In David Tong's lecture "Quantum Field Theory" - Lecture 2, he said that
"In Quantum mechanics, position is the dynamical degree of the particle which get changed into an operator but in Quantum Field Theory, the position will be just considered as label"
So, I was wondering: Why?
I would call it a parameter. You can expect to see that in QFT because of at least two different reasons:
In a relativistic theory, you want space and time to have the same status. In quantum mechanics t is a parameter and position is an operator. In QFT they are all parameters (an alternative would be to transform time into an operator, but my understanding is that this seems to be mathematically impossible, or at least extremely difficult).
QFT is a theory of fields. In classical mechanics, when you describe fields in a Lagrangian, position is no longer a degree of freedom, but rather, the fields become the degrees of freedom.That is, particles have a position that can change, fields do not change position, rater they change themselves, likely in different ways at different positions (I am not sure if you have studied classical field theory). So you would expect that in QFT the position operator will also disappear and be replaced by the fields.
Classical particle mechanics is to classical field theory as quantum mechanics is to quantum field theory.
In both classical particle physics and quantum mechanics we describe particles which have dynamical positions that change over time. In both classical and quantum field theory we have fields that take on values throughout space time.
Strictly speaking, quantum field theory is still quantum mechanics. For systems involving a fixed number of particles, you can still try to keep track of the dynamics of each one of these particles by defining $\hat{x}_i,\hat{p}_i$ for all of them. Then you can write down the Hamiltonian $H(\hat{x}_i,\hat{p}_i)=\sum_i \frac{\hat{p}_i^2}{2m}+V(\hat{x}_1,...,\hat{x}_N)$, where the $V$ could be potential or/and particle-particle interactions. Then you can try to solve for the spectrum and eigenstates, which is in fact notoriously difficult to do. Even for a single particle, we are only able to write down the exactly solution for a handful of cases (e.g. infinite potential well, harmonic trap etc.).
The advantages of QFT include, but not limited to: (1) We can forget about the detailed dynamics of each individual particle, and focus on the creation and annihilation of this type of particle (since they are all identical if we only have one species) at each spacetime point using $\hat{\psi}^\dagger(t,x),\hat{\psi}(t,x)$. It's like looking at the waves on a lake instead of tracking down each water molecule. (2) When special relativity is incorporated, we have relativistic QFT, which can deal with cases where particles can pop up from vacuum as long as you give it enough energy ($E=mc^2$). (3) Perturbation theory can also be very systematically done using Feynman diagrams.
To summarize, using $x$ as an operator or label, there is no contradiction. It's a matter of convenience.
Edit: Based on comments below from Tobias and Prox. I realized that in writing the above answer I overlooked the subtlety of defining position operator in relativistic quantum theories. I also found another well-explained post here along the same line with the comments.
