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Homework Help: Conditions for the X dot P expectation to be constant?

  1. Oct 4, 2015 #1
    1. The problem statement, all variables and given/known data

    Under what conditions is [itex] \left\langle{{\mathbf{x} \cdot \mathbf{p}}}\right\rangle [/itex] a constant.

    A proof of the quantum virial theorem starts with the computation of the commutator of [itex] \left[{\mathbf{x} \cdot \mathbf{p}},{H}\right] [/itex]. Using that one can show for Heisenberg picture operators [itex] \mathbf{x} [/itex] and [itex] \mathbf{p} [/itex], and expectations relative to stationary states (i.e. states that are not time dependent), and [itex] H = \mathbf{p}^2/2m + V(\mathbf{x}) [/itex], we have

    [tex]\frac{d}{dt} \left\langle{{\mathbf{x} \cdot \mathbf{p}}}\right\rangle=\left\langle{{\frac{\mathbf{p}}{m}}}\right\rangle + \left\langle{{ \mathbf{x} \cdot \boldsymbol{\nabla} V}}\right\rangle.[/tex]

    Getting that far is mostly just algebra. I'm asked when is the LHS zero. That's clearly true when the expectation is a constant, but it is not clear to me, for general time dependent Heisenberg picture operators, when that would be.

    2. Relevant equations

    1D SHO position and momentum operator representation in the Heisenberg picture are

    [tex]x = x(0) \cos(\omega t) + \frac{p(0)}{m \omega} \sin(\omega t)[/tex]
    [tex]p = p(0) \cos(\omega t) - m \omega x(0) \sin(\omega t)[/tex]

    3. The attempt at a solution

    I was hoping the rationale for this zero time derivative would be clear if I worked an example, so I calculated such an expectation for the 1D SHO Heisenberg operators. For that specific case, this expectation does in fact vanish. For one of the stationary states [itex] {\left\lvert n \right\rangle} [/itex] I calculate

    [tex]\begin{align*}\left\langle x p \right\rangle &= \left\langle n \right\rvert x p \left\lvert n \right\rangle \\ &= \cdots \\ &= \frac{i \hbar}{2} \left( \cos^2(\omega t) + \sin^2(\omega t) \right) - \left( n + 1/2 \right) \sin(\omega t) \cos(\omega t) \left( \frac{\hbar^2}{x_0^2 m \omega } - m \omega x_0^2 \right) \\ &= \frac{i \hbar}{2}.\end{align*}[/tex]

    So, for this particular Hamiltonian, it's possible to show that the expectation of [itex] \mathbf{x} \cdot \mathbf{p} [/itex] is constant, despite the time dependence of the operators themselves. Exactly what principles would justify extending this to the general case is not obvious to me.

    I could probably include a hand-waving argument, stating that this LHS is zero for stationary states, but this doesn't seem adequate to me.

    It's not hard to find other quantum virial theorem treatments:

    1. http://www.physicspages.com/2012/10/09/virial-theorem/
    2. http://www7b.biglobe.ne.jp/~kcy05t/viriproof.html
    3. https://www.univie.ac.at/physikwiki/images/a/a0/T2_Skript_final.pdf

    In the first, I don't follow the authors argument (using <p> and d<x>/dt).
    In the second the author just states that the LHS is zero for stationary states, without any elaboration, despite the fact that both [itex] \mathbf{x} [/itex] and [itex] \mathbf{p} [/itex] are time dependent in their Heisenberg representation.
    In the third, the author states that ``Finally we assume stationary states which satisfy [itex] 0 = \frac{d}{dt} \left\langle \mathbf{x} \cdot \mathbf{p} \right\rangle [/itex]''. He also states in a footnote, ``This condition can also be regarded as a form of Hamilton's principle, since the product of position and momentum has the dimension of an action, whose variation is required to vanish''. That footnote doesn't clarify things for me, since I don't see what the connection between the action principle and this expectation value is.

    Another way to potentially show this might be to use Hamilton's equations of motion. I note that for a stationary state [itex] \left\lvert \psi \right\rangle [/itex], we have

    \frac{d}{dt} \left\langle \mathbf{x} \cdot \mathbf{p} \right\rangle
    &= \frac{d}{dt} \left\langle \psi \right\rvert \mathbf{x} \cdot \mathbf{p} \left\lvert \psi \right\rangle \\
    &= \left\langle \psi \right\rvert \frac{d \mathbf{x}}{dt} \cdot \mathbf{p} + \mathbf{x} \cdot \frac{d\mathbf{p}}{dt} \left\lvert \psi \right\rangle \\
    &= \left\langle \psi \right\rvert \frac{\partial H}{\partial \mathbf{p}} \cdot \mathbf{p} - \mathbf{x} \cdot \frac{\partial H}{\partial \mathbf{x}} \left\lvert \psi \right\rangle

    I suspect that there might be a way to show that this is zero, at least for certain Hamiltonians, but am not sure what that would be. I am also guessing this might be related to the action minimization comments in one of the references above.
  2. jcsd
  3. Oct 4, 2015 #2
    Note that I've got the sign wrong on the gradient above, but that doesn't change my question, since I want to be able to justify setting the LHS to zero.
  4. Oct 4, 2015 #3


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    I wonder what the asker would consider an appropriate answer. After all, ##\frac{d}{dt}\langle\mathbf{X\cdot P}\rangle=0## is itself an equation in ##t## that specifies when it is. One can manipulate that to get different equations, some more explicit than others. Does the asker perhaps want a formula that gives t in terms of ##\psi##?

    I am profoundly ignorant of the virial theorem, so I just tried manipulating ##\langle\mathbf{X\cdot P}\rangle##, first expanding that as
    $$\left\langle\psi\left|\sum_{k=1}^3 X_kP_k\right|\psi\right\rangle$$

    After some messing about I ended up with the requirement being that

    $$\sum_{k=1}^3\int_{-\infty}^\infty x_k\frac{\partial}{\partial t}\left[\psi(\mathbf{x},t)^*\ \frac{d\partial}{\partial x_k}\psi(\mathbf{x},t)\right]dx_k=0$$

    where ##\psi(\mathbf{x},t)\equiv \langle\mathbf{x}\ |\ \psi(t)\rangle## is the representation of ##\psi(t)## in the position basis. But that didn't seem to be any more informative than what we started with, and I couldn't see any promising directions to head from there.
  5. Oct 4, 2015 #4
    The expansion you've done isn't quite valid in this case, because [itex]X_k[/itex] and [itex]P_k[/itex] operators are Heisenberg picture operators

    $$\begin{align*}X_{k,H}(t) &= U^\dagger(t) X_k U(t) \\ P_{k,H}(t) &= U^\dagger(t) P_k U(t).\end{align*}$$

    Also note that the state vectors in question (say, [itex]\left\lvert \psi \right\rangle[/itex] is not time dependent), since the time dependence is in the operators themselves.

    If the Hamiltonian isn't explicitly time dependent, that time evolution operator is [itex]U(t) = e^{-i H t/\hbar}[/itex]. If I try plugging that in I just get to the starting point of the virial theorem derivation (i.e. evaluate the commutator of [itex]\mathbf{x} \cdot \mathbf{p}[/itex] with the Hamiltonian)

    $$\begin{align*}\frac{d}{dt} \left\langle \psi \right\rvert X_{k,H} P_{k,H} \left\lvert \psi \right\rangle &=\frac{d}{dt} \left\langle \psi \right\rvert U^\dagger X_{k} P_{k} U \left\lvert \psi \right\rangle \\ &= \left\langle \psi \right\rvert \left( U^\dagger X_{k} P_{k} \frac{dU}{dt}+\frac{d U^\dagger}{dt} X_{k} P_{k} U \right) \left\lvert \psi \right\rangle \\ &= \left\langle \psi \right\rvert \left( -U^\dagger X_{k} P_{k} \frac{i H}{\hbar} U + U^\dagger \frac{i H}{\hbar} X_{k} P_{k} U \right) \left\lvert \psi \right\rangle \\ &=\frac{1}{i\hbar} \left\langle \psi \right\rvert U^\dagger \left( X_{k} P_{k} H- H X_{k} P_{k} \right) U \left\lvert \psi \right\rangle\\ &=\frac{1}{i\hbar} \left\langle \psi \right\rvert \left[ X_{k,H} P_{k,H} , H\right] \left\lvert \psi \right\rangle.\end{align*}$$

    This is just the Heisenberg picture time evolution equation for a Heisenberg picture operator [itex]A_H[/itex]

    $$\frac{d A_H}{dt} = \frac{1}{i \hbar} [ A_H, H ].$$

    Expanding those commutators will give you

    $$\frac{d}{dt} \left\langle \mathbf{x} \cdot \mathbf{p} \right\rangle = \left\langle \frac{\mathbf{p}^2}{m} \right\rangle - \left\langle \mathbf{x} \cdot \boldsymbol{\nabla} V \right\rangle,$$

    but that's doesn't help figuring out when the LHS is zero.

    Note that the virial theorem is just a statement that provided the conditions for the LHS to be zero are satisfied, one is left with a relation between twice the kinetic energy and the potential:

    $$\left\langle \frac{\mathbf{p}^2}{m} \right\rangle = \left\langle \mathbf{x} \cdot \boldsymbol{\nabla} V \right\rangle.$$
  6. Oct 12, 2015 #5
    Working some subsequent parts of the assignment provide a clue. We are asked to compute the following for the 3D SHO Hamiltonian:

    $$\left\langle \frac{\mathbf{p}^2}{m} \right\rangle - \left\langle \mathbf{x} \cdot \boldsymbol{\nabla} V \right\rangle$$

    (a) for the eigenstates ##\left\langle n_x, n_y, n_z \right\vert##
    (b) for the state ##\left\langle 0,0,0 \right\rvert + \left\langle 0,0,2 \right\vert##

    For (a) the results of each of the expectations are time independent and identical, whereas for (b) they are time dependent and the signs of the time dependent portions differ, resulting in a non-zero value for the difference. It seemed like the answer that was sought was the LHS time derivative of ##\left\langle \mathbf{x} \cdot \boldsymbol{p} \right\rangle## expectation is zero when the expectation is taken with respect to eigenstates.

    Is there a way to show, in general, that ##\left\langle A \right\rangle## is not time dependent when that expectation is with respect to eigenstates of that operator ##A##?
  7. Oct 12, 2015 #6


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    Isn't the way you come to eigenstates that the TDSE is separable and ##<A> = <\Psi|A|\Psi> = a <\Psi|\Psi> ## hence time independent ?
  8. Oct 12, 2015 #7
    But in this case the eigenstates were for the Hamiltonian, not for the operator itself?
  9. Oct 12, 2015 #8


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    o:) Yes. (Sorry, I'm living in the Schroedinger picture far too much).
  10. Mar 14, 2018 #9
    I know this is an old post but I got stuck on this for a little while too. I don't know the general answer but to prove that it is true for stationary states is fairly straightforward. Sticking in the Schrodinger picture:
    $$\langle \mathbf{x} \cdot \mathbf{p} \rangle = \langle \psi(x,t) | \mathbf{x} \cdot \mathbf{p} | \psi(x,t) \rangle = \langle \psi(x,0) | e^{i\hat{H}t} (\mathbf{x} \cdot \mathbf{p}) e^{-i\hat{H}t} | \psi(x,0) \rangle.$$.
    Stationary states are eigenvectors of the Hamiltonian and so, for such states, this equals
    $$\langle \psi(x,0) | e^{iEt} (\mathbf{x} \cdot \mathbf{p}) e^{-iEt} | \psi(x,0) \rangle = \langle \psi(x,0) | \mathbf{x} \cdot \mathbf{p} | \psi(x,0) \rangle,$$
    because the exponential term becomes a number that is not affected by the operators and so it cancels. The term that we end up with is manifestly constant. Note that this result would be true for any function of position and momentum operators, provided we are dealing with stationary states.
  11. Mar 14, 2018 #10


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    You will get the same result for any time-independent observable ##\hat{A}##. If the state is stationary and the observable time-independent, then nothing changes with time!

    For a potential of the form ##V(x) = \alpha x^n##, one has
    [\hat{H}, \hat{x} \hat{p}] = i \hbar \left( \alpha n x^n - \frac{\hat{p}^2}{m} \right)
    so a potential for which ##\langle \hat{x} \hat{p} \rangle = 0## will not be a trivial potential, if it exists at all.
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