How Does the Second Quantized Field Operator Act on a Two-Fermion Wave Function?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 2K views
Einj
Messages
464
Reaction score
59
I have a doubt on the second quantization formalism. Suppose that we have two spin-1/2 fermions which can have just two possible quantum number, 1 and 2. Consider the wave function:
$$
\psi(r_1,r_2)=\frac{1}{\sqrt{2}}\left(\psi_1(r_1)\psi_2(r_2)-\psi_1(r_2)\psi_2(r_1)\right).
$$
The second quantized field is defined as:
$$
\Psi(r)=\sum_{k=1}^2a_k\psi_k(r),
$$
where [itex]a_k[/itex] are the annihilation operators for fermions, i.e. anti-commuting with each other.

What's the action of [itex]\Psi(r)[/itex] on the wave function [itex]\psi(r_1,r_2)[/itex]?
 
Physics news on Phys.org
Einj said:
I have a doubt on the second quantization formalism. Suppose that we have two spin-1/2 fermions which can have just two possible quantum number, 1 and 2. Consider the wave function:
$$
\psi(r_1,r_2)=\frac{1}{\sqrt{2}}\left(\psi_1(r_1)\psi_2(r_2)-\psi_1(r_2)\psi_2(r_1)\right).
$$
The second quantized field is defined as:
$$
\Psi(r)=\sum_{k=1}^2a_k\psi_k(r),
$$
where [itex]a_k[/itex] are the annihilation operators for fermions, i.e. anti-commuting with each other.

What's the action of [itex]\Psi(r)[/itex] on the wave function [itex]\psi(r_1,r_2)[/itex]?
In second quantization, the field operator doesn't act on a wavefunction at all, it acts on an abstract state vector in Fock space.

Be careful to keep straight the meaning of the subscripts in the two examples. In your first-quantized wavefunction, the subscripts refer to particle 1 or particle 2. But in the expression for the second-quantized field operator, the subscript k is used to indicate a single-particle state, not a particle.