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The time evolution of any wave function is deterministically specified by the Schrödinger equation (unless measured).

However, particle creation is not allowed in quantum mechanics.

As I am unfamiliar with quantum field theory, what would be a brief outline of how QFT extends QM to allow for particle creation?

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    $\begingroup$ Have you tried to read the wikipedia entry? $\endgroup$ Dec 9, 2022 at 7:31
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    $\begingroup$ Then the short answer is: particles in QFT are analogous to excitation modes in regular QM. They are created by creation operators and annihilated by annihilation operators. $\endgroup$ Dec 9, 2022 at 7:35
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    $\begingroup$ No, a quantum field is not a wavefunction. Yes, there is one electron-positron field throughout the universe and all electrons and positrons are its quanta (loosely, its “excitations”). $\endgroup$
    – Ghoster
    Dec 9, 2022 at 7:43
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    $\begingroup$ A value of a quantum field at a point is an operator, not a complex number like for a wavefunction. $\endgroup$
    – Ghoster
    Dec 9, 2022 at 7:48
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    $\begingroup$ One first step toward QFT is to learn the simple harmonic oscillator in QM, especially the approach using “ladder operators” between energy eigenstates. Similar-looking operators will re-appear in QFT as operators that create and destroy particles, rather than turning an energy eigenstate of a single oscillating particle into a higher or lower one. $\endgroup$
    – Ghoster
    Dec 9, 2022 at 7:55

6 Answers 6

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In what ways QFT generalizes QM
The answer depends on what is understood by Schrödinger equation: if it is $$i\partial_t|\psi\rangle = H|\psi\rangle,$$ where $H$ could be any Hamiltonian operator, then this is also true in quantum field theory (QFT). However, in quantum mechanics (QM) the Hamiltonian $H$ is usually understood as non-relativistic, corresponding to what is classically particles (mostly electrons and protons), often one-particle or with interactions limited to electromagnetic ones. Thus, QFT extends QM in the following ways (the list is by no means comprehensive):

  • relativistic quantum theory
  • treating fermions and bosons on equal footing, as fields (in this sense QFT is also an extension of the classical field theory - aka electromagnetism)
  • providing coherent treatment of interactions for arbitrary number of particles and for electromagnetic, strong and weak interactions (I am not sure about the current status of the gravity)

Importantly, QFT is a quantum theory built from fundamental principles - like the universal symmetries - from which one naturally obtains many things that in QM had to be introduced ad-hoc - e.g., electron spin or various new particles.

Last but not least, QFT has introduced many new methods (diagrammatic expansions, path integrals, renormalization group) which turned out to be extremely useful for dealing with what is essentially "quantum-mechanical" problems - like those in solid state physics.

In what ways QFT does not generalize QM
What QFT doesn't do, is it doesn't question the postulates of quantum theory, which are routinely referred to as postulates of quantum mechanics - this is an example of equivocation, i.e., the use of term quantum mechanics to mean both non-relativistic quantum theory and quantum theory in general. In other words, it would be correct to say that QM is a non-relativistic limit of QFT, which itself is a sub-domain (or application) of general quantum theory.

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    $\begingroup$ Some observables of QFT are localised in spacetime regions and this localisation plays a crucial role in several contexts (locally covariant renormalisation for instace). The fact the the field operators are smeared with (compactly supported) functions implements these ideas. That is a novelty with respect to QM. $\endgroup$ Dec 9, 2022 at 10:04
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    $\begingroup$ @ValterMoretti this is a bit technical for me to discuss myself - it could be interesting, if you expand this in an answer. $\endgroup$
    – Roger V.
    Dec 9, 2022 at 12:53
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    $\begingroup$ Quantum field theory, in practice, does change the postulates of quantum mechanics, because relativistic QFTs require renormalization. This means that the Born Rule for determining probabilities can only be applied to external, states. Intermediate states do not have a probabilistic interpretation. $\endgroup$
    – Buzz
    Dec 10, 2022 at 3:27
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    $\begingroup$ @Buzz you have a point there... Not sure though what you mean by "intermediate states" - those in perturbative expansion? I think they are not measurable in QM either... $\endgroup$
    – Roger V.
    Dec 10, 2022 at 7:08
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    $\begingroup$ The Schrodinger evolved wavefunctional between the asymptotic times could be called the "intermediate state". These are measurable in QM. But idk why Buzz would call them "non-measurable" in QFT. They are non-measurable in scattering because of the interaction time being tiny. They are measurable in, say, bound state experiments. $\endgroup$
    – Ryder Rude
    Dec 11, 2022 at 6:14
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There have historically been a couple of ways to motivate QFT from QM, but one useful framework is that of first vs. second quantization.

First quantization

This is standard quantum mechanics. If we wish to analyze a physical system, we write down its Hamiltonian and then apply the various quantization substitutions. So we have the usual steps of, e.g., $p_x \to -i \hbar \frac{d}{dx}$. Then we use this quantized Hamiltonian in the Schrödinger equation and solve for the time evolution of the wavefunction

Conceptually, in first quantization, we have taken the idea of a particle in classical mechanics and substituted it for the idea of a wavefunction. In classical mechanics, a particle can take on any possible state in phase space. But in quantum mechanics, if the eigenbasis of the wavefunction is discrete, then the wavefunction will collapse into one of a set of quantum states when it is observed --- a particle cannot be found just anywhere in phase space. So the particle has become quantized.

However, crucially, in first quantization, we still treat the fields that the particle interacts with the same way we did classically. If the particle is in a quadratic potential, for example, the potential is treated as continuous.

Second quantization

The step that quantum field theory takes is that now the fields that a particle interacts with are quantized as well. As an example, let's consider the electron in a hydrogen atom. In standard QM we were content to just write down the classical Hamiltonian as:

$$ H = \frac{\textbf{p}^2}{2m} - \frac{e^2}{r} $$

and then just make the substitution of $\textbf{p} \to -i \hbar \nabla$

But in QFT, we now need to consider how the electron actually interacts with the electromagnetic field. In standard QM, the electron is just interacting with a continuous electromagnetic field. But in QFT, the electron can only interact with the field in discrete interactions. This is because the field is no longer continuous, but, just like the electron itself, can only exist in discrete states (what we call photons). The quantization of the electromagnetic field produces a new effect, the Lamb shift.

So QFT is a more principled kind of theory than standard QM. In standard QM, we have two kinds of things: wavefunctions and potentials. But in QFT, all you have is fields. These fields can only occupy quantized states, and we call a quantized state of a field a "particle." There are four special fields which will interact with other fields, and we call those special fields "forces." So conceptually, it's actually quite simple. The mathematical details, of course, are a different story....

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There are two things people can mean by "Quantum Field Theory".

One is that it's the theory of quantum physics of fields. The second is that Quantum Field Theory (often capitalised) = particle physics (where relativity also becomes important). The latter is an ancient notion that was a result of generations of physicists being raised on outdated texts like Peskin and Schroeder. Unfortunately, this view is still quite common, especially among high-energy physicists.

Being "Relativistic" is just saying the theory's symmetry group contain the Lorentz Group. Of course, there are quantum field theories that are non-relativistic, as one encounters all the time in condensed matter physics. Moreover, the question of "particles" vs "fields" is often really just a question of ontology that is often given to theories posthoc (ironically, by positivists). What is referred to as "Quantum Mechanics" can be viewed as a 1-dimensional quantum field theory (for a field $x(t)$ where you interpret the dimension as "time" and the value of the field as the "position".). Nothing stops you from giving it also a perturbative treatment and drawing Feynman diagrams. In this sense, quantum field theory generalises quantum mechanics.

Neither is renormalisation anything unique to "Quantum Field Theory". The often-told story of Renormalisation being about "infinities of Quantum Field Theory" is also terribly outdated. It is now much better understood to occur due to a historical premature assumption taken at the time that the QFTs that appeared in particle physics are well-defined at all energy scales. To see how it can also occur in Quantum Mechanics, see section 4.5 of David Tong's lecture notes 1. The modern understanding of these issues is in terms of the Renormalisation Group and effective field theories.

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Unfortunately the idea of particle creation is really just a terribly confusing way to describe quantum field theory. Quantum Field theory doesn’t have particles. It has (guess it…) fields. The amplitudes of those fields are quantized, and, for historical reasons people have chosen to call (often delocalized, basis-dependent, and non-conserved) excitations in those fields as “particles”. This has led to tons of confusion and misconceptions.

So quantum field theory extends standard quantum mechanics in the same way classical field theory (electromagnetism, fluid dynamics) extends classical particle dynamics (billiards balls, point charges).

Particle theories are theories that predict the space time trajectories of localized (often infinitesimal in the model) particles. Field theories are theories that predict the amplitudes in space time of delocalized (often infinite in spatial extent) fields.

What is often call “particle creation and annihilation” in qft is better thought of as “the amplitude of a field changing”. In qft field amplitudes change due to interactions between different fields eg the electron and photon fields in quantum electrodynamics (qed).

For more on this viewpoint see https://arxiv.org/abs/1204.4616 "There are no particles, there are only fields" by Art Hobson

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  • $\begingroup$ But what does a fermion field mean? $\endgroup$ Dec 10, 2022 at 22:44
  • $\begingroup$ @AimanAl-Eryani It's a good question and I don't have a complete answer and it's something I wonder about myself. Here is what I have so far: Classically, it is possible to define a "Grassman odd bispinor field". I don't know what Grassman odd means but I think it is related to the Pauli exclusion principle. My understanding is that such a field would have an object at each point in space which is a Grassman odd bispinor. for QFT this "Grassman odd bispinor" valued field gets promoted to a "Grassman odd bispinor operator" valued field. It will behave similarly to the classical object but $\endgroup$
    – Jagerber48
    Dec 10, 2022 at 23:10
  • $\begingroup$ may now have a quantized range of values which it can take on, it can be in a superposition of those values, and it can participate in entanglement. That is to say quantum field theory is richer than classical field theory exactly because the fields can be in superposition states and participate in entanglement. $\endgroup$
    – Jagerber48
    Dec 10, 2022 at 23:11
  • $\begingroup$ OK, here's a challenge. Can you represent using "Grassman valued fields" two different fermionic field configurations? $\endgroup$ Dec 11, 2022 at 7:45
  • $\begingroup$ @AimanAl-Eryani Like I said, I don't really know what "Grassman" means so that will be quite a challenge for me. Maybe I'll learn one day though. Are you trying to imply with your questions that it is not possible to think of fermions as "something"-valued fields? i.e. fields which assign objects of type "something" to each point in space in generalization to classical tensor fields? $\endgroup$
    – Jagerber48
    Dec 11, 2022 at 16:21
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However, particle creation is not allowed in quantum mechanics.

what would be a brief outline of how QFT extends QM to allow for particle creation?

There is nothing inherent about QFT that allows for particle creation: instead the important point is that when QFT is mentioned, people are often implicitly referring to relatvistic QFT and it's the introduction of relativity that allows for particle creation.

In textbook applications, QM enjoys Galilean symmetries which imply that mass is a conserved quantity and this fact forbids particle creation. In textbook applications, QFT instead enjoys the symmetries of special relativity, in which mass is no longer conserved and particle creation is kinematically possible.

This is not a general property of QFTs, though. There are many examples of non-relativistic QFTs which again forbid particle creation due to the same symmetry constraints which forbid it in QM.

The above is my answer to your direct question. I can't help but also write about how QFT extends QM in general, below.

General Relationship Between QFT and QM

QFT is a pretty straightforward generalization/extension of QM, despite the fact that it is not often presented this way in textbooks.

In QM, we are tracking degrees of freedom which depend on a single variable: time. For instance, point particles can be described by their positions $\vec{q}(t)$, from which we can form Hamiltonians $H(\vec{q}, \vec{p})$ and create states $|\Psi\rangle$ which evolve under $H(\vec{q}, \vec{p})$ according to Schrodinger's equation. $|\Psi\rangle$ predicts the probability of finding the system with any particular value of $q$.

In QFT, things are nearly the same. The only difference is that we are now tracking fields, $\phi(t, x)$ which are functions of both time and space. The rest proceeds as before: from $\phi(t,x)$ we can form Hamiltonians $H(\phi, \pi_\phi)$ and create states $|\Psi\rangle$ which evolve under $H(\phi, \pi_\phi)$ according to Schrodinger's equation. $|\Psi\rangle$ predicts the probability of finding the system with any particular value of $\phi$ for every point in space.

The main reason that QFT is not typically taught this way in textbooks is that the answers to questions of interest in QFT are usually more efficiently computed by other means, i.e. you don't need all the scaffolding of wavefunctions and Schrodinger's equation. Action- and path-integral-based methods are typically more convenient. For instance, using the wavefunction is a terrible way to compute the $S$-matrix which describes scattering events.

The one area I know of where so-called "Schrodinger-picture field theory" is used and the computation look roughly like their QM counterparts is cosmology where it falls under the (awesomely) named "wavefunction of the universe" method; see the Hartle-Hawking paper of the same name.

I think this viewpoint is briefly mentioned in Weinberg somewhere (as is almost all of QFT), but I couldn't find the section at the moment. I know it is discussed fairly extensively in this book by Hatfield.

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  • $\begingroup$ thank you vm. i see now that the reason particle creation is not allowed in QM is for conservation of mass. Suppose though we are dealing with a modeling problem like "the wavefunction of a hydrogen atom", which we know cannot have been the complete picture (because there are other things beside this hydrogen atom in the universe), then is introducing new particles ad hoc into the wavefunction allowed practically, because we know the wavefunction in analysis is just a limited model anyway, or will it lead to some catastrophic failure of certain properties of this original model? $\endgroup$
    – James
    Dec 11, 2022 at 3:06
  • $\begingroup$ I think you need to more precisely define the procedure you have in mind for me to give a useful answer. $\endgroup$
    – user26866
    Dec 11, 2022 at 13:41
  • $\begingroup$ How about we have a single hydrogen atom wavefunction evolving stably in time. We send a localized wavepacket some distance away. This is just around 10 "excitation" localized numbers on the wavefunction which can be inputed manually. We place this wavepacket some distance away so it will not mess up any of the existing hydrogen wavefunction structure. The wavepacket propagate automatically per Schrodinger equation. We aim it to hit right into the middle of the hydrogen wavefunction. The waves collide, new waves disperse all over the place... then we see what is the new stable state from this? $\endgroup$
    – James
    Dec 11, 2022 at 14:38
  • $\begingroup$ That sounds like some complicated setup, but not a problematic one. It's just a scattering event. The results depend on the details of the setup. $\endgroup$
    – user26866
    Dec 11, 2022 at 22:48
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Quantum field theory explains why the particles of quantum mechanics behave as they do .For example it details why electric charges attract or repel each other .

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