# The classical limit of quantum mechanics; Ehrenfest’s theorem

Required math: calculus

Required physics: 3-d Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 4.40.

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercise 13.1.5.

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We’ve seen the virial theorem in one dimension, which states:

$\displaystyle 2\langle T\rangle=\left\langle x\frac{dV}{dx}\right\rangle \ \ \ \ \ (1)$

where ${T}$ is the kinetic energy.

We can derive the 3-d version of the virial theorem using a similar method. From the formula for the rate of change of an observable, we have,

$\displaystyle \frac{d}{dt}\langle\mathbf{r}\cdot\mathbf{p}\rangle=\frac{i}{\hbar}\langle[\hat{H},\mathbf{r}\cdot\mathbf{p}]\rangle \ \ \ \ \ (2)$

assuming that the potential is time-independent. (This is what Shankar refers to as Ehrenfest’s theorem.) In three dimensions, we have

 $\displaystyle \mathbf{r}\cdot\mathbf{p}$ $\displaystyle =$ $\displaystyle -i\hbar x\frac{\partial}{\partial x}-i\hbar y\frac{\partial}{\partial y}-i\hbar z\frac{\partial}{\partial z}\ \ \ \ \ (3)$ $\displaystyle \hat{H}$ $\displaystyle =$ $\displaystyle T+V\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right)+V \ \ \ \ \ (5)$

Since each term in the commutator (except for the potential ${V}$) contains only one of the three spatial coordinates, any derivative term commutes with any other derivative term that contains a different variable. The remaining three non-zero commutators, one for each coordinate, can be calculated in the same way as in one dimension. We are therefore left with a simple generalization of the result for one dimension.

 $\displaystyle \frac{i}{\hbar}[\hat{H},\mathbf{r}\cdot\mathbf{p}]$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{m}\left(\frac{\partial^{2}}{\partial x^{2}}+\frac{\partial^{2}}{\partial y^{2}}+\frac{\partial^{2}}{\partial z^{2}}\right)-x\frac{\partial V}{\partial x}-y\frac{\partial V}{\partial y}-z\frac{\partial V}{\partial z}\ \ \ \ \ (6)$ $\displaystyle \frac{d}{dt}\langle\mathbf{r}\cdot\mathbf{p}\rangle$ $\displaystyle =$ $\displaystyle 2\langle T\rangle-\left\langle \mathbf{r}\cdot\nabla V\right\rangle \ \ \ \ \ (7)$

For stationary states the time derivative is zero, so

$\displaystyle 2\langle T\rangle=\left\langle \mathbf{r}\cdot\nabla V\right\rangle \ \ \ \ \ (8)$

For hydrogen,

$\displaystyle V=-\frac{e^{2}}{4\pi\epsilon_{0}}\frac{1}{r} \ \ \ \ \ (9)$

so since ${r=\sqrt{x^{2}+y^{2}+z^{2}}}$,

 $\displaystyle \frac{\partial V}{\partial x}$ $\displaystyle =$ $\displaystyle \frac{e^{2}}{4\pi\epsilon_{0}}\frac{x}{r^{3}}\ \ \ \ \ (10)$ $\displaystyle \frac{\partial V}{\partial y}$ $\displaystyle =$ $\displaystyle \frac{e^{2}}{4\pi\epsilon_{0}}\frac{y}{r^{3}}\ \ \ \ \ (11)$ $\displaystyle \frac{\partial V}{\partial z}$ $\displaystyle =$ $\displaystyle \frac{e^{2}}{4\pi\epsilon_{0}}\frac{z}{r^{3}}\ \ \ \ \ (12)$ $\displaystyle \mathbf{r}\cdot\nabla V$ $\displaystyle =$ $\displaystyle \frac{e^{2}}{4\pi\epsilon_{0}}\frac{x^{2}+y^{2}+z^{2}}{r^{3}}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{e^{2}}{4\pi\epsilon_{0}}\frac{1}{r}\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -V \ \ \ \ \ (15)$

Thus we have

$\displaystyle 2\langle T\rangle=-\left\langle V\right\rangle$

But we know that the total energy for the hydrogen atom in quantum state ${n}$ is ${E_{n}=\langle T\rangle+\langle V\rangle=\langle T\rangle-2\langle T\rangle=-\langle T\rangle}$ so we get ${\left\langle T\right\rangle =-E_{n}}$ and ${\left\langle V\right\rangle =2E_{n}}$.

For the 3-d harmonic oscillator

$\displaystyle V=\frac{1}{2}m\omega^{2}r^{2} \ \ \ \ \ (16)$

so

 $\displaystyle \nabla V$ $\displaystyle =$ $\displaystyle m\omega^{2}\mathbf{r}\ \ \ \ \ (17)$ $\displaystyle \mathbf{r}\cdot\nabla V$ $\displaystyle =$ $\displaystyle m\omega^{2}r^{2}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2V \ \ \ \ \ (19)$

The total energy in state ${n}$ is ${E_{n}=\langle T\rangle+\langle V\rangle=\frac{1}{2}(2\left\langle V\right\rangle )+\left\langle V\right\rangle =2\left\langle V\right\rangle }$ so ${\left\langle V\right\rangle =E_{n}/2=\left\langle T\right\rangle }$.