Uncertainty and particle in a box

[hokhani]

TL;DR

More confinement doesn't result in more uncertainty!

As far as I know about the position -momentum uncertainty if a quantum particle is more confined, we expect its momentum to be more uncertain. However, I think, as a counterexample one may take the particle in a box. Each sin wave function (which is the solution of particle in the box) is always a combination of the two opposite momenta no matter how much the particle is confined. So, it seems that the uncertainty in momentum doesn't change by confinement. I appreciate any explanation.

 

[anuttarasammyak]

TL;DR: More confinement doesn't result in more uncertainty!

Each sin wave function (which is the solution of particle in the box) is always a combination of the two opposite momenta no matter how much the particle is confined.

These solution wave functions must be zero outside the box. They are not simple sinusoids functions. Their momentum wave functions do not give simple two opposite values. Ref. https://en.wikipedia.org/wiki/Particle_in_a_box

 

[Morbert]

The wavefunction is zero outside the box, so you don't strictly have two exact but opposite momenta. Also, these two momenta are further from the mean the smaller the box is, increasing uncertainty.

 

[hokhani]

These solution wave functions must be zero outside the box. They are not simple sinusoids functions. Their momentum wave functions do not give simple two opposite values. Ref. https://en.wikipedia.org/wiki/Particle_in_a_box

Right, but there are two opposite momentum values for the particle in the box. So, the momentum uncertainty in the box is within two uncertain momentum values. However, it seems that uncertainty outside the box increases. Doesn't it?

 

[hokhani]

The wavefunction is zero outside the box, so you don't strictly have two exact but opposite momenta. Also, these two momenta are further from the mean the smaller the box is, increasing uncertainty.

Here, do you also consider the deviation from the mean momentum (which is zero) a criterion of uncertainty?

 

[Morbert]

Here, do you also consider the deviation from the mean momentum (which is zero) a criterion of uncertainty?

The momentum uncertainty is just $\sqrt{\langle p^2\rangle - \langle p \rangle^2}$.

 

[weirdoguy]

Right, but there are two opposite momentum values for the particle in the box.

No, as is explicitly stated in wiki article: https://en.wikipedia.org/wiki/Particle_in_a_box#Momentum_wave_function

And also was explicitly said in what you quoted