Wave function of hydrogen atom including spin of nucleus

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.

• The eigen numbers of orbital and internal momenta are not additive (unlike the operators), i.e. $F \neq J + I$; $\hat{\mathbf{F}} = \hat{\mathbf{J}} + \hat{\mathbf{I}}$; $F_3 = J_3 + I_3$ Feb 1, 2012 at 17:31
• Does "the wave function" mean "the wave functions of energy eigenstates"? Or does it mean "any wave function"? Feb 2, 2012 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 $<2P_{\frac{1}{2}},F',F'_{3}|$ Feb 3, 2012 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? Feb 12, 2012 at 11:23
• No $\hat{J}$ is the total angular momentum of the electron and nucleus have zero angular momentum. Feb 12, 2012 at 20:48

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 $2I+1$.
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 $2 \times (2I+1)$.
• Does the non additivity of $\hat{F}$ is the consequence of not using Born Oppenheimer approximation? Feb 1, 2012 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 $\hat{\mathbf{F}}$. Feb 2, 2012 at 8:38