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 $$\right \left| l,m_l,m_s\rangle$$ )

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

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.