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Why ignore the potential term in the quantum Hamiltonian?

  1. Oct 5, 2011 #1


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    I am reading about the recovery of some classical rules from quantum mechanics.

    My text (Shankar) considers a Hamiltonian operator in a one-dimensional space
    H = P^2 / 2m + V(X)
    where P and X are the momentum and position operators respectively.

    It then asserts that [X,H] = [X,P^2/2m]

    That is, it has discarded the potential term of the Hamiltonian without comment or explanation.

    How is that justified? I would have thought that if, as indicated, V is a function of X, the hamiltonian operator should be expressed in terms of that function.
    For example, if V is a gravitational potential V(X) = -k/X, I would expect the above commutator to be

    [X,H] = [X,P^2/2m-k/X] = [X,P] - k[X,1/X]

    Why does Shankar discard the second term?
    If one didn't discard it, what would [X,1/X] mean? Is there any way to handle the reciprocal of an operator?

    Thanks for any help.
  2. jcsd
  3. Oct 5, 2011 #2


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    The commutator of an operator with any function of itself is 0. That's why it's discarded. [x,f(x)]=0 for any f.
  4. Oct 5, 2011 #3


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    You can see that by letting act V(x) and x act on a test function:

    [x, V(x)] f(x) = xVf - Vxf = 0

    (simply b/c multiplication of functions is commutative)

    This does no longer hold for p wich acts as a derivative; using the stanbdard rules for derivatives w.r.t. x one finds

    [p, V(x)] f(x) = pVf - Vpf = (pV)f + V(pf) - V (pf) ~ V'f + Vf' - Vf' = V'f
  5. Oct 5, 2011 #4
    Think of it in terms of linear algebra. Parameter x -> operator X; parameter 1/x -> operator X-1. Then you have XX-1=X-1X=I, (the identity operator), because inverses commute (otherwise you wouldn't get I regardless).

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