Yes the period the same. But The direction of the angular momentum is parallel to the direction of the magnetic field, then the electron orbital momentum vector should point to the opposite direction?
I though the periodic time and the radius of the rotation are affected by the relativistic motion, so the periodic time from the point view of the electron will be different from the point view of the proton.
He said he is going to use a process of successive approximations, I am not familiar with the "asymptotic expansion" so I can not tell if this what he did
In equation [9.15] ##c_a(0) = 1## this is before the perturbation (at time ##t= 0##) no transition happened yet and the particle still in the upper state ##a##. But equation [9.17] after perturbation at time t and ca(1)(t) = 1 is the first-order correction. I understand that the zeroth-order...
He used a process of successive approximations, so for this two particle system the particle starts at state ##a##, then at time t=0: ca(0)=1 and cb(0)=0. If there were no perturbation the system will stay there forever, so we can say the zeroth- order terms are: ca(t)=1 and cb(t)=0. To...
I was reading in the Book: Introduction to Quantum Mechanics by David J. Griffiths. In chapter Time-Dependent Perturbation Theory, Section 9.12. I could not understand that why he put the first order correction ca(1)(t)=1 while it equals a constant.
I only studied two spin half particles from the book Introduction to Quantum Mechanics by David J. Griffiths. I do not know if this enough For example see the attached picture.
The problem is I am not familiar with this type of product using this symbol: \otimes ! So will it be wrong if I used the way I described above a got the correct matrices?
I have this homework: consider the case of two spin half particles. Use the basis: |++>, |+->, |-+>, |--> to find the matrices representing the operators S^2 and S_z.
My idea for the solution for S_z is: S_z=S_z(1)+S_z(2) where S_z(1) is the operator for the first particle ... etc
So I...
It worked now, but I equated the the first order of ##\epsilon## (in the left side) with the zeroth order of ##\epsilon ## (in the right side) because the right side is multiplied by ##H^{'}## which is small (of the same order of ##\epsilon##). For example for the first order of ## \epsilon## i...
I was reading in the Book: Introduction to Quantum Mechanics by David J. Griffiths. In chapter Time-Dependent Perturbation Theory, Section: Two-level system. Every thing was fine till He said He will solve this equation:
by a process of successive approximations. I have no idea what this...
I was reading in the Book: Introduction to Quantum Mechanics by David J. Griffiths. In chapter Time-independent Perturbation Theory, Section: Spin -Orbit Coupling. I understood that the spin–orbit coupling in Hydrogen atom arises from the interaction between the electron’s spin magnetic moment...