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I'm having a rather technical question concerning the calculation of the lifetime of a quasiparticle. The reference I use is: http://web.rosario-conicet.gov.ar/IFIR/manuel/condmat/cursos/notes/daneses.pdf on page 244 in the document itself.

I don't really see where the factors in Eq. (14.45) for the scattering into the state k and for scattering out of state k are coming from. I suppose that at the beginning you sum over k,k',q,$\sigma$,$\sigma'$ and then use an argument similar to the one of Equation 2.40 in the same document.

I do understand the theory of second quantization in principle, however for some reason I'm really feeling uneasy when I have to do actual calculation myself. For example I'm not really sure where the $(1-n_{k+q})$, etc.- factors are coming from. I suppose that they could come from the kronecker-deltas since $n_{k}$ is either one or zero for fermions but still I'm not quite sure.

I'd be more than happy if you could provide me with some insight into how this calculation works and how to deal with actual calculations based on second quantization in general.

Thanks in advance. Stan.

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1 Answer 1

You are asking about a formula that is confusing only because it is not really giving a lifetime in the sense of quantum field theoretic decays, like the lifetime of the muon, but a master equation element. These things are different, although related. The master equation is telling you the rate at which probability flows in and out of a given state.

What the master equation is describing is the rate of decay of the occupied state into an occupied state of wavenumber k. The formula doesn't consider it a decay if the electron in the state k gets scattered out and later a new electron takes its place. The latter process is in principle not the same as doing nothing, because of the loss of phase coherence. But for purposes of getting the classical occupation probability $n_k$ it is the same as doing nothing.

The only quantum mechanical part of the formula is the previous equation, which is an application of Fermi's golden rule (the asymptotic contribution of perturbative transition amplitudes to the probability transition rate for times long enough for energy to be conserved):

$$\Gamma_{k+q,k'-q;k',k} = |W(q)|^2 (2\pi) \delta(E_f - E_i)$$

I rearranged the formula slightly to place the $2\pi$ factor next to the $\delta$ function where it belongs. Once you have this differential transition rate, you can integrate it over the final states to find the true transition rate.

This integration is complicated by the fact thatsome of the final k-states are occupied, and so unavailable. Further, there are also transitions into the k-state from other k states.

The master equation for the probability of leaving state k then becomes the first expression the book writes down:

$$2\sum_{k',q} |W(q)|^2 2\pi\delta(E_f-E_i) n_k n_k' (1-n_{k+q})(1-n_{k'-q})$$

The probability of having k-mode occupied is leaking out at this rate, which is just summing the Fermi golden rule contribution over all the available phase space, and over the second particle involved in the scattering, assuming everything is thermally distributed.

The second expression subtracted from it is the Fermi golden rule contribution to entering the state. This tells you how the probability in the state decays, and the factors involved are a purely classical probability argument, no quantum mechanics at all.

There are some assumptions in this. First, that the scattering process is described by Fermi's Golden rule--- this assumes that you can consider the scattering process completed independently to the asymptotic energy-conserving limit, so that no second later quantum process is happening in intermediate stages of the decay, when energy is not conserved. This assumption is reasonable considering that the lifetime is long and the metal is thermalizing.

But there is a confusing later statement about using the formula to compute the decay of a state which is forced to be occupied. In this case, the "in" rate vanishes, since n_k is not the thermal n_k, but 1. The out rate is also modified from what you would naively compute because the occupation probability is 1 at t=0 in this k-state. As n_k evolves with time according to the decay, it relaxes according to the changing rate given by the formula.

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