# Ehrenfest theorem for kinetic energy

• cartatum
In summary, the time derivative of the kinetic energy of a particle in classical mechanics is given by the particle velocity multiplied by the force. This is independent of the picture used, but the proof is simpler in the Heisenberg picture. Using the Heisenberg algebra, we can deduce the covariant time derivative of the momentum and the Hamiltonian, which leads to the result that the time derivative of the kinetic energy is equal to the negative gradient of the potential energy. This result is also supported by Ehrenfest's theorem, which states that the equations of motion for the average values of observables are not always identical to the classical equations of motion, except for specific cases such as the linear oscillator or motion in a constant force field.
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...

Try with Heisenberg picture!

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:
some correction is needed in fourth line above.

andrien said:
some correction is needed in fourth line above.

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

vanhees71 said:
It's of course independent of the picture.
The final result is of course independent, but the proof is simpler in the Heisenberg picture.

vanhees71 said:
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.
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.

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).

## 1. What is the Ehrenfest theorem for kinetic energy?

The Ehrenfest theorem for kinetic energy is a fundamental principle in quantum mechanics that states that the time derivative of the expectation value of kinetic energy is equal to the expectation value of the force multiplied by the expectation value of the velocity.

## 2. Who developed the Ehrenfest theorem for kinetic energy?

The Ehrenfest theorem for kinetic energy was developed by Paul Ehrenfest, a Dutch physicist, in 1907.

## 3. How is the Ehrenfest theorem for kinetic energy used in quantum mechanics?

The Ehrenfest theorem for kinetic energy is used to describe the behavior of quantum mechanical systems and to calculate the time evolution of their physical properties.

## 4. Can the Ehrenfest theorem for kinetic energy be applied to classical systems?

Yes, the Ehrenfest theorem for kinetic energy has a classical counterpart and can be applied to classical systems as well as quantum systems.

## 5. What is the significance of the Ehrenfest theorem for kinetic energy in quantum mechanics?

The Ehrenfest theorem for kinetic energy is important in quantum mechanics because it relates the time evolution of the expectation values of kinetic energy, force, and velocity, which are essential physical quantities in understanding the behavior of quantum systems.

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