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In our lecture today, we introduced two kinds of creation and annihilation operators.

I want to restrict myself to the antisymmetric case:

The first operator $a_k^{\dagger}$ creates a state $|k\rangle$ and the action is given by

$a_k^{\dagger} |1,...,N \rangle = \sqrt{N+1} |k,1,...,N \rangle$ , where $|1,...,N \rangle$ were antisymmetrized states(!) ( slater determinants).

Now a couple of minutes later, our lecturer switched to the notion of occupation numbers and introduced for $n_j$ being the occupation of state $k$

$a_k^{\dagger} |n_1,..,n_k,.,.. \rangle = (-1)^{N_{\beta}} \sqrt{ n_{\beta}+1} |n_1,...,n_{k}+1,.. \rangle,$ where $|n_1,..,n_k,.,.. \rangle$ is supposed to be an element of the Fock space and $N_{k}:= \sum_{i=1}^{k-1} n_i$.

Now my problem with this is, that both operators apparently create a new state k, but I don't see why they act so differently( the first one gets me a $\sqrt{N+1}$, whereas the second one does something completely different, although both of them just create a new state $k$.

I mean, I think both of them look somehow plausible, but somehow they seem to contradict each other.

If anything is unclear, please let me know.

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I think you either misunderstood your lecturer or he used a very bad and uncommon notation. I would claim the expression $a_k^{\dagger} |1,...,N \rangle = \sqrt{N+1} |k,1,...,N \rangle$ is simply wrong when understood within the usual notations.

It is unclear what the state $|1,...,N \rangle$ denotes. Now my best guess is that it is in 'first quantization' notation, and that the modes $1, 2,\dots, N$ are all occupied by one fermion (since you mention antisymmetric). But if it is fermionic, it would only be physically allowed if you additionally antisymmetrize it. $S_- |1,...,N \rangle$. This is cumbersome and implicit when written in second quantization in the occupation number representation, which is the main reason why this notation is useful.

One more problem is the action of the creation operator $a_k^\dagger$ on your state. If it is fermionic, the expression is also generally not true, since the state is annihilated if a particle already occupies this mode, i.e. if $l \leq N$. Furthermore, the factor $\sqrt{N+1}$ (where $N$ now smells like a occupation number and not a mode index) can be omitted for fermions, as it is always either 0 or 1. The sign also has to be taken care of for fermions (as you correctly write in the second paragraph) when writing such an expression with a ladder operator.

For you all of your fellow student's sake I hope that this is just an accumulation of many typos, since otherwise somebody coming up with this did not understand some fundamental principles.

Concerning the second paragraph: this is somewhat better, but the correct expression would be $$a_k^{\dagger} |n_1,..,n_k,.,.. \rangle = (-1)^{N_{k}} (1 - n_k )|n_1,...,n_{k}+1,.. \rangle$$ for fermions. In this way the prefactor $ (1 - n_k )$ takes care of annihilating the state if $n_k=1$ before acting on it.

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  • $\begingroup$ You said that this is only defined if we additionally antisymmetrize, but as I tried to say in my question: These states are indeed antisymmetrized states. Does the first equation now make more sense to you? $\endgroup$ – Tokoyo Nov 12 '14 at 8:33
  • $\begingroup$ The state may be defined, but not a valid fermionic state without antisymmetrization. Unfortunately, it does still not make sense to me, since I don't know what the state $|1,...,N \rangle$ is supposed to denote. $\endgroup$ – ulf Nov 12 '14 at 15:14
  • $\begingroup$ Usually, for a many-body state in first quantization, one starts with a set of $M$ single particle orbitals $|1 \rangle, \ldots, |M \rangle$ (each of which is a valid single-particle state). A basis of states for $N$ particles is then given by $|i_1 \rangle, \ldots, |i_N \rangle$, where each $i_n$ is a state index for the $n$-th particle. Thereafter you can antisymmetrize for fermions. $\endgroup$ – ulf Nov 12 '14 at 15:24

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