Expectation value of uncertainty

[capandbells]

Homework Statement

Show that, if [H,A] = 0 and dA/dt = 0, then <&Delta;A> is constant in time.

Homework Equations

d<A>/dt = <i/ℏ[H,A] + dA/dt>

The Attempt at a Solution

I am trying to use the above equation to show that d<&Delta;A>/dt is 0, and I can get to d&Delta;A/dt = 0, but I can't figure out how to compute [H,&Delta;A]. The only thing I can think of is that since &Delta; A is just a function of A and A commutes with H, then &Delta; A also commutes with H, but I can't find a theorem that says that.

 

[dextercioby]

What's the definition of $\delta A$ ? Then differentiate it wrt time and use Heisenberg's equation of motion.

 

[capandbells]

Sorry, I don't know Heisenberg's equation of motion. We haven't gone over it in class and it doesn't show up in my book until much later.

 

[dextercioby]

What is the name of the equation you placed under <Relevant equations> ?

 

[paranormal]

I would first review the commutation relation [H,A] = 0 and understand its implications. This relation means that the operators H and A commute with each other, which implies that they share a set of common eigenstates. This also means that the values of H and A can be simultaneously measured with arbitrary precision, as they do not affect each other's measurements.

Next, I would consider the expectation value of the uncertainty, <&Delta;A>, which is a measure of the spread or uncertainty in the measurement of A. If the commutation relation [H,A] = 0 holds, then the uncertainty in the measurement of A should remain constant over time. This is because the commutation relation implies that the operator A does not change in time, as given by dA/dt = 0. Therefore, the expectation value of the uncertainty, <&Delta;A>, should also remain constant over time.

To prove this mathematically, we can use the equation d<&Delta;A>/dt = <i/ℏ[H,A] + dA/dt> and plug in the values of [H,A] = 0 and dA/dt = 0. This yields d<&Delta;A>/dt = 0, which shows that the uncertainty in the measurement of A is indeed constant over time.

In conclusion, the commutation relation [H,A] = 0 and the fact that dA/dt = 0 imply that <&Delta;A> is constant in time. This is a significant result in quantum mechanics, as it shows that certain observables can be measured with a constant level of precision over time.