2. A particle is trapped inside a finite square well (left) and a finite tilted
ID: 1768952 • Letter: 2
Question
2. A particle is trapped inside a finite square well (left) and a finite tilted well (right). The depth of the square well is the same as the average of the tilted well. (a) Draw the particle's wave functions in both weds at the ground state and the first two excited states (n =1,2, and 3). For the tilted well, make sure you depict clearly the asymmetry in wavelength and amplitude in each wave function and how the wave function penetrates the left and right sides of the well. Mark any shifts in the nodal positions. (b)Compare the values of En in the square well and En' in the tilted well and give your reasons.Explanation / Answer
A particle of mass m is in the ground state of a one-dimensional infinite square well with walls at x=0 and x=a. ?1(x)=2a--vsin(pxa) , E1= h2p22ma2 * What is the initial wave function ?(x,0)? * h is supposed to be h bar, I just couldn't find it) 2. Relevant equations 3. The attempt at a solution My attempt: If the general solution is a superposition of all stationary states, ?(x,t)=?cn?ne-iEnth , at t=0, ?(x,0)=?cn?n . Also, at this time, the particle is in the ground state (n=1), so: ?(x,0)=c1?1 . Do I assume c1=1 at this point, because the wave function "collapses" once the energy becomes known? I'm just not sure if I understand exactly what happens when the known data is given. The solution itself is supposed to be ?(x,0)=?1(x)=2a--vsin(pxa) .
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