Consider the Hilbert space $L^2(\mathbb{R}^d)$ and a self-adjoint, positive operator $H$ (Hamiltonian).
Let $\psi_t$ be a solution to the Schrödinger equation (with $\psi_0$ the initial condition), then consider the function
$$t \mapsto \int_{\mathbb{R}} \overline{f(\boldsymbol{x})} e^{-iHt} \psi_0(\boldsymbol{x}) \ d^dx.$$
Assume that $\psi_o$ is compactly supported in some ball of radius $r>0$ centred at the origin.
I am supposed to be observing how quantum mechanics violates special relativity by showing that the support of $\psi_t$ grows faster than the speed of light, but I'm lost as to where to go with this.
How to proceed? Thanks.
In Peskin and Schroeder, the argument goes $$ U(t) = \langle x\mid e^{-i(p^2/2m)t}\mid x_0\rangle\\ = \int\frac{d^3p}{(2\pi)^3}\langle x\mid e^{-i(p^2/2m)t}\mid p\rangle\langle p\mid x_0\rangle\\ = \frac{1}{(2\pi)^3}\int dp^3 e^{-i(p^2/2m)t}e^{ip(x-x_0)}\\ =\left(\frac{m}{2\pi i t}\right)^{3/2}e^{im(x-x_0)^2/2t}. $$
That the propagator does not vanish for space-like separated $x$, $x_0$ implies violation of causality.
So instead of a generic positive Hamiltonian, he's assuming a free nonrelativistic Hamiltonian $H=\frac{p^2}{2m}$. And instead of assuming an initial state that has compact support, I think he's assuming here a Dirac delta function. For a generic function of compact support, I think just insert its Fourier transform, and proceed as above.
Then they talk about the Hamiltonian $H=\sqrt{p^2+m^2}$, and how the problems with this Hamiltonian are the same, via an asymptotic argument through Bessel functions only vaguely alluded to, you wind up with $U(t)\sim e^{-m\sqrt{x^2-t^2}}.$ Note that this shows, contrary to what I said in the comments, using a relativistic Hamiltonian does not enforce causality (but switching from a quantum mechanics to a quantum field theory does).
The general case, with an arbitrary self-adjoint positive Hamiltonian, may proceed similarly, but the details were too vague for me to follow.