Quantum time corr: expectation value of particle motion in Schro. pic

[pfnewt]

Homework Statement

The expectation value of motion of a particle over a time interval t-to is

C(t,to) = <0|x(t)x(to)|0>
(product of position operators in Heisenberg representation for ground state harmonic oscillator)

Homework Equations

Schrodinger picture:
<ψ(t)|Ω|ψ(t)> = <ψ(to)|U⁺(t,to) Ω U(t,to)|ψ(to)>

I believe:
U(t,to) = exp[-iωt/h]
U⁺(t,to) = exp[iωt/h]

Heisenberg rep: time-indep states, time-dep operators
Schrodinger rep: time-dep states, time-indep operators

The Attempt at a Solution

I'm confused about how to convert x(t) in the Heisenberg representation to the Schrodinger representation and how to incorporate it into C(t,to).

Here is what I scribbled:
<x(t)> = <ψ(t)|x(to)|ψ(t)> = <ψ(to)|exp[iωt/h] x(to) exp[-iωt/h]|ψ(to)>
= <0|exp[iωt/h] x(to) exp[-iωt/h]|0>
Is the first equality even true?

The question seems like it should be simple, but I am sincerely confused. Thanks for your time and for any help!

 

[Thoros]

In the Heisenberg picture all time dependence is attributed to the operators, the states carry no time dependence. In the Schrödinger picture this is reversed, the operators are constants but the states change in time.

The connection between the pictures is established by requiring physics to be unchanged. That is expectations values must give the same numbers in both cases.

I believe:
U(t,to) = exp[-iωt/h]
U+(t,to) = exp[iωt/h]

This is a little wrong, why do you write ωt/h?
In the Schrödinger picture you can formally solve the Schrödinger equation by means of the evolution operator U(t,t0) = exp[-iHt/h], where H is the Hamiltonian of the system. So the states will be given by ψ(t)> = U(t,to)|ψ(to)>. To go to the Heisenberg picture you simply have to get rid of the time dependence so you can write |ψ(Heisen.)> = U+(t,to) U(t,to)|ψ(to)> = ψ(to, Schr.)> due to the unitarity of the evolution operator.

Now you can simply apply the same thing in any of your expectation value equations and find how your operators must change in order for the expectation values to be invariant.

 

[pfnewt]

Hello and thanks for responding. The omega represents energy and the h is supposed to be h bar. I could see how that would look strange actually, so I won't use omega anymore.

Anyway, here is another attempt:

<0|x(t)x(to)|0> = <0|exp[iHt/h]exp[-iHt/h] x(to)x(to)exp [-iHt/h]exp[iHt/h]|0>
= <0|x²|0> = h/(2mω)

This doesn't match my answer for solving it another way :/. I think I understand what to do conceptually, but not mathematically. Are the exponentials supposed to cancel? Aren't those what make the states time-dependent?

I hope I am making some amount of sense. Thanks again.

 

[Thoros]

<0|x(t)x(to)|0> = <0|exp[iHt/h]exp[-iHt/h] x(to)x(to)exp [-iHt/h]exp[iHt/h]|0>
= <0|x²|0> = h/(2mω)

How did you get exp[iHt/h]exp[-iHt/h] on either side?

The correct way should be:

In the Heisenberg picture you have <0|x(t)x(to)|0>
So using the evolution operator to switch the states to the Schrödinger picture you get:
<0|exp[iHt/h] x(t) x(to) exp[-iHt/h]|0> = <0|U+ x(t) x(to) U |0>

So in order for the expectation value to be invariant you must write
<0|U+ U x(t) U+ U x(to) U+ U |0> = <0|x(t)x(to)|0> due to unitarity of U.

So the operators must change like U x(t) U+

Hope this helps. Note the ambiguity when i write x(t) even in the Schrödinger picture.

 

[paranormal]

Great question! The expectation value of the particle's motion in Schroedinger picture is a fundamental concept in quantum mechanics. In order to convert the position operator from the Heisenberg representation to the Schroedinger representation, we need to use the time evolution operator, U(t,to), which relates the two representations. As you correctly wrote, U(t,to) = exp[-iωt/h] and U+(t,to) = exp[iωt/h].

To incorporate this into C(t,to), we can use the following steps:

1. Rewrite the position operator in the Heisenberg representation as a function of the time evolution operator:

x(t) = U(t,to) x(to) U+(t,to)

2. Substitute this into the expectation value expression:

C(t,to) = <0|x(t)x(to)|0> = <0|U(t,to) x(to) U+(t,to) x(to)|0>

3. Use the fact that the ground state, |0>, is an eigenstate of the position operator, x(to)|0> = x(to)|0>, and that U(t,to)|0> = |0> (since the ground state is time-independent).

C(t,to) = <0|U(t,to) x(to) x(to) U+(t,to)|0> = <0|U(t,to) x(to) U+(t,to)|0>

4. Finally, using the properties of the time evolution operator, we can simplify the expression further:

C(t,to) = <0|U(t,to) x(to) U+(t,to)|0> = <0|x(to)|0>

So the final expression for the expectation value of the particle's motion in the Schroedinger picture is:

C(t,to) = <0|x(to)|0>

which is simply the expectation value of the position operator in the ground state.

I hope this helps clarify the conversion process and how to incorporate it into the expectation value expression. Keep up the good work in your studies!