# Fock Space, Propagator

1. Is it correct to say that a classical time-varying wave function is a map $\mathbb{R}\times\mathbb{R}^3\xrightarrow{\psi}\mathbb{C}$ (saatisfying certain conditions at infinity)?
2. Is it correct to say a quantum field is a map $\mathbb{R}\times\mathbb{R}^3\xrightarrow{\phi}\mathbb{F}^\mathbb{F}$ (satisfying certain conditions at infinity, $\mathbb{F}=\bigoplus_n\mathbb{H}_n$, and $\mathbb{H}_n$ is a Hilbert space)?
3. In his book "Quantum Field Theory and the Standard Model," Matthew D. Schwartz on page 30 remarks that "kinetic terms are bilinear, meaning they have exactly two fields." As an example he mentions $\frac{1}{2}\phi\Box\phi$. Is it not standard notation practice in QFT to elide the composition operator $\circ$, so that this example could be written $\frac{1}{2}\phi\,\circ\,\Box\phi$? If $\Box\phi$ is derived from $\phi$, then how is this an example of "exactly two fields"?
4. In his Lecture notes, Chapter 3, Feynman Calculus, Professor Etingof explains that calculation of coefficients "$A_i$ reduces to calculation of integrals of the form $\int_VP(x)e^{-B(x,x)/2}dx$ where $P$ is a polynomial and $B$ is a positive definite bilinear form (in fact, $B(v,u)=\partial_v\partial_u(c)$)." Should this not be "nondegenerate" positive definite bilinear form? In his statement of Wick's theorem he writes, "Let $B^{-1}$ denote the inverse form on $V^\ast$." Does this bilinear form relate directly to Schwartz's remark in my question 3.? If $B^{-1}(l_i,l_{\sigma(i)})$ in his Theorem 3.1 is indeed a (Feynman) propagator, how exactly does it relate to Schwartz's $\Pi=-\frac{1}{\Box}$ on page 41?

1. Is it correct to say that a classical time-varying wave function is a map $\mathbf{R}\times\mathbf{R}^3\xrightarrow{\psi}\mathbf{C}$ (satisfying certain conditions at infinity)?

It could be modeled that way, sure. But this describes a single wave function. If we were interested in the space of wave functions, we would be looking at some subspace of $L^{2}(\mathbb{R})$.

1. Is it correct to say a quantum field is a map $\mathbf{R}\times\mathbf{R}^3\xrightarrow{\phi}\mathbf{F}^\mathbf{F}$ (satisfying certain conditions at infinity, $\mathbf{F}=\bigoplus_n\mathbf{H}_n$, and $\mathbf{H}_n$ is a Hilbert space)?

Hmmm...this is misleading, I think because quantum fields are operator densities, whereas I'm not certain $\mathbf{F}^\mathbf{F}$ captures this aspect to them. (Now, I may be ignorant of various theorems mathematical analysis which supply the necessary & sufficient conditions for an operator density to be represented by a function, so just be aware of that caveat.)

When I say "operator density", I mean -- physically -- we ought not ask the question "What's the value of the field at this point?" But it is meaningful to ask "In this region, what is the average value of the field?"

I keep arguing back and forth whether $\mathbf{F}^\mathbf{F}$ provides the right domain for this. The number operator lives there, but it's also implicitly defined over a region of spacetime. The creation and annihilation operators live there too, and they're operator densities.

There's also subtleties surrounding interacting field theories...

1. In his book "Quantum Field Theory and the Standard Model," Matthew D. Schwartz on page 30 remarks that "kinetic terms are bilinear, meaning they have exactly two fields." As an example he mentions $\frac{1}{2}\phi\Box\phi$. Is it not standard notation practice in QFT to elide the composition operator $\circ$, so that this example could be written $\frac{1}{2}\phi\,\circ\,\Box\phi$? If $\Box\phi$ is derived from $\phi$, then how is this an example of "exactly two fields"?

Well, two things:

First, Schwartz is a little sloppy here. By they "have exactly two fields" he means that there are two factors of the same field in a given kinetic term, or derivatives of the same field. So there are two factors in this kinetic term: the field $\phi$ and the D'Alembertian of the field $\Box\phi$.

Second, remember composing functions $g\circ f$ requires the codomain of $f$ to be the domain of $g$. We cannot write $\phi\circ\Box\phi$ because $\phi\colon\mathcal{M}\to\mathrm{something"}$ a field is a "function" which takes in a point in spacetime $\mathcal{M}$ and produces "some value"...not a point in spacetime. So composing $\phi\circ\Box\phi$ is a "meaningless" (wrong) statement.

• (1) I should not have used the word "classical" in my question. (2) Is an "operator density" like a probability density? (3) If f is a real-valued function of a real variable, then f' is also a real-valued function of a real variable, Therefore, f and f' can be multiplied, and they can be composed. Multiplication is commutative, but composition not necessarily. – Ellis D Cooper Aug 5 '17 at 15:40
• @EllisDCooper (2) Well, "operator valued distribution" might be a better phrase, not related at all to probability theory. (3) But a quantum field is not a real-valued function of a real variable (remember: $q$-numbers are not $c$-numbers), but even classically your reasoning breaks down when talking about functions of $\mathbb{R}^{n}$ unless it's an $\mathbb{R}^{n}$-valued function...and even then, there's subtleties (confusing spacetime with the internal space, etc.). – Alex Nelson Aug 6 '17 at 20:45
• @EllisDCooper Also, composing functions is not the same as multiplying them. If $f(x)=x^{2}$ and $g(x)=\sin(x)$, then $(f\circ g)(x) = f(g(x)) = \sin^{2}(x)$ whereas $f(x)g(x)=x^{2}\sin(x)\neq\sin^{2}(x)$. – Alex Nelson Aug 6 '17 at 21:59
1. It is unclear what is meant by "classical time-varying wave function". For example:

• The quantum state of a particle without spin in three dimensions is a function $\mathbb{R}^3\to \mathbb{C}$. Its time evolution gives a map $\mathbb{R}\times\mathbb{R}^3\to \mathbb{C}$. But this isn't "classical".
• A classical complex scalar field in $\mathbb{R}\times\mathbb{R}^3$ is also a function $\mathbb{R}\times\mathbb{R}^3\to\mathbb{C}$. But this isn't a "wave function" (at least, in its usual meaning: some kind of quantum state).
2. It is true that a quantum field is (roughly) a map from space-time ($\mathbb{R}\times\mathbb{R}^3$) to the endomorphisms of a Hilbert space $F$. However, if the $H_n$ are the Hilbert spaces of $n$-particle states, then it isn't true in general that $F$ is decomposed as their direct sum. That is only true for the free theory.

3. The kinetic term is $E(\phi,\phi)$, where $E$ is the bilinear map given by $E(\phi_1,\phi_2)=\phi_1\circ\square\phi_2$. This is the sense in which the kinetic term is bilinear and "has exactly two fields". Perhaps it's clearer to say that the kinetic term is quadratic in $\phi$.

4. $B(\phi, \phi)$ is the quadratic part of the action. That includes the kinetic term plus possibly a mass term $\sim m^2 \phi^2$. For a massless theory, the quadratic part equals the kinetic term. We can rewrite a bilinear form $B$ as a linear transformation $\tilde{B}$ followed by a scalar product: $$B(\phi_1, \phi_2)=\left<\phi_1,\tilde{B} \phi_2\right>.$$ The inverse form is $B^{-1}(\phi_1,\phi_2)=\left<\phi_1,\tilde{B}^{-1} \phi_2\right>$. In the case in which $B(\phi_1,\phi_2)=\int d^4x \phi_1\square\phi_2$, we have $\tilde{B}=\square$ and therefore the propagator is $\tilde{B}^{-1}=\frac{1}{\square}$.

• (1) Of course I ought not have used the word "classical." (2) The elements of A for a map A x X ---> X are operators on X. I would say a map E ---> A is a field of operators on E. (3) & (4). I understand that a bilinear form B : V x V ---> F on a vector space V over a field F corresponds to a matrix relative to a basis for V, and that B is non-degenerate iff the matrix is invertible. My real question is to understand exactly the abstraction in Professor Etingof's version of Wick's Theorem relative to the concrete version of Wick's Theorem in, say, Peskin-Schroeder pp.88-90. – Ellis D Cooper Aug 5 '17 at 15:40
• @EllisDCooper About (2): I agree with you, but I don't ser how what you say in the comment is related to what I say in the answer. I'm stating that the space $X$ ($F$ in the concrete case of the question and answer) is the Fock space of $n$-particle states only in the free case – coconut Aug 5 '17 at 20:09
• About(2): The minor point is that whatever F is, a direct sum or a quotient or whatever, a quantum field is a field of operators defined on spacetime E which are endomaps of F. (I would note that lectures by Scott Glasgow on creation and annihilation operators define bosonic Fock space as a direct sum of Hilbert spaces. So, a reference to the Fock space for the interacting case would be important to me.) Regarding (3)-(4): I am still befuddled by Etingof's definition of B as the Hessian of a function S. And, he says it is the Hessian evaluated at a minimum of S. I have yet to see. – Ellis D Cooper Aug 6 '17 at 17:19
• @EllisDCooper About (2): I don't know how to say this in different way: first, a field of operators on $E$ is (roughly) a map from $E$ to $F^F$ (and I don't see how what we say about this differs in any way); second, $F$ doesn't need to be a Fock space (but it clearly is in the free case) – coconut Aug 6 '17 at 17:41
• @EllisDCooper About the Hessian thing: roughly speaking again, a function $S$ of several variables is Taylor expanded around a minimum as $S(\vec{x})=b_{ij}x_ix_j+\cdots$. The coefficient of the quadratic term $b_{ij}$ is the Hessian evaluated at the minimum. – coconut Aug 6 '17 at 17:52

Pretty much everything I was trying to understand on this question is handled by A. Zee, "Quantum Theory in a Nutshell." Basically, my confusion was not appreciating that there are multiple approaches to understanding what a Feynman Diagram is. The "canonical" approach versus the "path integral" approach, for me, loses out in terms of my personal "principle of least thought." This is basically that I seek understanding in terms of the least number of steps, each of which is no greater than my intuition can handle.