
Also, we will see that the method can be used to find energy levels in confining quantum wells.

Finally, it can be used to obtain approximate solutions to complex potential profiles.

E1

E2

etc


1 2 3 4 5

i

rt

1 2 3 4 5

A2

B2

C2

D2

A3

B3

C3

D3

A4

B4

C4

D4

i

rt

In a multi-layer problem, the difficulties come in handling the reflections at all of the interfaces. However, if we can determine how the plane waves relate from one side of a constant potential region to the other, including the effects of scattering at the interfaces, then we can relate the transmitted amplitude to the incident amplitude (or reflected to incident) of the overall system.

For each region, we’ll try to write a matrix of the form:

where [Mn] is a 2x2 matrix describing the nth region.

�An�1

Bn�1

�= [Mn]

�Cn

Dn

�


1 2 3 4 5

A2

B2

C2

D2

A3

B3

C3

D3

A4

B4

C4

D4

i

rt

1 2 3 4 5

i

rt

= [M2]

A3

B3

�A2

B2

�= [M2]

C2

D2

�

= [M2] [M3]

C3

D3

�= [M2] [M3]

A4

B4

�

= [M2] [M3] [M4]

C4

D4

�

ir

�= [IL] [M2] [M3] [M4] [IR]

t0

�


transmission:

reflection:

Finding reflection and transmission coefficients is a process of multiplying individual layer matrices to obtain an overall “system” matrix. The reflection and transmission coefficient ratios are found from the elements of the system matrix.

ir

�= [IL] [M2] [M3] [M4] [IR]

t0

�

ir

�=

M11 M12

M21 M22

� t0

�

r

i=

M21

M11

t

i=

1

M11


Unfortunately, our approach won’t be quite as tidy as the previous slide would imply. Since the electron waves are scattered at interfaces between layers, we need a matrix to describe what happens at each interface. Of course, an interface is not a property of single layer, but depends on the layer properties on either side of the interface.

Also, the waves change phase as they propagate across regions where E > Un, or they grow and decay exponentially across regions where E < Un. We will need propagation matrices to describe these changes.

Intuitively then, each layer leads to two matrices to be included in the sequence, one propagation matrix and one interface matrix.

However, once we have the form of the propagation and interface matrices, we’ll see that we can combine them in the right way to obtain a simple, one-matrix description of each layer. However, getting to this point is not intuitive, so we’ll take a round-about approach, but one that will hopefully give a clearer picture of what we’re doing.


Interface matrices

Case 1 - E > U on both side of the interface.

We assume the interface occurs at x = xn. For x < xn, the potential is Un, and for x > xn, it is Un+1. Since E > Un and E > Un+1, we use plane-wave solutions on both sides of the interface.

E

Un

Un+1

xn

�Q ([) = $ exp [LNQ ([ � [Q)] + % exp [�LNQ ([ � [Q)]

�Q+� ([) = & exp [LNQ+� ([ � [Q)] + ' exp [�LNQ+� ([ � [Q)]

$+ % = &+'

LNQ$� LNQ% = LNQ+�&� LNQ+�'

Apply the boundary conditions


Note that the matrix elements in this case are all real.

AB

�=

I11 I12I21 I22

� CD

�

$ =��

��+

NQ+�NQ

�&+

��

�� NQ+�

NQ

�'

% =��

�� NQ+�

NQ

�&+

��

��+

NQ+�NQ

�'

Use the two equations to write A and B in terms of C and D.

�$%

�=

�

�

�� + NQ+�

�NQ

� �� NQ+�

�NQ

�

�� NQ+�

�NQ

� �� + NQ+�

�NQ

�

�

��&'

�In matrix form


Case 2 - E > U on the left and E < U on the right.

As before, we assume the interface occurs at x = xn. For x < xn, the potential is Un, and for x > xn, it is Un+1. For x < xn, we need plane wave solutions and for x > xn, we use growing and decaying exponentials.

Using the connection rules at the interface:

E

Un

Un+1

xn

�Q ([) = $ exp [LNQ ([ � [Q)] + % exp [�LNQ ([ � [Q)]

�Q+� ([) = & exp [�Q+� ([� [Q)] +' exp [��Q+� ([� [Q)]

$+ % = &+'

LNQ$� LNQ% = �Q+�&� �Q+�'


Solving for An and Bn in terms of Cn+1 and Dn+1:

�An

Bn

�=

�

�

�12 � i�n+1

2kn

� �12 + i�n+1

2kn

�

�12 + i�n+1

2kn

� �12 � i�n+1

2kn

�

�

��Cn+1

Dn+1

�

$Q =��

�� L�Q+�

NQ

�&Q+� +

��

��+ L�Q+�

NQ

�'Q+�

%Q =��

��+ L�Q+�

NQ

�&Q+� +

��

�� L�Q+�

NQ

�'Q+�

Note that the matrix elements in this case are all complex.

In matrix form


Case 3 - E < U on the left and E > U on the right.

Same song, third verse. For x < xn, the potential is Un, and for x > xn, it is Un+1. For x < xn, we need growing and decaying exponentials and for x > xn, we use plane wave solutions.

Using the connection rules at the interface:

E

Un

Un+1

xn

�Q ([) = $Q exp [� ([� [Q)] + %Q exp [�� ([� [Q)]

$Q + %Q = &Q+� +'Q+�

�Q+� ([) = &Q+� exp [LNQ+� ([ � [Q)] + 'Q+� exp [�NQ+� ([ � [Q)]

�Q$Q � �Q%Q = LNQ+�&Q+� � LNQ+�'Q+�



The matrix elements are again all complex.

In matrix form

$Q =��

��+ L NQ+�

�Q

�&Q+� +

��

�� L NQ+�

�Q

�'Q+�

%Q =��

�� L NQ+�

�Q

�&Q+� +

��

��+ L NQ+�

�Q

�'Q+�

�An

Bn

�=

�

�

�12 + ikn+1

2�n

� �12 � ikn+1

2�n

�

�12 � ikn+1

2�n

� �12 + ikn+1

2�n

�

�

��Cn+1

Dn+1

�


Case 4 - E < U on the left and E < U on the right.

Final stanza. Everything is growing and decaying exponentials.

E

Un Un+1

xn

�Q ([) = $Q exp [�Q ([� [Q)] + %Q exp [��Q ([� [Q)]

�Q+� ([) = &Q+� exp [�Q+� ([� [Q)] +'Q+� exp [��Q+� ([� [Q)]

$Q + %Q = &Q+� +'Q+�

�Q$Q � �Q%Q = �Q+�&Q+� � �Q+�'Q+�

Applying the boundary conditions one last time:



$Q =��

��+

�Q+��Q

�&Q+� +

��

�� Q+�

�Q

�'Q+�

%Q =��

�� Q+�

�Q

�&Q+� +

��

��+

�Q+��Q

�'Q+�

�An

Bn

�=

�

�

�12 + �n+1

2�n

� �12 � �n+1

2�n

�

�12 � �n+1

2�n

� �12 + �n+1

2�n

�

�

��Cn+1

Dn+1

�

The matrix elements are again all real.

In matrix form


It’s pretty easy to see that k should be replaced with -iα in regions where E < U.

E > Un

E > Un+1

E > Un

E < Un+1

E < Un

E > Un+1

E < Un

E < Un+1

Summary:I11 I12I21 I22

�

�

�

�12 + �n+1

2�n

� �12 � �n+1

2�n

�

�12 � �n+1

2�n

� �12 + �n+1

2�n

�

�

�

�

�

�12 + ikn+1

2�n

� �12 � ikn+1

2�n

�

�12 � ikn+1

2�n

� �12 + ikn+1

2�n

�

�

�

�

�

�12 � i�n+1

2kn

� �12 + i�n+1

2kn

�

�12 + i�n+1

2kn

� �12 � i�n+1

2kn

�

�

�

�

�

�12 + kn+1

2kn

� �12 � kn+1

2kn

�

�12 � kn+1

2kn

� �12 + kn+1

2kn

�

�

�


As a traveling wave crosses a region where E > Un, it changes phase.

Propagation matrices

C = A exp(iknLn)

For a wave traveling the -x direction

D = B exp(�iknLn)

Expressed in matrix form:

AB

�=

exp(�iknLn) 0

0 exp(iknLn)

� CD

�

E

Un

xn xn+1

kn

A C

B D

For a wave traveling in the +x direction, we can write by inspection:

AB

�=

P11 P12

P21 P22

� CD

�


For an evanescent wave in a region where E < Un, there is no phase change, but the amplitude must change exponentially.

C = A exp(↵nLn)

D = B exp(�↵nLn)

In this case, the “propagation” matrix is

AB

�=

exp(�↵nLn) 0

0 exp(↵nLn)

� CD

�

E

Un

xn xn+1

αn

A C

B D


Looking again at the potential posed on the first slide, we see that we need a whole string of interface and propagation matrices.

U4

U1

U2

U3

U5

incident

reflected

transmitted

L2 L3 L4

t

i=

1

M11

r

i=

M21

M11

ir

�= [I12] [P2] [I23] [P3] [I34] [P4] [I45]

t0

�=

M11 M12

M21 M22

� t0

�

T =k5k1

|t|2

|i|2 =k5k1

1

|M11|2

R =k1k1

|r|2

|i|2 =|M21|2

|M11|2


The matrix technique lends itself well to programming in Matlab or some other language. However, handling the interfaces is a bit unwieldy since the interface matrix involve properties of two layers. It would be nice if everything about a given layer could be included in one matrix. Can this be done? Look at the form of an interface matrix.

The matrix with kn represents the left side of the interface. The matrix with kn+1 represents the right side of the interface.

Mathematically, it can be split in two:

Alternatively, the matrix with kn represents the right end of the nth layer. The matrix with kn+1 represents the left end of the (n+1)st layer.

In,n+1 =1

2

2

4

⇣1 + kn+1

kn

⌘ ⇣1� kn+1

kn

⌘

⇣1� kn+1

kn

⌘ ⇣1 + kn+1

kn

⌘

3

5

In,n+1 =1p2

2

41

⇣1kn

⌘

1⇣� 1

kn

⌘

3

5 · 1p2

1 1

(kn+1) (�kn+1)

�

[,Q,Q+�] = [/Q] [5Q+�]


Look a section of the sequence of matrices from our original problem.

[0] = [,��] [3�] [,��] [3�] [,��] [3�] [,��]

Split the interface matrices

[0] = [/�] [5�] [3�] [/�] [5�] [3�] [/�] [5�] [3�] [/�] [5�]

= [/�] [5�] [3�] [/�] [5�] [3�] [/�] [5�] [3�] [/�] [5�]

= [/�] [0�] [0�] [0�] [5�]

[0Q] = [5Q] [3Q] [/Q]

where

The layer matrix [Mn] contains all of the information about a particular layer. The parameters for layer n show up only in that particular matrix. This makes it easier to specify and compute the matrices in a program.


In the layer, if E > Un (propagating wave)

[0Q] = [5Q] [3Q] [/Q]

If E < Un (evanescent wave)

[Mn] =

�

�

�1�2

� �1�2

�

�� i�n�

2

� �i�n�

2

�

�

� ·�exp (�i�nLn) 0

0 exp (�nLn)

�·

�

�

�1�2

� �i�2�n

�

�1�2

� ��i�2�n

�

�

�

[Mn] =

�

�

�1�2

� �1�2

�

�kn�

2

� �� kn�

2

�

�

� ·�exp (�iknLn) 0

0 exp (iknLn)

�·

�

�

�1�2

� �1�2kn

�

�1�2

� �� 1�

2kn

�

�

�

[Mn] =

�cos (knLn) � i

knsin (knLn)

�ikn sin (knLn) cos (knLn)

�

[Mn] =

�cosh (�nLn) i

�nsinh (�nLn)

�i�n sinh (�nLn) cosh (�nLn)

�


Example - tunneling through a square barrier (redux)

incident

reflected

transmitted

U = Uo

U = 0

x = 0 x = L

1 2 3

We’ve done this before and know the result. This may a good test for our matrix approach.

There is an electron incident from the left in region 1 (where U = 0), so we need a left half matrix for region 1 at x = 0. We need a layer matrix (of the E < U variety) for the barrier. Finally, we must have a right-half matrix for region 3 at x = L. Since k1 = k3 = k and region 2 is characterized by α, we can dispense with the subscripts.

Now comes tedious algebra to get to the answer. Note that to the find the transmission probability, we only need M11.

[M ] =

�

�

�1�2

� �1�2k

�

�1�2

� �� 1�

2k

�

�

� ·�

cosh (�L) i� sinh (�L)

�i� sinh (�L) cosh (�L)

�·

�

�

�1�2

� �1�2

�

�� k�

2

� �� k�

2

�

�

�


M11 = cosh (↵L) +i

2

✓k

↵� ↵

k

◆sinh (↵L)

T =k3k1

1

|M11|2=

1

|M11|2

⇡16E (U

o

� E)

U2o

�exp (�2↵L)

As we saw earlier when we first looked at tunneling.

= �+ sinh� (�/) +��

��N�

�� +

��

N��

�sinh� (�/)

= �+��

��N�

��+ �+

��

N��

�sinh� (�/)

|0��|� = cosh� (�/) +��

�N�

� �

N

��sinh� (�/)

= �+��

�N�

+�

N

��sinh� (�/)

= �+

�8�R

�( (8R � ()

��sinh� (�/)

=

��

��+

�8�R

�( (8R � ()

��sinh� (�/)

��

�

��


Bound statesCan the matrix method be used to learn something about bound states?

It requires a slightly different approach, since a bound state does not “propagate” and so we will not calculate transmission or reflection probabilities.

A bound state is characterized by the requirement that ψ( x → ±∞ ) → 0.

This requirement means that, in the “input” and “output” regions, the wave function must be in the form of a decaying exponential.

�0B

�= [M ]

�C0

�

M11 = 0

So the matrix procedure would be to find the total matrix description for the problem, and then finds the roots of the M11 matrix element.


Example - finite height square well (redux)

U = 0

U = Uo

+L/2

0 x

–L/2

U = Uo

12 3

To give a quantitative comparison, use Uo = 1 eV and L = 2 nm.

Using the even / odd approach with solving the transcendental characteristic equation repeatedly give four solutions:

w1 ≈ 1.31 → E1 = 0.066 eV

w3 ≈ 3.86 → E3 = 0.569 eV

w2 ≈ 2.608 → E2 = 0.260 eV

w4 ≈ 4.96 → E3 = 0.939 eV


U = 0

U = Uo

+L/2

0 x

–L/2

U = Uo

12 3

[07] = [/�] [0�] [5�]

=

�

�

�1�2

� �i�2�

�

�1�2

� �� i�

2�

�

�

� ·�

cos (kL) �ik sin (kL)

�ik sin (kL) cos (kL)

�·

�

�

�1�2

� �1�2

�

��i��

2

� �i��2

�

�

�

M11 = cos (kL) +1

2

��

k� k

�

�sin (kL)


To find the bound states, set M11 = 0 and find the roots.

cos (kL) +1

2

��

k� k

�

�sin (kL) = 0

Of course, k and α depend on energy, so we will be finding particular energies for which the above equation goes to 0. An easy way to see what is going on is to make a plot.

-2!

-1!

0!

1!

2!

3!

4!

5!

6!

0! 0.1! 0.2! 0.3! 0.4! 0.5! 0.6! 0.7! 0.8! 0.9! 1!

Energy (eV)!

M11!

Just from the plot, we see that the approximate energies are:

E1 ≈ 0.06 eV

E2 ≈ 0.25 eV

E3 ≈ 0.56 eV

E4 ≈ 0.94 eV

With just a bit of effort, the numbers can be made more precise.

1-d matrix method - gary tuttle's isu web site

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