# Fermi's Golden Rule and Density of States

1. Sep 15, 2008

### yaychemistry

So we all know, or can look up Fermi's golden rule to be something like:
$$\sigma\left(E\right) \propto \rho\left(E\right) |\langle \psi_i | \mu | \psi_f\left(E\right) \rangle|^2$$

Sigma is the cross section (or can be the transition rate as well), rho is the density of states of the final state and mu is the dipole operator (or other transition operator).

My question relates to the density of states, and what exactly are we to use as the "final" state.

Suppose we are calculating a photodetachment cross section for H and that the electron the S-angular momentum state ground state. Our final state is then a P-wave free electron state, as that is the only final state with a nonzero transition dipole matrix element.

In literature, the formulas for the cross section always use

$$\rho\left(E\right) \propto \sqrt{E}$$

which one can derive from plane waves in a 3-D box (disregarding the fact that the positive energy solutions of the schrodinger equation aren't in fact plane waves).
But, if I limit myself to only l = 1 (e.g. P) angular momentum spherical waves, haven't I essentially put myself into a 1-D box? In that case

$$\rho\left(E\right) \propto 1/\sqrt{E}$$.

So which density of states should I use? Literature picks the 3-D box but I don't see why, since the distribution of l = 1 energy eigenvalues follows the 1-D box results.

2. Sep 15, 2008

### ZapperZ

Staff Emeritus
This depends on the problem that you have. If you are talking about atomic transition, the final state can be anything, but it doesn't mean that the matrix element is non-zero. This is because there is a symmetry consideration when you calculate each of the matrix element as you run the indices f over all possible basis.

If you do this carefully, you'll find out that f must be a state that differs from the initial state by 1, i.e. f = i+1 or i-1. This comes right out of the symmetry of the dipole matrix element.

Zz.

3. Sep 15, 2008

### yaychemistry

First! Thanks for the reply and for your help. I apologize in advance if I seem agitated, but this problem has been bugging me for a while now.

I understand that the matrix element depends on the symmetry of the initial and final states. But I'm not sure how that dictates which density of states to use. Suppose I choose my final state to be a P-wave, since I know that will be the only symmetry-allowed state. Then I will choose $$\rho \propto 1/\sqrt{E}$$.

However, what if I choose my final state to be a plane-wave? Well, then the density of states follows that of plane waves with $$\rho \propto \sqrt{E}$$. However, only the component of the plane wave which has P-symmetry will contribute, so the dipole matrix element will be the same as the previous case, although possibly with a different constant factor out front.

But that's when I hit a quandary. The energy dependence of the cross section for the first case will go as $$\sigma\left(E\right) \propto 1/\sqrt{E}|\mu_P|^2$$
and in the second case it will go as
$$\sigma\left(R\right) \propto \sqrt{E}|\mu_P|^2$$.
This means the answer depends what I consider to be my outgoing wave, but the underlying physics of the problem hasn't changed! In either case I'm shooting photons at the H atom and counting the number of ejected electrons.

4. Sep 15, 2008

### ZapperZ

Staff Emeritus
You can't choose whatever basis states you like. It depends entirely on the problem, like I said earlier.

If this is an atomic energy level for a hydrogenic atom, then your basis functions will be the wave function of the hydrogenic atom. If this is a tunneling problem, then your wavefunction will be the plane wave states on both sides of the barrier, etc. In other words, you just can't use any wavefunction that you like. It has to be the wavefunction of THAT particular problem.

Zz.

5. Sep 15, 2008

### yaychemistry

Ok, so I guess what it boils down to is that I'm unsure what to use as my final state. In the case of photodetachment of H atom, both plane waves and spherical l-waves are asymptotic solutions to the Schrodinger equation for positive energy (neglecting the infinite range of the Coulomb attraction, but this should hold for high enough kinetic energy of the free electron).

With two equally valid solutions of the Schrodinger equation to use, which is the correct one? Do boundary conditions specify plane waves over spherical waves, and if so how? I'm sure it probably has something to do with detectors and such, but I can't figure it out right now. Perhaps I should be calculating the differential cross section and "aiming" an outgoing plane wave at it, and then integrating over all solid angles to get the total cross section?

6. Sep 16, 2008

### olgranpappy

Hi. It might help people (e.g., me) to understand the question if you could explain once more what you want to calculate. For example, what is "photodetachment of H atom"? Presumably,
we shoot in light ("photo") at a neutral hydrogen atom and... what? What are we detecting?--Again, I'm just guessing but presumably the photon gets absorbed and the electron gets detected ("detached").

Also, what operator are we assuming causes the transition? Etc?

If we are detecting free electrons (that got kicked off a H-atom), I would probably start off by calculating the
following rate:

$$R_{1s\to k_f}=\frac{(2\pi)}{\hbar}{|\langle{\bf k_f}|\frac{e}{mc}\vec\epsilon_0\cdot\hat\vec p|1s\rangle|}^2\delta(E_{1s}+\hbar \omega-E_{k_f})\;,$$
where e is the electron charge, m the electron mass, c the speed of light, epsilon_0 the incident polarization, hbar the reduced Planck constant, E's are energies, omega is the incident angular
frequency of the light, p is the momentum operator of the electron, 1s is the initial state and, k_f is the plane wave final state.

7. Sep 17, 2008

### Marty

The way you've set the problem up, I think you should use the spherical waves in your calculation. Because if you try to use plane waves, you will have to integrate over different angles where the coupling between the states varies according to the angle: strongest when it is perpendicular to the incident light. And it is really just this integral over weighted plane waves that gives you the spherical wave.

8. Sep 17, 2008

### yaychemistry

Right, so I think I have the solution and it has everything to do with the proper normalization of basis states.

But to address olgranpappy: You can think of the H atom photodetachment experiment in that that we fire a photon of a specific frequency at an H atom such that the electron is ejected (or detached) e.g.

$$\textrm{H} + h\nu \rightarrow \textrm{H}^+ + e^{-}\left(\vec{k}\right)$$

Where $$\vec{k}$$ is the wavevector of the free electron, and $$k^2/2m = h\nu - 13.6\textrm{eV}$$ where 13.6 eV is the binding energy of H atom. For the transition operator we are assuming the dipole approximation holds here, for better or worse.

One of the strategies of dealing with the delta function in the rate formula provided by olgranpappy is that we integrate over energy: e.g. we acknowledge that our detector can only determine the electron energy to within a small uncertainty, thus we must include all of the states in a small dE of energy, which leads to the density of states between E and dE ($$\rho\left(E\right)$$) in the formula I first provided. The cross section is proportional to the rate of ejecting an electron, and what I'm interested is the shape of the energy dependence of the cross section, so I'm not too worried about which constants, etc, live out front of the equation.

So, to calculate the rate we need both the final state and its density of states. As it turns out both plane wave states and spherical wave angular momentum states are perfectly valid and will provide the same answer as long as we carry out things in a correct manner.

First, the spherical wave states:
We have to specify what our outgoing states look like asymptotically. Also, to properly normalize we must operate in a spherical box of radius R and let the size of the box go to infinity. Thus the proper basis of final states look like:
$$|f^\textrm{sph}_l \left(k\right)\rangle = 1/\sqrt{N}\frac{\sin\left(kr + l\pi/2\right)}{r}$$
where the normalization N depends on R but not on k, and the superscript "sph" specifies a spherical wave basis.
The correct density of states for these states is $$\rho\left(E\right) \propto 1/\sqrt{E}$$ as stated earlier. And the transition dipole matrix element is
$$\mu^{\textrm{sph}} = \langle i | \mu | f^\textrm{sph}_\textrm{p}\rangle$$
where we note that selection rules only allow P-waves (l = 1).
Thus our transition rate (or cross section) goes as
$$\sigma \propto 1/\sqrt{E} |\mu^{\textrm{sph}}|^2$$.

For the plane waves we can calculate the rate of electrons going in one direction (i.e. into one final plane wave state) but we have to integrate that over all directions, like Marty stated, in order to get the complete answer, since the plane waves are a complete set.
For the plane waves the density of states from my first post $$\rho\left(E\right) \propto \sqrt{E}$$ is correct. However, we must ask 1) what is the proper normalization and 2) what does $$\mu^\textrm{pw}$$ look like ("pw" means plane wave basis)?
Again, we can go into a box normalization so our plane waves look like
$$|f^{\textrm{pw}}\left(k\right)\rangle = 1/\sqrt{N}\exp\left(i\vec{k}\cdot\vec{r}\right)$$
where again N only depends on the size of the box, but not on k. But here's the tricky part. Like I noted the selection rules of photons say that l can only change by one, so the dipole matrix element only connects with the portion of the plane wave that contains p-like waves. That being said, the dipole matrix element for the plane wave, $$\mu^\textrm{pw} \ne \mu^\textrm{sph}$$ as I originally thought, even after averaging over all angles. This is what lead me into the original problem.
If I had taken the time to expand plane waves in a basis of spherical waves I would have realized that
$$\exp\left(i\vec{k}\cdot\vec{r}\right) = \sum_{l,m} a_{l,m}Y_{l,m}j_l\left(kr\right)$$
where Ylm are the spherical harmonics and $$j_l\left(kr\right)$$ is a spherical bessel function AND asymptotically one should note that
$$j_l\left(kr\right) \rightarrow \frac{\sin\left(kr + l\ip/2\right)}{kr} \propto 1/k|f^\textrm{sph}\rangle$$
and that last proportional relationship is the crux of the problem and leads to the fact that
$$|\mu^\textrm{pw}|^2 \propto 1/k^2|\mu^\textrm{sph}|^2 \propto 1/E|\mu^\textrm{sph}|^2$$.
Thus the final cross section/golden rule rate in the plane wave basis goes as
$$\sigma \propto \sqrt{E}|\mu^\textrm{pw}|^2 \propto \sqrt{E}/E|\mu^\textrm{sph}|^2 = 1/\sqrt{E}|\mu^\textrm{sph}|^2$$
which is the same as the spherical basis cross section!
So it ALL has to do with normalization!!

Thanks everyone for your input and help!

9. Sep 17, 2008

### olgranpappy

nice. I'm glad you figured it out. Cheers.