- #1
Peeter
- 305
- 3
Homework Statement
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.
Homework 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]
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:
- http://www.physicspages.com/2012/10/09/virial-theorem/
- http://www7b.biglobe.ne.jp/~kcy05t/viriproof.html
- 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
[tex]
\begin{align*}
\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
\end{align*}
[/tex]
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.