Why do we use second quantization in condensed matter?
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It's just a convention to write the Dirac Hamiltonian in terms of electron and positron operators. The operator bs†pbps†creates a "hole", i.e. a positron, which is the same as annihilating an electron. So we could just as well define csp=bs†pcps=bps† and write the Hamiltonian in terms of aa and ccinstead of aa and bb. Then everything is expressed in terms of electron operators.
So we are reduced to asking why you need two sets of operators aa and cc. The reason is simply that the dispersion relation of the Dirac equation is E=±p2+m2−−−−−−−√E=±p2+m2, so at any given momentum there are two bands. Hence, you need one operator that creates/annihilates an electron in the top band, and one which creates/annihilates an electron in the bottom band. This would equally well be true in a condensed matter system treated in momentum space, if there are multiple bands.
(As for why this convention is chosen: in relativistic quantum theory, if you write everything in terms of electrons, you are forced to conclude that the vacuum |0⟩|0⟩has all the negative-energy states occupied, since cs†p|0⟩=0cps†|0⟩=0. This was, in fact, Dirac's original picture, but conceptually it's easier to imagine that the vacuum contains no particles. On the other hand, in condensed matter physics, we are used to thinking about our solid-state materials as containing a lot of electrons, so it's actually less confusing to take the opposite point of view.)
hope it will help you dear
up vote12down voteaccepted
It's just a convention to write the Dirac Hamiltonian in terms of electron and positron operators. The operator bs†pbps†creates a "hole", i.e. a positron, which is the same as annihilating an electron. So we could just as well define csp=bs†pcps=bps† and write the Hamiltonian in terms of aa and ccinstead of aa and bb. Then everything is expressed in terms of electron operators.
So we are reduced to asking why you need two sets of operators aa and cc. The reason is simply that the dispersion relation of the Dirac equation is E=±p2+m2−−−−−−−√E=±p2+m2, so at any given momentum there are two bands. Hence, you need one operator that creates/annihilates an electron in the top band, and one which creates/annihilates an electron in the bottom band. This would equally well be true in a condensed matter system treated in momentum space, if there are multiple bands.
(As for why this convention is chosen: in relativistic quantum theory, if you write everything in terms of electrons, you are forced to conclude that the vacuum |0⟩|0⟩has all the negative-energy states occupied, since cs†p|0⟩=0cps†|0⟩=0. This was, in fact, Dirac's original picture, but conceptually it's easier to imagine that the vacuum contains no particles. On the other hand, in condensed matter physics, we are used to thinking about our solid-state materials as containing a lot of electrons, so it's actually less confusing to take the opposite point of view.)
hope it will help you dear
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