1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
    [tex]=x\left(-\frac{\hbar^2}{2m}\frac{\delta^2}{\delta{x^2}}+V\right)\Psi-\left(-\frac{\hbar^2}{2m}\frac{\delta^2}{\delta{x^2}}+V\right)x\psi[/tex]

    [tex]x\left(-\frac{\hbar^2}{2m}\frac{{\delta^2}\psi}{\delta{x^2}}+V\psi\right)+\frac{\hbar^2}{2m}\frac{{\delta^2}(x\psi)}{\delta{x^2}}-Vx\psi[/tex]

    [tex]x\left(-\frac{\hbar^2}{2m}\frac{{\delta^2}}{\delta{x^2}}\right)+\frac{\hbar^2}{2m}\frac{\delta}{\delta{x}}\left(\psi+x\frac{\delta\psi}{\delta{x}}\right)[/tex]

    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

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    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.
    AB
     
  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?!

    AB
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook