# Ehrenfest theorem for kinetic energy

1. Feb 7, 2013

### cartatum

In classical mechanics, the time derivative of the kinetic energy of a particle is given by the particle velocity multiplied by the force. Specifically, in one dimension, this translates into:
(d/dt)(p^2/2m)= −p/m ∂V/∂x

I'm unsure how to prove this....

2. Feb 8, 2013

### Demystifier

Try with Heisenberg picture!

3. Feb 8, 2013

### vanhees71

It's of course independent of the picture. The covariant time derivative of operators, representing observables, is given by
$$\frac{\mathrm{D} \hat{A}}{\mathrm{D} t}=\mathring{\hat{A}}=\frac{1}{\mathrm{i} \hbar} [\hat{A},\hat{H}],$$
where I have assumed that there is no explicit time dependence.

For a Hamiltonian that is not explicitly time dependent you thus find
$$\mathring{\hat{H}}=\frac{1}{\mathrm{i} \hbar} [\hat{H},\hat{H}]=0.$$
If, further, you have the motion of a particle in an external (time-independent) potential, you have
$$\hat{H}=\frac{\hat{\vec{p}}^2}{2m} + V(\hat{\vec{x}}),$$
and from the Heisenberg algebra, you can easily deduce
$$\mathring{\hat{\vec{p}}}= \frac{1}{\mathrm{i} \hbar} \left [\hat{\vec{p}},V(\hat{\vec{x}}) \right]=-\vec{\nabla} V(\hat{\vec{x}}).$$
From this it follows
$$\frac{\mathrm{D}}{\mathrm{D} t} \frac{\hat{\vec{p}}^2}{2m} = \frac{1}{\mathrm{i} \hbar 2m} \left (\hat{\vec{p}} \cdot [\hat{\vec{p}},V(\hat{\vec{x}})]+[\vec{p},V(\hat{\vec{x}})] \cdot \hat{\vec{p}} \right ) = -\frac{1}{2m} \left (\hat{\vec{p}} \cdot \vec{\nabla} V(\hat{\vec{x}}) + \vec{\nabla} V(\hat{\vec{x}}) \cdot \hat{\vec{p}} \right).$$

Ehrenfests theorem tells you that for any state $\hat{R}$ in any picture of time evolution you have
$$\frac{\mathrm{d}}{\mathrm{d} t} \langle A \rangle=\frac{\mathrm{d}}{\mathrm{d} t} \mathrm{Tr} (\hat{R} \hat{A})=\mathrm{Tr} (\hat{R} \mathring{\hat{A}}).$$
Note that the Averaging goes over $\mathring{\hat{A}}$, and the equations of motion for the averages are in general not identical with the classical equations of motion for the average values. That's the case only for the linear oscillator or for the motion in a constant force field, because only then $\langle \vec{\nabla} V(\vec{x}) \rangle=\vec{\nabla} V(\langle \vec{x} \rangle)$.

Last edited: Feb 8, 2013
4. Feb 8, 2013

### andrien

some correction is needed in fourth line above.

5. Feb 8, 2013

### vanhees71

Argh! Thanks, I've done the corrections. I hope, it's right now.

6. Feb 8, 2013

### Demystifier

The final result is of course independent, but the proof is simpler in the Heisenberg picture.

In the Heisenberg picture this is correct. But in the Schrodinger picture, $$\hat{A}$$ wouldn't depend on time even implicitly, so the left-hand side would be zero.

7. Feb 8, 2013

### vanhees71

No! Note that I defined the covariant derivative of an observable operator as the commutator with the Hamiltonian. In general, it's not the time derivative of the operator in the sense of a differential quotient of an operator valued function of time. That's the case only in the Heisenberg picture. As you say, in the Schrödinger picture the "mathematical time derivative" would indeed be 0 (except for explicitly time dependent operators, but this I don't consider here, because this would make the whole issue even more complicated).