# In what way does QFT combine particles' wavefunctions to form fields?

If I understand correctly, in terms of the math of QFT, saying that particles are excitations in fields is mathematically equivalent to saying that a field is a representation of all possible properties the relevant particle can have. If that's the case, it makes me think the fields of QFT must be able to be "built-up" from the wavefunctions of QM. I also asked a related question a few weeks ago (What is the mathematical relationship between the wave functions of QM and the fields in QFT?) and the responses also gave me that impression.

So now my question is, how exactly are the wavefunctions combined? (I know it's some kind of integral transform, judging from the responses to my other question, but I found a lot of the notation used in those responses difficult to understand, even having looked up the meanings of individual symbols, so I'm not clear on the specific operations being done inside the integral.) Is it a Fourier transform? A tensor product? Something else entirely?

If reasonably possible, please limit mathematical notation to that likely to be understood by someone who's taken in undergrad courses on linear algebra, differential equations, and introductory quantum mechanics, as well as the standard three semesters of university physics, in addition to a bunch of electrical engineering courses, but no courses on QFT or advanced quantum theory.

• the fields of QFT must be able to be "built-up" from the wavefunctions of QM That’s not correct. What kind of Schrodinger-style wavefunction could describe a particle that starts or stops existing? The concept of particle creation and annihilation doesn’t even exist in the QM of particles. QFT can describe processes that QM cannot. Feb 1 at 6:30
• @Ghoster, from what I understood from the responses to my previous question, it seemed like the creation and annihilation operators were applied to the wavefunctions themselves in some manner. Feb 1 at 12:13
• There is a rather new book that should meet your needs: John Donoghue, Lorenzo Sorbo, A Prelude to Quantum Field Theory, Princeton University Press, 2022. From the preface: "Quantum Field Theory is the ultimate way to understand quantum physics. ... After teaching it for many years, we have found that the primary hurdle is to make the transition from the way a student thinks of quantum mechanics to the way we think in field theory. ... This book is dedicated to helping students make this transition. ..." Feb 1 at 14:12
• " but no courses on QFT or advanced quantum theory"... But that's the point... Unless you appreciate the repackaging of an infinity of mutually coupled oscillators involved in QFT, and how localized matrix elements of those contrast to a localized coherent state of just one QM oscillator, you won't get an answer to your question... Feb 1 at 17:25
• @Ghoster The Schrödinger wave function describing a system of uncertain particle count is defined on a Fock space. There's nothing about the mathematical framework of undergrad QM that prevents modeling systems with particle creation; it's just not usually covered. It's not much different from modeling spin ½ particles with a phase space that's effectively a direct sum of $2^n$ copies of the spin-0 phase space. Feb 3 at 1:12

## 1 Answer

OK, I'll sketch the trailmap for you, but it's all math, of the most abstract kind, since it asks you (and all students) to suspend/abrogate their suppositious intuition from QM when transitioning to QFT.

Instead of transitioning by thinking about wavefunctions, I will, perhaps perversely, beg you to completely ignore any and all wavefunctions, which I have often found confuse students (and chemists) more than illuminate them (with illusory intuition).

I'll just use the standard Heisenberg/Dirac matrix mechanics picture that provides all answers for the oscillator in matrix mechanics, based on operators and their matrix elements. Non-dimensionalize ℏ=1 throughout, for simplicity, and work in one space dimension.

$$[a,a^\dagger ]=1, \qquad H=\omega ( a^\dagger a +1/2), \leadsto \\ [H, a^\dagger]=\omega a^\dagger, \qquad a|0\rangle, \qquad (a^\dagger)^n|0\rangle = |n\rangle ,$$ etc... The spectrum, the time evolution , and all matrix elements, etc, can be derived without any mention of wavefunctions, whose subordinate features are given in the above link; but you never really need to know about Hermite functions and stuff to do anything. (In fact, these (eigen)wavefunctions, $$\psi_n(x)=\langle x| n\rangle$$, are but a mere formal basis in building up your states; but, as you must appreciate, Schrödinger's wavepacket states that truly remind you of a classical oscillator, the so-called coherent states, are something much simpler than superpositions of confounded Hermite functions!)

• QFT (in 1D for simplicity, but easy to generalize to 3D):

Starting from classical field theory (classical strings that you simulate as an infinity of coupled oscillators, whose uncoupled normal modes you find by Fourier transformation to momentum space), you are impelled to consider the quantum linear combination of an infinity of decoupled oscillator generators, $$a_p$$ and $$a^\dagger_p$$, $$[a_p,a_p'^\dagger]=2\pi \delta(p-p'), \qquad \int \!\! dp~~{\omega_p \over 2\pi} (a_p^\dagger a_p + [a_p, a_p^\dagger]/2 ),\\ [H,a_p^\dagger]=\omega_p a_p^\dagger, \qquad ...$$ where $$\omega^2_p\equiv m^2 +p^2$$, and the commutator term in the hamiltonian is an infinite constant $$\propto \delta (0)$$, ignored, like all such constants in such systems, where we are only interested in energy differences.

• This is but an infinity of oscillators with idiosyncratic energies.

The actual quantized (real, scalar) field operator is $$\phi(x)=\int\!\!{ dp\over 2\pi \omega_p}(a_p e^{ipx}+ a_p^\dagger e^{-ipx})\\ = \int\!\!{ dp\over 2\pi \omega_p}(a_p + a_{-p}^\dagger )e^{ipx},$$ and so on; it satisfies interesting commutation relations of its own, and produces all the answers acting on $$|0\rangle$$, thus generating the Fock space; but this is not the place to understand QFT and its computational techniques. Time dependence trivially follows QΜ: $$\phi(x,t)=e^{iHt}\phi(x)e^{-iHt}$$. This is a counterintuitive operator whose functions' matrix elements in vacuum should not be thought of as wavefunctions, even when localized in a peculiar way—see linked question in the comments and below.

The point of the non-answer to your misbegotten question is to illustrate that you really do not need the treacherous intuition of wavefunctions to appreciate the essence of the QM $$\mapsto$$ QFT transition.

There are lots of murky wavefunctions' illustrations in the market, and you could concoct your own, given the links above; but they almost always misdirect you rather than help you.

• QFT runs on decoupled oscillator operators in momentum space. Each state $$a^\dagger_p|0\rangle$$ represents a particle of momentum p. The state $$(a^\dagger_p)^2/\sqrt{2}|0\rangle$$ two particles of momentum p, etc... No need for wavefunctions.
• Clear enough now? Feb 2 at 23:58
• not really, but I think it's because I need to learn the linear algebra formulation of standard QM. Most of the mathematics I've learned in regards to QM has been all about solving the Schrödinger equation and working with wavefunctions. I had started watching some videos going into the linear algebra formulation though, so I think I'll go back to those videos and then reread your answer later on. Feb 3 at 1:48
• The wikipedia article is a start. The book by Sakurai & Napolitano is exceptionally good. Feb 3 at 1:57