Quantum Field Operators for Bosons

[knowwhatyoudontknow]

TL;DR

Trying to better understand Quantum Field Operators

Consider the field creation operator ψ^†(x) = ∫d³p a_p^†exp(-ip.x)

My understanding is that this operator does not add particles from a particular momentum state. Rather it coherently (in-phase) adds a particle created from |0> expanded as a superposition of momentum eigenstates states, exp(-ip.x), at x, to a particle (if it exists) expanded as a superposition of basis states, exp(ip.x') at x'. The probability amplitude at x is then:

∫d³p exp(-ip.x)/√2π exp(ip.x')/√2π = δ⁽³⁾(x - x') which is an eigenvalue of position

Is this the correct interpretation of how things work? Sorry, if my question is a little redundant, but I am just starting out with QFT.

 

[vanhees71]

If you are talking about non-relativistic free particles, it's correct. Here $\hat{\psi}(x)$ annihilates a particle at position $x$.

 

[knowwhatyoudontknow]

Thanks. If a particle exists at x₁ and another is added at the same point is it also fair to say that the resulting state is the ⊗ product of the 2 wavefunctions?

 

[vanhees71]

Be careful! The field operators are distribution valued operators due to the equal-time canonical commutator (bosons) relation
$$[\hat{\psi}(t,\vec{x}),\hat{\psi}^{\dagger}(t,\vec{y})]=\mathrm{i} \delta^{(3)}(\vec{x}-\vec{y}).$$
That's why squaring an operator at the same point is ambigous. It's of course the same singularity as in
$$\langle \vec{x}|\vec{y} \rangle = \delta^{(3)}(\vec{x}-\vec{y}).$$
There's a cure for that: Use "smeared" position states, i.e., true square integrable functions, i.e., instead of a position eigenstate $|\vec{y} \rangle$ you can use a Gaussian wave packet $|\phi_{\vec{y}} \rangle$ with
$$\langle \vec{x}|\phi_{\vec{y}} \rangle=N \exp\left [-\frac{(\vec{x}-\vec{y})^2}{4 \sigma^2} \right].$$
Then you deal with realistic "position eigenstates", i.e., with a particle in a well-defined small region around the point $\vec{y}$. Then you can have of course easily two bosons "at the same place" in this "smeared" sense, i.e., taking into account the necessarily finite accuracy you can localize any particle.

 

[knowwhatyoudontknow]

Understood. I have read about the difficulties of using δ functions in this context. Smearing makes sense. I have also heard about using 'rigged' Hilbert space as a possible alternative.