As I understand, the wave-function in QFT becomes a wave-functional dependent on fields. I have heard this told by Sean Carroll in his Biggest Ideas in the Universe lectures. Srednicki gives the $n$-particle state as $$ |\psi,t\rangle = \int d^3x_1\cdots d^3x_n\psi(x_1,\cdots,x_n,t)a^\dagger(x_1)\cdots a^\dagger(x_n)|0\rangle $$ (chapter 1). In QM, with a particle state $|\psi\rangle$, we can obtain the wave-function by acting with $\langle x|$ (i.e., $\psi(x) = \langle x|\psi\rangle$). Does that mean the QFT $n$-particle wave-functional is given by $$ \psi[\phi(x)] = \langle\phi(x)|\psi,t\rangle, $$ with $\phi(x)$ some field configuration? When we measure the wave-functional, will it collaps to a certain field configuration, which we then interpret as some collection of particles with certain momenta?
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1Linked; see this outstanding review by Jackiw. – Cosmas Zachos Aug 21 '23 at 14:21
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The mathematically inclined often go for this here. Is this what you are after? – Cosmas Zachos Aug 21 '23 at 18:37
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Why did you skip powers of oscillator creators for some sites, e.g. $^\dagger (_1)^5$ ? Wavefunctional states certainly don't exclude those. See links. – Cosmas Zachos Aug 21 '23 at 23:52
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1@CosmasZachos As long as we have at least 5 $a^\dagger(x)$'s, and we integrate over all $x$'s independently, won't we also have the case $a^\dagger(x)^5$ included in the expression? At any rate, I just wrote it like Srednicki does. – Depenau Aug 23 '23 at 10:16
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Check. You have the entire Fock space, indeed. – Cosmas Zachos Aug 23 '23 at 10:38
1 Answers
When we measure the wave-functional, will it collapse to a certain field configuration, which we then interpret as some collection of particles with certain momenta?
Basically/ontologically yes, but one never executes this in practice, and the devil is in the details of translating from it to what one actually does. The point is that free QFT is an artful, tortured, repackaging of an infinity of decoupled momentum space oscillators, and the Fourier transform of these normal modes couples them together to something nonlocal and hard to visualize; it took me a while to appreciate the expression you wrote involves n excitations of an infinity of normal mode oscillators in momentum space, the Fock space of QFT. I have some trouble visualizing multiple simultaneous nonlocal Fourier transforms. M Lüscher has it better. I will stick to p78 et seq of R Jackiw's review lecture or, better, Ch 10 of the fine book by Hatfield here.
Indeed, the dense set of states you wrote is equivalent to the Fock space of states of QFT. More commonly, one reconfigures it to the space of eigenfunctionals of the quantum field $\hat \Phi(x)$ with eigenvalue a classical field $\phi(x)$, so, as you state, $$ \Psi[\phi] = \langle\phi(x)|\psi\rangle, $$ where, crucially, in the QFT Schrödinger representation, $$ \hat\Phi(x) \Psi[\phi] = \phi(x) \Psi[\phi]. $$ In the textbook of M Schwartz, the above is (14.19), cf. problem 14.4, underlying the extension of the QM path integral to the QFT one.
This setup is but the analog of the functorial extension (E Nelson) $q\longrightarrow \phi(x)$, where x is a variable label, not a dynamical variable, encompassing both x and t. Just as the QM wavefunction accomodates all qs, just so the QFT wavefunctional encompasses all classical field functions $\phi(x)$, eigenvalues of the field operator. There are questions on this site constructing such complete eigenfunctional bases, but I am not sure you are focussed on that.
This formalism is adequate in yielding measurable matrix elements, such as Schwinger-Dyson equations, and not abstractly interpreting collapsed wavefunctionals; still, your vision is broadly not unsound.
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