# Position field and momentum field in QFT

Im new in QFT (quantum field theory)

We know tha in QM we have operators:

Position $\hat{x} \psi(x) = x\psi(x)$

Momenutum ${\mathbf {\hat {p}}}=-i\hbar \nabla$

How are defined the operator field in QFT?

Position field $\hat{\phi}(x)$

Momentum field $\hat{\pi}(x)$

QFT is also known as "second quantization". "Why not first quantization?" you may ask. To put it explicitly, "first quantization", or "Quantum Mechanics" as more familiarly known, is the procedure of quantizing a Langrangian $L[x,p]$ that describes one (or more) point particle(s) characterized by a position $\vec{x}$ and a conjugate momentum $\vec{p}=\frac{\partial L}{\partial \dot{\vec{x}}}$. The quantizing procedure is:

1) Solve the classical equations of motion to find $\vec{x}(t)$ and, consequently, $\vec{p}(t)$

2) Promote the position $\vec{x}$ (which are the degrees of freedom of the problem) and conjugate momentum $\vec{p}$ to operators that satisfy the equal-time fundamental commutation relations, \begin{equation} [\hat{x}_{i}(t),\hat{p}_{j}(t)]=i\hbar \delta_{ij} \end{equation}

Eventually, what you get are some ladder operators $\hat{a}_{\alpha}$ and $\hat{a}^{\dagger}_{\alpha}$ that act on the vacuum state $|0\rangle$ to raise the eigenvalue of the discrete observable $\alpha$, such as the energy ($\alpha=E$) for the case of the Harmonic oscillator or the z-component of the angular momentum ($\alpha=L_{z}$) for the case of spherical harmonics. As a result, point particles are no longer described by a deterministic trajectory, but rather by a wavefunction. To be more accurate, the general state, \begin{equation} |\psi(t)\rangle = \sum_{n=0}^{\infty}c_{n}(t)\frac{(a^{\dagger}_{\alpha})^{n}}{\sqrt{n!}} |0\rangle \end{equation} becomes a wavefunction only after sandwiching it with the position bra, \begin{equation} \psi(\vec{x},t)=\langle\vec{x}|\psi(t)\rangle \end{equation}

Similirarily, "second quantization", or "Quantum Field Theory", goes a step further to say "ok, forget that we are dealing with point particles. What if we were dealing with fields?". Then the dynamics of the system would be described by a Lagrngian density $\mathscr{L}[\phi,\partial \phi]$ with the fundamental field $\phi(t,\vec{x})$ taking the place of the position vector $\vec{x}(t)$. The associated conjugate momenta $\pi(t\vec{x})=\frac{\partial \mathscr{L}}{\partial_{t}\phi}$ are really conjugate momenta densities. The corresponding procedure of quantizing a field theory is known as "canonical quantization" and the steps are practically the same:

1) Solve the classical equations of motion to find $\phi(t,\vec{x})$ and, consequently, $\pi(t,\vec{x})$.

2) Promote the fundamental field $\phi$ (which are the degrees of freedom of the problem) and conjugate momentum $\pi$ to operators that satisfy the equal-time fundamental commutation/anti-commutation relations, \begin{equation} [\hat{\phi}(t,\vec{x}),\hat{\pi}(t,\vec{y})]_{\pm}=i\hbar \delta^3(\vec{x}-\vec{y}) \end{equation} with $[A,B]_{-}\equiv[A,B]=AB-BA$ and $[A,B]_{+}\equiv\{A,B\}=AB+BA$

To finally answer your question, all operators in quantum field theory are constructed from the fundamental fields $\hat{\phi}$ and the conjugate momenta densities $\hat{\pi}$ just as all operators in quantum mechanics are constructed from the fundamental degrees of freedom $\hat{\vec{x}}$ and the conjugate momenta $\hat{\vec{p}}$.

The construction of the operators of the fundamental fields them selves is done by solving the classical equations of motion and promoting the "parameters" involved in the classical solution to operators. These "parameters" are just the classical integration constants that you would normally fix via some initial and boundary conditions and become eventually ladder operators. For example, the real Klein-Gordon field has the general classical solution, \begin{equation} \phi(t,\vec{x}) = \int \frac{d^{3}\vec{p}}{(2\pi)^{\frac{3}{2}}} \; \left(a_{\vec{p}} e^{i(\omega t -\vec{p}\cdot\vec{x})} + a^{\dagger}_{\vec{p}} e^{-i(\omega t -\vec{p}\cdot\vec{x})} \right) \end{equation} with $a_{\vec{p}}$ and $a_{\vec{p}}^{\dagger}$ the integration constants. Quantizing the theory means to promote these integration constants to operators and the commutation relations eventually lead to the identification of these integration constants/operators as ladder operators that create and destroy particles with momentum $\vec{p}$.

• This is wrong: you don't need to solve the classical equations of motion to quantise the theory. In general, we don't even know how to solve them; but canonical quantisation works anyway. – AccidentalFourierTransform Apr 9 '18 at 21:18
• In order to construct the operators for the fundamental fields and their conjugate momenta through ladder operators, you need to know the classical solutions. After all what you eventually want is to compute correlation functions of the fundamental fields and to do that, you need to know how they act on the vacuum state. – Panos C. Apr 10 '18 at 9:04
• Thank you for your answer. You said: "all operators in quantum field theory are constructed from the fundamental fields $\hat{ϕ}$ and the conjugate momenta densitie"- "Quantizing the theory means to promote these integration constants to operators". The crucial point of my problem is this: the promoting is completly arbitrary? Do I use a particular formula to transform the fundamental fields or the constants fields $a_p$ in operators? In QM position and momentum have a precise formula, In QFT I can use any formula with only conditions of commutation/anti-commutation relations? – asv Apr 10 '18 at 11:33
• The promoting is not arbitrary. It is such so that the commutatation/anti-commutation relations are satisfied. The exact same thing holds in QM. The operators for the position and the momentum are not always what you wrote but they are always such that satisfy the fundamental commutation relation. What you wrote is just the position and momentum operator in the position representation. In the momentum representation, for example, the position operator becomes a derivative over momentum and the momentum operator just multiplies with a constant. – Panos C. Apr 10 '18 at 15:56
• Eventually, both the position and the momentum operator are constructed from some ladder operators as well, but such a formulation is just too formal in QM and is, thus, taught only as a special tool for the harmonic oscillator even though it is mystically applied in every problem in QM. – Panos C. Apr 10 '18 at 16:00