Quantum version of Larmor precession

[Bobs]

Homework Statement

Homework Equations

I didn't get what this actually means.

The Attempt at a Solution

I'm not sure whether it is correct. Could you take a look?
Regards!

 

[TSny]

Hello. I haven't gone through all of the details of your calculations. But, I don't see where you have taken into account the evolution of the system between $t = 0$ and $t = T$. It appears to me that you are assuming that the system is in state $\chi_+^{(x)}$ at time $t = T$.

 

[Bobs]

Do you see any mistakes at the final answer? And yes, I'm assuming that the system in state $\chi_+^{(x)}$ at time $t=T$. Staying tuned for your sincerely reply!

 

[TSny]

Maybe we are interpreting the statement of the problem differently. The way I read it, at time $t = 0$ the system is in state $|\psi(0) \rangle = \chi_+^{(x)}$. It evolves for time $T$ with $\mathbf B$ in the z direction. Then, in the time interval $T \leq t \leq 2T$ it evolves with $\mathbf B$ in the y direction. So, you will need to find $|\psi(T) \rangle$ before you can determine the state for $t > T$.

Your final result for the probability appears to be a complex number. But probabilities are real numbers. Also, shouldn't the answer depend on the magnetic field strength B?

 

[Chandra Prayaga]

It looks like this post has appeared twice.

Do you see any mistakes at the final answer? And yes, I'm assuming that the system in state at time $t=T$. Staying tuned for your sincerely reply!

The systaem will not be in state $\chi_+^{(x)}$ at t = T. As TSny points out in the above post, the evolution between t = 0 and t = T is dictated by the magnetic field only in the z direction. You should calculate |ψ(T)> from that, and then let that evolve from t = T to t = 2T with the magnetic field only in the y direction.

 

[Bobs]

I found the probability as $\dfrac{B^2}{2}$ is that correct?

 

[vela]

I found the probability as $\dfrac{B^2}{2}$ is that correct?

That answer can't possibly be correct. Probabilities are unitless.

 

[Bobs]

Finally, I've found the probability as $0$ from $P = \dfrac{1}{4}exp \biggr( \dfrac{-2t\mu BB}{\hbar}\biggr ) \biggr | (1-1) \biggr |^2 = 0$ Does that seem correct now? Thanks in advance.

 

[Chandra Prayaga]

What steps did you follow to arrive at that answer?

 

[Bobs]

Is that correct? Or what would be the correct probability we're looking for? I'm receiving too many different answers.

 

[vela]

No. The argument of the exponential should be unitless.