gptejms said:
If the probability is not 100%,I don't see why it should follow a different law/transformation.
I agree with Fra on this point:
Fra said:
... I agree on the fact that probabilities can't just be transformed like a lorentz scalar. ... probability is not an objective thing, it's not an object that everyone agress upong. It's intrinsically relative to the observer. What is probably for me, need not be probably to you for the simple reason that we devise our ensembles differently.
I think that your transformation law \psi(x) \to \psi(\Lambda x) is not obvious at all. If you think it is true, you should be able to prove it, rather that give a vague reference to analogies.
gptejms said:
I asked you to complete your calculation in post #252 with the time coordinate included.
This is done most easily in the momentum representation. If \psi(\mathbf{p}) is the wave function at time t=0 from the point of view of observer at rest O, then
e^{\frac{it}{\hbar} E_p}\psi(\mathbf{p})
is the wavefunction at time t from the point of view of O
\sqrt{\frac{E_{\Lambda p}}{E_p}}\psi(\Lambda \mathbf{p})
is the wavefunction from the point of view of moving observer O' at time t=0 (by her clock)
e^{\frac{it'}{\hbar} E_{\Lambda p}} \sqrt{\frac{E_{\Lambda p}}{E_p}}\psi(\Lambda \mathbf{p})
is the wavefunction seen by O' at time t' (by her clock). You can get position-space representations of these wavefunctions by making usual Fourier transformations. I don't think these integrals can be written in a nice analytical form.
The above formulas follow from the general fact that boost transformations are realized by unitary operators \exp(-\frac{ic}{\hbar} K \theta), where K is the generator of boosts; and time translations are realized by unitary operators \exp(\frac{i}{\hbar} Ht), where H is the Hamiltonian.
gptejms said:
Anyway---if the wavefunction in one frame is \exp{\iota p_\mu x^\mu}(where \hbar=1),then you can see that the transformed wavefunction is going to be not much different.
This is true, but the wavefunction you wrote is just a specific case of a plane wave, i.e., a state with definite momentum, where our formulas coincide. I was talking about transformations of general wavefunctions.
meopemuk said:
It is possible to show (I will not do it here, but I am ready to reproduce calculations, if you are interested) that if we
1) take the classical limit of the above theory and consider trajectories of two non-interacting particles that intersect at the space-time point (x,t) from the point of view of observer O and
2) transform these trajectories according to commutators between the Newton-Wigner operator and generators of boost transformations U_g,
then
3) we will find that from the point of view of O', the intersection of trajectories occurs at point (x',t') which is related to (x,t) by usual Lorentz formulas (1) and (2).
gptejms said:
OK. To make things simple, I'll consider a one-dimensional problem in which the movement of the particles and the reference frame occur only along the x-axis. I will denote P, H, and K the generators of the Poincare group corresponding to space translations, time translations, and boosts, respectively. In the case of spinless particle, the Newton-Wigner position operator is R = -c^2 \frac{K}{H} (In fact, the correct quantum expression is R = -\frac{c^2}{2} (KH^{-1} + H^{-1}K), however, in the end we will take the classical limit anyway, so it is justified to ignore the non-commutativity of operators here.)
From the point of view of observer at rest O, the time dependence of R is
R(0,t) = \exp(\frac{i}{\hbar}Ht) R \exp(-\frac{i}{\hbar}Ht) = R + Vt
where V = Pc^2/H is velocity (Calculations performed in this post can be easily done with the knowledge of commutators between P, H, K, R, and V. I'll skip the detailed calculations, but, if you like, I can justify all steps)
From the point of view of moving observer O', the position at t'=0 (where time t' is measured by the clock of observer O') is given by
R(\theta,0) = \exp(-\frac{ic}{\hbar}K \theta) R \exp(\frac{ic}{\hbar}K \theta) = \beta \frac{R}{\cosh \theta}
(Here I will use a convenient notation \cosh \theta = (1-v^2/c^2)^{-1/2} and \sinh \theta = v/c (1-v^2/c^2)^{-1/2} and
\beta = (1 - Vv/c^2)^{-1}, where v is the velocity of the reference frame O')
The time dependence of position from the point of view of O' is
R(\theta,t') = \exp(\frac{i}{\hbar}H't') R (\theta,0)\exp(-\frac{i}{\hbar}H't') = \beta (\frac{R}{\cosh \theta} + (V-v)t'} ) (1)
where
H' = \exp(-\frac{ic}{\hbar}K \theta) H \exp(\frac{ic}{\hbar}K \theta)
is the Hamiltonian in the reference frame O'.
Eq. (1) is the transformation of position obtained by quantum-mechanical Poincare-group analysis. On the other hand, standard Lorentz transformations of special relativity would predict
R(\theta,t') = R (0,t)\cosh \theta - ct \sinh \theta (2)
t' = t \cosh \theta - \frac{R (0,t)}{c} \sinh \theta (3)
Formulas (1) and (2) do not look identical. However, if we take their difference, replace t' by eq. (3) and use R (0,t) = R + Vt, we will get exactly zero.
This means that space-time coordinates of free classical particles transform exactly by Lorentz formulas within relativistic quantum mechanics. This also means that if there is a localized event defined as intersection of trajectories of two non-interacting particles, then space-time coordinates of this event will transform by Lorentz formulas as well.
One important remark. Here we obtained Lorentz transformation formulas, which are usually interpreted as an evidence of equivalence of space and time coordinates. However, there is no such symmetry in our approach: Position is described by a Hermitian operator, but time is just a classical parameter that labels reference frames.
Another important point is that by this method we will not be able to prove Lorentz formulas (2) and (3) for the point of collision if the two particles interact with each other.
Eugene.
