Spin 1/2 planced in a magnetic field at x-direction

[M-st]

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 S_x

where S_x = [STRIKE]h[/STRIKE]/2 * pauli x-matrix. We have absorbed all couplings into an effective magnetic field B>0 so that B has dimension inverse time.

1. Find the energy eigenvalues and corresponding eigenstates of H.

the spin is known to be in its lowest energy state. Then a measurement (M1) of the spin-component in the x-direction is carried out.

2. Calculate the probability of finding the value h-bar /2 in measurement M1.

The Attempt at a Solution

Many attempts but, don't have a clue where to start from.

 

[arkajad]

To start with: Can you write S_x$ as a matrix? What matrix would it be? Can you find eigenvectors and eigenvalues of this matrix?

 

[M-st]

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

 

[arkajad]

Can you find eigenvectors and eigenvalues of this matrix?

 

[M-st]

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?

 

[arkajad]

So, which will be "the lowest energy eigenstate"?

 

[M-st]

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

 

[arkajad]

Alright, the minus one is lower. Now, if you want to calculate probabilities, you must normalize the eigenvector. What will be the normalized (norm one) eigenvector?

 

[M-st]

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

 

[arkajad]

Now, be careful. You have H=-BSx. Eigenvalues of Sx are hbar/2 and -hbar/2. When Sx has eigenvalue hbar/2, H has eigenvalue -Bhbar/2 - which is the lowest. In general, the probability is given by the expression of the form

|<u|v>|^2

But here |u> and |v> are the same: on your (normalized) eigenvector u=v= <br /> \frac{1}{\sqrt{2}}<br />[1,1] H has the lowest eigenvalue and Sx has eigenvalue hbar/2. So?

 

[M-st]

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