# Zero momentum distribution and consequences on uncertainty principle

1. Mar 5, 2009

### 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?

Last edited: Mar 5, 2009
2. Mar 6, 2009

### 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.

3. Mar 6, 2009

### alxm

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

4. Mar 6, 2009

### CompuChip

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

5. Mar 6, 2009

### alxm

It never is for a solution to the Schrödinger equation.

6. Mar 13, 2009