# Hydrogen 2p doublet

1. Jun 7, 2010

### oddiseas

I am trying to better understand this concept.

for an n=2 state, l=1 we have three values for the magnetic quantum number, -1,0,1.
Now in the absence of an "external" field we still have the orbital magnetic moment.So we can say that the magnetic field that the electron craete by its orbit interacts with the spin magnetic moment and gives it a "potential" energy.

For the afformentioned state we will thus get the energies:{E(0)=initial energy of 2p state,omega =larmour frequancy}

E=E(0)-h(omega), E(0) ,E(0)+h(omega)

So i know that in a porbital there are six possible states.
For j=3/2 we get 4 states and for j=1/2 we get two states.My book says that this orbital-spin interaction has removed the degenaracy of the p state.

My questions are:
1) does the l=1, m=0 state cease to exist?

2)And considering that there are two energy levels and j=3/2 has 4 quantum states and j=1/2 ,2 quantum states,are they not degenerate "because" each now has a unique total angular momentum J?

2. Jun 8, 2010

### DrDu

To question 1) the state l=1 and m=0 still exists but it is no longer an eigenstate of the atomic hamiltonian. These eigenstates are now mixtures of e.g. l=1, m=0 and s=1/2, m_s=1/2 and of l=1, m=1 and s=1/2 and m_s=-1/2. From these states one can construct eigenstates with l=1, s=1/2 and J=3/2 M_J=1/2 and l=1, s=1/2, J=1/2 and M_J=1/2. The coefficients of this mixings are called Clebsch Gordan coefficients and you can either look them up or find them by acting on e.g. l=1, m=1, s=1/2, m_s=1/2 with an angular momentum lowering operator.

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