Consider a particle of mass m and energy E incident from the left on a potential

Question

Consider a particle of mass m and energy E incident from the left on a potential step of height U located at x = 0. Assume that E > U. The wavefunction for x 0 is psi L (x) = A exp (ikLx) + B exp (-ikLx), where Aexp(ikLx) represents the incident beam and Bexp(-ikLx) the reflected one. On the right of the step, i.e. for x > 0, the wavefunction is phi R(x) = C exp (ikRx). Prove that phi L(x) (phi R(x)) is a solution of the Schrodinger equation for x 0 (x > 0) and determine kL and kR in terms of m, E and U. Determine B as a function of A by imposing the boundary conditions at x = 0: phi L(0) = phi R (0) and d phi L / dx |x=0 = d phi R / dx |x=0.

Explanation / Answer

A)

schrodinger Equation is (1-D)

E(x)=-((h/2)2/2m)d2/dx2+U(x)

for x0

U=0

=>

  E(x)=-((h/2)2/2m)d2/dx2

the standard & regular solution for this type of differential equation is exponential form as its derivate is also same form with a constant in multiplication.

so, the solution is of type

(x)=Aexp(iklx)+Bexp(-iklx)

in cross verification we can find it as true with kl2=2mE/(h/2)2

for x>0

schrodinger Equation is (1-D)

E(x)=-((h/2)2/2m)d2/dx2+U(x)

the solution is of type

(x)=Cexp(ikRx)

with

kr2= 2m(E-U)/(h/2)2

B) from first condition

A+B=c

dL/dx=Aiklexp(iklx)-Biklexp(-iklx)

dR/dx=CikRexp(ikRx)

from condition 2

(A-B)kl=Ckr

=> (A-B/A+B)=kr/kl

=> 2A/2B=(kr+kl/kl-kr)

=> B=A(kl-kr/kr+kl)

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