QM Lecture 6: Quantum Tunneling

**By: Tao Steven Zheng (鄭濤)**

*This article was originally posted on Brilliant.org by the same author.*

## **Transmission and Reflection**

My past lectures have been largely theoretical and mathematically intensive. The reason was to introduce the foundations or "language" in which quantum mechanics works. Now it is time to look into actual physical phenomena, which will lead us into the wonderful applications that have helped us forge the modern world. One important and astounding application of quantum mechanics is **quantum tunnelling**, a quantum mechanical phenomenon whereby a wavefunction can propagate through a potential energy barrier. The existence of this phenomenon may be explained in terms of the Heisenberg uncertainty principle! In quantum mechanics, an object can exist as a wave or as a particle; therefore, the uncertainty in the exact location of (very light) particles allows these particles to break the rules of classical mechanics and move in space without passing over the potential energy barrier. To begin our investigation of this quantum tunnelling, consider the **rectangular barrier** problem. The objective of this problem is to determine the mathematical expressions for computing the probability of a particle-wave moving with energy

E
<
V

{\displaystyle E<V }

 (moving from left to right) being transmitted or reflected as it interacts with a rectangular potential barrier with potential energy

V

{\displaystyle V}

.

Let us separate the picture into three regions:

(
1
)

Region 1:

−
∞
<
x
<
0

(
2
)

Region 2:

0
≤
x
≤
L

(
3
)

Region 3:

L
<
x
<
∞

{\displaystyle
\begin{align}
&(1) \; \textbf{Region 1:} \;-\infty < x<0 \\[2 pt]
&(2) \; \textbf{Region 2:} \; 0 \le x \le L \\[2 pt]
&(3) \; \textbf{Region 3:} \;L < x< \infty \\
\end{align}}

In **Region 1**, the potential is zero. A moving wave thus has energy that is greater than the potential. This is also true in **Region 3**. However, in **Region 2**, the energy of the wave is either less than or equal to the potential. Thus, from the Schrödinger equation, we find two differential equations:

**Region 1 and Region 3**

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
−
E
)

ℏ

2

{\displaystyle \frac{{d}^{2}\psi}{d{x}^{2}}={\kappa}^{2}\psi, \quad \kappa=\sqrt{\frac{2m(V-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

F
e

i
k
x

Region 3

{\displaystyle \psi (x)=\begin{cases}
{ Ae }^{ ikx }+{ Be }^{ -ikx } &\textbf{Region 1} \\
{ Ce }^{ \kappa x }+{ De }^{ -\kappa x } &\textbf{Region 2} \\
{ Fe }^{ ikx } &\textbf{Region 3} \end{cases} }

If you were wondering where

G
e

−
i
k
x

{\displaystyle {Ge}^{-ikx} }

 went in the solution of the above differential equations, you should know what the exponents mean. Plane-waves that travel to the right are of the form

e

i
k
x

{\displaystyle {e}^{ikx} }

 and plane-waves that travel to the left are of the form

e

−
i
k
x

{\displaystyle {e}^{-ikx} }

. In quantum tunnelling, there are waves that get reflected and transmitted, so in **Region 1** we find waves travelling to the left (heading towards the barrier and transmitted) and waves travelling to the left (reflected by the barrier); whereas, the wave travels solely to the right in **Region 3**.

The continuity of wavefunctions and its derivative wavefunctions allows us to solve this problem. From the continuity condition we get:

(
1
)

A
# B
C
D

(
2
)

C
e

κ
L

D
e

−
κ
L

=

F
e

i
k
L

(
3
)

A
i
k
−
B
i
# k
C
κ
−
D
κ

(
4
)

C
κ
e

κ
L

−

D
κ
e

−
κ
L

=

F
i
k
e

i
k
L

{\displaystyle \begin{align}
&(1) \; A+B=C+D \\[2 pt]
&(2) \; {Ce}^{\kappa L}+{De}^{-\kappa L}= {Fe}^{ikL} \\[2 pt]
&(3) \; Aik-Bik=C\kappa-D\kappa \\[2 pt]
&(4) \; {C\kappa e}^{\kappa L}-{D\kappa e}^{-\kappa L}= {Fike}^{ikL}
\end{align} }

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

(
5
)

2
# A

(

1

κ

i
k

)

C

(

1
−

κ

i
k

)

D

(
6
)

2
C

e

κ
L

=

(

1

i
k

κ

)

F

e

i
k
L

(
7
)

2
D

e

−
κ
L

=

(

1
−

i
k

κ

)

F

e

i
k
L

{\displaystyle \begin{align}
&(5) \; 2A=\left(1+\frac{\kappa}{ik}\right)C+\left(1-\frac{\kappa}{ik}\right)D \\[2 pt]
&(6) \; 2C{e}^{\kappa L}=\left(1+\frac{ik}{\kappa}\right)F{e}^{ikL} \\[2 pt]
&(7) \; 2D{e}^{-\kappa L}=\left(1-\frac{ik}{\kappa}\right)F{e}^{ikL} \\[2 pt]
\end{align} }

Combining these three equations yields

2
# A

(

1

κ

i
k

)

(

1

i
k

κ

)

F

e

i
k
L

e

−
κ
L

2

(

1
−

κ

i
k

)

(

1
−

i
k

κ

)

F

e

i
k
L

e

κ
L

2

{\displaystyle 2A=\left(1+\frac{\kappa}{ik}\right)\left(1+\frac{ik}{\kappa}\right)\frac{F{e}^{ikL}{e}^{-\kappa L}}{2}+\left(1-\frac{\kappa}{ik}\right)\left(1-\frac{ik}{\kappa}\right)\frac{F{e}^{ikL}{e}^{\kappa L}}{2} }

which can be simplified as

A

e

−
i
k
L

F

=
cosh
⁡

(
κ
L
)

i

(

κ

2

−

k

2

2
k
κ

)

sinh
⁡

(
κ
L
)

{\displaystyle \frac{A{e}^{-ikL}}{F}=\cosh{(\kappa L)}+i\left(\frac{{\kappa}^{2}-{k}^{2}}{2k\kappa}\right)\sinh{(\kappa L)} }

Now the probability of a wave to tunnel through the barrier is equal to the probability of the wavefunction in **Region 3** divided by the probability of the wavefunction in **Region 1**. Thus, the probability of transmission is quantified by

T

=

F

∗

F

A

∗

A

=

[

cosh

2

(
κ
L
)

(

κ

2

−

k

2

2
k
κ

)

2

sinh

2

(
κ
L
)

]

−
1

{\displaystyle \begin{align}
T &= \left|\frac{{F}^{\*}F}{{A}^{\*}A}\right| \\[5 pt]
&={\left[{{\cosh}^{2}(\kappa L)+\left(\frac{{\kappa}^{2}-{k}^{2}}{2k\kappa}\right)}^{2}{\sinh}^{2}(\kappa L)\right]}^{-1}
\end{align} }

(remember probabilities are defined by the complex conjugate product).

Since

cosh

2

(
x
)
−

sinh

2

(
x
# )
1

{\displaystyle {\cosh}^{2}(x)-{\sinh}^{2}(x)=1 }

, we get

# T

[

1

(

k

2

κ

2

2
k
κ

)

sinh

2

(
κ
L
)

]

−
1

.

{\displaystyle T={\left[1+\left(\frac{{k}^{2}+{\kappa}^{2}}{2k\kappa}\right){\sinh}^{2}(\kappa L)\right]}^{-1}. }

If we let

# β

(

k

2

κ

2

2
k
κ

)

{\displaystyle \beta=\left(\frac{{k}^{2}+{\kappa}^{2}}{2k\kappa}\right) }

, we find

# T

1

1

β

2

sinh

2

(
κ
L
)

.

{\displaystyle T=\frac{1}{1+{\beta}^{2}{\sinh}^{2}(\kappa L)}. }

Naturally, the probability of reflection is

1
−
T

{\displaystyle 1-T }

; hence,

# R

β

2

sinh

2

(
κ
L
)

1

β

2

sinh

2

(
κ
L
)

.

{\displaystyle R=\frac{{\beta}^{2}{\sinh}^{2}(\kappa L)}{1+{\beta}^{2}{\sinh}^{2}(\kappa L)}. }

In our macroscopic experience, objects colliding against a wall will be deflected. This is analogous to the reflection probability being 100% and transmission probability being 0%. The above example shows that it is possible for matter waves to "go through walls" with some probability, given that a matter wave has sufficient energy or the barrier being sufficiently narrow (small L). Check out this video

## **Applications of Quantum tunnelling**

Quantum tunnelling is responsible for many physical phenomena that baffled scientists in the early 20th century. The first would be radioactivity. The most common type of radioactive decay is caused by an electron being captured by atomic nuclei. Contrary to the Bohr model of the atom, electrons exist not in shells but in "clouds" of probability densities. Because quantum tunnelling is common at the atomic and subatomic levels, this electron capture is nothing strange. Ever since, quantum tunnelling has played a vital role in our capability of manipulating electrons.

Other applications of quantum tunnelling include:

**Scanning Tunnelling Microscope**

**P-N junction**

**Tunnel junction**

## **Problems**

### **Problem 1: Step Potential**

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.

Solution