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Potential commutes with position?

  1. May 16, 2008 #1
    Why does the potential energy (operator?) commute with the position operator? Is it because we deal only with real potentials?
     
  2. jcsd
  3. May 16, 2008 #2

    Dr Transport

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    The potential operator commutes with the position operator when the potential operator is a function of position only, it will not commute if it is a function of the momenta as is seen in nuclear physics.
     
  4. May 16, 2008 #3

    dx

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    You can think of the potential operator and the position operator as measurements. If the potential doesnt depend on the particles motion, then it wouldnt matter in which order you measure them. But if the potential depends on the momentum of the particle, then they dont commute because the measurement of the particles position affects its momentum.
     
  5. May 16, 2008 #4
    They commute because they are both multiplication operators (assuming potential depends only on position), and multiplication operators always commute (unless you have some more exotic algebra). It does not matter if the multiplication operators are real or complex.

    For given measurable function [itex]V:\mathbb{R}^3\to\mathbb{R}[/itex] the potential operator is

    [tex]
    \hat{V}:D(\hat{V})\to L^2(\mathbb{R}^3),\quad (\hat{V}\psi)(x)=V(x)\psi(x)\quad\forall x\in\mathbb{R}^3
    [/tex]

    and the position operator (or a component of it, k=1,2,3), is

    [tex]
    \hat{x}_k:D(\hat{x}_k)\to L^2(\mathbb{R}^3),\quad
    (\hat{x}_k\psi)(x)=x_k\psi(x)\quad\forall x\in\mathbb{R}^3
    [/tex]

    Here the domains [itex]D(\hat{V})[/itex] and [itex]D(\hat{x}_k)[/itex] are some subsets of [itex]L^2(\mathbb{R}^3)[/itex]. For all [itex]x\in\mathbb{R}^3[/itex]

    [tex]
    (\hat{V}\hat{x}_k\psi)(x)\;=\;V(x)(\hat{x}_k\psi)(x)\;=\;V(x)x_k\psi(x)\;=\;x_kV(x)\psi(x)\; =\;x_k(\hat{V}\psi)(x)\;=\;(\hat{x}_k\hat{V}\psi)(x).
    [/tex]

    The key step is V(x)x=xV(x), which is justified because x and V(x) are merely numbers. Therefore

    [tex]
    \hat{V}\hat{x}_k\psi = \hat{x}_k\hat{V}\psi,
    [/tex]

    whenever [itex]\psi[/itex] is in domain of these composition maps.
     
    Last edited: May 16, 2008
  6. May 17, 2008 #5

    samalkhaiat

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    Last edited: May 17, 2008
  7. May 17, 2008 #6

    reilly

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    The momentum and position representations are connected by a unitary transformation, which always preserves canonical or any commutation relations. So, if V is a function of x, then [x,V]=0 holds necessarily in the momentum representation, or any representation. Hint; write out V in terms of the momentum rep of x.
    Regards,
    Reilly Atkinson
     
  8. May 17, 2008 #7
    (Let us remain extremely civilized and scientific. I'm concerned with the content)

    This was a mistake. The operator, which is multiplication operator [tex]\hat{V}[/tex] in the spatial space, becomes

    [tex]
    \hat{\psi}\mapsto \mathcal{F}\hat{V}\mathcal{F}^{-1}\hat{\psi},
    [/tex]

    [tex]
    (\mathcal{F}\hat{V}\mathcal{F}^{-1}\hat{\psi})(p) = \frac{1}{2\pi} \int dx\; d\bar{p}\; e^{ix(p-\bar{p})} V(x) \hat{\psi}(\bar{p})
    [/tex]

    in the momentum space. Usually you cannot write this as an integral operator with an ordinary function as the kernel. For example, if we have [itex]V(x)=x^2[/itex], and insist on writing the potential operator as an integral operator, we need derivatives of the delta function to the kernel.

    [tex]
    \frac{1}{2\pi} \int dx\; d\bar{p}\; e^{ix(p-\bar{p})} x^2 \hat{\psi}(\bar{p}) = \cdots =
    \int d\bar{p}\;\big(-\partial_p^2 \delta(p-\bar{p})\big) \hat{\psi}(\bar{p})
    [/tex]
     
  9. May 18, 2008 #8
    It could I complained about a small thing. If we accept

    [tex]
    \frac{1}{2\pi} \int dx\; e^{-ix(p-\bar{p})} x^2 = -\partial_p^2\delta(p-\bar{p}),
    [/tex]

    then this

    is fine. However, here

    you seem to assume that V was a multiplication operator in the momentum space too. If you calculate the commutator with the integral operator representation of the potential, the commutator vanishes again.
     
  10. May 18, 2008 #9

    reilly

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    Operators in Hilbert Space can be multiplied by other operators, and can multiply states, as in O|p>.

    !. As is discussed in almost all QM texts, a commutator is an operator, and any operator is independent of basis. So, if an operator is zero in any representation, it is zero in all representations.

    2. Define W = xV(x) = V(x)x. W commutes with itself. Show this is so in your approach, and you will find your error.(This is most easily done with Dirac notation.)
    Regards,
    Reilly Atkinson
     
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