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Heisenberg's uncertainty relation not invariant ? 
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#1
Dec706, 02:07 PM

P: 185

I'm not sure whether this has been discussed before. If one just looks at Heisenberg's uncertainty relation (energytime), 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 ?



#2
Dec706, 02:33 PM

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Sci Advisor
P: 7,594

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. 


#3
Dec706, 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 wellforces could be very far from the static usual toymodel ! 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 xp 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 4vector, 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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