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:

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

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

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

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

<br /> \Delta x = \sqrt{\langle (\Delta x)^{2} \rangle} = \sqrt{\langle x^{2} \rangle - \langle x \rangle^{2}}<br />

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