Suppose Hamiltonian of the system in 2nd quantised notation is $H= t\sum_{x=1}^{N}d_{x}^{\dagger}d_{x}$.
Does this mean that eigenstates of the Hamiltonian is $N$-fold degenerate with energy $t$?
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$\begingroup$ Could you clarify if $d$ is bosonic or fermionic? That is, which of $[d^\dagger,d]=1$ or $\{d^\dagger,d\}=1$ holds? $\endgroup$– jacob1729Commented May 12, 2021 at 12:45
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$\begingroup$ 'd' is a fermionic operator. $\endgroup$– RakeshCommented May 12, 2021 at 13:02
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2 Answers
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Yes, indeed, there are $N$ identical states, each having energy $t$.
Remark:
Note that usually $t$ is used for a hopping integral, as in
$$
H= \epsilon_0\sum_{x=1}^Nd_x^\dagger d_x + t\sum_{x=1}^N\left(d_x^\dagger d_{x+1} + h.c.\right)
$$
Remark
In view of the comments and the other answer, it is necessary to note that what we mean here depends on the context. Specifically:
- one may speak about a single-particle Hamiltonian in second quantized notation (or simply the fact that the system is composed of $N$ identical orbitals), which si how I understood the OP
- one may also speak about many-body states, as suggested by the use of the second quantization.
Note however that second quantization is widely used to deal with one-particle problems, so one cannot reasonably claim that only one interpretation is correct.
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$\begingroup$ This is misleading. Assuming $d$ to be fermionic in OP's post there are $2^N$ states. The phrase 'identical states' doesn't have any particular meaning. Of these $2^N$ states, there are $N$ different energy levels which have degeneracies $\binom{N}{r}$, thus only the first excited energy level is $N$ fold degenerate. $\endgroup$ Commented May 12, 2021 at 12:47
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$\begingroup$ @jacob1729 I see what you mean. It depends on how one interprets the question. $\endgroup$– Roger V.Commented May 12, 2021 at 12:54
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$\begingroup$ I don't know of any interpretation of 'degeneracy' that would be correct to say it is $N$-fold unless you very specifically restrict to only states on energies $0,t$. $\endgroup$ Commented May 12, 2021 at 12:55
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$\begingroup$ @jacob1729 Perhaps you are used to this form of the Hamiltonian in a specific context. It could be just a one-particle Hamiltonian in second-quantized notation - not uncommon when studying, e.g., scattering problems in solid state. $\endgroup$– Roger V.Commented May 12, 2021 at 13:00
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1$\begingroup$ I think the above Hamiltonian "H" gives single particle eigenstates , which have same energy "t". The number of such eigenstates is "N". From this we can construct multiparticle states, by filling various states with a fermionic particle. There will be degeneracy in these multi-particle states as well. $\endgroup$– RakeshCommented May 12, 2021 at 13:16
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The degeneracy of the state with energy $nt$ is the number of ways you can make the integer $n$ out of $N$ numbers eg if $N=3$ and then we can get $4t$ as $$ 4+0+0\\ 0+4+0\\ \vdots $$ etc, and $$ 3+1+1\\ 1+3+1\\ \vdots $$ or $$ 2+0+2\\ 0+2+2\\ \vdots $$ I leave it to you to compute the degeracy of the level $E=nt$!
answered May 12, 2021 at 12:07
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$\begingroup$ Remark that this answer assumes the system is a collection of harmonic oscillators (ie bosons). The notation in the OP is ambiguous but I would tend towards assuming it means fermions, in which case obviously $4t$ isn't allowed by Pauli. $\endgroup$ Commented May 12, 2021 at 12:52
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$\begingroup$ (But this is a fairly simple change that the OP should be able to adapt to their case. The fermion combinatorics are simply a bit easier.) $\endgroup$ Commented May 12, 2021 at 12:54