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Homework Help: Commutator maths problem

  1. Jan 14, 2010 #1
    1. The problem statement, all variables and given/known data
    Determine [tex]\left[\hat{x},\hat{H}\right][/tex]

    2. Relevant equations

    3. The attempt at a solution



    How do you get from the 2nd line to the 3rd? is it an identity of partial differentials? can someone explain how to split the d squared term into two separate parts?
  2. jcsd
  3. Jan 14, 2010 #2


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    Hi 8614smith! :smile:

    (use "\partial" not "\delta" :wink:)

    Hint: ∂2/∂x2(xψ) = (∂2x/∂x2)ψ + 2(∂x/∂x)(∂ψ/∂x) + x∂2/∂x2(ψ) :wink:
  4. Jan 14, 2010 #3
    And one more hint: (∂2x/∂x2)ψ = 0.
  5. Jan 14, 2010 #4
    why does this = 0? i would have thought it would equal [tex]\psi[/tex]

    assuming the above quote is correct and the 1st term = 0, I have got to the 3rd line - except that i now have an extra 2 in the equation, how do i get rid of this?

    and has your identity worked the same way as expanding expanding ( )^2?
  6. Jan 14, 2010 #5
    But I got the right answer. Let's make it look better: [tex]\frac{\hbar^2}{2m}\frac{{\partial^2} (x\psi)}{\partial{x^2}}=\frac{\hbar^2}{2m}\frac{{\partial}}{\partial{x}}\left(\frac{{\partial} (x\psi)}{\partial{x}}\right)[/tex]. Now use the simple rule (xy)'=x'y+y'x and see what happens.
  7. Jan 14, 2010 #6
    It would be like that if it were (∂x/∂x)ψ.
  8. Jan 14, 2010 #7
    if i use the product rule on both sides then i just get the same thing on both sides. meaning i just have 0? or have i done that wrong?

    The answer quoted for this problem is [tex]\frac{\hbar^2}{m}\frac{\partial}{\partial{x}}[/tex]
  9. Jan 14, 2010 #8
    It means that you have done it right. Let's take the final step:

    [tex]\frac{\hbar^2}{2m}\frac{{\partial^2} (x\psi)}{\partial{x^2}}=
    \frac{\hbar^2}{2m}\frac{{\partial}}{\partial{x}}\left(\frac{{\partial} (x\psi)}{\partial{x}}\right)=
    \frac{\hbar^2}{2m}\frac{{\partial}}{\partial{x}}\left(\frac{{\partial} (x)}{\partial{x}}\psi+x\frac{{\partial} (\psi)}{\partial{x}}\right)=
    \frac{\hbar^2}{2m}\frac{{\partial}}{\partial{x}}\left(\psi+x\frac{{\partial} (\psi)}{\partial{x}}\right)[/tex]. So we ended up getting the expression in the parenthesis of desired answer. You see it, right?!

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