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How do I write the wave function of hydrogen atom taking into consideration of nucleus spin? For example consider $2S_{\frac{1}{2}}$ state with nucleus spin $I$, then wave function $\psi=\langle2S_{\frac{1}{2}},F,F_{3}|$ where $F$ is the total angular momentum of hydrogen atom $F=J+I$ and $F_{3}$ is the projection of it along z axis. Now what will be the explicit form of the wave function? Thanks in advance.

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The eigen numbers of orbital and internal momenta are not additive (unlike the operators), i.e. $F \neq J + I$; $\hat{\mathbf{F}} = \hat{\mathbf{J}} + \hat{\mathbf{I}}$; $F_3 = J_3 + I_3$ – Maksim Zholudev Feb 1 '12 at 17:31
Does "the wave function" mean "the wave functions of energy eigenstates"? Or does it mean "any wave function"? – Steve B Feb 2 '12 at 18:54
wave functions of energy eigenstates.More precisely it's the parity violating hamiltonian of hydrogen atom,I need to calculate the matrix element between the states of $2S_{\frac{1}{2}}$ and $2P_{\frac{1}{2}}$ states The eigenstates are $<2S_{\frac{1}{2}},F,F_{3}|$ and $<2P_{\frac{1}{2}},F',F'_{3}|$ – WInterfell Feb 3 '12 at 21:34
What is $\hat{\mathbf{J}}$ here? Is it the orbital momentum of the electron, nucleus or whole atom? Does the nucleus have zero orbital momentum? – Maksim Zholudev Feb 12 '12 at 11:23
No $\hat{J}$ is the total angular momentum of the electron and nucleus have zero angular momentum. – WInterfell Feb 12 '12 at 20:48

1 Answer

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The easy way

If we do not take into account the dependence of the electron state on the spin state of the nucleus, the wavefunction is just a product of electron and nucleus wavefunctions: $$ \psi = \psi_e(\mathbf{r} - \mathbf{R}) \psi_n(\mathbf{R}) $$ Both are spinors of rank 1 (columns of functions). The spinor $\psi_e$ consists of two components. The number of the components of $\psi_n$ depends on the total spin of the nucleus $I$ and is equal to $2I+1$.

The hard way

If the spin of the nucleus affects the electron state, then the total wavefunction is a spinor of rank 2 i.e. a table of functions with dimensions $2 \times (2I+1)$.

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Does the non additivity of $\hat{F}$ is the consequence of not using Born Oppenheimer approximation? – WInterfell Feb 1 '12 at 23:00
@user7429, the operators are additive. The numbers $J$ and $I$ are not. The eigen number of $\hat{\mathbf{F}}^2$ is $F(F+1)$. Here is one of the definitions of $F$. The same for $J$ and $I$. So the sum $J+I$ is not $F$ even though the sum of operators $\hat{\mathbf{J}}$ and $\hat{\mathbf{I}}$ is the operator $\hat{\mathbf{F}}$. – Maksim Zholudev Feb 2 '12 at 8:38

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