Spin, let me make sure I have this straight

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So if you measures the spin of an electron in a Hydrogen atom (which I understand requires a magnetic field to eliminate the degeneracy of energy levels) in an arbitrary direction

x^i

you would measure plus or minus

\frac{eh}{2Mc}B_i

in Joules. Where B_i is the magnetic field component in the direction x^i

I understand you never measure anything between this.

What about the magnetic field components of the other 2 orthogonal dimensions? Is the spin in those directions also

\frac{eh}{2Mc}B_i

That would give a total intrinsic angular momentum of 3/2.
 
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Well it's been a couple years since I took my quals, so someone correct me if I'm wrong. But I believe the Hamiltonian for this system is:

H=\dfrac{e\hbar}{2mc}B_z\sigma_z

But if you want to find the possibile energies that you'd get by making a measurement of the x-component of the spin, you're basically finding the eigenvalues of the x spinor. So you'd do,

\dfrac{e\hbar}{2mc}B_z\sigma_z \dfrac{1}{\sqrt{2}} \left(\begin{array}{cc}1\\1\end{array}\right) = E \dfrac{1}{\sqrt{2}}\left(\begin{array}{cc}1\\1\end{array}\right)

So assuming I got that right, now you just need to do the eigenvalue problem.
 
You can't sum up the components; the operators don't commute for different directions.
They do commute with S2 and so you get an expectation value S^2 = 3\hbar^2/4 and more generally S = \sqrt{s(s+1)}

So the total spin angular momentum is S = \hbar\frac{\sqrt{3}}{2}
 
So what does it mean to say the spin of an electron is l=1/2 where l is the total angular momentum.

Edit:

I think I understand my mistake, l is the total angular momentum quantum number, not the energy eigenvalue.

So then am I correct in saying that the absolute value of the intrinsic angular momentum of a particle is equal in any direction a magnetic field of equal intensity is applied?
 
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