Ehrenfest theorem allows us to see how physical quantities evolve through time in terms of other physical quantities. In quantum mechanics, physical quantities, like momentum or position, are represented by operators. To get the average value of a physical quantity of a system, we act the operator on a wavefunction. The wavefunction represents the physical system’s probabilistic behavior. The average of a physical quantity is called an “expectation value”.

The expectation value of an operator \mathscr{O} is:
\left< \mathscr{O} \right> = \left< \Psi \left| \mathscr{O} \right| \Psi \right> = \int dx \Psi^{\dagger} \mathscr{O} \Psi
where \Psi is a wavefunction \Psi(x).

This definition allows us to take a time derivative of the expectation value.
\frac{d}{dt}\left< \mathscr{O} \right> = \frac{d}{dt} \left< \Psi \left| \mathscr{O} \right| \Psi \right> = \frac{d}{dt} \int dx \Psi^{\dagger} \mathscr{O} \Psi
The total derivative moves inside the integral as a partial derivative, where we use the product rule to differentiate:
= \int dx (\frac{\partial}{\partial t} \Psi^{\dagger} \mathscr{O} \Psi + \Psi^{\dagger} \frac{\partial}{\partial t} \mathscr{O} \Psi + \Psi^{\dagger} \mathscr{O} \frac{\partial}{\partial t} \Psi)

A time derivative acting on a wave function is equivalent to a Hamiltonian operator acting on a wave function (with a factor of \frac{1}{i \hbar}):
\frac{\partial}{\partial t} \Psi = \frac{1}{i \hbar} H \Psi
Take the complex conjugate:
\frac{\partial}{\partial t} \Psi^{\dagger} = \frac{-1}{i \hbar} \Psi^{\dagger} H

We can then swap the time derivatives for Hamiltonians:
\frac{d}{dt}\left< \mathscr{O} \right> = \int dx ( \frac{-1}{i \hbar} \Psi^{\dagger} H \mathscr{O} \Psi + \Psi^{\dagger} \frac{\partial}{\partial t} \mathscr{O} \Psi + \frac{1}{i \hbar} \Psi^{\dagger} \mathscr{O} H \Psi)
= \frac{-1}{i \hbar} \left< \Psi \left| H \mathscr{O} \right| \Psi \right> + \left< \Psi \left| \frac{\partial}{\partial t} \mathscr{O} \right| \Psi \right> + \frac{1}{i \hbar} \left< \Psi \left| \mathscr{O} H \right| \Psi \right>
The first and last terms can be combined using a commutator:
= \left< \Psi \left| \frac{\partial}{\partial t} \mathscr{O} \right| \Psi \right> + \frac{1}{i \hbar} \left< \Psi \left| [\mathscr{O},H] \right| \Psi \right>

So then we have Ehrenfest Theorem, relating the time derivative of an expectation value to the expectation value of a time derivative:
\frac{d}{dt}\left< \mathscr{O} \right> = \left< \frac{\partial}{\partial t} \mathscr{O} \right> + \frac{1}{i\hbar}\left< [\mathscr{O},H] \right>

The general form of the Hamiltonian H, has a momentum-dependent kinetic energy term, and a position-dependent potential energy term V(x).
H = \frac{p^2}{2m} + V(x)

As an example, let’s see how changes in momentum over time can be expressed.
\frac{d}{dt}\left< p \right> = \left< \frac{\partial}{\partial t} p \right> + \frac{1}{i\hbar}\left< [p,H] \right>
Momentum doesn’t have an explicit time-dependence, so the first term is zero. Further, operators commute with themselves: [p,p^2] = 0, so the Hamiltonian reduces to just the potential energy term. So we’re left with:
\frac{d}{dt}\left< p \right> = \frac{1}{i\hbar}\left< [p,V(x)] \right>

To see what this remaining commutator of operators reduces to, we will have to use a little calculus. Momentum expressed in terms of position is essentially the derivative operator:
p = -i\hbar \frac{d}{dx}. Keep in mind that potential energy is a function of position which is why momentum does not commute with it.
Since we are dealing with operators, we need them to act on something: we’ll use a dummy wavefunction \Phi(x). Then just remember the product rule for derivatives:
[p,V]\Phi = -i\hbar[\frac{d}{dx},V]\Phi = -i\hbar(\frac{d}{dx}V\Phi - V\frac{d}{dx}\Phi) = -i\hbar(\frac{d}{dx}V)\Phi
So then we can see that the change in the expectation value of momentum with respect to time is:
\frac{d}{dt}\left< p \right> =-\left< \frac{d}{dx}V(x) \right>
On the left you have impulse over change in time, and on the right you have change in potential energy over change is position. Both are ways of measuring force in classical physics. In fact, this Newton’s second law!
\frac{d}{dt}\left< p \right> = \left< F \right>

If we do the same for change in expectation value of position with respect to time, we get an equation for velocity:
\frac{d}{dt}\left< x \right> = \left< \frac{\partial}{\partial t} x \right> + \frac{1}{i\hbar}\left< [x,H] \right>
Again, position has no explicit time-dependence, so the first term is zero. However, now position commutes with potential energy (since it’s just a function of position), and position does not commute with kinetic energy (momentum). So we’re left with:
\frac{d}{dt}\left< x \right> = \frac{1}{i\hbar} \frac{1}{2m}\left< [x,p^2] \right>
If we go through the dummy wavefunction process we’ll arrive at:
\frac{d}{dt}\left< x \right> = \frac{\left< p \right>}{m}

What if we try a much more complication operator, like the translation operator? I covered how this operator works in a previous blog entry. A translation operator shifts a wavefunction’s position by some distance a:
T(a)\Psi(x) = \Psi(x+a)
What does the change in this expectation value of this operator look like?
\frac{d}{dt}\left< T(a) \right> = \left< \frac{\partial}{\partial t} T(a) \right> + \frac{1}{i\hbar}\left< [T(a),H] \right>
There’s no explicit time dependence here, so the first term is zero. The commutator term is all that is left. We must now consider the explicit form of T(a).
T(a) = e^{a \frac{d}{dx}}
It only contains x derivative operators (momentum operators) and so it commutes with the kinetic energy term in the Hamiltonian, but not the potential energy term.
\frac{d}{dt}\left< T(a) \right> = \frac{1}{i\hbar}\left< [e^{a \frac{d}{dx}},V(x)] \right>
How do we evaluate something like this with a derivative operator in the exponent? This is what Taylor expansions are for!
e^{a \frac{d}{dx}} = 1+ \sum\limits_{n=1}^{\infty} \frac{a^n}{n!} (\frac{d}{dx})^n
After using a dummy wavefunction and Pascal’s triangle a bit, you get:
\frac{d}{dt}\left< T(a) \right> = \frac{1}{i\hbar}\sum\limits_{n=1}^{\infty} \frac{a^n}{n!} \sum\limits_{m=0}^{n-1}  \frac{n!}{m!(n-m)!} \left< ((\frac{d}{dx})^{n-m}V) (\frac{d}{dx})^m \right>
which can be expressed as a sum of products of force derivatives and momentum.
-\sum\limits_{n=1}^{\infty} \sum\limits_{m=0}^{n-1}  \frac{1}{(i\hbar)^{m+1}}\frac{a^n}{m!(n-m)!} \left< ((\frac{d}{dx})^{n-m-1}F) p^m  \right>
This tells us the change in a translation operator with respect to time can be quantified with this particular series of force and momentum products. This result looks messy, but it seems intuitive that force would be involved here.


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