- #1

Quiablo

- 9

- 0

Hi everyone,

I am really stuck here, and I would really appreciate if someone could help me out. The statement of the problem is attached as a image. Some equations refered by the problem are also attached in another image. I will try to explain how I have reasoned using more words than formulas, since its so darn difficult to write expressions using this editors. Here we go:

First I inverted eq. 14.4.35, multiplying both sides by the same exponential shown without the minus sign. This way we get an expression for psi(t) in terms of psi

(i * hbar (d/dt) + gama*S*B) * exp(iwt * S

Then i applyied the derivative using the product rule (which I am not 100% sure if can be done with operators like exp (iwt * S

i * hbar ( iw * S

Multiplying everything by exp(-iwt * S

- w * S

Which is the same as:

i * hbar (d/dt) psi(t) = (w * S

Now if we compare this to the form of 14.4.34, we see that the Hamiltonian is indeed indepent of time, as expected, but I am almost 100% positive that this is the wrong Hamiltonian, for if we carry on the computations we dont get to the result shown in 14.4.36. I am pretty positive that the term w * S

I am really stuck here, and I would really appreciate if someone could help me out. The statement of the problem is attached as a image. Some equations refered by the problem are also attached in another image. I will try to explain how I have reasoned using more words than formulas, since its so darn difficult to write expressions using this editors. Here we go:

First I inverted eq. 14.4.35, multiplying both sides by the same exponential shown without the minus sign. This way we get an expression for psi(t) in terms of psi

_{r}(t). Substituting that in 14.3.34, and considering that H = - gama * S * B, i got:(i * hbar (d/dt) + gama*S*B) * exp(iwt * S

_{z}/ hbar) * psi(t) = 0Then i applyied the derivative using the product rule (which I am not 100% sure if can be done with operators like exp (iwt * S

_{z}/ hbar) ) and got:i * hbar ( iw * S

_{z}/ hbar) * exp(iwt * S_{z}/ hbar) * psi(t) + i * hbar * exp(iwt * S_{z}/ hbar) (d/dt) psi(t) = - gama*S*B * exp(iwt * S_{z}/ hbar) * psi(t)Multiplying everything by exp(-iwt * S

_{z}/ hbar) to the left side, and considering that: exp(-iwt * S_{z}/ hbar) is the operator that rotates the spinor around the z axis; exp(-iwt * S_{z}/ hbar) * S_{z}* exp(iwt * S_{z}/ hbar) = S_{z}(rotates counterclockwise and clockwise around the same axis); B * exp(iwt * S_{z}/ hbar) = B_{r}, which is the (static) B field in the rotating frame; exp(-iwt * S_{z}/ hbar) * S equals an operator I called S_{r}, which is the version of S in the rotating frame (static relative to that rotating frame), i got to the expression:- w * S

_{z}* psi(t) + i * hbar (d/dt) psi(t) = - gama*S_{r}*B_{r}* psi(t)Which is the same as:

i * hbar (d/dt) psi(t) = (w * S

_{z}* - gama*S_{r}*B_{r}) * psi(t)Now if we compare this to the form of 14.4.34, we see that the Hamiltonian is indeed indepent of time, as expected, but I am almost 100% positive that this is the wrong Hamiltonian, for if we carry on the computations we dont get to the result shown in 14.4.36. I am pretty positive that the term w * S

_{z}SHOULD NOT be in this expression, for it it weren't there, the right answer would be obtained. Can anyone tell me what operation was done incorrectly by me in the steps above?#### Attachments

Last edited: