This question comes in consequence of another one.

I want to stress a problem that none of the answers addressed it. For making my problem more understandable let me first remind a well-known state, the spin-singlet of spin $1/2$ fermions,

$$|S\rangle = \frac {|\uparrow\rangle |\downarrow\rangle - |\downarrow\rangle |\uparrow\rangle}{\sqrt {2}}. \tag{1}$$

The two fermions are completely identical, though, we try to keep track of their identity by writing in each product first the fermion 1 and second the fermion 2. The position of the fermion in the product keeps track of the identity of the fermion. If we interchange the fermions,

$$|S\rangle = -\frac {|\downarrow\rangle |\uparrow\rangle - |\uparrow\rangle |\downarrow\rangle}{\sqrt {2}}, \tag{2}$$

the state changes sign.

Now, to my problem. Let's begin with a state comprising a single fermion with spin up

$$\psi_0 = |\uparrow \rangle \tag{3}$$

A) To this state I add a new fermion. From now on, I will refer to the old fermion as fermion 1, and to the newly added fermion as fermion 2. If the twob fermions were distinguishable, their positions in any product should be according to their names, $a_{\downarrow}^{\ \dagger} \psi_0 = |\uparrow \rangle |\downarrow\rangle$. But they are not independent. They arrange themselves anti-symmetrically, in the state $|S\rangle$. None of them has anymore a well-defined spin-projection, we can't say anymore that the fermion 1 (old fermion) is in the spin-up state.

In continuation I apply the annihilation operator $a_{\uparrow}$. Note again, it is not clear which one of the fermions will be destroyed, 1 or 2. So the state we get, rigorously speaking, is something like this

$$|\psi_1\rangle = \frac {|0_{\uparrow}\rangle |\downarrow\rangle - |\downarrow\rangle |0_{\uparrow}\rangle}{\sqrt {2}}. \tag{4}$$

There is no such Fock state, the Fock state is

$$|\psi_1\rangle = |\downarrow\rangle \tag{5}$$.

The form $(4)$ just stresses that we don't know which one of the fermions was destroyed.

B) However, starting from the state $\psi_0$ and applying the two operations in inverse order, the situation is totally different. First of all one creates vacuum,

$$a_{\uparrow} \psi_0 = |0\rangle . \tag{6}$$

So, fermion 1 is destroyed. Applying the creation operator $a_{\downarrow}^{\ \dagger}$ to the vacuum, one obtains the fermion 2.

$$ |\psi_2\rangle = a_{\downarrow}^{\ \dagger}|0\rangle = |\downarrow\rangle . \tag{7}$$

Obviously, it's a different situation because here we have full knowledge of which one of the fermions was present and when. Also, it's not clear where from should appear the minus sign in $(7)$ to the difference from $(5)$.

What can be wrong here?

Please notice that I tried to stick to the phenomenology. I am against introducing manipulations with vacuum which are not involved in the sequence of events described above. Also, please don't send me other questions about fermions.

  • 3
    $\begingroup$ Creation/annihilation operators act in the second quantized form, so your equation (3) does not make sense. They do not create a certain fermion, but a fermion in a certain mode in an antisymmetric state. $\endgroup$ Feb 19, 2015 at 18:27
  • $\begingroup$ @NorbertSchuch did you read my discussion about the singlet state? Please read it, and then we can talk. The state |S> has asymmetry between whom and whom? What means interchanging the fermions? $\endgroup$
    – Sofia
    Feb 19, 2015 at 18:35
  • $\begingroup$ I read it, and I still say that eq. (4) is wrong. On the RHS, there will be $\lvert S\rangle$. That's how creation operators (in second quantization!) are defined. They produce Slater determinants. $\endgroup$ Feb 19, 2015 at 18:37
  • $\begingroup$ @NorbertSchuch you argue with me about something on which we agree. I say in the text that the state becomes (5). (And please let aside Slater determinants, for two particles I don't need them.) About how creation and annihilation operators are defined, definitions have to be consistent. I repeat, the old fermion was spin-up. After adding a fermion in spin-down state, the state becomes antisymmetrical, i.e. each one of the fermions may become up or down. But the antisymmetry is at interchange between fermion 1 and fermion 2. $\endgroup$
    – Sofia
    Feb 19, 2015 at 18:58
  • 2
    $\begingroup$ But they are identical! You cannot pretend they are not! You obtain $\lvert\psi_1\rangle=\lvert\psi_2\rangle$, i.e., the same state. There is no contradiction. $\endgroup$ Feb 19, 2015 at 22:30

2 Answers 2


I will formalise my comments into an answer, and try to introduce a notation that might avoid some of the confusions in the OP. The correct way to think about creation and annihilation operators, Fock spaces and so on, is in the occupation-number representation, where we write the basis states in terms of the number of quanta $n_i$ carrying a given value of all possible quantum numbers, denoted collectively as $i$. These states are written as: $$ \lvert n_a n_b n_c \cdots \rangle = \left(\prod_i \frac{1}{\sqrt{n_i!}} \right)(c_a^\dagger)^{n_a} (c_b^\dagger)^{n_b} (c_c^\dagger)^{n_c} \cdots \lvert 0\rangle. \tag{1}$$ For bosons, the numbers $n_a$ can take any value, in the absence of other constraints. When one is dealing with fermions, the Pauli principle tells you that $n_a = 0,1$ for all quantum numbers $a, b, c, \ldots$. The exchange symmetry is enforced by defining the ladder operators to obey canonical (anti-)commutation relations $[c_a,c_b^\dagger] = \delta_{ab}$ ($\{c_a,c_b^\dagger\} = \delta_{ab}$). Some people like to call this representation "second quantisation". It differs from the "first quantised" representation, where one writes the states as $$ \lvert n_a n_b n_c \cdots \rangle = \mathcal{S}_\pm \left[\lvert a\rangle^{\otimes n_a} \otimes \lvert b\rangle^{\otimes n_b} \otimes\lvert c\rangle^{\otimes n_c} \otimes \cdots \right], \tag{2}$$ where each factor in the tensor product corresponds to a distinguished particle, and we thence apply a combined (anti-)symmetrisation and subsequent normalisation operation $\mathcal{S}_\pm$ to get the correct exchange statistics for identical particles. The advantage of representation (1) over representation (2) becomes crystal clear as soon as you try to write down any non-trivial many-body state.

Now, to the main part of the question. The definition of the Fermi operators $c_a \lvert 0 \rangle = 0$, combined with the canonical anti-commutation relations, imply that these operators do not simply "annihilate" particles, unlike the bosonic case. There are also some tricky phase factors which must be tracked carefully. This is taken care of automatically using the form on the right-hand side of (1). However, if you insist on using the form on the left-hand side, you should find that $$ c_d\lvert n_a \ldots n_d \ldots \rangle = \delta_{n_d,1}(-1)^{\sum_{i<d} n_i} \lvert n_a \ldots (n_d - 1) \ldots \rangle, \tag{3}$$ where $i< d$ simply means all indices which appear to the left of $d$. You can prove this easily from the definition on the right-hand side of (1).

Specifying to to the example from the OP, one would like to "prove" the canonical anti-commutation relations (although they are actually part of the definition of $c_a$), by considering the action of $\{c_\uparrow,c_\downarrow^\dagger\}$ on the state $\lvert \uparrow \rangle = |0_\downarrow 1_\uparrow \rangle$ (these are the "first" and "second quantised" representations, respectively). These manipulations will go via the intermediate states $(\lvert\downarrow \rangle \lvert\uparrow\rangle - \lvert\uparrow \rangle \lvert\downarrow\rangle)/\sqrt{2} = \lvert 1_\downarrow 1_\uparrow \rangle $ and $\lvert \downarrow \rangle = |1_\downarrow 0_\uparrow \rangle$ . We have $$c_\uparrow c_\downarrow^\dagger \lvert 0_\downarrow 1_\uparrow \rangle = c_\uparrow|1_\downarrow 1_\uparrow \rangle = -|1_\downarrow 0_\uparrow \rangle, $$ where in the last step we used (3). On the other hand, one has $$ c_\downarrow^\dagger c_\uparrow \lvert 0_\downarrow 1_\uparrow \rangle = c_\downarrow^\dagger \lvert 0_\downarrow 0_\uparrow \rangle = \lvert 1_\downarrow 0_\uparrow \rangle. $$ Therefore, we have confirmed that $\{c_\uparrow,c_\downarrow^\dagger\} = 0$ in this case.

  • $\begingroup$ @Mark, a minor remark: For bosons, Eq. (1) is missing a prefactor (or your states are not normalized). $\endgroup$ Feb 20, 2015 at 17:17
  • 1
    $\begingroup$ @NorbertSchuch Thanks, you are right of course. I have updated the answer with (hopefully) the correct normalisation. $\endgroup$ Feb 20, 2015 at 19:57

$\newcommand{\ket}[1]{\lvert #1 \rangle}\newcommand{\up}{\ket{\uparrow}}\newcommand{\down}{\ket{\downarrow}}$Let us write your starting state as $\psi_0 = a^\dagger_\uparrow\ket{0} = \up$.

You claim that $a_\downarrow^\dagger\psi_0 = \up\down$. This is false, since

$$a^\dagger_\downarrow\psi_0 = a_\downarrow^\dagger a_\uparrow^\dagger \ket{0} \overset{\text{CAR}}{=} - a_\uparrow^\dagger a_\downarrow^\dagger \ket{0} = -a^\dagger_\uparrow \down \tag{i}$$

where we would write the RHS as $\down\up$ by your logic, but $\down\up \neq - \up\down$. The state created by the action of $a_\uparrow^\dagger a_\downarrow^\dagger = - a_\downarrow^\dagger a_\uparrow^\dagger$ is already your $\ket{S}$.

Such errors continue throughout the post - for example, $(4)$ is nonsense, because $\ket{0_\uparrow}$ is not a state at all - the state in which there is only one fermion left is just $\up$ or $\down$ or a linear combination of these. You have to use the proper definition of the Fock space instead of "intuitively" concluding things like $(4)$ because "it is not clear which fermion is destroyed". $\ket{0_\uparrow}\down$ is not a state in the Fock space. You have to use the Fock space with the canonical (anti-)commutation relations of the creators and annihilators to get correct results in such a "second-quantized" description.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – David Z
    Feb 20, 2015 at 0:00

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