \phi - \hat{p} \hat{x} \phi

    = -i\hbar x \frac{d}{dx} \phi -( -i\hbar \frac{d}{dx} (x \phi))

    = -i\hbar x \frac{d}{dx} \phi -( -i\hbar \phi -i\hbar x \frac{d}{dx} \phi ) (Look Out! Your Carelessness Happens Here!)

    = i\hbar \phi

    Just as the textbook gives.
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