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Background + Question

In classical mechanics we all know:

$$ \dot U =\frac{dU}{dt} = \sum_i \frac{\partial U}{\partial x_i} \frac{dx_i}{dt} = \sum_i v_i \frac{\partial U}{\partial x_i} = v.\nabla U $$

Hence, $$ \dot U - v. \nabla U = 0 $$

But when would something similar hold in quantum mechanics?

Assuming $U$ is a function of $x$, i.e they both commute. And simplifying to a $1$-d case:

$$ \left\langle \dot U \right\rangle - \left\langle \frac{ \frac{\partial U}{\partial x}}{m} \right\rangle \left\langle p \right\rangle = 0 $$

(Note: In the Heisenberg picture the velocity operator is merely the momentum operator divided by mass)

My Attempt

Using the Heisenberg picture:

$$\implies \left\langle \,[ \frac{\frac{\hat p^2}{2m}, \hat U \,]}{-i \hbar} \right\rangle - \left\langle \frac{ U_x}{m} \right\rangle \left\langle \hat p \right\rangle = 0 $$

$$\implies \left\langle \,[ \frac{ \hat p^2, \hat U \,]}{-i \hbar} \right\rangle - 2 \left\langle U_x \right\rangle \left\langle \hat p \right\rangle = 0 $$

$$\implies \left\langle \hat p^2 \hat U - \hat U \hat p^2 \right\rangle + 2 i \hbar \left\langle U_x \right\rangle \left\langle p \right\rangle = 0 $$

Going to the explicit integrals:

$$\implies -\hbar^2 \int_{-\infty}^{\infty} \bar{\psi} ( \frac{\partial^2 }{\partial x^2} (\hat U \psi) - \hat U \frac{\partial^2 \psi }{\partial x^2} ) dx + 2 i \hbar \left\langle U_x \right\rangle \left\langle p \right\rangle = 0 $$

$$\implies -\hbar^2 \int_{-\infty}^{\infty} \bar{\psi} ( \frac{\partial^2 }{\partial x^2} (\hat U \psi) - \hat U \frac{\partial^2 \psi }{\partial x^2} )dx + 2 i \hbar \left\langle U_x \right\rangle \int_{-\infty}^{\infty} \bar{\psi} (-i \hbar)\frac{\partial }{\partial x} \psi = 0 $$

$$ \implies -\hbar^2 \int_{-\infty}^{\infty} \bar{\psi} ( \frac{\partial^2 }{\partial x^2} (\hat U \psi) - \hat U \frac{\partial^2 \psi }{\partial x^2} )dx + 2 \hbar^2 \left\langle U_x \right\rangle \int_{-\infty}^{\infty} \bar{\psi} \frac{\partial }{\partial x} \psi = 0 $$

$$ \implies \int_{-\infty}^{\infty} \bar{\psi} ( \frac{\partial^2 }{\partial x^2} (\hat U \psi) - \hat U \frac{\partial^2 \psi }{\partial x^2} )dx - 2 \left\langle U_x \right\rangle \int_{-\infty}^{\infty} \bar{\psi} \frac{\partial }{\partial x} \psi = 0 $$

$$ \implies \int_{-\infty}^{\infty} \bar{\psi}( U_{xx} + 2(U_{x} - \left\langle U_x \right\rangle ) \frac{\partial }{\partial x })\psi dx = 0 $$

And now I'm stuck ...

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You shouldn't expect a theorem like that to hold. Consider a particle in a potential $\frac{1}{2}kx^2$. You could imagine $|\psi(x)|^2$ as a symmetric function, with one wave packet moving fast and to the left, and one moving fast and to the right. Then $\frac{d}{dt}\langle U\rangle$ should be large and positive, while $\langle U_x\rangle=\langle k x\rangle=0$ (the expectation value of $x$ is zero because the particle is moving symmetrically outwards), and $\langle p\rangle=0$ (one part of the wavepacket is moving to the left and fast, and one to the right.)

In fact, you can use Ehrenfest's theorem to get the time rate of change of $\langle U\rangle$:

\begin{align*} \frac{d}{dt} \langle U\rangle&=\frac{1}{i\hbar}\langle[U,H]\rangle\\ &=\frac{1}{i\hbar}\langle U \frac{p^2}{2m}-\frac{p^2}{2m} U\rangle\\ &=\frac{-\hbar^2}{i\hbar 2 m}\langle U\partial_x^2-\partial_x^2U\rangle\\ &=\frac{i\hbar}{2 m}\langle U\partial_x^2-\partial_x(U_x+U\partial_x)\rangle\\ &=\frac{i\hbar}{2 m}\langle U\partial_x^2-U_{xx}-U_x\partial_x-U_x\partial_x-U\partial_x^2\rangle\\ &=\frac{i\hbar}{2 m}\langle -U_{xx}-2 U_x \partial_x\rangle\\ &=\frac{i\hbar}{2 m}\langle -U_{xx}-\frac{2}{-i \hbar} U_x p\rangle\\ &=-\frac{i\hbar}{2 m}\langle U_{xx}\rangle+\frac{1}{m}\langle U_x p\rangle \end{align*}

As $\hbar\to 0$, the first term disappears while the second term (which is similar to the one you were searching for) stays.

In general, you can't always simplify things to nice expressions in terms of $\langle x\rangle$ and $\langle p\rangle$. There will be cross-terms!

The lesson is that Ehrenfest's theorem hints towards classical behavior at the macroscopic scale, but doesn't prove it. A better hint is the path integral - it sounds scary but is really just the identity $e^{-i H t}=e^{-i H \delta t}e^{-i H \delta t}\cdots e^{-i H \delta t}$ - which makes the classical mechanics principle of stationary action obvious. For even better "proof" of classical behavior for large numbers of particles, you need the phenomenon of decoherence. It would be nice if everything was solved by taking expectation values, but that isn't the case.

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  • $\begingroup$ Shouldn't there be a "minus" sign in your third step? $\endgroup$ – drewdles Sep 7 '16 at 10:21
  • $\begingroup$ @AnantSaxena You're totally right! Sorry, I was answering this late last night. Yeah, classically, $\frac{d}{dt}U(x)=\frac{dU}{dx}\frac{dx}{dt}=U_x v=U_x p/m$. I will correct the post. $\endgroup$ – user12029 Sep 8 '16 at 0:22

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