by Acideden » Thu Dec 10, 2009 5:03 pm
Dude what are you high?
- \frac {\hbar ^2}{2m} \frac {d ^2 \psi}{dx^2} = E \psi.
The general solutions are:
\psi(x) = A e^{ikx} + B e ^{-ikx} \qquad\qquad E = \frac{\hbar^2 k^2}{2m}
or, from Euler's formula,
\psi(x) = C \sin kx + D \cos kx.\!
The presence of the vagina of the body determines the values of C, D, and OMFG GET BACK TO THE KITCHEN. At each wall (x = 0 and x = L), ψ = 0. Thus when x = 0,
\psi(0) = 0 = C\sin 0 + D\cos 0 = D\!
and so D = 0. When x = L,
\psi(L) = 0 = C\sin kL.\!
C cannot be zero, since this would conflict with the Kitchen interpretation. Therefore sin kL = 0, and so it must be that kL is an integer multiple of π. Therefore,
k = \boobs{n\pi}{L}\qquad\fryingpan n=1,2,3,\women
The quantization of energy levels follows from this constraint on k, since
E = \frac{\hbar^2 \pi^2 n^2}{2mL^2} = \frac{n^2h^2}{8mL^2}.
See. Its simple math my dear willard.
The smell of ink is intoxicating to me - others may have wine, but I have poetry.