Expectation of position and momentum at time t, pictures

[ma18]

Homework Statement

Consider a particle, with mass m, charge q, moving in a uniform e-field with magnitude E and direction X_1.

The Hamiltonian is (where X, P, and X_1 are operators):

The initial expectation of position and momentum are <X(0)> = 0 and <P(0)>=0

Calculate the expectation of position and momentum oprator in the Schrodinger picture is X (t) = (X_1, X_2, X,3)

2. The attempt at a solution

I know the Hamiltonian and the initial condition. I am doing this in the schrodinger picture, and that with the time evolution operator I can represent the ket as

|phi(t)> = U (t,0) |phi(0)>

Is H is time-independent? As P is squared and is the magnitude of the P? If so the time evolution operator can be represented as

U (t,0) = exp (-i*t*H/hbar)

Then

I can represent phi(t) in terms of those terms and phi(0) and proceed but I am not sure if that is correct.

I know that P = mv = m dx/dt so I think I can use the derivative of the expectation value of X when I get that, correct. According to the equation:

d/dt <O>_t = i/h <[H,O]>_t + <dO/dt>_t

but I have to find the expectation of the position first.

Any help would be much appreciated!

 

[DrClaude]

I can represent phi(t) in terms of those terms and phi(0) and proceed but I am not sure if that is correct.

You don't need to know φ(t).

I know that P = mv = m dx/dt so I think I can use the derivative of the expectation value of X when I get that, correct.

That should not be necessary.

d/dt <O>_t = i/h <[H,O]>_t + <dO/dt>_t

Start with that equation, and see what you get for P and X.

 

[ma18]

You don't need to know φ(t).That should not be necessary.Start with that equation, and see what you get for P and X.

Okay, plugging X into that equation (get rid of _t for legibility) I get

d/dt <X> = i/h <[H,X]> + <dX/dt>
= i/h <[(P^2/2m - E*q*X_1), X]>+<dX/dt>
= i/h (-E*q) <X,X1>+<dX/dt>

and I'm not sure where to go from here...

 

[ma18]

What if I just did:

U (t,0) = exp (-i*t*H/hbar)
= exp (-i*t*(P^2/2m - E*q*X_1)/hbar)

|phi (t)> = exp (-i*t*(P^2/2m - E*q*X_1)/hbar) |phi(0)>

<phi (t)|X|phi(t)> = <phi(0) exp (i*t*(P^2/2m - E*q*X_1/hbar) | X | exp (-i*t*(P^2/2m - E*q*X_1)/hbar) |phi(0)>

but I don't know where I would go next, could I just multiply it in? But they are operators.

 

[DrClaude]

d/dt <X> = i/h <[H,X]> + <dX/dt>
= i/h <[(P^2/2m - E*q*X_1), X]>+<dX/dt>
= i/h (-E*q) <X,X1>+<dX/dt>

What is the commutator of X and P? What can you say about the time-dependence of an operator (not explicitly time dependent) in the Schrödinger picture?

What if I just did:

U (t,0) = exp (-i*t*H/hbar)
= exp (-i*t*(P^2/2m - E*q*X_1)/hbar)

|phi (t)> = exp (-i*t*(P^2/2m - E*q*X_1)/hbar) |phi(0)>

<phi (t)|X|phi(t)> = <phi(0) exp (i*t*(P^2/2m - E*q*X_1/hbar) | X | exp (-i*t*(P^2/2m - E*q*X_1)/hbar) |phi(0)>

but I don't know where I would go next, could I just multiply it in? But they are operators.

You can't go anywhere with that. Note also that you do not know what |φ(0)>, only that <φ(0)|X|φ(0)> = <φ(0)|P|φ(0)> = 0.

 

[ma18]

I know that the commutator of x and p is i*hbar and that in the Schrödinger picture the operators are time independent

 

[DrClaude]

I know that the commutator of x and p is i*hbar and that in the Schrödinger picture the operators are not time dependent. However here I have x_1 and p^2

X₁ is only one of the spatial dimensions (if I understand correctly "X (t) = (X_1, X_2, X,3)"), and p² is simply pp, so
$$
\begin{align*}
[p^2,x] &= p^2x-xp^2 \\&= ppx - xpp \\
&= ppx - xpp - pxp + pxp \\
&= ppx - pxp + pxp - xpp \\
&= p[p,x] + [p,x]p
\end{align*}
$$

 

[ma18]

X₁ is only one of the spatial dimensions (if I understand correctly "X (t) = (X_1, X_2, X,3)"), and p² is simply pp, so
$$
\begin{align*}
[p^2,x] &= p^2x-xp^2 \\&= ppx - xpp \\
&= ppx - xpp - pxp + pxp \\
&= ppx - pxp + pxp - xpp \\
&= p[p,x] + [p,x]p
\end{align*}
$$

Ah ok so then, assuming that the commutator of x and x_1 is 0 I would get:

d/dt <X> = i/h <[H,X]> + <dX/dt>
= i/h <[(P^2/2m - E*q*X_1), X]>+<dX/dt>
= i/h <[X*P^2/2m -X*P^2/2m]>+<dX/dt>
= i/(2mh) <[X,P^2]>+<dX/dt>
= i/(2mh) <2*i*h*p>+<dX/dt>
= i/(2mh)*(2*i*h) <p>+<dX/dt>
= -1/m<p> +<dX/dt>

==>
<p> = m (d/dt <X> - <dX/dt>)

But that still has a lot of unknows for a final answer as I have to find out <X> as well

 

[ma18]

If I did the same procedure for P then it would be d/dt <P> = i/h <[H,P]> + <dP/dt>
= i/h <[(P^2/2m - E*q*X_1), P]>+<dP/dt>
= i/h <[P*-E*q*X_1 -E*q*X_1*P]>+<dP/dt>
= i(-E*q)/h <[P,X_1]>+<dP/dt>
= i/(2mh) <i*h>+<dP/dt>

But then if [P,X_1] = ih then this doesn't really make sense...

 

[blue_leaf77]

Sorry for butting in, just a short intervention to remind the OP that dX/dt = dp/dt = 0.

 

[ma18]

Sorry for butting in, just a short intervention to remind the OP that dX/dt = dp/dt = 0.

Ah, thank you.

So then

<p> = m (d/dt <X>)

and

d/dt <P>= i(-E*q)/h <[P,X_1]>

Is that the final answer?
The second portion of the question is to, with the same Hamiltonian, knowing that in the Heisenberg Picture , X_H = (X_1H,X_2H,X_3H), find the commutator

[X_aH (t), X_bH (0)]

where all the X are operators

I know that in the Heisenberg picture the operators evolve with time as U^+ (t,0) O U(t,0)

 

[ma18]

So then for the Heisenberg picture, because e^(0) = 1 and thus for X_bH (0), U would equal 1. I would get (where U_d represents U dagger)

U_d X_aH U X_b - X_b U_d X_aH U

then I don't know what to do

 

[blue_leaf77]

Is that the final answer?

No, you still have to solve those two coupled differential equations.

 

[ma18]

No, you still have to solve those two coupled differential equations.

ah, okay. Plus I still need to find <[P,X_1]> right. Could you help me with that commutator?

Thanks

 

[blue_leaf77]

Plus I still need to find <[P,X_1]> right.

Is P actually a vector in your notation? Is it actually $\mathbf{P} = P_1\hat{i} + P_2\hat{j} + P_3\hat{k}$?

 

[ma18]

It doesn't specify but I would assume it is a vector

 

[blue_leaf77]

In that case, it will be $[P_1\hat{i} + P_2\hat{j} + P_3\hat{k},X_1] = [P_1,X_1]\hat{i} + [P_2,X_1]\hat{j} + [P_3,X_1]\hat{k}$. You know the value of each term, don't you?

 

[ma18]

Yes so then it would just be [P,X_1] = ih ?

Then the equation would be, as I put in post 9

d/t <P> = = i/(2mh) <i*h>

 

[blue_leaf77]

Yes so then it would just be [P,X_1] = ih ?

Don't forget the unit vector, $[P,X_1] = i\hbar\hat{i}$.

d/t <P> = = i/(2mh) <i*h>

That equation will then disperse into three equations, each for the momentum components and it should be easily solvable.

 

[ma18]

As ih is a constant wouldn't <ih> just be equal to ih since the exponentials of the U term would become 1 and the phi(0) would as well.

And then I could sub in the first equation to get the value for <x>?

 

[blue_leaf77]

As ih is a constant wouldn't <ih> just be equal to ih

Yes, it is.
The equation for $\langle \mathbf{X} \rangle$ in post #8 also contains three further equations for each components of the position, so you need all values of $\langle P_i \rangle$'s.

 

[ma18]

But the whole thing has been done with the whole X anyway and the only reason the X_1 is being used is because it is in the Hamiltonian so isn't it fine?

Then

d/dt <p> = Eq i_hat

<p> = Eq t i_hat + C

using the first equation then:

d/dt <x> = (Eq t i_hat + C)/m

<x> = t(Eq t i_hat + C)/m + C_1

 

[blue_leaf77]

<p> = Eq t i_hat + C

Find C from the initial conditions for $\mathbf{P}$.

<x> = t(Eq t i_hat + C)/m + C_1

Find C_1 from the initial conditions for $\mathbf{X}$, also check again if any constant number is missing in the first term proportional to $t^2$.

 

[ma18]

From the initial conditions

<p(0)> = 0 = C = 0

<x(0) = 0 = C_1 = 0

so

<p> = Eqt i_hat

<x> = t (Eq t i_hat)/m = <p> *t/m

 

[blue_leaf77]

<x> = t (Eq t i_hat)/m

You did the integration wrong, $d<\mathbf{X}>/dt = -<\mathbf{P}>/m = -1Eqt/m \hat{i}$. How do you integrate $- Eqt/m$ over $t$?

 

[ma18]

Ah of course, silly mistake

d/dt <x> = (Eq t i_hat + C)/m
<x> = (E q t^2 i_hat )/(2m) + C_1

and then C_1 is zero from the IC so

<x> = (E q t^2 i_hat )/(2m)

 

[blue_leaf77]

Yes, I think that's the correct answer, only that remember that $\langle \mathbf{X} \rangle = \langle X_1 \rangle\hat{i} + \langle X_2 \rangle\hat{j} + \langle X_3 \rangle\hat{k}$.

 

[ma18]

Thank you for your help!

Onto the Heisenberg picture portion of the question :)

 

[ma18]

I feel like there is a way to manipulate the terms so as to put them back into a known relationship but I am not seeing it

I know that

[X_aH (t), X_bH (0)] = [U_d (t,0) X(t)_aH U (t,0),U_d (0,0) X (0)_bH U (0,0)] = U_d (t,0) X(t)_aH U (t,0) X (0)_b - X (0)_b U (t,0)_d X (t)_aH U (t,0)

Where
U (t,0) = exp (-i*t*H/hbar)
= exp (-i*t*(P^2/2m - E*q*X_1)/hbar)