0
$\begingroup$

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/larmor.html

Consider the process of Larmor precession of the magnetic moment associated with the spin angular momentum of an electron inside a hydrogen atom (ref. above link).

It is written that there is an "....angular frequency associated with the spin transition involving an energy change...".

But angular frequency is usually defined for rotating bodies. How can it be associated with the process of an energy transition, just as in this case? I can't imagine a rotating body in a spin-transition process!

Also, the energy difference and the angular frequency are related to each other using - $E = \hbar \omega$. Where did this come from?

$\endgroup$
1
  • $\begingroup$ in the hyperphysics link you attached while it is spinning up in the same direction as the magnetic field it is in the low energy state. While it is spinning facing down against the magnetic field it is in the high energy state. There are only 2 spin states for up and down but they are coupled when excited so that the up and down cancel but the sides do not cancel because they are coupled and facing same direction this explains T1 T2 relaxation phenomenon $\endgroup$
    – ChemEng
    Feb 13 '20 at 7:05
0
$\begingroup$

When you have two levels separated by energy $\Delta E$, you can induce a transition between them by introducing a time-dependent term to the hamiltonian (for example exciting it with a periodic electromagnetic field). This term must have a time dependence of the form $\cos(\omega t)$ with $\omega=\frac{\Delta E}{\hbar}$. The angular frequency being referred to is $\omega$, there are no rotating bodies here.

In order to understand why time-dependent perturbations of a certain frequency induce transitions between certain levels, you need to read up on time-dependent perturbation theory. Both Cohen and Sakurai are great sources for this.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.