Question

Is it possible to construct a pure Bosonic creation operator in scalar QFT?

Answer 1

Is it possible to construct a pure Bosonic creation operator in scalar QFT?

❱ In theoretical physics, quantum field theory (QFT) is the theoretical framework for constructing quantum mechanical models of subatomic particles in particle physics.

Answer 2

1

I've been working on this since, and the answer I've come up with is a qualified no. The qualification is that you can construct an operator that increases particle number and maintains normalization, but it doesn't maintain that property when superposed. One exception is when the state being operated on is an eigenstate of the particle number operator.

The reasoning is this. We construct a QFT as an infinite number of d=1 simple harmonic oscillators based on a momentum space grid ki. Instead of using sine and cosine transforms and separating by parity, it's easier to use a multi-demensional Hartley transform

ϕ~(k)≡∫dd−1x(2π)(d−1)/2ϕ(x)2–√cos(k⋅x−π4),

which is basically saying ϕ~≡ϕ~++ϕ~−.

On this grid, the Hamiltonian becomes

H=∑im2+k2i−−−−−−−√a†iai.

At each site we can define a pure creation operatorn as

A†i≡[a†iai]+−−−−−−√a†i

where [a†iai]+ is the Moore-Penrose pseudoinverse of [a†iai] (basically invert the non-zero eigenvalues). These operators have the properties

AiA†jA†iAj=δij+(1−δij)A†jAi=δijΠ0¯,i+(1−δij)AjA†i,

where Π0¯,i is the projection operator onto the complement of the ground state of the ith site.

In order to transition to the continuum field limit, we make the substitutions ai→a(k)d3k−−−√ and Ai→A(k)d3k−−−√, which gives

A(k)A†(k′)A†(k)A(k′)=δ(k−k′)+1k≠k′A†(k′)A(k)=δ(k−k′)Π0¯(k)+1k≠k′A(k′)A†(k)

The above identities are enough to show that if

|f,ψ⟩⟨f,ψ|f,ψ⟩≡∫dd−1kf~(k)A†(k)|ψ⟩, then=∫dd−1kdd−1k′f~∗(k)f~(k′)⟨ψ|A(k)A†(k′)|ψ⟩=∫dd−1kdd−1k′f~∗(k)f~(k′)⟨ψ|δ(k−k′)+1k≠k′A†(k′)A(k)|ψ⟩=∫dd−1kf∗(k)f(k)+∫k≠k′dd−1kdd−1k′f~∗(k)f~(k′)⟨ψ|A†(k′)A(k)|ψ