# Quantum version of Larmor precession

## 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!

#### Attachments

TSny
Homework Helper
Gold Member
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##. #### Attachments

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
Homework Helper
Gold Member
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.

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

vela
Staff Emeritus
Homework Helper
I found the probability as ##\dfrac{B^2}{2}## is that correct?
That answer can't possibly be correct. Probabilities are unitless.

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