There is a method of finding propagators in external field called Fock-Schwinger proper time method.

It is designed to solve the equation of the form (2-133).

(The image was taken from the book QUANTUM FIELD THEORY, C. ITZYKSON J.-B. ZUBER)

enter image description here

The very first thing they do (2-134) is not clear to me at all. What those $\left|x\right>$ vectors are? What is their explicit form? Please don't tell me that they are abstract things, corresponding to the position state blah blah. In functional analysis (which deals with abstract vector spaces and operators) one can study abstract things in the sense of generalization of points, functions, arrows to a single notion of a vector, but when it comes to solving equations those abstract things each time are very concrete. For example, you may study Schrödinger equation from the point of view of abstract vector spaces and write $i \frac{d}{dt}\left|f\right> = H\left|f\right>$, but you have in mind that $\left|f\right>$ is a good old wave-function. And so $\frac{d}{dt}\left|f\right>$ means to you a very special operation, i.e. a limit $$ \frac{d}{dt}f(t) = \lim_{dt\rightarrow 0} \frac{f(t+dt)-f(t)}{dt} $$

Therefore, my question is... what is this entity $\left|x\right>$ here? We are dealing with differential equations, so those $\left|x\right>$ things should respect differentiation etc..., but I can't see how. Also there is the inner product of $\left|x^{\prime}\right>$ and $\left|x\right>$. What is its explicit form? For example, in QM $$ \left<f^{\prime}|f\right> = \int{f^{\ast}(x) f(x) dx}. $$

Anticipating the answer "there is no explicit form neither for $\left|x\right>$ nor for the inner product", I will rephrase my question. If not explicit form is available, then what are the mathematical grounds of such behaviour of the authors? In maths everything can be brought back to explicit basic logical blocks. What are they here?


UPDATE After thinking about it for a bit, I think whats tripping you is that you are hoping for explicit functional forms $f(x)$ that describe a system. You may have learnt quantum mechanics with wavefunctions $\Psi(x)$ describing the state of a quantum system followed by the explicit wavefunctions that describe the states of a harmonic oscillator or the hydrogen atom, etc. In many cases, this is enough. However, at the level of QM presented in the question, you seem to have reached a stage where you should train yourself to think more abstractly which does not mean -- as the phrasing of the question seems to indicate -- less explicitly.

In this abstract construction of QM, there is a vector space (a Hilbert space) ${\cal H}$. We denote the elements of this vector space by $|\Psi\rangle$. On this vector space, there are at least two Hermitian linear operators $x$ and $p$, satisfying $[{\hat x},{\hat p}]=1$. Since ${\cal H}$ is a vector space, one can choose any basis of vectors and then expand every element of ${\cal H}$ in this basis. A convenient choice if often the basis of position eigenstates which satisfy $$ {\hat x} |x\rangle = x |x\rangle \, , \qquad \langle x | x'\rangle = \delta ( x - x' ) \, . $$ Every state in ${\cal H}$ is then expanded as $$ | \Psi \rangle = \int dx | x \rangle \Psi(x) \, , \qquad \Psi(x) = \langle x | \Psi \rangle $$ The function $\Psi(x)$ is known as the wave-function that you are possibly already familiar with. But note that it appears here only because I chose to work in the $x$-basis. There was no need for me to do so. I could have chosen any other basis I want. For instance, another basis one can work with is the $p$-basis. Yet another (important) one is the energy eigen-basis (where we choose the basis states to be eigenstates of the Hamiltonian operator ${\hat H}$). Anyway, the point is that wavefunctions are good objects, but they force you to work in the position basis, something that may not always be the most convenient choice.

If you wish for an "explicit representation of $|x\rangle$ maybe you want my to describe its wavefunction which is $$ \Psi_x(x') = \delta (x-x') $$ Here, $x$ is to be thought of as the "parameter" that labels the state and $x'$ is the position eigenstate that I am projecting upon.

I'll leave the rest of the answer as is.

We are interested in a solution to $$ H(x, i \partial_x) G(x,x') = \delta^4(x-x') $$ Denote eigenstates of $x$ by $| x\rangle$ so that $$ {\hat x}^\mu |x\rangle = x^\mu | x \rangle \,, \qquad \langle x | x' \rangle = \delta^4(x-x') $$ I don't know what you mean next by "abstract things". The above is as concrete as one can get. Maybe you're talking about having the "explicit wave-function" in which case - if we denote the wavefunction corresponding to $|x\rangle$ by $\Psi_x(x')$, then by definition $$ \Psi_x(x') = \langle x | x' \rangle = \delta^4(x-x') $$ One can talk about differentiating the wavefunction, $$ \partial_{x'^\mu} \Psi_x(x') = \partial_{x'^\mu} \delta^4(x-x') \, . $$ You could differential these states if you wish $$ a^\mu \partial_{x^\mu} | x \rangle = \lim_{\epsilon \to 0 } \frac{1}{\epsilon} \left[ | x + \epsilon a \rangle - | x \rangle \right] \, . $$ This operation is also as concrete as one can get.

Maybe what is tripping you is that the Dirac delta function that appear as wavefunctions here are not functions at all. This is true and strictly speaking, the Hilbert space of the theory only contains normalizable superpositions of these eigenstates such as $$ |f\rangle = \int d^4 x | x \rangle f(x) $$ The wave-function of the state $|f\rangle$ is $f(x)$ by definition. It is taken to be $L^2$ normalizable, which means $$ \int d^4 x |f(x)|^2 < \infty \, . $$ We also have the inner product $$ \langle f | g \rangle = \int d^4 x f(x)^* g(x) \, . $$

Note that $|f\rangle$ is in no sense a wavefunction. It is a state. A wavefunction is the inner product of a state with $|x\rangle$. A wavefunction is just that - a function.

PS - I have to say, I am totally unclear what you are asking since everything you have written is literally as explicit as you can get - after all, in QM you are dealing with abstract Hilbert space of states along with an abstract space if operators. I have made a few comments hoping that some of them can help you understand this better. If not, please leave a comment and I will clarify myself further.

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  • $\begingroup$ I'm not confused with abstract states in general, for example, in QFT. What I was saying is that I don't like the "physicist's approach" of solving this kind of PDE. Let's say we're mathematicians (btw, I'm physicist) and we don't know about physics at all. How do we solve that PDE? We should only rely on proved mathematical stuff without looking for the unclear "physical sense". So all those bra's and ket's in mathematical sense should be justified by the appropriate theorems. $\endgroup$ – Pingguoren Nov 22 '17 at 8:54
  • $\begingroup$ What I'm looking for is a piece of mathematics that justifies this approach. 1) It should justify that we have a right to "invent" such a Hilbert space out of nothing. 2) It also should justify that all the operations in that space are not forbidden and all the properties of Hilbert spaces are fulfilled (convergence of sequences, comleteness etc) 3) This piece of maths should also give some baby examples in order to, so to say, believe in this approach. In physical textbooks neither justifications nor the baby examples are given. $\endgroup$ – Pingguoren Nov 22 '17 at 9:00
  • $\begingroup$ What PDE are you talking about?? If you're talking about (2-135), you'll need to provide me with a Hamiltonian first. Once you do it's a simple differential equation. For instance if it's the non-relativistic free particle $H = \frac{p^2}{2m}$ and the equation is $i \partial_\tau U(x,x',\tau) = - \frac{1}{2m} \partial_x^2U(x,x',\tau)$. Then, you just solve this. $\endgroup$ – Prahar Nov 22 '17 at 9:03
  • $\begingroup$ @Pingguoren - You have a right to invent whatever you want. As is done in all of mathematics, you start with some set of assumptions and then follow them to their logical conclusion. Whether these set of assumptions implies things that are reproduced in nature or not then tells us the authenticity of those assumptions. What justifies us using Hilbert spaces is the simple fact that they lead to the right predictions. 1) and 2) are taken to be the basic axioms of QM. For 3) have you not studied the harmonic oscillator or the hydrogen atom for baby examples? $\endgroup$ – Prahar Nov 22 '17 at 9:06
  • $\begingroup$ What PDE are you talking about?? I'm talkin about (2-133). "Right predictions" is a physical justification of the method. I'm looking for a mathematical one, without "physical sense". As I've said we have a PDE (2-133) and no nothing about where it came from. Neither hydrogen atom nor harmonic oscillator require introducing Hilbert spaces out of nothing. $\endgroup$ – Pingguoren Nov 22 '17 at 9:32

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