Spin state function of a beam of particles in terms of eigenfunctions.

Mithra
Messages
16
Reaction score
0
Hi there, I apologise that I should probably know this/its a stupid question but I seem to have forgotten all physics over the holiday and so any help would be great!

I have been told that there is a beam of atoms with spin quantum number 1/2 and zero orbital angular momentum, with spin +1/2 along the x axis. I am then asked what the spin state function of this beam is in terms of eigenfunctions of \hat{S}^2 and \hat{S}_z, being the kets |1/2 , m_{s_z} = 1/2> and |1/2 , m_{s_z} = -1/2>

Would I be right in assuming that the answer is simply
\phi_s = \frac{1}{\sqrt{2}}|1/2 , m_{s_z} = 1/2> + \frac{1}{\sqrt{2}}|1/2 , m_{s_z} = 1/2>
because knowing the spin in the x direction doesn't tell you about the z direction (as they are incompatible observables) or is it more complex than that? Any help/advice greatly appreciated, thanks.
 
Physics news on Phys.org
You want to find the eigenstates of Sx in terms of the eigenstates of Sz. Your answer is a good guess, but there could be a different relative phase between the two terms.
 
Thanks for the help :).

I got the eigenstates of S_x in terms of S_z as

\frac{1}{\sqrt{2}} (|\frac{1}{2},\frac{1}{2}> + |\frac{1}{2},\frac{-1}{2}>
and
\frac{1}{\sqrt{2}} (|\frac{1}{2},\frac{1}{2}> - |\frac{1}{2},\frac{-1}{2}>

With the first one corresponding to the same eigenvalue as the positive spin in the x direction given by S_x. Is that anything like what the correct answer would be? (I'm guessing that it should be similar to if the beam is not polarised as the following questions suggest that they would not be differentiated by a Stern-Gerlach experiment with the magnetic field along z.).

I don't know how much you're actually allowed to say "yes that's the correct answer" but thanks for the help anyway ;).
 
Mithra said:
Thanks for the help :).

I got the eigenstates of S_x in terms of S_z as

\frac{1}{\sqrt{2}} (|\frac{1}{2},\frac{1}{2}> + |\frac{1}{2},\frac{-1}{2}>
and
\frac{1}{\sqrt{2}} (|\frac{1}{2},\frac{1}{2}> - |\frac{1}{2},\frac{-1}{2}>

With the first one corresponding to the same eigenvalue as the positive spin in the x direction given by S_x. Is that anything like what the correct answer would be?
Yup, that's it.
(I'm guessing that it should be similar to if the beam is not polarised as the following questions suggest that they would not be differentiated by a Stern-Gerlach experiment with the magnetic field along z.).

I don't know how much you're actually allowed to say "yes that's the correct answer" but thanks for the help anyway ;).
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top