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I'm trying to understand the connection between the wavefunction and a quantum field, so I wanted to know if my reasoning was in the right track (this is a more mathematical/conceptual approach, and it could be useful for others studying QFT)

Let's consider, for example, a system of N particles in 3D space, and examine the differences between the description by a wavefunction and a quantum field.

Wavefunction

¤ We have a total of 3N arguments (3 for each direction) that characterize the value of the wavefunction in position space $\psi(\vec{x})$

¤ The wavefunction is, mathematically, just a continuous function with (at least) first order derivatives that can satisfy boundary conditions.

¤ Given a definite momentum $\vec{p}$ of one particle, we cannot specify the value of the wavefunction for each point in space (because of the uncertainty principle).

¤ The wavefunction completely describes the evolution of the system and allows the calculation of observables that come from operators applied to it

Quantum field

¤ We have infinite degrees of freedom, so we technically have infinite arguments to describe the system (I'm not sure if this is true or we also have 3N arguments).

¤ The quantum field can be seen as an operator-valued distribution (in terms of creation and anhiquilation operators); each particle corresponds to a Fourier mode of oscillation of the quantum field.

¤ Given a particle with momentum $\vec{p}$ we can specify the value of the field $\phi(x)$ in each point (since we can just do a Fourier transform of a given oscillation mode). Does the uncertainty principle apply here?

¤ The quantum field doesn't describe one particle per-se, but the excitations in each point of some "fluid" that permeates all the space, each excitation corresponding to one possible particle.

Do you think these differences are correct? Should I add anything else?

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    $\begingroup$ Related post $\endgroup$ – DanielSank Sep 27 '18 at 7:06
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    $\begingroup$ By the way, do you think the title is useful so other people who're also studying this topic can find it easily? $\endgroup$ – Charlie Sep 27 '18 at 16:30
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We have a total of $3N$ arguments (3 for each direction) that characterize the value of the wavefunction in position space $\psi(\mathbf{x})$

For a single particle, it easy to think of the (coordinate space) wavefunction $\psi(\mathbf{x})$ as 'living' in position space but for $N \gt 1$ particles, it's clear that the wavefunction $\psi(\mathbf{x}_1,\mathbf{x}_2,\cdots,\mathbf{x}_n)$ lives in a $3N$ dimensional configuration space

Given a definite momentum p⃗ of one particle, we cannot specify the value of the wavefunction for each point in space (because of the uncertainty principle).

This isn't correct. For example, the wavefunction for a single particle with definite momentum $\mathbf{p}$ is $\psi_\mathbf{p}(\mathbf{x}) = e^{\frac{i}{\hbar}\mathbf{p}\cdot\mathbf{x}}$ which has a definite value at each point in space.

We have infinite degrees of freedom, so we technically have infinite arguments to describe the system (I'm not sure if this is true or we also have $3N$ arguments).

A quantum field is a function of spacetime, i.e., it assigns an operator to each event in spacetime so the are four arguments, the coordinates of an event

each particle corresponds to a Fourier mode of oscillation of the quantum field.

For a free (non-interacting) field, a particle with definite momentum is the quanta of a single Fourier mode. But there are one particle states which are a superposition of one-particle states from different modes

Given a particle with momentum p⃗ we can specify the value of the field $\phi(\mathbf{x})$ in each point (since we can just do a Fourier transform of a given oscillation mode). Does the uncertainty principle apply here?

The quantum field $\phi(\mathbf{x})$ doesn't have a value at each point. Remember, $\phi(\mathbf{x})$ is a field of operators. The operator $a^\dagger(\mathbf{p})$ creates a particle with definite momentum $\mathbf{p}$ and we can express $a^\dagger(\mathbf{p})$ in terms of operator field $\phi(\mathbf{x})$ via an inverse Fourier transform

The quantum field doesn't describe one particle per-se, but the excitations in each point of some "fluid" that permeates all the space, each excitation corresponding to one possible particle.

The ontology of quantum fields isn't settled

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  • $\begingroup$ Could you elaborate on why you mean by assigning an operator to an "event" in spacetime? For example, in the classical (relativistic) case for the EM field is the "event" something like: The electric field took on value $10 \frac{V}{m}$ at point $(t,x,y,z)$? Whereas the quantum case is maybe written as: the photon field took on value $\hat{a} at point $(t,x,y,z)$? What specifically do you mean by event here? $\endgroup$ – jgerber Sep 27 '18 at 6:02
  • $\begingroup$ @jgerber, 'points' in spacetime are events with 1 temporal and 3 spatial coordinates, e.g., $(t,x,y,z)$. The classical four-potential $A^\mu$ assigns a number to each event in spacetime. The quantum (gauge) field $A^\mu$ assigns an operator to each event in spacetime. $\endgroup$ – Alfred Centauri Sep 27 '18 at 10:29
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    $\begingroup$ I see, I think I was confusing the classical definition of field with the quantum one, hence why I was still assigning points of space to the quantum field. Thanks also for the other answers, it helped me clarify better the differences, so I might as well summarize this information in another answer for easy reference. $\endgroup$ – Charlie Sep 27 '18 at 16:29

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