Step Potential

## **Problem**

A beam of particles coming from very far away from the left

(
# x
−
∞
)

{\displaystyle (x= -\infty)}

 meets a step potential barrier described by

V
(
x
# )

V
0

{\displaystyle V (x) = V\_0}

, where

V
0

>
0

{\displaystyle V\_0 > 0}

, at

# x
0

{\displaystyle x = 0}

. The incident beam of particles have energy

0
<
E
<

V
0

{\displaystyle 0<E<V\_0}

. Determine the wavefunction left and right of the step potential barrier as well as the transmission and reflection coefficients for this potential.

(
1
)

Region 1:

−
∞
<
x
<
0

(
2
)

Region 2:

0
<
x
<
∞

{\displaystyle
\begin{align}
&(1) \; \textbf{Region 1:} \;-\infty < x<0 \\[2 pt]
&(2) \; \textbf{Region 2:} \; 0 < x < \infty \\[2 pt]
\end{align}}

## **Solution**

A plot of the potential is shown above (Figure 1). The energy of the incoming particles

(
0
<
E
<

V

0

)

{\displaystyle (0<E<V\_{0})}

 is indicated by the thick dashed line.

**Region 1**

d

2

ψ

d

x

2

=

k

2

ψ
,

# k

−
2
m
E

ℏ

2

{\displaystyle \frac{{d}^{2}\psi}{d{x}^{2}}={k}^{2}\psi, \quad k=\sqrt{\frac{-2mE}{{\hbar}^{2}}} }

**Region 2**

d

2

ψ

d

x

2

=

κ

2

ψ
,

# κ

2
m
(

V

0

−
E
)

ℏ

2

{\displaystyle {\frac {{d}^{2}\psi }{d{x}^{2}}}={\kappa }^{2}\psi ,\quad \kappa ={\sqrt {\frac {2m(V\_{0}-E)}{{\hbar }^{2}}}}}

Hence, we find that the wavefunctions take the form:

ψ
(
x
# )

{

A
e

i
k
x

B
e

−
i
k
x

Region 1

C
e

κ
x

D
e

−
κ
x

Region 2

{\displaystyle \psi (x)=\begin{cases}
{ Ae }^{ ikx }+{ Be }^{ -ikx } &\textbf{Region 1} \\
{ Ce }^{ \kappa x }+{ De }^{ -\kappa x } &\textbf{Region 2} \\
\end{cases} }

For this problem, the incident wavefunction (moving to the right) is

ψ

i
n
c

(
x
# )

A
e

i
k
x

{\displaystyle \psi\_{inc} (x)= {Ae }^{ ikx } }

 and the reflected (moving to the left) is

ψ

r
e
f

(
x
# )

B
e

−
i
k
x

{\displaystyle \psi\_{ref} (x)= {Be }^{-ikx } }

. The complete wavefunction for **Region 1** is the sum of the incident and reflected wavefunction; therefore,

ψ

1

(
x
# )

A
e

i
k
x

B
e

−
i
k
x

.

{\displaystyle {\psi}\_{1} (x)= { Ae }^{ ikx }+{ Be }^{ -ikx }.
 }

We reject the first solution because it blows up as

x
→
∞

{\displaystyle x\to\infty}

 and is physically unacceptable, so,

ψ

t
r
a
n
s

(
x
# )

D
e

−
κ
x

{\displaystyle {\psi}\_{trans} (x) = { De }^{ -\kappa x } }

.

Therefore, in **Region 2**, we get

ψ

2

(
x
# )

D
e

−
κ
x

.

{\displaystyle {\psi}\_{2} (x) = { De }^{ -\kappa x }. }

The wavefunction must obey two conditions at the boundary

# x
0

{\displaystyle x = 0}

:

ψ
(
x
)

{\displaystyle \psi(x) }

 must be continuous
2. The first derivative

d
ψ

d
x

{\displaystyle \frac{d\psi}{dx} }

 must be continuous
Matching up the wavefunctions

ψ

1

(
x
)
,

ψ

2

(
x
)

{\displaystyle {\psi}\_{1}(x), {\psi}\_{2}(x)
 }

at

# x
0

{\displaystyle x = 0}

 and the derivatives

d

ψ

1

d
x

,

d

ψ

2

d
x

{\displaystyle \frac{d{\psi}\_{1}}{dx}, \frac{d{\psi}\_{2}}{dx}
 }

gives the conditions:

(
1
)

A
# B
D

(
2
)

i
k
(
A
−
B
# )
κ
D

{\displaystyle \begin{align}
&(1) \; A+B=D \\[2 pt]
&(2) \; ik(A-B)= \kappa D \\[2 pt]
\end{align} }

Therefore, by making combinations of the above equations (try this yourself), we find

(
5
)

i
k
(
A
−
B
# )
κ
(
A
B
)

{\displaystyle \begin{align}
&(5) \; ik(A-B)=\kappa (A+B)
\end{align} }

Solving for

B

{\displaystyle B}

 and

D

{\displaystyle D}

 in terms of

A

{\displaystyle A}

 gives

B

=

i
k
κ

i
k
−
κ

A

D

=

2
i
k

i
k
−
κ

A

{\displaystyle \begin{align}
B &= \frac{ik + \kappa}{ik - \kappa} A \\[5 pt]
D &= \frac{2ik}{ik - \kappa} A \\
\end{align} }

The probability of reflection is calculated by

R

=

B

∗

B

A

∗

A

=

i
k
−
κ

i
k
κ

⋅

−
i
k
−
κ

−
i
k
κ

=

k

2

κ

2

k

2

κ

2

=
1

{\displaystyle {\begin{aligned}R&=\left|{\frac {{B}^{\*}B}{{A}^{\*}A}}\right|\\[5pt]\\&={\frac {ik-\kappa }{ik+\kappa }}\cdot {\frac {-ik-\kappa }{-ik+\kappa }}\\[5pt]&={\frac {k^{2}+\kappa ^{2}}{k^{2}+\kappa ^{2}}}\\[5pt]&=1\\\end{aligned}}}

Naturally, the probability of reflection is

1
−
R

{\displaystyle 1-R }

; therefore,

# T
0

{\displaystyle T = 0}

.