Momentum and uncertainty in the 1D infinite square well

[Hyperreality]

(a) In the infinite one-dimensional well, what is $$p_{av}$$?

(b) What is $$(p^2)_{av}$$?

(c) What is $$\Delta p = \sqrt{(p^2)_av - (p_av)^2}$$?

(d) Compute $$\Delta p \Delta x$$, and compare with the Heisenberg uncertainty relationship.

Here's my working:

(a) $$p_{av}=0$$.

I'm not so sure about this bit
(b)$$(\frac{p^2}{2m})_{av} = E_{n} = \frac{\hbar^2\pi^2n^2}{2mL^2}$$.
There fore $$(p^2)_{av}=(\frac{\hbar\pi^2n^2}{L})^2$$

(c)Therefore,
$$\Delta p = \frac{\hbar\pi n}{L}$$.

(d)$$\Delta p\Delta x = \frac{\hbar}{2}\sqrt{2n^2\pi^2 -1}$$

Part (d) it seems the most suspicious, that is, the uncertainty increases with n^2. Have I done anything wrong?

 

[dextercioby]

Actually it increases with "n"...You have square root form a "n^{2}"...It looks okay...Though you didn't show the calculations leading to $\Delta x$...

Daniel.

 

[wais]

Your working for parts (a) and (b) is correct. For part (c), you seem to have forgotten to square the term inside the square root, so it should be \Delta p = \frac{\hbar\pi n}{L}. For part (d), you are correct that the uncertainty increases with n^2. This is a fundamental aspect of the Heisenberg uncertainty principle, which states that the product of uncertainties in position and momentum must be greater than or equal to a certain value (in this case, \frac{\hbar}{2}). So as n increases, the uncertainty in momentum also increases, in order to satisfy this relationship. This shows the inherent uncertainty in the position and momentum of a particle in the infinite one-dimensional well.