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Commutation (Ehrenfest?) relations

  1. Jun 21, 2005 #1
    I'm following a derivation (p85 of Symmetry Principles in Quantum Physics by Fonda & Ghirardi, for anyone who has it) in which the following assertion is made:

    "...we have
    [tex]\left[\mathcal{G}_p,\mathbf{r}_i\right] &=& \mathbf{v}_0t\mathcal{G}_p, [/tex]
    [tex]\left[\mathcal{G}_r,\mathbf{p}_i\right] &=& \mathbf{v}_0m_i\mathcal{G}_r,[/tex]
    that is,
    [tex]-i\hbar\frac{\partial\mathcal{G}_p}{\partial p_{ik}} = v_{0k}t\mathcal{G}_p,[/tex]
    [tex]i\hbar\frac{\partial\mathcal{G}_r}{\partial r_{ik}} = v_{0k}m_i\mathcal{G}_r.[/tex]"
    In the book's notation, [tex]\mathcal{G}_p[/tex] and [tex]\mathcal{G}_r[/tex] are unknown unitary operators which depend only on the momentum and position respectively; [tex]\mathbf{r}_i[/tex], [tex]\mathbf{p}_i[/tex] and [tex]m_i[/tex] are the position, momentum and mass of the [tex]i[/tex]th particle in the system; [tex]\mathbf{v}_0[/tex] is a constant velocity; and [tex]t[/tex] is the time.

    I follow the derivation up to this point, and continue to follow it afterwards, but do not understand how the second pair of equations follows from the first. It clearly seems to assert that
    [tex]-i\hbar\frac{\partial\mathcal{G}_p}{\partial \mathbf{p}_{i}} = \left[\mathcal{G}_p,\mathbf{r}_i\right], [/tex]
    which reminds me of the Ehrenfest relation, and I also note that the derivatives are with respect to the canonical conjugates of the quantities in the commutators, but I'm not sure if that's just a coincidence or how to show that last equality. I have a feeling that it may be something rather trivial that I've just overlooked, but I'm out of ideas right now... Can anybody explain this step, or point me to a theorem that may help?

  2. jcsd
  3. Jun 21, 2005 #2


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    For the life in me,i haven't sen these before.I mean the RHS in those is unknown to me.Anyway,since i don't have the book,can u give another source where i can look them up ?

    Go the the book's bibliography and see where this chapter was inspired from.

  4. Jun 21, 2005 #3
    This section is in Chapter 2, Geometrical Symmetries in Ordinary Quantum Mechanics. Unfortunately there are references for the chapter for sections 2.2 (Transformation Laws of Coordinates and Momenta in Nonrelativistic Classical Mechanics), 2.4 (Space Rotations) and 2.6 (Space Reflection) but the section I'm looking at here is 2.5 (Special Galilei Transformations). A similar statement is made in section 2.3 (Space Translations) when they say:

    From [tex]D_\ell^\dagger\mathbf{r}_i D_\ell=\mathbf{r}_i-\mathbf{\ell}[/tex], using the unitarity of [tex]D_\ell[/tex], we get

    [tex]\left[D_\ell\left(\mathbf{p}_1, \mathbf{p}_2,\ldots, \mathbf{p}_N\right), \mathbf{r}_i\right] = D_\ell\left( \mathbf{p}_1, \mathbf{p}_2,\ldots, \mathbf{p}_N\right)\mathbf{\ell},[/tex]

    which gives immediately the following system of equations:

    -i\hbar\frac{\partial D_\ell}{\partial p_{xi}} = D_\ell\ell_x,
    -i\hbar\frac{\partial D_\ell}{\partial p_{yi}} = D_\ell\ell_y,
    -i\hbar\frac{\partial D_\ell}{\partial p_{zi}} = D_\ell\ell_z.

    Edit:In this case, two observers are in reference frames differing only by a translation. [tex]D_\ell[/tex] is the unitary operator that relates the two observers and [tex]\ell[/tex] is the displacement vector between the two systems.

    But again, there are no references for that section of the text.

    The motivation of the section I'm looking at is to find the unitary operator [tex]\mathcal{G}[/tex] which allows us to perform Galilean transformations, i.e. if we have an operator [tex]\hat{A}[/tex] in one inertial frame and want to transform it (non-relativistically of course) to a frame moving at constant velocity [tex]\mathbf{v}_0[/tex] relative to the first (coincident at t=0), we calculate [tex]\hat{A}' = \mathcal{G}^\dagger\hat{A}\mathcal{G}[/tex].

    As far as I can tell, when the authors go from a commutator to a partial derivative the step is not related to the motivation of this particular section, although I haven't found a similar step in the first ~100 pages of the book other than the two places I've now mentioned.
    Last edited: Jun 21, 2005
  5. Jun 21, 2005 #4
    I think I have figured out what is happening... sort of. Much earlier in the text (p22) the authors state that

    [tex]\mathcal{U}^\dagger\left[\alpha_j, \alpha_k\right]\mathcal{U} = \mathcal{U}^\dagger i\hbar\left[\alpha_j, \alpha_k\right]_{\text{QPB}}\mathcal{U}[/tex]

    where QPB means Quantum Poisson Bracket, but the left-hand side is a commutator. Then for [tex]\mathcal{U}[/tex] unitary this gives us

    [tex]\left[\alpha_j, \alpha_k\right] = i\hbar\left[\alpha_j, \alpha_k\right]_{\text{QPB}}[/tex].

    Then in the specific case of my first post, we find e.g.

    [tex] \left[\mathcal{G}_p, \mathbf{r}_i\right] = i\hbar\left[\mathcal{G}_p, \mathbf{r}_i\right]_{\text{QPB}} = i\hbar\sum_{k=1}^N\left(\frac{\partial \mathcal{G}_p}{\partial x_{ik}} \frac{\partial \mathbf{r}_i}{\partial p_{ik}} - \frac{\partial \mathcal{G}_p}{\partial p_{ik}} \frac{\partial \mathbf{r}_i}{\partial r_{ik}}\right)[/tex]

    But by assumption [tex]\mathcal{G}_p[/tex] is independent of the [tex]r_{ik}[/tex], and

    [tex]\frac{\partial \mathbf{r}_i}{\partial r_{ik}} = \hat{e}_k,[/tex]

    the kth unit vector, so we're left with

    [tex] \left[\mathcal{G}_p, \mathbf{r}_i\right] = -i\hbar\sum_{k=1}^N\frac{\partial \mathcal{G}_p}{\partial p_{ik}}\hat{e}_k [/tex]

    which, taken component-wise, yields the original equation I was after. So this shows how the equations in my first post were derived, as long as the assumption that a commutator is equal to [tex]i\hbar[/tex] times the Poisson bracket holds. I know that to move from a classical Poisson bracket to a quantum one the factor of [tex]i\hbar[/tex] appears, but do you move from a 'classical' quantum commutator to the QPB in the same way?
    Last edited: Jun 21, 2005
  6. Jun 21, 2005 #5

    Kane O'Donnell

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    Seems like a fairly odd way of going about finding these translations. If I remember correctly, for the free-particle non-relativistic hamiltonian the generators of the translation are just the components of the momentum operator, you can show this in a fairly straightforward manner by using the part of Stone's theorem that asserts

    [tex] iT\psi = \lim_{s \to 0}\frac{1}{s}(U_{s}\psi - \psi) [/tex]​

    where T is the (self-adjoint) generator of the unitary operator U_s, and s is the parameter for the strongly-continuous one-parameter group we're talking about generating.
  7. Jun 21, 2005 #6

    Kane O'Donnell

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    Whoops, don't mind me, you're after a different kind of translation. :smile:

    Still, the above method does seem rather complex for what isn't a particularly difficult problem - if I remember correctly you can do it crudely by transforming the Schroedinger equation and then finding a phase change for the wavefunction which makes the equation return to the same form - this is perfectly ok since our probabilities are invariant under phase changes.
  8. Jun 23, 2005 #7

    George Jones

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    I think you're making this harder than it actually is. For simplicity, and because I know almost nothing about tex, consider one particle in one dimension.

    [G , p] psi = -i hbar G dpsi/dx + i hbar d/dx (G psi)

    = (i hbar dG/dx) psi.

    Now let there be many particles and dimensions. For the top equation, remember that in momentum space, r is a differentail operator.

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