Heisenberg Uncertainty Principle question

[jap33]

given 2 unnormalized wave functions:

Y1(x)=e^i(x/m)

Y2(x)=1/2*[e^2i(x/m) + e^3i(x/m) + e^-2i(x/m) + e^-3i(x/m)]

if the positions of the particles were measured, which would be found to be more localized in space? (that is, which has a position known more precisely?)

to my understanding, i understand the principle if you know position specifically, then you know nothing about the momentum, etc.

 

[Dickfore]

Hint: Take the x-axis to be of finite extent ( $-L/2 \le x \le L/2$). this will make all your integrals converge. Then evaluate:

$$A = \int_{-L/2}^{L/2}{|\psi(x)|^{2} \, dx}$$

$$A \, \langle x \rangle = \int_{-L/2}^{L/2}{x \, |\psi(x)|^{2} \, dx}$$

$$A \, \langle x^{2} \rangle = \int_{-L/2}^{L/2}{x^{2} \, |\psi(x)|^{2} \, dx}$$

Then, once you have calcualated $\langle x \rangle$ and $\langle x^{2} \rangle$, you can evaluate the uncertainty in the position by:

$$\Delta x = \sqrt{\langle (\Delta x)^{2} \rangle} = \sqrt{\langle x^{2} \rangle - \langle x \rangle^{2}}$$

What happens to this number if you set $L \rightarrow \infty$?