Effect of spins on hydrogen atom ground state energy

[forcefield]

What do you think about Feynman's description (http://feynmanlectures.caltech.edu/III_12.html#Ch12-S3) ? It seems to be inconsistent with hyperphysics (http://hyperphysics.phy-astr.gsu.edu/hbase/quantum/h21.html).

 

[blue_leaf77]

The hyperphysics link does not work in my computer. Just an advice to make your post able to attract more comments, try to summarize or point out which part of those long texts that you are concerned with.

 

[forcefield]

try to summarize or point out which part of those long texts that you are concerned with.

The inconsistency is that hyperphysics says that the energy difference is due to spins going from parallel (higher energy) to antiparallel (lower energy) whereas according to Feynman antiparallel spins can be associated also with the higher energy. I don't see how hyperphysics can be correct if one considers also Zeeman splitting like Feynman does in the next section.

 

[blue_leaf77]

It seems that the link with hyperphysics is using a hand-waving argument to say that the upper level in the ground state of hydrogen is solely associated to parallel spins between proton and electron. The perturbation term in the one-electron hyperfine splitting is diagonalized by the eigenstates of the total angular momentum operator $\mathbf{F}$, which is equal to the sum of the nucleus and electron individual angular momenta, $\mathbf{F} = \mathbf{I}+\mathbf{L}+\mathbf{S}$. Therefore, each level in the hyperfine structure is assigned with the total angular momentum quantum number. For ground state, $L=0$ and $\mathbf{F} = \mathbf{I}+\mathbf{S}$. As both proton and electron have spin one-half, the possible value of $F$ in the ground state is $F=0,1$. $F=1$ is the upper level and $F=0$ is the lower level. The upper level with $F=1$ is further three-fold degenerate with $m_F = -1,0,1$ as usual. The state associated with $|F=1,m_F=1\rangle$ has both proton and electron in spin up, $|F=1,m_F=-1\rangle$ has both proton and electron in spin down, and $|F=1,m_F=0\rangle$ is a symmetric combination between electron spin up-proton spin down and the opposite. I guess it's this last state which that hyperphysics link has omitted and which Feynman's lecture referred to as "antiparallel spins can be associated also with the higher energy". The lower level with $|F=0,m_F=0\rangle$, however, is not degenerate and is an anti-symmetric combination between electron spin up-proton spin down and the opposite.

 

[forcefield]

$|F=1,m_F=0\rangle$ is a symmetric combination between electron spin up-proton spin down and the opposite. I guess it's this last state which that hyperphysics link has omitted and which Feynman's lecture referred to as "antiparallel spins can be associated also with the higher energy".

Thanks, I think your description is consistent with Feynman's although you are using a slightly different notation. Also, your quoted text above is my summary of what Feynman actually said.