Recent content by M-st

i got it now. then the probability is 1. thanks

The normalized eigenvectors: \frac{1}{\sqrt{2}} [1 1]. And the probability for getting hbar/2 is 0?

Since the hamiltonian has two eigenvalues, so the one with lowest energy will be E = (- Hbar*B)/2 ? or is it the plus one. ?

yes. the eigenvalues are +- (h-bar/ 2). and the eigenvectors are [1 1] and [1 -1] So the eigenvalues of Hamiltonian become :\lambda = \pm \frac{B hbar}{2} the eigenvectors are the same as written above. what about the propability now?

yes. the S_x matrix is (h-bar)/2 * pauli x- matrix. I have written it in the question above.

spin 1/2 placed in a magnetic field at x-direction Hello guys I have been a regular reader of this forum. This is my first thread. Homework Statement A spin-1/2 is place in a magnetic field which points in the x-direction. The Hamiltonian is \hat{H} = - B Sx where Sx = h/2 * pauli...