Finite universe, quantized linear momentum, and the HUP....

[asimov42]

Hi all,

Yet another question: if the universe is finite, then linear momentum should be quantized (I assume in a similar manner to an infinite potential well since there are boundary conditions). My question is, then, if one computes a value for $\Delta p$ (momentum variance), is the variance also quantized? Due to the uncertainty principle, the variance cannot be zero - but I'm unsure about the implications of quantized momentum.

 

[Demystifier]

The variance is not quantized. The variance comes from superposition principle, and coefficients of superposition can take any values (constrained only by the overall normalization condition).

 

[asimov42]

So there would be a finite number of momentum states, but the probability distribution over superpositions of those states (combinations) is continuous, hence continuous values for variance?

 

[dextercioby]

There is no quantized momentum in a finite universe.

 

[asimov42]

Is it not the case that momentum would have to be quantized? Since the universe has finite size, don't the boundary conditions enforce quantization (albeit with extremely small separation for a universe of any reasonable size).

 

[Orodruin]

Is it not the case that momentum would have to be quantized? Since the universe has finite size, don't the boundary conditions enforce quantization (albeit with extremely small separation for a universe of any reasonable size).

If the Universe has a boundary, there is no momentum operator whose eigenvalues can be quantised. The presence of the boundary breaks translational invariance.

You might consider a compact universe (such as considering a particle living on a circle) where something like momentum exists (actually, more like angular momentum, but let's skip that for now).

 

[asimov42]

Thanks @Orodruin! Are there implications then for a universe with a boundary? If translational invariance is broken (something I hadn't thought about), then this also break momentum conservation (at the boundary), correct? Presumably this does not have implications for the uncertainty principle?

 

[asimov42]

Are there implications then for a universe with a boundary? If translational invariance is broken (something I hadn't thought about), then this also break momentum conservation (at the boundary), correct? Presumably this does not have implications for the uncertainty principle?

Also, with a boundary in place, the wavefunction would not be differentiable at the boundary itself - how is this dealt with? (since both the wavefunction and its derivative should be continuous)

 

[Orodruin]

Also, with a boundary in place, the wavefunction would not be differentiable at the boundary itself - how is this dealt with? (since both the wavefunction and its derivative should be continuous)

It is differentiable and continuous. If you are thinking "wave function = 0" outside: It is not. The outside is not part of the domain of the wave function.

Also, the "continuous and differentiable" is a truth with modification that makes some assumptions about the potential. Consider what happens in a delta function potential.

 

[asimov42]

Ah I see - what about at the exact boundary itself? (this would be the set of all points that actually are 'on' the boundary, if one can specify such a thing) If I understand correctly, you're saying there is no discontinuity in the derivative because all points are 'inside' the universe? So the analogy of an infinite square well isn't really correct.

 

[Demystifier]

So there would be a finite number of momentum states, but the probability distribution over superpositions of those states (combinations) is continuous, hence continuous values for variance?

Yes.