jostpuur said:
The Klein-Gordon equation is particularly useful, because we know it to be Lorentz invariant, and know that its solutions don't lead to causality paradoxes. Immediate consequence is, that the solutions of the relativistic Shrodinger's equation don't lead to causality paradoxes either.
This is how the proof goes:
Solutions of relativistic SE are also solutions of the KGE. Because solutions of KGE are know to be paradox free, we conclude that so are solutions of relativistic SE paradox free too.
Can you say what's wrong in this?
What is your definition of "Lorentz invariance"? For example, why do you say that the Scroedinger equation
i \hbar \frac{\partial}{\partial t} \psi(x,t) = \sqrt{-\hbar^2 c^2 \frac{\partial}{\partial t} + m^2 c^4} \psi(x,t)... (1)
is not Lorentz invariant while the Klein-Gordon equation
-\hbar^2 \frac{\partial^2}{\partial t^2} \psi(x,t) = (-\hbar^2 c^2 \frac{\partial}{\partial t} + m^2 c^4) \psi(x,t)...(2)
is Lorentz invariant?
What do you mean by "solutions of KGE are know to be paradox free"? How exactly you are using this statement for proving that solutions of (1) cannot exhibit superluminal propagation?
jostpuur said:
Here,
F. Strocchi, "Relativistic quantum mechanics and field theory", Found. Phys. 34 (2004), 501;
http://www.arxiv.org/abs/hep-th/0401143
I didn't even understand how the proof was supposed to work.
In his proof Strocci uses a (supposedly well-known) lemma which states that the Fourier transform of an analytical function has a compact support (i.e., it is non-zero only in a finite region of its argument). And inversely, the Fourier transform of a function with compact support is analytical. I don't know how these statements are proved. However, intuitively, they make sense. An analytical function is supposed to be differentiable infinite number of times. So, its Fourier spectrum should not contain infinite frequencies, because they usually correspond to discontinuities of the function.
Once we established this, the Strocci's proof becomes simple. Suppose that the wave function \psi(x,0) has compact support (i.e., localized) at time t=0. If we assume that the spreading cannot be superluminal, we conclude that \psi(x,t) for finite t > 0 also has a compact support (the support at t=0 can expand only by ct, so it remains compact). Then, the time derivative \partial \psi(x,t)/ \partial t at t=0 also has a compact support. Now we can take the Fourier transform of both sides of the Schroedinger equation (1) at t=0
i \hbar \frac{\partial}{\partial t} \psi(p,t) = \sqrt{p^2c^2 + m^2 c^4} \psi(p,t)....(3)
where (according to the Lemma) i \hbar \partial \psi(p,t) / \partial t and \psi(p,t) are both analytical functions of p. However, \sqrt{p^2c^2 + m^2 c^4} is
not an analytical function of p. So, there cannot be equality between the left and right hand sides of (3). This controversy demonstrates that our assumption (that \psi(x,t) propagates with a finite speed) was wrong.
I agree with parts of what you write but now I wonder if we are arguing past each other? I got the impression that you questioned what you now seem to argue in favour of.
I intended the other thread "propagation speeds in literature" to be about the history of this stuff.)
