Post on 24-Feb-2018
7/24/2019 FinteWell
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Finite Well Potential Unbound States
V(x) =
0 for x < a
V0 for a < x < a0 for x > a
This potential has unbounded states for E >0. There are solutions of the form:
(x) =
Beikx for x < aCeilx + Deilx for a < x < aF eikx for x > a
only for these energies:
E+ V0= n2 22
2m
(2a
)
2
For these energies, the energy eigenstates are:
(x) =
Beikx for x < a
Bkl
2l
eia(k+l)eilx + B
k+l2l
eia(k+l)eilx for a < x < a
Be2aikeikx for x > a
To derive these expressions,in addition to using the continuity of the wave function andthe derivative at x = a we use use the relation
2al= n e4ail = 1, e2ail = 1
At the edges of the well, the wave function is
(a) = Beika
(a) = Beika
These wave functions do not vanish.One can also compute the currents using the expression we gave in class:
j(x, t) = i2m
x
x
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7/24/2019 FinteWell
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jI
= km|B|2 jI
= 0 x < a
jII = lm |B|2kl2l2
jII = lm |B|2k+l2l2
a < x < a
jII I
= km|B|2 jII I
= 0 x > a
That is, in the intermediate region, region II, there are currents in both directions. Butthe currents are conserved, as expected at x= a
jI
+ jI
= jII
+ jII
jII
+ jII
=jII I
+jII I
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