# What fraction of electrons is captured in semiconductor defects?

I'm having trouble with the following exercise:

Some point defects (impurities, holes, etc.) in semiconductors can trap an electron in a localized state with energy $E_{1}$ and spin $-1/2$. A second electron captured by the same defect would face the coulombian repulsion of the first electron and its energy would be $E_{2}>E_{1}$. If $E_{2}>0$ the second electron is repelled; if $E_{2}<0$, the electron can be bound to the defect.

Prove that the probability that an electron is captured by a simple defect is

$$\frac{1}{\frac{1}{2}\exp\left(E_{1}/kT\right)+1}$$

and that the mean number of electrons in a double defect ($E_{2}<0$) will be

$$2\frac{\exp\left(-\beta\left(E_{1}-\mu\right)\right)\left(1+4\exp\left(-\beta\left(E_{2}-2\mu\right)\right)\right)}{1+2\exp\left(-\beta\left(E_{1}-\mu\right)\right)+4\exp\left(-\beta\left(E_{1}+E_{2}-2\mu\right)\right)}$$

where $\mu$ is the chemical potential of the electrons, $\beta = 1/kT$, $T$ is the temperature, and $k$ is Boltzmann's constant.

I'll post what I have done in an answer (this one). The expressions I obtain are very different from the ones suggested by the exercise and I think there's something wrong with it, so I wanted to get confirmation.

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The complicated formula is correct, if you assume there are 2 and 4 spin states for 1 and 2 electrons respectively. the simple formula is meaningless and wrong, but you can figure out what the author was thinking. Both exercises are badly written, and you should get a better book. –  Ron Maimon May 18 '12 at 17:32

You are almost right, but you made two mistakes:

• (trivial error) you didn't take the doubled electrons in the defect into account-- you need a 2 in front of the second term in the numerator in your expression for $\langle n_1\rangle$
• you ignored the possible spin states of the bound electrons, and the book wants you to take this into account.

The easy way to get your result is to consider one defect to be a subsystem of the whole thing, capable of holding 0,1,2 spinless electrons, and then the grand-canonical partition function is

$$\Phi = (1 + e^{-\beta(E_1 - \mu)} + e^{-\beta(E_1 + E_2 -2\mu})$$

To get the full grand canonical partition function, you just consider n independent single wells, reproducing the combinatorics of your answer. The expected number of electrons is found by averaging 0,1,2 using the weights above, to get

$$\langle n_1 \rangle = {e^{-\beta(E_1-\mu)} + 2 e^{-\beta(E_1 + E_2 - 2 \mu)} \over 1 + e^{-\beta(E_1 -\mu)} + e^{-\beta(E_1 + E_2 - 2\mu)}}$$

The two being required because there are 2 electrons in the 2 electron state. This is different from your answer. The answer factorizes as follows

$$\langle n_1 \rangle = { e^{-\beta(E_1-\mu)} (1 + 2 e^{-\beta(E_2-\mu)}) \over 1+ e^{-\beta(E_1-\mu)} + e^{-\beta(E_1+E_2 -2 \mu)}}$$

Now you can see what the book is assuming:

• There are two different degenerate electronic spin states for the one-bound electron problem $E_1$ (this is sensible)
• There are four different degenerate electronic spin states for the two-bound electron states of energy $E_1 + E_2$ (this is physically dubious)

If you make the book's unstated assumptions, there are really 7 states for the single defect subsystem: no electrons bound, two different E_1 states with one electron of each spin bound, four different E_1+E_2 states with two electrons bound, each of their spins independent.

The grand canonical sum for the single defect

$$\Phi = 1 + 2 e^{-\beta(E_1 - \mu)} + 4 e^{-\beta(E_1+E_2 -2\mu)}$$

While the numerator now has 2 and 4 decorating factors

$${1*2e^{-\beta(E_1 -\mu)} + 2*4 e^{-\beta(E_1 + E_2 - 2\mu)}\over \Phi}$$

Where the factor 1*2 means two different (spin) states with one electron, and 2*4 is four different (spin) states, each containing 2 electrons. There is only one state with no electrons. The spin degree of freedom makes occupation more probable at a given chemical potential, because there are more occupied states. Factoring this reproduces the book's answer.

In a real system, I would have expected the double occupied state to contain two electrons of necessarily opposite spin, since the most likely way for a microscopic defect to house two electrons is in an overlapping spatial wavefunction, which requires opposite spins. This would split the triplet from the singlet in the two-electron problem, and would change the answer. But the book doesn't assume this, and this is why the book got the answer that it did. You must take the spin degeneracy of a single electron into account, but in a real-world problem, opposite spins tend to have much lower energies in a localized configuration as same-spins, and the answer in this case is that the multiplicities of the states are 0 electrons/1-state, 1 electron/2-states, 2-electrons/1-state, and the expected value would be

$$\langle n_1\rangle = {2e^{-\beta(E_1-\mu)} + 2 e^{-\beta(E_1 + E_2 -2\mu)} \over 1+ 2^{-\beta(E_1-\mu)} + e^{-\beta(E_1 + E_2 - 2\mu)}}$$

As for "prove the probability that an electron is captured by a single defect is ..." this is a terrible question--- the probability of it being captured under what conditions? If you have one electron in a crystal, it isn't going to be captured by a defect, and the concept of temperature is ridiculous.

The meaning of this is something else--- it means, consider an electron which is at temperature t, and can be in exactly 3 states, unbound, bound to the defect with energy $E_1$ and spin up, bound to the defect with energy $E_1$ with spin down. Then what is the probability that the electron is bound? In this problem, you reproduce the book's answer. But this problem is insanely artificial--- why should the unbound electron have only one state it can be in? This is revealing the author's confusion between grand-canonical systems, where you can think about the states being occupied or unoccupied separately, with a chemical potential cost, and single-particle pictures, where you need to know how many states there are overall to figure out the probability to be in a given state (which is tantamout to figuring out the effective chemical potential corresponding to a given number of electrons).

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I'm sorry I missed the bounty deadline. I was away for the weekend. Now I can only reward you 200 points (instead of the original 100), which would severely amputate my reputation, but would only add 0.8% to yours. Hope you understand ;) Thanks a lot for the reply! –  becko May 20 '12 at 18:51
@becko: It's no big deal, I was just happy I could get an answer to you in time, you seemed to really want to know. For some reason, the first nonsense question really threw me off regarding the second sensible question, so I couldn't answer immediately. –  Ron Maimon May 20 '12 at 19:40

Let there be $n$ defects in the semiconductor; let $n_{1}$ be the number of defects occupied by a single electron and let $n_{2}$ be the number of defects occupied by two electrons. The number of trapped electrons is then $n_{1}+2n_{2}$ and their energy is $n_{1}E_{1}+n_{2}\left(E_{1}+E_{2}\right)$. Considering the system of trapped electrons as an open subsystem of the semiconductor, the grand-partition function is given by

$$\Phi=\sum_{n_{1},n_{2}}M\left(n_{1},n_{2}\right)e^{-\beta\left(n_{1}E_{1}+n_{2}\left(E_{1}+E_{2}\right)-\mu\left(n_{1}+2n_{2}\right)\right)}$$

where $M\left(n_{1},n_{2}\right)$ is the multiplicity of states, given by:

$$M\left(n_{1},n_{2}\right)=\frac{n!}{n_{1}!n_{2}!\left(n-n_{1}-n_{2}\right)!}$$

because there are $M\left(n_{1},n_{2}\right)$ ways of distributing the defects between simple-occupied, double-occupied, and empty. Performing the sum:

$$\Phi=\left(1+e^{-\beta\left(E_{1}-\mu\right)}+e^{-\beta\left(E_{1}+E_{2}-2\mu\right)}\right)^{n}$$

The expected number of electrons in single defects is then

$$\left\langle n_{1}\right\rangle =n\frac{e^{-\beta\left(E_{1}-\mu\right)}+e^{-\beta\left(E_{1}+E_{2}-2\mu\right)}}{1+e^{-\beta\left(E_{1}-\mu\right)}+e^{-\beta\left(E_{1}+E_{2}-2\mu\right)}}$$

This is obviously different from the suggestion in the exercise. Who's right?

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