Heisenberg's uncertainty relation not invariant ?


by notknowing
Tags: heisenberg, invariant, relation, uncertainty
notknowing
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#1
Dec7-06, 02:07 PM
P: 185
I'm not sure whether this has been discussed before. If one just looks at Heisenberg's uncertainty relation (energy-time), one easily sees that this is not Lorentz invariant. Even very simple results, such as the energy of a particle in a potential well seems not to transform according to the Lorentz transformation. How can it be that such simple and basic results from quantum physics are in such striking contradiction with special relativity ?
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pervect
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#2
Dec7-06, 02:33 PM
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Quote Quote by notknowing View Post
I'm not sure whether this has been discussed before. If one just looks at Heisenberg's uncertainty relation (energy-time), one easily sees that this is not Lorentz invariant. Even very simple results, such as the energy of a particle in a potential well seems not to transform according to the Lorentz transformation. How can it be that such simple and basic results from quantum physics are in such striking contradiction with special relativity ?
I think there was a short thread on this a while back. The short answer is that quantum mechanics, in the form of quantum field theory, is fully covariant, i.e. compatible with special relativity, and this includes the Heisenbereg relationships.

The interesting thing is that there are some issues with quantum mechanics as a single particle theory that disappear when you go to the mulitple particle theory of QFT. I'm not sure exactly where this is discussed, the Wikipedia has a short discussion of some of these aspects, which isn't totally satisfactory and unfortunatley doesn't cite it's sources.

The second problem arises when trying to reconcile the Schrödinger equation with special relativity. It is possible to modify the Schrödinger equation to include the rest energy of a particle, resulting in the Klein-Gordon equation or the Dirac equation. However, these equations have many unsatisfactory qualities; for instance, they possess energy eigenvalues which extend to –∞, so that there seems to be no easy definition of a ground state. Such inconsistencies occur because these equations neglect the possibility of dynamically creating or destroying particles, which is a crucial aspect of relativity. Einstein's famous mass-energy relation predicts that sufficiently massive particles can decay into several lighter particles, and sufficiently energetic particles can combine to form massive particles. For example, an electron and a positron can annihilate each other to create photons. Such processes must be accounted for in a truly relativistic quantum theory. This problem brings to the fore the notion that a consistent relativistic quantum theory, even of a single particle, must be a many particle theory.
However, this isn't really a "quantum mechanics is incompatible with SR" sort of deal. It should be understood that in the form of QFT, quantum mechanics is fully covariant, i.e. fully compatible with special relativity.
lalbatros
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#3
Dec7-06, 03:49 PM
P: 1,235
notknowing,

Let's consider the potential well. If the forces included in this model are not invariant under Lorentz transformation one should not expect invariant results. However, in the frame attached to the well, the model could be excellent. Therefore, if you want to built a relativistic potential well model, you need to include fields equations for the forces in the well and these fields equations need to be relativistic. Note that such well-forces could be very far from the static usual toy-model ! But maybe it is simpler than what I imagine.

I guess this is how -in principle- an atom could be modelled in quantum mechanics. Here the electromagnetic forces could be described in a fully relativistic way, this is known stuff. At least one knows the hamiltonian (lagrangian). However, if things are simpler in the frame attached to the atom, why should we make them more complicated with another frame?

By the way, where could I find the full expression for the fully relativistic equations for an hydrogen atom?

Concerning the x-p uncertainty relations, I feel already more confortable when I remember that the phase of a plane wave is a relativistic invariant. The wavevector is a 4-vector, just like the coordinates. The uncertainty relation is related to the spread of a wavefunction and its fourier transform. I don't think there should be any problem there. But I need to think a bit more to express that clrearly. Maybe you can be faster than me on that point.

Michel


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