For Boson creation and annihilation operators, b|n>=\sqrt{n}|n-1> and b^\dag=\sqrt{n+1}|n+1>, in the Fock space. I know for Fermion systems, n=1,0. However, does any general relationship exist for fermions in the same way as bosons? In short, c|n>=? and c^\dag|n>=?, where, c and c^\dag are Fermionic creation and annihilation operators obeying Fermi anti-commutation rules. If I have interpreted the analogy in a wrong way, please rectify me.
What has been described here is the quantum mechanics situation, i.e. the harmonic oscillator in one dimension. In this case, there is only one creation operator, and one annihilation operator, and thus - up to rescaling - just one state of "level n", denoted |n>. In QFT (or other situations such as many particle QM) the setting is more involved. Here you have many (usually infinitely many) states at "level n". For example, one has to deal with a single particle space of dimension greater than 1 to accomodate the single particle states of a system with several degrees of freedom (think of wave functions of one variable).
As a consequence, one has to introduce not just one, but many creation/annihilation operators, labelled by the one particle states of the system. That is, you have one annihilator c(f) and creator c(f)* per vector f in the one-particle space. In the Fermionic case, both these operators square to zero as an expression of the Pauli principle. But still, for each number n, there are infinitely many states of n particles, described by antisymmetric tensor products
g_1 x ... x g_n,
where the g_k are single particle states. These states replace the single |n> from the harmonic oscillator. One can then wonder how the creation/ann. operators act on them, and finds
c(f)* g_1 x .... x g_n = \sqrt{n+1} f x g_1 x ... x g_n,
c(f) g_1 x ... x g_n = \sqrt{n} \sum_{k=1}^n (-1)^{k-1} g_1 x ... (g_k omitted) ... x g_n
Here the scalar product on the one particle space.
As you can see, they go up/down one level, together with the familiar sqrt{n}, sqrt{n+1} factors, and furthermore satisfy the commutation relations
c(f)c(g) = - c(g)c(f)
c(f)*c(g)* = - c(g)*c(f)*
c(f)c(g)* = - c(g)*c(f) + 1
for arbitrary f,g in the single particle space. In particular, for f=g you see from the first equation c(f)^2=0.
If you work on the Bose Fock space instead, the situation is quite similar in many ways: Then x denotes the symmetric tensor product, you have to omit the factor (-1)^{k-1} in the equation for the annihilation operators, and all minus signs in the commutation relations. Then you get the corresponding formulas for the bosonic creation/annihilation operators.
As far as I understand |n> is a state with n identical particles. For the case of fermions you only have the states |0> and |1>. So the relations would be a^\dagger |0>=|1>, a |1>=|0> and a^\dagger |1>=a |0>=0.
Dear Serodio,
I thank you for confirming this! It is right.
What has been described here is the quantum mechanics situation, i.e. the harmonic oscillator in one dimension. In this case, there is only one creation operator, and one annihilation operator, and thus - up to rescaling - just one state of "level n", denoted |n>. In QFT (or other situations such as many particle QM) the setting is more involved. Here you have many (usually infinitely many) states at "level n". For example, one has to deal with a single particle space of dimension greater than 1 to accomodate the single particle states of a system with several degrees of freedom (think of wave functions of one variable).
As a consequence, one has to introduce not just one, but many creation/annihilation operators, labelled by the one particle states of the system. That is, you have one annihilator c(f) and creator c(f)* per vector f in the one-particle space. In the Fermionic case, both these operators square to zero as an expression of the Pauli principle. But still, for each number n, there are infinitely many states of n particles, described by antisymmetric tensor products
g_1 x ... x g_n,
where the g_k are single particle states. These states replace the single |n> from the harmonic oscillator. One can then wonder how the creation/ann. operators act on them, and finds
c(f)* g_1 x .... x g_n = \sqrt{n+1} f x g_1 x ... x g_n,
c(f) g_1 x ... x g_n = \sqrt{n} \sum_{k=1}^n (-1)^{k-1} g_1 x ... (g_k omitted) ... x g_n
Here the scalar product on the one particle space.
As you can see, they go up/down one level, together with the familiar sqrt{n}, sqrt{n+1} factors, and furthermore satisfy the commutation relations
c(f)c(g) = - c(g)c(f)
c(f)*c(g)* = - c(g)*c(f)*
c(f)c(g)* = - c(g)*c(f) + 1
for arbitrary f,g in the single particle space. In particular, for f=g you see from the first equation c(f)^2=0.
If you work on the Bose Fock space instead, the situation is quite similar in many ways: Then x denotes the symmetric tensor product, you have to omit the factor (-1)^{k-1} in the equation for the annihilation operators, and all minus signs in the commutation relations. Then you get the corresponding formulas for the bosonic creation/annihilation operators.
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