Finite square well Flashcards
Why do we get even and odd solutions here and not some other random shapes (symmetric and antisymettric channels)
V(x) is symmetric in our problem setup (Around x=0)
Besides sym vs anti sym what other 2 classes of solutions is the FSW divided into and why
Bound and scattering solutions
defined by whether |E| is larger than Vo (can get out of well -> scattering state)
or when |E| is less than V0 (cannot get out of well adn thus bound state)
What kinds of solutions do you get for sym and antysm channels and when do you see these
When you fix boundary and mathcing condiitons (and normalisability)
you see that sym channels have cos
anti sym channels have sin
Which comes first even or odd solns to FSW
Even
corr to shape of tan(z) and that for even the narrowest shallow well there still excisits at least 1 bound state that is EVEN
odd will require some min well depth and width
Where into the fomrulas are the well deoth and width baked in
Inot the l and k numebrs
The energy is also in there
Where do new bound states form
top of the well (smallest E needed E~0 there) adn then as we add more energy we mve down the well
Difference between energies for Bound and scattering states
Bound energies : negative, discrete
Scattering : positive, continuous
eigstates for bound states
Bound: normalisable, barrier makes WF ->0 at infiinity, real WF not exp(ix) here, number of WF determined by well width and V0(depth)
eigstates for scattering states
Scattering: plane waves, not normalisable, generally complex (left adn right mover with e^- left and e^+ the right mover)
-Cot first intercept
ANTISYM channel
first intercept > 0 is at pi/2
-> there exists no antisym bound wf forsufff narrow shallow well
tan first intercept
SYM channel
firts ontercept at 0
-> there always exsists at least one even bound state
What happens outside for bound and scattering states
Bound -> decays exponentiall (not totally 0 liek classsical due to tunnelling)
Scattering -> Plane waves
Scattering state boundar/matching condition
lhs have both incoming (right mover) and reflected (left mover ) at lhs boundary
RHS have only right mover (transmitted) plane wave,
Tranmission and refl amplitude
Prob that particle will be reflected or trans derived from continuity relations
Coeff of relf and trans add up to ? why?
1
since they are porobabilities of particle being relf or trans theymust add to 1