# Hydrogen 2p3/2 -> 1s1/2 transition polarisation and angular distribution

Could you please help me. I have to calculate the intensity angular and polarisation distribution in hydrogen electric dipole transition $\text{2p}_{3/2}\rightarrow \text{1s}_{1/2}$. To do this I have to calculate the transition dipole moment $\langle \text{out}|q_e\hat{\vec{r}}|\text{in}\rangle$ But nothing is said about the quantum number $j_z = m_j$. Does it mean that I have 8 (3, in fact, but still not 1) problems for each transition $|n=2,l=1,j=3/2, m_j=\pm3/2\rangle \rightarrow |n=1,l=0,j=1/2, m_j=\pm1/2 \rangle$ $|n=2,l=1,j=3/2, m_j=\pm3/2\rangle \rightarrow |n=1,l=0,j=1/2, m_j=\mp1/2 \rangle$(forbidden) $|n=2,l=1,j=3/2, m_j=\pm1/2\rangle \rightarrow |n=1,l=0,j=1/2, m_j=\pm1/2 \rangle$ $|n=2,l=1,j=3/2, m_j=\pm1/2\rangle \rightarrow |n=1,l=0,j=1/2, m_j=\mp1/2 \rangle$

Than I can write those wavefunctions through general hydrogen wavefunctions and spin part using Clebsch–Gordan coefficients:

$\lvert l=1,j=3/2, m_j=3/2\rangle = \lvert l=1,m_l=1\rangle\lvert\uparrow\rangle$ $\lvert l=1,j=3/2, m_j=1/2\rangle = \sqrt{\frac{2}{3}}\lvert l=1,m_l=0\rangle\lvert\uparrow\rangle+ \sqrt{\frac{1}{3}}\lvert l=1,m_l=1\rangle\lvert\downarrow\rangle$ $\lvert l=0,j=1/2, m_j=1/2 \rangle = \lvert l=0,m_l=0\rangle\lvert\uparrow\rangle$

analogues for $m_j<0$ with only change of $m_l$ and spin to contrary.

Then I can calculate transition dipole moments (charge dropped, any bra has $\langle n=2\rvert$, any ket - $\lvert n=1\rangle$):

$\langle l=0,j=1/2, m_j=1/2\rvert \hat{\vec{r}}\lvert l=1,j=3/2,m_j=3/2 \rangle=$ $=\langle l=0,m_l=0 \rvert \hat{\vec{r}} \lvert l=1,m_l=1 \rangle=$ $=\text{assuming the exact configurational representation}= \frac{2^7}{3^5}a_0 \overrightarrow{\{1,i,0\}}=:\vec{a}$ $\langle l=0,j=1/2, m_j=-1/2\rvert \hat{\vec{r}}\lvert l=1,j=3/2,m_j=-3/2 \rangle=\vec{a}^{*}$ $\langle l=0,j=1/2, m_j=1/2 \rvert \hat{\vec{r}}\lvert l=1,j=3/2,m_j=1/2\rangle=$ $= \left(\langle l=0,m_l=0\rvert\langle\uparrow\rvert\right) \hat{\vec{r}} \left(\sqrt{\frac{2}{3}}\lvert l=1,m_l=0\rangle\lvert\uparrow\rangle+ \sqrt{\frac{1}{3}}\lvert l=1,m_l=1\rangle\lvert\downarrow\rangle \right) = 0$, what told me @Sofia, but I still do not clearly understand, why this is obvious. $\langle l=0,j=1/2, m_j=-1/2 \rvert \hat{\vec{r}}\lvert l=1,j=3/2,m_j=-1/2\rangle=0$ $\langle l=0,j=1/2, m_j=-1/2 \rvert \hat{\vec{r}}\lvert l=1,j=3/2,m_j=1/2\rangle=$ $= \left(\langle l=0,m_l=0\rvert\langle\downarrow\rvert\right) \hat{\vec{r}} \left(\sqrt{\frac{2}{3}}\lvert l=1,m_l=0\rangle\lvert\uparrow\rangle+ \sqrt{\frac{1}{3}}\lvert l=1,m_l=1\rangle\lvert\downarrow\rangle \right) = \sqrt{\frac{1}{3}}\vec{a}$ $\langle l=0,j=1/2, m_j=1/2 \rvert \hat{\vec{r}}\lvert l=1,j=3/2,m_j=-1/2\rangle=\sqrt{\frac{1}{3}}\vec{a}^{*}$ $\langle l=0,j=1/2, m_j=-1/2 \rvert \hat{\vec{r}}\lvert l=1,j=3/2,m_j=3/2\rangle=$ $=\left(\langle l=0,m_l=0\rvert\langle\downarrow\rvert\right)\hat{\vec{r}}\left(\lvert l=1,m_l=1 \rangle\lvert\uparrow\rangle\right) =0$ (forbidden) $\langle l=0,j=1/2, m_j=1/2 \rvert \hat{\vec{r}}\lvert l=1,j=3/2,m_j=-3/2\rangle=0$

So, I analysed all cases and now can write $I(p=\{L,R\}, \theta,\varphi)\propto \left|\vec{a} \vec{e}_p(\theta,\varphi)\right|^2$, were $\vec{e}_p(\theta,\varphi)$ is polarisation ort ( $\vec{e}_R(\theta,\varphi)=\{-\cos\theta \cos\varphi, -\cos\theta \sin\varphi, \sin\theta\},$ $\vec{e}_L(\theta,\varphi)=\{\sin\varphi, -\cos\varphi, 0\}$). But doing this I will receive asymmetric distributions (there is dependency on $\theta$ and no symmetry $L\leftrightarrow R$): $I(R,\theta,\varphi)=I_0\cos^2\theta,\text{ }I(L,\theta,\varphi)=I_0$ that is strange as the problems looks totally symmetric.

Is it true, that we can't determin the intensity angular and polarisation distribution in hydrogen for transition $\text{2p}_{3/2}\rightarrow \text{1s}_{1/2}$ in general? What will be observed in the experiment?

• Unless I have forgotten something trivial about the hydrogen problem I can't see why one could not calculate the angular distribution and polarization. Is the symmetry broken by a magnetic field? I think so, so you may have to deal with that one by one. But since I have forgotten almost everything about spin-orbital angular momentum coupling, I have to pass on real help with the problem. – CuriousOne Dec 27 '14 at 22:03
• @Basil Some transitions that you mention are impossible: from +1/2 to +1/2 and from -1/2 to -1/2. The photon is considered as having angular momentum $+\hbar$ or $- \hbar$. There remain 4 possibilities. But I believe that there has to be symmetry between the cases $3/2 \to 1/2$ and $-3/2 \to -1/2$. Also between $1/2 \to -1/2$ and $-1/2 \to 1/2$. – Sofia Dec 27 '14 at 22:05
• @Basil : so you have to calculate the angular intensity for only 2 cases. As to polarization, it is circular in all these cases. From bigger $m_z$ to lower you have right polarization, including the case $1/2 \to -1/2$, and from lower to bigger, left circular polarization, including the case from $-1/2 \to 1/2$. – Sofia Dec 27 '14 at 22:14
• @Sofia Thank you for answer. Sure, there is a symmetry in cases $3/2→1/2$ and $−3/2→−1/2$ and also between $1/2→−1/2$ and $−1/2→1/2$, I mean this by writing $\pm$. But I thought that transition with $\Delta m_j = 0$ is not forbidden and corresponds to linear polarization (please note, that $j$ is the total moment ($\hat{\vec{J}} = \hat{\vec{L}}+\hat{\vec{S}}$ )? I understand, that due to the symmetry $m_j$ shold not be important, but for some reason I need it in my calculations. Should it be given several answers depending on the transition? What will be observed in the experiment? – Basil Dec 27 '14 at 22:36
• Probably, nothing is said about $m_z$ because no magnetic field is present. – Sofia Dec 27 '14 at 22:37