A simple doubt regarding the propagator of the free particle

Question

I start by saying that I'm not a physicist so forgive me if this question is somehow trivial.

I am studying the White noise approach to Feynman integrals, from Hida et al.'s "White noise analysis".

They start from the simpler case, i.e. the free particle, at a certain point the authors arrive to the following expression:

$$\frac{1}{\sqrt{2\pi i}}e^{i/2 (x-x_0)^2}$$ and they claim that this is the solution of the propagator for the free particle.

As far as I understand this is the solution of the Schrödinger equation where the initial condition is a delta function. From my very limited contact with QM I've learned that the solution's of the aforementioned equation are wavefunctions and the product with its complex conjugate gives a probability density.

The fact is that in this case it is clear that

$$\int_{\mathbb R} \bigg|\frac{1}{\sqrt{2\pi i}}e^{i/2 (x-x_0)^2}\bigg|^2dx\neq 1$$

and actually

$$\bigg|\int_{\mathbb R} \frac{1}{\sqrt{2\pi i}}e^{i/2 (x-x_0)^2}dx\bigg|^2=1.$$

What am I missing?

Thanks in advance for bearing with me.

Answer 1

I am missing time $t$ in your expression. Looking at formula (3.10) of the paper by Hida & Streit it seems that their Feynman propagator of the free particle in ordinary quantum theory is the same as what everybody else is using (for example Peskin & Schroeder p. 14): $$ G_0(y_1,y_0,t)=(\frac{m}{2\pi i\hbar t})^{1/2}\exp\Big(\frac{im(y_1-y_0)^2}{2\hbar t}\Big)\,. $$ It is obvious, and answers to related questions confirm that this solution of the Schroedinger equation cannot be normalized in the ordinary sense $$ \int |G_0(y_1,y_0,t)|^2\,dy_1=+\infty. $$ Apart from that, Peskin & Schroeder show nicely that it violates causality because the particle can propagate between any two points in arbitrarily short time. They take this as the motivation for field theory where, as we know, the Schroedinger equation is not used, and a much better Feynman propagator does not have these deficiencies.