I have the same question as the person here: Action of Fermionic Creation and Annihilation Operators

The question actually wasn't anwered, because using anticommutation relations between creation $c_\nu^\dagger$ and annihilation $c_\nu$ operators means knowing their explicit form.

Books, that I read, just give the following definitions without any remark: $$c_\nu^\dagger\left | n_1,n_2,\dots,n_\nu,\dots \right \rangle=\left ( -1 \right )^{\sum_{i<\nu}n_i}(1-n_\nu)\left | n_1,n_2,\dots,n_\nu+1,\dots \right \rangle$$ $$c_\nu\left | n_1,n_2,\dots,n_\nu,\dots \right \rangle=\left ( -1 \right )^{\sum_{i<\nu}n_i}n_\nu\left | n_1,n_2,\dots,n_\nu-1,\dots \right \rangle$$

The sense of the multipliers $(1-n_\nu)$ and $n_\nu$ is clear: there can be only $1$ or $0$ fermions in the $\nu$-th one-particle state.

The question is how to obtain the factor $\left ( -1 \right )^{\sum_{i<\nu}n_i}$ without knowing anticommutation relations between $c_\nu$ and $c_\nu^\dagger$?

I found a hint in the book "Feynman Diagram Techniques in Condensed Matter Physics" of Radi A. Jishi (chapter 3.4.1). He uses many-particle state representation by the Slater determinant $\left | \phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu_N}\right \rangle$, where $N$ is current number of fermions and $\phi_{\nu_i}\in\left \{ \phi_1,\phi_2,\dots,\phi_\nu,\dots \right \}$ is a one-particle state of the $i$-th fermion. He claims that: $$c_\nu^\dagger\left | \phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu_N}\right \rangle=\left |\phi_{\nu}\phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu_N}\right \rangle\quad \text{if $\nu_i\neq\nu\quad \forall i$}$$ $$c_\nu\left |\phi_{\nu} \phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu_N}\right \rangle=\left |\phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu_N}\right \rangle$$ (a one-particle state on the new fermion $\phi_{\nu_{N+1}}=\phi_{\nu}$)

It means that $c_\nu^\dagger$ adds the first row to the Slater determinant with the fucntion $\phi_{\nu}$, and $c_\nu$ removes the first row from the Slater determinant if this row contains $\phi_{\nu}$. Then it's clear why the factor $\left ( -1 \right )^{\sum_{i<\nu}n_i}$ appears: $$\left | n_1,n_2,\dots,n_\nu,\dots,n_N,\dots \right \rangle=\left |\phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu}\dots\phi_{\nu_N}\right \rangle=\left ( -1 \right )^{\sum_{i<\nu}n_i}\left |\phi_{\nu}\phi_{\nu_1}\phi_{\nu_2}\dots\phi_{\nu_N}\right \rangle$$ But then I don't understand, why a one-particle state of a created or annihilated fermion has to be in the first row of the Slater determinant.

Anyway, I'll accept your answer even if it doesn't use the Slater determinant

  • $\begingroup$ Can you clarify your starting point? The Slater determinant only makes sense if you've already decided that the many-particle states need to be totally antisymmetric. Are you asking how the total antisymmetry of many particle states implies the anticommutation relations and yields the minus sign in question? $\endgroup$
    – J. Murray
    Dec 11, 2019 at 16:36
  • $\begingroup$ @J.Murray, yes, indeed, using the Slater determinant means using an antysymmetric wavefucntion. I want to understand, how to obtain expressions for operators knowing only antisymmetric property of the fermionic many-particle state. And there is an additional question, which I didn't mention hoping to understand this after the answer on the written question: how are anticommutation relations derived? $\endgroup$
    – Olexot
    Dec 11, 2019 at 17:43

1 Answer 1


But then I don't understand, why a one-particle state of a created or annihilated fermion has to be in the first row of the Slater determinant.

It doesn't have to be. The order of the rows or columns in a Slater determinant is not physically meaningful; it's only important that you set it up consistently. This is easy to see if you remember that exchanging any two rows or columns of a determinant only changes its sign, not its value. So you could also formulate it so that your creation operator appends a new row at the bottom, and the annihilation operator removes it. Same difference, save for a possible sign change if you compare it to the "top" version.

With this in mind, the meaning of the $(-1)^{\sum...}$ prefactor can be explained in very simple terms as well; you've already hinted at it in your question. It's the sign change from the successive row swaps that you need to do in order to move a state from somewhere in the middle of the determinant to the position where your operators act on it. Again, whether that be the first or last row, or just any other position in between, is up to you. I assume it's always done to the first row because it makes your manipulations more easy to read when writing the wavefunction in the "linear" form $|\phi_\nu ... \rangle$, because then all the interesting stuff happens at the front of that term instead of its back.

  • 1
    $\begingroup$ Thank you for the answer! Indeed, swapping the rows positions means changing the sign, that's why I wrote the last formula. Unfortunately, I don't understand why it's necessary to operate with a certain row. Why can't we say that $c_\nu^\dagger$ creates the state exactly there where it should be (regarding occupation numbers representation)? For example, if we have one fermion in $\phi_3$ (the first row) and another fermion in $\phi_8$ (the second row), then $c_5^\dagger$ adds the row between the first and the second rows. But if the 2nd fermion in $\phi_4$, then $c_5^\dagger$ adds to the end $\endgroup$
    – Olexot
    Dec 11, 2019 at 18:02
  • $\begingroup$ @Olexot, "Why can't we say that $c^{\dagger}_\nu$ creates the state exactly there where it should be". Fair enough, and accordingly we have $\left |1, 1 \right \rangle = c^{\dagger}_1c^{\dagger}_2\left |0, 0 \right \rangle$ and also $\left |1, 1\right \rangle = c^{\dagger}_2c^{\dagger}_1\left |0, 0 \right \rangle$, thus you proved that $c^{\dagger}_1c^{\dagger}_2= c^{\dagger}_2c^{\dagger}_1$. $\endgroup$
    – MadMax
    Dec 11, 2019 at 18:16
  • $\begingroup$ @MadMax, good counterexample. But how to obtain the known result? Imagine, we're developing new field in physics. Then we don't have a straightforward answer whether $\left \{ c_\nu^\dagger,c_\mu^\dagger \right \}=0$ or $\left [ c_\nu^\dagger,c_\mu^\dagger \right ]=0$. At the moment we have only antisymmetric property of the fermionic wave function. Then, there is something in my definition that contradicts the antisymmetricity. It's not clear for me what exactly is, because the total antisymmetricity is kept $\endgroup$
    – Olexot
    Dec 11, 2019 at 18:36
  • $\begingroup$ In this case, we obtain the known result because we know that fermionic wavefunctions must be antisymmetric. In a way, you could say that the whole (fermionic) second quantization formalism that we are discussing here was built around that principle. And having the operators always act on the same position in the determinant ties in with that antisymmetry requirement. $\endgroup$
    – Antimon
    Dec 11, 2019 at 20:37

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