Emission in hydrogen atom: recoil and photon properties

[deccard]

I have been trying to picture the whole process of a photon emission by a atom. So to have good understanding what is going on, I have came up with following experimental setup. A single hydrogen atom in excited state ^2\!P_{1/2}, which has been orientated with a magnetic field so that precession of its total angular momentum is pointing to +z axis. The hydrogen atom is located at rest in the middle of a hollow spherical detector with radius r=1cm.

Now, without of the presence of any external magnetic field, the excited state is relaxed to the ground state in the dipole transition ^2\!P_{1/2}\rightarrow^2\!\!S_{1/2} with an emission of a photon with wavelength \lambda=1 215.674\textrm{\AA}.

Okay now to the questions:

1.
As the emitted photon has momentum p=h/\lambda, we get that the hydrogen atom is given recoil p=h/\lambda=mv. So the time between the detection of the photon and detection of the hydrogen atom at the detector is

t=\frac{r\lambda m}{h}=3.1\textrm{ms},

where m is the mass of hydrogen atom.

right?

2.
Because J=j=l+s=1/2, the state ^2\!P_{1/2} can have following configurations with notation\right \left| l,m_l,m_s\rangle
\right \left| 1,+1,-1/2\rangle
\right \left| 1,-1,+1/2\rangle
\right \left| 1,0,+1/2\rangle
\right \left| 1,0,-1/2\rangle

and so, as the spin doesn't change, the transition can be one of these

\right \left| 1,+1,-1/2\rangle \rightarrow \right \left| 0,0,-1/2\rangle
\right \left| 1,-1,+1/2\rangle \rightarrow \right \left| 0,0,+1/2\rangle
\right \left| 1,0,+1/2\rangle \rightarrow \right \left| 0,0,+1/2\rangle
\right \left| 1,0,-1/2\rangle \rightarrow \right \left| 0,0,-1/2\rangle

right?

3.
Is the reason for the unchanging spin the need of \Delta j=\pm1? Photon has spin \pm1 and in dipole transition we must have \Delta l=\pm1. Which would mean that we are left with \Delta s=0.

I will continue my questions when these are answered.

 

[alxm]

1. Seems fine with me. Kind of a contrived situation, though.

2. Seems fine.

3. Right. (assuming LS-coupling)

 

[deccard]

Now, I am little bit confused. I have started to doubt the permissibility of the two transitions below. (With \right \left| l,m_l,m_s\rangle )

<br /> \right \left| 1,0,+1/2\rangle \rightarrow \right \left| 0,0,+1/2\rangle<br />
<br /> \right \left| 1,0,-1/2\rangle \rightarrow \right \left| 0,0,-1/2\rangle<br />

In the transition ^2\!P_{1/2}\rightarrow^2\!\!S_{1/2} we have \Delta J=0, so a transition which has \Delta m_{j}=0 cannot occur. Can I calculate \Delta m_{j} as

\Delta m_{j} = m_{ji}-m_{jf}=m_{li}+m_{si}-(m_{lf}+m_{sf})

where indices i f represent initial and final states respectively?

If I can, this would give for the transitions \Delta m_{j}=0, which would then make the transitions forbidden.

On the other hand in the transition we have \Delta l=1 and \Delta m_{l}=0, which should be enough to make the transition allowed.

So... If some one could elaborate this little more.