The hydrogen atom is a bound state of two spin ½ particles. Its Hamiltonian may be
written as *H=H _{0}+W_{f}+W_{hf}, *with

The eigenfunctions of *H _{0}* may be chosen as {|

= *W _{mv} + W_{D} +W_{so}* .

.

*W _{mv} *and

*W*, a relativistic correction term:_{mv}where

Given the stationary state wavefunctions of the hydrogen atom we find

,

Therefore

with

In general,

*W*, the Darwin term:_{D}(integration by parts)

Therefore

For

*l¹0*, |*R*|_{nl}(0)^{2}=0, so*W*is non zero only for_{D}*l=0*states.

*W*, the spin-orbit term:_{so}For the spin-orbit interaction we have

*W*is a scalar operator. It does not operate on the proton spin variables and commutes with_{so}. Its eigenvalues are therefore independent of**J***m*._{j}*W*also commutes with_{so}*L*and matrix elements between states with different values of^{2}*l*are zero.The eigenfunctions of

*H*may also be chosen as {|_{0}*n l s j m*>}. These basis functions are eigenfunctions of_{j }i m_{i}*W*._{so}If

*l=0*then*j=s*and*W*. If_{so}=0*l¹0,*thenWith

*j=l*±1/2 we have*W*partially remove the six fold degeneracy of the_{so}*2p*level. The*2p*level splits into a 2 fold degenerate*2p*level and a 4 fold degenerate_{1/2}*2p*level._{3/2}

**The fine structure of the hydrogen atom**:

*W _{f} = W_{mv} + W_{D} + W_{so}*.

is the general expression.

are degenerate.
This is an
accidental degeneracy, and it remains in the exact solution of the Dirac equation
neglecting the proton spin. However, QED corrections raise the *2s _{1/2}*
level with respect to the

If we do neglect the proton spin, then the 1s level is 4-fold degenerate in the state
space of two spin 1/2 particles. Can *W _{hf}* remove this 4 fold
degeneracy?

With and and we have

We choose {|*n l s j i f m _{f}*>} as the eigenbasis of

(a) Describe the energy levels for *B=*0.

(b) Describe the energy levels for weak magnetic fields (Zeeman effect). What are the Lande' g-factors?

(c) Describe the energy levels for large magnetic fields (Paschen-Back effect).

- Solution:
(a) If

*B=0*, we have just the spin orbit interaction,*W*_{so}µ**L×S.**The eigenstates of

*H*are the eigenstates of_{0}*J*, and^{2}, L^{2}*S*and they may be denotes by {^{2}*|nls;jm*}._{j}>We have

*l*=1,*j*=1/2 or*j*=3/2.For the

(b) Consider the case of a*p*level we have_{1/2}*<W*. For the_{so}>=-A*p*level we have_{3/2}*<W*. The_{so}>=(1/2)A*p*level lies (3/2)_{3/2}*A*above the*p*level._{1/2}**weak field**. The splitting introduced by*B*is much smaller than the splitting between the*p*level and the_{1/2}*p*level. This is called the_{3/2}**Zeeman effect**. We considera perturbation to

The eigenstates of

*H*are the_{0}’*|E*. The corresponding eigenvalues are_{0}ls;jm_{j}>*2j+1*fold degenerate.**L**+2**S**=**V***k*,*j*) we haveusing the projection theorem. Choose your coordinate system such that

points in the z-direction. Then the matrix elements of**B***L*are proportional to the matrix elements of_{z}+2S_{z}*J*. To evaluate the proportional constant we use_{z}For the

*p*level we have_{1/2}*g*=2/3. For the*p*level we have_{3/2}*g*=4/3.The

(c) Consider the case of a*B*field completely removes the degeneracy of each level. The energy differences are proportional to*B*and a constant*g*, the_{J}**Landé g - factor**.**strong field**. The splitting introduced by*B*is much larger than the splitting introduced by the spin orbit interaction. This is called the**Paschen-Back effect**. Assume the spin orbit interaction can be ignored. We consider*V’*a perturbation to*H*. The eigenstates of_{0}*H*are_{0}*|nlm;sm*. They are_{s}>*(2l+*1*)(2s+*1*)*fold degenerate.*V’*is diagonal in this basis.The possible values for

*m*are -1, 0, 1, and the possible value for*2m*are -1, 1. Therefore the possible values for_{s}*(m+2m*are -2, -1, 0, 1, 2. The value 0 is two fold degenerate._{s})

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