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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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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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