Is There a Superluminal Signal When Turning Off a Harmonic Potential?

[kof9595995]

Consider we initially have a ground state particle of a harmonic oscillator:
\psi = \exp ( - a{x^2}) (neglecting nomalization)
And we have a detector far in space monitoring the local probability of finding a particle. Now if we suddenly turn off the harmonic potential, the wavefunction will evolve as free particle, and
\Psi (x,t) = \frac{{\exp (\frac{{ - a{x^2}}}{{1 + 2iat/m}})}}{{\sqrt {1 + 2iat/m} }}.
We see no matter how far the detector is, the local probability will start to change immediately after we turn off the potential. So will the detector record a change in number of particles detected? If so, it seems there's a superluminal signal transmitted since the detector could be very far from the origin .

 

[The_Duck]

Nonrelativistic quantum mechanics does indeed allow particles to travel arbitrarily fast. This tells you that it's just a low-energy approximation. To account for relativistic effects completely you need quantum field theory.

 

[kof9595995]

True, but in my thought experiment the group velocity is 0, so I don't think it's the crux of the problem.

 

[The_Duck]

Note that the mere fact that the wavefunction changes everywhere isn't surprising. What would clearly violate relativity is if, say, the expectation value of the absolute value of position increase faster than light. This would indicate that the particle, initially localized around the origin, heads away from it faster than light.

This can indeed happen if "a" is large enough, meaning that the wave function is initially localized very sharply around the origin. Then by the uncertainty principle there is a very high uncertainty in momentum, and the wave function will spread out extremely rapidly--even faster than light, if you take the nonrelativistic Schrodinger equation at face value, since it doesn't know about relativity.

You can expand the wave function in momentum eigenstates. Though they average out to an expectation value of zero momentum, some of them have arbitrarily high momenta and thus arbitrarily high speeds. It's the presence of these impossibly high-speed modes that allows the particle to flee the origin faster than light. In a relativistic theory, by contrast, momentum is not proportional to speed but rather high-momentum particles have speeds approaching that of light. In a relativistic theory, the presence of the high-momentum modes does not allow the wave function to spread faster than light.

 

[kof9595995]

But it seems the relativistic, let say Klein-Gordon propagator, is also non-zero outside the light cone. Then if I modify my argument with KG propagator would it still be a paradox?

 

[kof9595995]

Emm, I know in QFT to discuss causality they use commutator instead of propagator itself. However in my thought experiment I can't see how commutator of propagator is involved, it seems only propagator itself matters.

 

[kof9595995]

I see where I was wrong, the propagator is indeed 0 in spacelike region. I was thinking
< 0|\phi (x)\phi (y)|0 > as the propagator, but it should be
< 0|[\phi (x),\phi (y)]|0 >, which is 0 for spacelike separation.