# Time Evolution of Operators

• bananabandana
In summary, the given conversation discusses the definition of a general operator and its corresponding energy eigenstates, as well as the derivation of equations involving the operator and the Hamiltonian. However, there are some discrepancies in the equations presented, particularly in the last equation which seems to be a result of Ehrenfest's theorem but does not fully apply in this context. More clarification is needed to resolve these issues.
bananabandana

## Homework Statement

[/B]

For a general operator ## \hat{O}##, let ##\hat{O}_{mn}(t)## be defined as:
$$\hat{O_{mn}}(t) = \int u^{*}_{m}(x,t) \hat{O} u_{n}(x,t)$$
and
$$\hat{O_{mn}} = \int u^{*}_{m}(x) \hat{O} u_{n}(x)$$
##u_{m}## and ##u_{n}## are energy eigenstates with corresponding energy ##E_{n}## and ##E_{m}##.
From this it can be shown that: (sticking in ## \hat{O} = [\hat{A},\hat{H}] ##
$$d\frac{[\hat{A},\hat{H}]}{dt} =(E_{n}-E_{m}) exp\bigg[-i\frac{(E_{n}-E_{m})t}{\hbar}\bigg] \bigg( -\frac{i \hat{A_{mn}} (E_{n}-E_{m})}{\hbar} + \frac{d \hat{A_{mn}}}{dt}\bigg)$$
But simultaneously, it is also possible to prove that: (as we'd expect from Ehrenfest anyhow)
$$\frac{d\hat{A_{mn}}}{dt} = -\frac{i}{\hbar} [\hat{A_{mn}},\hat{H}]$$
Where ## \hat{H}## is the Hamiltonian, and ##\hat{A}## some general Hermitian operator.

This is an issue, since if ## [\hat{A},\hat{H}] =0 ## we have simultaneously:
$$\frac{d \hat{A_{mn}}}{dt} = 0$$
and
$$\frac{d\hat{A_{mn}}}{dt} = i \frac{\hat{A_{mn}}}{\hbar} (E_{n}-E_{m})$$

## The Attempt at a Solution

This is a problem I made up and got stuck with! I think the resolution should be fairly trivial, but I can't see it - I'd bet that there is probably a simpler way of looking at this using Dirac notation, but I don't really know that very well...

Thanks!

bananabandana said:
$$d\frac{[\hat{A},\hat{H}]}{dt} =(E_{n}-E_{m}) exp\bigg[-i\frac{(E_{n}-E_{m})t}{\hbar}\bigg] \bigg( -\frac{i \hat{A_{mn}} (E_{n}-E_{m})}{\hbar} + \frac{d \hat{A_{mn}}}{dt}\bigg)$$
The LHS is an operator while the RHS is a scalar function, is such an equation possible? I would like to know how you obtained it in the first place?

Ah, ## [\hat{A},\hat{H}]## should in fact read ## [\hat{A}, \hat{H}]_{mn}## But otherwise:

$$[\hat{A},\hat{H}]_{mn} = exp\bigg( i \frac{(E_{m} -E_{n})t}{\hbar} \bigg) \int u^{*}_{m}(x) [\hat{A},\hat{H}] u_{n}(x) \ dx$$
$$= exp\bigg( i \frac{(E_{m} -E_{n})t}{\hbar} \bigg) (E_{n}-E_{m}) \int u^{*}_{m}\hat{A}u_{n}$$
Where the second line follows from that Hermitian nature of the operators, ( ## \int \phi^{*} \hat{H}\hat{A} \psi \ dx = \int (\hat{A}\hat{H}\phi)^{*}\psi dx ##)

Differentiating that under the integral sign ( right phrase? I mean - ## \frac{d}{dt}\int \rightarrow \int \frac{\partial}{\partial t} ##) with respect to ##t## gets the result shown.

About the ##[\hat{A},\hat{H}]## vs. ## [\hat{A},\hat{H}]_{mn}## I think, given that any arbitrary wave function can be expanded out in terms of the eigenstates of energy, we can still get the result in general for the expectation commutator? I mean, if we do:

$$\int \psi^{*}(x,t) [\hat{A},\hat{H} ] \psi(x,t) dx$$

and write $$\psi = \sum c_{n}u_{n}(x)exp(-\frac{E_{n}t}{\hbar})$$
Then, if the commutator ## [\hat{A},\hat{H}] = 0## then the integral must be zero, meaning that each term in $n$ individually yields zero, which gives us a sum of exponentials [Setting third eq'n in first posting = 0]...

Does that make sense?

Thanks!

This equation
$$\frac{d\hat{A_{mn}}}{dt} = -\frac{i}{\hbar} [\hat{A_{mn}},\hat{H}]$$
doesn't look right to me. ##A_{mn}## which appears in the derivative depends on ##t## internally, e.g. oscillating electric field, hence it does not need to obey the Heisenberg equation of motion. Given Heisenberg equation of motion for a time-independent operator ##O##
$$\frac{d}{dt}O_H(t) = -\frac{i}{\hbar} [O,\hat{H}]$$
where ##O_H(t) = U^\dagger O U##. On the other hand, ##A_{mn}(t)## in the first equation clearly is not defined this way because the dependence on time is already inherent in the physical quantity represented by the operator itself.

blue_leaf77 said:
This equation
$$\frac{d\hat{A_{mn}}}{dt} = -\frac{i}{\hbar} [\hat{A_{mn}},\hat{H}]$$
doesn't look right to me. ##A_{mn}## which appears in the derivative depends on ##t## internally, e.g. oscillating electric field, hence it does not need to obey the Heisenberg equation of motion. Given Heisenberg equation of motion for a time-independent operator ##O##
$$\frac{d}{dt}O_H(t) = -\frac{i}{\hbar} [O,\hat{H}]$$
where ##O_H(t) = U^\dagger O U##. On the other hand, ##A_{mn}(t)## in the first equation clearly is not defined this way because the dependence on time is already inherent in the physical quantity represented by the operator itself.

Unfortunately, it turns out I got that result as a given from another question- I'm not sure where they derived it from exactly! I've double checked, and they are the same formula, so I'm not sure what to do.

One more strange thing, no matter how you look at it, as it stands now ##A_{mn}## is a scalar, therefore ##[A_{mn},H]=0##. Are you sure the other question you are talking about address exactly the same issue?

bananabandana said:

## Homework Statement

[/B]

For a general operator ## \hat{O}##, let ##\hat{O}_{mn}(t)## be defined as:
$$\hat{O_{mn}}(t) = \int u^{*}_{m}(x,t) \hat{O} u_{n}(x,t)$$
and
$$\hat{O_{mn}} = \int u^{*}_{m}(x) \hat{O} u_{n}(x)$$
##u_{m}## and ##u_{n}## are energy eigenstates with corresponding energy ##E_{n}## and ##E_{m}##.
From this it can be shown that: (sticking in ## \hat{O} = [\hat{A},\hat{H}] ##
$$d\frac{[\hat{A},\hat{H}]}{dt} =(E_{n}-E_{m}) exp\bigg[-i\frac{(E_{n}-E_{m})t}{\hbar}\bigg] \bigg( -\frac{i \hat{A_{mn}} (E_{n}-E_{m})}{\hbar} + \frac{d \hat{A_{mn}}}{dt}\bigg)$$
But simultaneously, it is also possible to prove that: (as we'd expect from Ehrenfest anyhow)
$$\frac{d\hat{A_{mn}}}{dt} = -\frac{i}{\hbar} [\hat{A_{mn}},\hat{H}]$$
Where did you get this last equation from? Ehrenfest's theorem applies to operators, not matrix elements (and there is in addition a partial derivative ##\partial \hat{A}/\partial t## in general).

nrqed said:
Where did you get this last equation from? Ehrenfest's theorem applies to operators, not matrix elements (and there is in addition a partial derivative ##\partial \hat{A}/\partial t## in general).
I also asked the same question and the OP said he/she got (or probably infer) it from another question (post #5).

## 1. What is meant by "time evolution of operators"?

Time evolution of operators refers to the change or transformation of a physical quantity or observable (represented by an operator) over time. In quantum mechanics, this is described by the Schrödinger equation, which governs how operators evolve in time.

## 2. How is the time evolution of operators related to quantum mechanics?

The time evolution of operators is a fundamental concept in quantum mechanics. It allows us to understand and predict the behavior of quantum systems over time, including the probabilities of different outcomes of measurements.

## 3. What are some examples of operators that undergo time evolution?

Some common examples of operators that undergo time evolution include the Hamiltonian operator, which represents the total energy of a system, and the position and momentum operators, which describe the position and momentum of a particle. Other operators can represent various physical quantities such as spin, angular momentum, or electric charge.

## 4. How does the time evolution of operators affect the state of a quantum system?

The time evolution of operators affects the state of a quantum system by changing the values of the operators and thus altering the probabilities of different outcomes. This allows us to study how a system evolves and changes over time.

## 5. Can the time evolution of operators be observed in experiments?

While the time evolution of operators is a fundamental concept in quantum mechanics, it is not directly observable in experiments. However, the effects of time evolution can be observed through measurements of physical quantities, such as energy or spin, which are represented by operators.

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