Quantum Mechanics - Infinite Potential Well

TheBaker
Messages
18
Reaction score
0

Homework Statement



A particle is trapped in an infinite potential well, with the infinite walls at ±a. At time t=0, the wavefunction of the particle is

\psi = \frac{1}{\sqrt{2a}}

between -a and a, and 0 otherwise.

Find the probability that the Energy of the particle is \frac{9 \bar{h}^2 \pi^2}{8ma^2}

Homework Equations



E_n = \frac{n^2\bar{h}^2\pi^2}{8ma}

\psi = A \cos{\frac{(2r+1) \pi x}{2a}} for |x| < a
\psi = 0 otherwise

The Attempt at a Solution



I've calculated the above equations, but I'm unsure how to get from them to the probability of the particle having a certain energy. This could be really simple and it's me just having a brain dead moment, but any help would be very much appreciated.
 
Physics news on Phys.org
btw there is also a sin solution with an argument (in your notation) 2r(pi)x/a

personally I prefer the notation n(pi)x/a n even

however the cos solution you wrote is the one you want with r=1, that 2r+1 thing is just a way of writing n so that n is always odd.

so just take the projection of psi at t=0 on your cos function and square the answer
 
The sin solution isn't valid because this well has even parity (i.e. it's symmetric).

How do I find A? Presumably I need to use the initial condition of Psi, but I found when doing that that A is x dependent, when it should be a constant.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top