quantum mechanics - Why Green's function will diverge at the same spacetime point?

In $d+1$ dimensional quantum field theory, the 2-point Green's function will diverge at the same spacetime point when $d\geq1$.

When $d=0$, $\phi(t)=q(t)$, that is the case of QM, and 2-point Green's function at the same spacetime point $\langle\Omega|T(q(t)q(t))|\Omega\rangle$ is well-defined.

While $d \geq1$, the 2-point Green's function at the same spacetime point $\langle\Omega|T(\phi(x)\phi(x))|\Omega\rangle$ will diverge.

So what's the physical or mathematical essence of this diverge. I especially want to know the physical picture in the path integral. Why the randomly walking of a particle will be different from that of a string?

Answer

The simple idea is the following. At short distances, you're probing the high-energy regime (UV) of the theory - you're sensitive to very short wavelengths. Take for instance the free boson in the Euclidean:

$$\langle \phi(x) \phi(0) \rangle \; = \int \frac{ d^dp}{(2\pi)^d} \frac{e^{ip\cdot x}}{p^2 + m^2}\,.$$ When $|x| > 0$, the oscillating factor $e^{ip \cdot x}$ modulates the integral, and there's essentially destructive interference when $|p| \gtrsim 1/|x|$. In other words, you don't expect that wavelengths $\lambda \sim 1/|p| \ll |x|$ contribute significantly. But in the strict limit $x \to 0$ the oscillating factor drops out, and the remaining integral $$\int^\Lambda \frac{ d^dp}{(2\pi)^d} \frac{1}{p^2 + m^2} \sim \Lambda^{d-2}$$ is just a measure of the number of modes at large $p$. And this is of course the big difference: the number of states at high energy in your Hilbert space is sensitive to $d$.

You can make a similar argument more rigorously also for an interacting theory, using the Källen-Lehmann spectral representation (see e.g. Peskin-Schroeder).