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Homework Help: Help with commutator question please!

  1. Dec 15, 2009 #1
    consider a general one dimensional potential v(x) drive an expression for the commutator [H,P] where h is hamiltonian operator and momentum operator. i keep getting zero and i dont think i should. since next part of hw question sais what condition must v(x) satisfy so that momentum will be constant of the motion recall that this is true if the momentum communtes with the hamiltonian. so i dont think it should zero. any help would be great trtying to do all hw problems to prepare for final!
  2. jcsd
  3. Dec 15, 2009 #2


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    [tex] [P^{2}/2m + V(x) , P] = [ V , P] = -i \frac{dV}{dx}[/tex]
  4. Dec 15, 2009 #3
    thats how i started but didnt arrive at that can u show a little bit more detail if u dont mind.
    thanks alot though
  5. Dec 15, 2009 #4
    Consider the action of the commutator on a wavefunction

    [H, P] psi(x)

    (so, we work in the position representation). Write out the commutator explicitely:

    H P psi - P H psi
  6. Dec 15, 2009 #5
    yes thats what im doing but im still getting zero
  7. Dec 15, 2009 #6
    Expand the potential as a Taylor series and use the commutation relation [tex][x,p]=i \hbar[/tex].
  8. Dec 15, 2009 #7


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    [tex][P^{2}/2m, P] + [V , P] = [V,P][/tex]

    because [P,P] = 0.


    [H,P] \phi(x) = [V(x) , id/dx] \phi(x) = iV \frac{d \phi}{dx} -i\frac{d}{dx} \{V(x)\phi(x)\} = \{-i dV/dx\} \phi(x)
  9. Dec 15, 2009 #8
    um is there an easier way? like the first respnse seems right but idk how to get there its an elementary problem that i have im sure
  10. Dec 15, 2009 #9
    samal where is the h bar term? is that a property of commutator i have never seen that before
  11. Dec 15, 2009 #10
    i think the problem is i dont know what to do when we have an additon in the commutator i dont know how to treat it
  12. Dec 15, 2009 #11
    The commutator is linear [A+B,C] = [A,C] + [B,C]. Just write it out and you'll see that.
  13. Dec 15, 2009 #12


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    I always set my h bar equal to 1. you can put it back in the equation. The first point you need to know is

    [A + B , C] = [A,C] + [B,C]

    the second point, an (differential) operator in the commutator acts on all the stuff on the right, so

    [tex][A , i\hbar d/dx] f(x) = A(i\hbar df(x)/dx) - i\hbar (d/dx)(Af(x))[/tex]
  14. Dec 15, 2009 #13
    oh got it thanks alot guys!
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