Spin, let me make sure I have this straight

[LostConjugate]

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.

 

[arunma]

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.

 

[alxm]

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 $$S = \hbar\frac{\sqrt{3}}{2}$$

 

[LostConjugate]

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?