Photon angular distribution in hydrogen 2P to 1S transition

[yxgao]

Hi, can someone help me with this problem?
A hydrogen atom is in the 2P state with n=2, l=1, m=1. It emits a photon and makes a transition to the 1S state (n=1, l=0, m=0).

The question is what is the probability distribution of the direction $$(\theta , \phi)$$ of the emitted photon?

The answer is:
$$P(\theta , \phi)=\frac{3}{16\pi} (1+(cos(\theta)^{2})$$

Thanks!






The solution is not very clear but the following facts may be helpful. Can someone please explain this problem clearly to me? I am having a very difficult time understanding this.

The electric field can be expressed in terms of creation and annihilation operators of photons in a box of volume V.

We know that the interaction is $$H_{I} = \vec{E} \cdot \vec{d} = -e \vec{E} \cdot \vec{r}$$ (d is electric dipole moment) so we know the matrix element of a photon is:
$$<i|H_{I}|f> = -e\sqrt{\frac{\hbar c k}{2 \epsilon_0 V}} <2P|\vec{\epsilon} \cdot \vec{r} | 1S>$$

From here, I don't understand what to do. If you can find a similar problem worked out online that would be helpful as well.

 

[dextercioby]

There's something divious with your problem:if the transition probability is given by the time-dependent perturbation matrix element squared:

$$P_{i\rightarrow f}=|\langle i|\hat{H}_{int}|f\rangle|^{2}$$
,then it shouldn't depend of the spherical angles.It should be a number between 0 and 1.After all,that integration,on normal basis,assumes integration wrt to all 3 spherical variables.
Maybe the problem asks for the matrix element computed only as integral over 'r' from 0 to infty and letting the spherical variables nonintegrated.In that case:
$$P_{i\rightarrow f}=\int d\Omega' P(\theta',\phi')$$

$$P_{i\rightarrow f}=|\langle i|\hat{H}_{int}|f\rangle|^{2}=\langle i|\hat{H}_{int}|f\rangle\langle f|\hat{H}_{int}|i\rangle$$
$$=[\int d\Omega' \int_{0}^{+\infty}dr' \psi_{i}^{*}(r',\theta',\phi') \hat{H}_{int} \psi_{f}(r',\theta',\phi')][\int d\Omega'' \int_{0}^{+\infty}dr'' \psi_{f}^{*}(r'',\theta'',\phi'') \hat{H}_{int} \psi_{i}(r'',\theta'',\phi'')]$$
$$=\int d\Omega' P(\theta',\phi')$$

U (actually I ) have found:
$$P(\theta',\phi')=[\int_{0}^{+\infty}dr' \psi_{i}^{*}(r',\theta',\phi') \hat{H}_{int} \psi_{f}(r',\theta',\phi')][\int d\Omega'' \int_{0}^{+\infty}dr'' \psi_{f}^{*}(r'',\theta'',\phi'') \hat{H}_{int} \psi_{i}(r'',\theta'',\phi'')]$$

Use the formula u have for the matrix element and treat those 2 vectors (one being the polarization vector of the emitted radiation,and the other the position vector of the electron) involved in the scalar product wrt to their spherical components.Make use of the fact that the polarization vector has modulus equal to unity.That scalar product should be 'r'.Then the angle dependence will be given only by the spherical harmonic of the initial state ('1s' orbital has no angle dependence).Hopefully u'll get the answer the problem gives u.

Daniel.

 

[Philcorp]

Hrmmm, I am not sure what to think about the reply to this topic. When I read the question, it seems to me that the answer looks like (sort of, I don't have them memorized) something to do with the squares of spherical harmonics. This makes some sense to me since you can write the solution to the simple model of the hydrogen atom as a product of a radial component and the spherical harmonics. Many third and fourth year undergraduates have been tortured by this kind of fun stuff. I don't really want to work through any details here, but i sugest looking towards this method of representing the wavefunctions of each level of the hydrogen atom. Not sure if this was any help...


Philcorp