# Recent content by ma18

1. ### Expectation of position and momentum at time t, pictures

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...
2. ### Expectation of position and momentum at time t, pictures

Thank you for your help! Onto the Heisenberg picture portion of the question :)
3. ### Expectation of position and momentum at time t, pictures

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)
4. ### Expectation of position and momentum at time t, pictures

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
5. ### Expectation of position and momentum at time t, pictures

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
6. ### Expectation of position and momentum at time t, pictures

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>?
7. ### Expectation of position and momentum at time t, pictures

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>
8. ### Expectation of position and momentum at time t, pictures

It doesn't specify but I would assume it is a vector
9. ### Expectation of position and momentum at time t, pictures

ah, okay. Plus I still need to find <[P,X_1]> right. Could you help me with that commutator? Thanks
10. ### Expectation of position and momentum at time t, pictures

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
11. ### Expectation of position and momentum at time t, pictures

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...
12. ### Expectation of position and momentum at time t, pictures

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...
13. ### Expectation of position and momentum at time t, pictures

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> ==>...
14. ### Expectation of position and momentum at time t, pictures

I know that the commutator of x and p is i*hbar and that in the Schrödinger picture the operators are time independent
15. ### Expectation of position and momentum at time t, pictures

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...