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Physical Interpretation of Field Operator in Quantum Field Theory and Mode Expansion

Physics Asked by Bentley Carr on January 16, 2021

I’m struggling to understand the physical interpretation behind the field operators $ phi(mathbf x)$ and $phi ^dagger (mathbf x)$ in quantum field theory. My understanding is $ phi ^dagger (mathbf x)$ is an operator which creates a particle at a position $mathbf x$. p37 of Quantum Field Theory for the Gifted Amateur (Lancaster & Blundell, OUP) says that from this definition of $phi ^dagger (mathbf x)$, we can then write
$$ phi ^dagger (mathbf x) propto sum_{mathbf{p}} a_{mathbf p}^dagger e^{-i mathbf p cdot mathbf x}.$$
However this is different to the approach in Peskin & Schroeder. They write out the mode expansion for $phi(mathbf x)$ for the Klein-Gordon field (eqn 2.25) as
$$phi(mathbf x) propto int frac{1}{left ( |mathbf p|^2 + m^2 right )^{1/4}} left ( a_{mathbf p} e^{i mathbf p cdot mathbf x} + a_{mathbf p} ^dagger e^{-i mathbf p cdot mathbf x} right ) d^3 p.$$
I understand that we have two terms since we require $phi(mathbf x)$ to be Hermitian. However does this not contradict the definition that $phi^dagger (mathbf x)$ creates a particle at a position $mathbf x$ above? Is it correct to interpret $phi(mathbf x)$ as an operator that creates a particle at $mathbf x$, or is it better to just ignore the interpretation and only think about the results the theory gives from performing measurements?

One Answer

One could define the field operator either way. The representation with only creation/annihilation operator $$ psi(x) sim sum_k a_ke^{ikx}, psi^dagger(x) sim sum_k a_k^dagger e^{-ikx} $$ has the interpretation of creating/removing a particle at a particular time point, which is a convenient interpretation, e.g., for electrons. Representations with sum or difference of creating and annihilation operators $$ psi(x)sim sum(b_k e^{ikx} + b_k^dagger e^{-ikx}), psi(x)sim sum i(b_k e^{ikx} - b_k^dagger e^{-ikx}) $$ are more convenient for bosonic fields, where in classical limit they reduce to the field strength (e.g., $E(x,t)$ for the electric field or polarization for phonons).

Answered by Vadim on January 16, 2021

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