Zero momentum distribution and consequences on uncertainty principle

[hoju]

What happens if the momentum distribution (sigma p) equals zero. Say the expectation value for the momentum (<p>) and <p^2> are zero. Then you will get 0>or=h/4Pi. How can this be possible? Or vice versa, what if sigma x equals zero?

 

[CompuChip]

That's not possible, indeed.
So the more accurately one measures the momentum, the less accurately one knows the position. When $\sigma p \to 0$, therefore, $\sigma x \to \infty$ and vice versa.
You can interpret this as saying that not only is it impossible to measure position or momentum with infinite precision, but even that the theory tells you so!
Or you can interpret this as saying that at some very small scale $\sigma x < L$ quantum mechanics breaks down and the uncertainty relation cannot be applied anymore anyway.

 

[alxm]

A zero expectation value doesn't imply a zero uncertainty.

 

[CompuChip]

No but if in addition <p^2> = 0 then the uncertainty $\sqrt{\langle p \rangle^2 - \langle p^2 \rangle}$ is zero.

 

[alxm]

No but if in addition <p^2> = 0 .

It never is for a solution to the Schrödinger equation.

 

[caduceus]

I appologize for my ignorance. I am a new learner as well, and just trying to figure out the meaning of quantum mechanics rather than just doing some algebra. However, what if a wave function is just a real constant? Will not we get a zero fluctuation in momentum?

 

[soothsayer]

However, what if a wave function is just a real constant?

I believe that's impossible, a real wavefunction couldn't satisfy the Schrodinger equation.