3.024 Electrical, Optical, and Magnetic Properties of Materials
Notes courtesy of Adam F. Hannon, used by permission.
Spring 2013
Piecewise Function/Continuity Review -Scattering from Step Potential
Piecewise Function/Continuity Review
Continuous piecewise functions are defined as follows:
𝑓! 𝑥 𝑥𝜖 (−∞, 𝑎! ]
𝑓 𝑥 𝑥𝜖 [𝑎! , 𝑎! ]
𝑓 𝑥 = !
⋮
⋮
𝑓! 𝑥 𝑥𝜖[𝑎! , ∞)
Here we note that the following continuity conditions must be in place for these functions
to be piecewise continuous:
𝑓! 𝑎! = 𝑓!!! 𝑎!
𝑑𝑓! 𝑎!
𝑑𝑓!!! 𝑎!
=
𝑑𝑥
𝑑𝑥
Wave functions must obey the same boundary conditions. Note however that the
potential function V(x) does not have to be piecewise continuous, just the wave function.
There are many problems of interest where V(x) is a piecewise function and not
necessarily continuous such as the particle in a 1D box and barrier and potential well
problems.
Example 1: Particle in a box – symmetric about x-axis
Consider the above system with the piecewise potential energy function V(x) for an
electron inside the infinite potential well.
𝑑
𝑑
𝑥𝜖 −∞, − ∪ , ∞
∞
2
2
𝑉 𝑥 =
0
𝑑 𝑑
𝑥𝜖 − ,
2 2
The wave function is 0 outside the middle region since there is 0 probability of finding
the electron in the infinite potential regions.
We write Schrodinger’s equation for the middle region.
𝐻𝜑 = 𝐸𝜑
1
3.024 Electrical, Optical, and Magnetic Properties of Materials
Notes by Adam Hannon
𝜕
−𝑖ℏ 𝜕𝑥
!
𝑝
𝜓 = 𝐸𝜓 →
2𝑚
!
𝜓 = 𝐸𝜓
2𝑚
𝜕!𝜓
𝜕𝑥 ! = 𝐸𝜓
2𝑚
−ℏ!
∴
The general solution of this equation is:
𝜓 𝑥 = 𝐴𝑒 !"# + 𝐵𝑒 !!"#
!!"
with 𝑘 ! = !
ℏ
The boundary conditions (BCs) for this problem are:
𝑑
𝑑
𝜓 −
= 0 and 𝜓
=0
2
2
The first BC gives:
𝐴𝑒
!"#
!
The second BC gives:
𝐴𝑒 !
Adding these two equations:
𝐴 𝑒
!"#
!
+ 𝑒!
+ 𝐵𝑒 !
!"#
!
!"#
!
+ 𝐵𝑒
!"#
!
=0
!"#
!
=0
+𝐵 𝑒
!"#
!
+ 𝑒!
!"#
!
=0
or
𝑘𝑑
𝑘𝑑
+ 2𝐵 cos
=0
2
2
𝑘𝑑
2 𝐴 + 𝐵 cos
=0
2
Subtracting the second equation from the first equation:
2𝐴 cos
𝐴 𝑒
!"#
!
− 𝑒!
!"#
!
or
−𝐵 𝑒
!"#
!
− 𝑒!
!"#
!
=0
𝑘𝑑
𝑘𝑑
− 2𝑖𝐵 sin
=0
2
2
𝑘𝑑
2𝑖 𝐴 − 𝐵 sin
=0
2
Thus these equations must simultaneously be true:
𝑘𝑑
2 𝐴 + 𝐵 cos
=0
2
𝑘𝑑
2 𝐴 − 𝐵 sin
=0
2
If A = B :
𝑘𝑑
cos
= 0 → 𝑘𝑑 = 𝑛𝜋 with 𝑛 odd
2
If A = -B :
𝑘𝑑
sin
= 0 → 𝑘𝑑 = 𝑛𝜋 with 𝑛 even > 0
2
2𝑖𝐴 sin
Spring 2013
3.024 Electrical, Optical, and Magnetic Properties of Materials
Notes by Adam Hannon
Thus the solution to the problem is:
𝜓! 𝑥 =
𝑐!"" 𝑒 !
!"#
!
+ 𝑒 !!
!"#
!
− 𝑒 !!
𝑐!"!# 𝑒 !
!"#
!
!"#
!
Such that:
𝑛𝜋𝑥
𝑑
=
𝑛𝜋𝑥
𝐶!"!# sin
𝑑
𝐶!"" cos
Spring 2013
𝑛 𝑜𝑑𝑑
𝑛 𝑒𝑣𝑒𝑛
𝐶!"" = 2𝑐!"" and 𝐶!"!# = 2𝑖𝑐!"!#
The constants C and D are found by the normalization condition for the total probability
of finding the particle:
!
!
1=
!
!
!
𝜓∗
! !
𝑥 𝜓! 𝑥 𝑑𝑥 =
!
𝜓! 𝑥 =
𝑛𝜋𝑥
𝑑𝑥 → 𝐶!"" =
𝑑
2
𝑑
!
𝐶!"!#
sin!
𝑛𝜋𝑥
𝑑𝑥 → 𝐶!"!# =
𝑑
2
𝑑
!
!
!
!
!
!
!
!
Therefore:
!
!
𝐶!"
" cos
2
𝑛𝜋𝑥
cos
𝑑
𝑑
𝑛 𝑜𝑑𝑑
2
𝑛𝜋𝑥
sin
𝑛 𝑒𝑣𝑒𝑛
𝑑
𝑑
The energies of the system are quantized such that:
!
! 𝑛𝜋
! !
ℏ
ℏ 𝑘
ℏ! 𝑛! 𝜋 !
ℎ! 𝑛!
𝑑
𝐸! =
=
=
=
2𝑚
2𝑚
2𝑚𝑑 !
8𝑚𝑑 !
3
3.024 Electrical, Optical, and Magnetic Properties of Materials
Notes by Adam Hannon
Spring 2013
Example 2: Scattering off a step potential.
Consider the following piecewise potential energy function V(x) for an electron traveling
incident from the left side with total energy E > V0.
𝑉 𝑥 =
0 𝑥𝜖 (−∞, 0)
𝑉!
𝑥𝜖 [0, ∞)
Find the general form of the wave functions for this potential energy and the transmission
and reflections coefficients for the incident electron, R and T.
First we write Schrodinger’s equation in the two regions.
𝐻𝜓 = 𝐸𝜓
𝜕 !
!
−𝑖ℏ
𝑝
𝜕𝑥 + 𝑉 𝜓 = 𝐸𝜓
+ 𝑉 𝜓 = 𝐸𝜓 →
2𝑚
2𝑚
In region I:
V=0
∴
In region II:
V = V0
∴
𝜕 ! 𝜓!
𝜕𝑥 ! = 𝐸𝜓
!
2𝑚
−ℏ!
𝜕 ! 𝜓!!
𝜕𝑥 ! + 𝑉 𝜓 = 𝐸𝜓
! !!
!!
2𝑚
−ℏ!
Since the wave function must be piecewise continuous, we have the following boundary
conditions (BCs).
BC are 𝜓! 0 = 𝜓!! 0 and !"!
!"
0 =
!"!!
!"
0 4
3.024 Electrical, Optical, and Magnetic Properties of Materials
Notes by Adam Hannon
Spring 2013
Now, to solve these we write the general solutions for the wave function in each region
and apply boundary conditions.
!!"
!!(!!! )
!
Let 𝑘 ! ≡ ! and 𝜌! ≡
ℏ
ℏ!
In region I:
𝜕 ! 𝜓!
𝑘 ! 𝜓! +
= 0 which has solutions of the form 𝜓! 𝑥 = 𝐴𝑒 !"# + 𝐵𝑒 !!!"
𝜕𝑥 !
In region II:
𝜕 ! 𝜓!!
!
𝜌 𝜓!! +
= 0 which has solutions of the form 𝜓!! 𝑥 = 𝐶𝑒 !!"! + 𝐷𝑒 !"!
𝜕𝑥 !
Since the electron is incident from the left, there can never be a rightward propagating
wave from the right side.
𝐶 = 0 → 𝜓!! 𝑥 = 𝐷𝑒 !"#
The coefficients R and T are simply related to the coefficients A, B¸and D such that A
corresponds to the incident electron, B the reflected electron, and D any transmission
electron.
The exact correspondence comes from the conservation of the flux of electrons from the
left equaling the flux of the electrons on the right.
The probability current/flux is simply the probability amplitudes times the velocity of the
electron.
𝑝 ℏ𝑘
𝑣= =
𝑚 𝑚
𝐹 = 𝐴! 𝑣
We have 3 probability currents/fluxes, incident, reflected, and transmitted.
𝐼+𝑅 =𝑇
ℏ𝜌
ℏ𝑘
ℏ𝑘
𝐴∗ 𝐴
+ 𝐵∗ 𝐵
= 𝐷∗ 𝐷
𝑚
𝑚
𝑚
Assuming I = 1, we can normalize this current/flux equation by 𝐴∗ 𝐴𝑘 and obtain the
following relations for R and T.
!∗!
!∗ ! !
We can thus write 𝑅 = ∗ ! and 𝑇 = ∗ ! ! .
!
!
Now, using the boundary conditions:
𝜓! 0 = 𝜓!! 0 → 𝐴𝑒 !"! + 𝐵𝑒 !!"! = 𝐷𝑒 !"! → 𝐴 + 𝐵 = 𝐷
𝜕𝜓!
𝜕𝜓!!
𝑖𝜌 0 =
0 → 𝑖𝑘(𝐴𝑒 !"! − 𝐵𝑒 !!"! ) = 𝑖𝜌𝐷𝑒 !"! → 𝐴 − 𝐵 =
𝐷
𝜕𝑥
𝜕𝑥
𝑖𝑘
!"
Subtracting the second equation from the 1st times we can find R:
!"
𝑖𝑘
𝑖𝑘
1−
𝐴+ 1+
𝐵=0
𝑖𝜌
𝑖𝜌
𝑖𝑘
1 − 𝑖𝜌
𝐵
𝑖𝜌 − 𝑖𝑘 𝜌 − 𝑘
=−
=
=
𝑖𝑘
𝐴
𝑖𝜌 + 𝑖𝑘 𝜌 + 𝑘
1 + 𝑖𝜌
5
3.024 Electrical, Optical, and Magnetic Properties of Materials
Notes by Adam Hannon
Spring 2013
𝜌−𝑘 !
𝐵∗ 𝐵
=
𝐴∗ 𝐴
𝜌+𝑘
𝜌−𝑘 !
𝑅=
𝜌+𝑘
2𝑚 𝐸 𝐸 − 𝑉!
2𝑚 𝐸 − 𝑉!
2𝑚𝐸
−2
+ !
!
!
ℏ
ℏ
ℏ
𝑅=
2𝑚 𝐸 𝐸 − 𝑉!
2𝑚 𝐸 − 𝑉!
2𝑚𝐸
+2
+ !
!
ℏ
ℏ!
ℏ
𝐸 − 𝑉! − 𝐸 𝐸 − 𝑉! + 𝐸
∴𝑅=
𝐸 − 𝑉! + 𝐸 𝐸 − 𝑉! + 𝐸
𝑅=
Adding the two equations we can find T:
𝑖𝜌
𝐷
𝑖𝑘
𝐷
2
2𝑖𝑘
2𝑘
=
=
=
𝑖𝜌
𝐴
𝑖𝑘 + 𝑖𝜌
𝑘+𝜌
1+
𝑖𝑘
2𝐴 = 1 +
2𝑘 ! 𝜌
4𝜌𝑘
𝐷∗ 𝐷 𝜌
=
=
𝑇= ∗
𝐴 𝐴𝑘
𝑘+𝜌 𝑘
𝑘+𝜌 !
2𝑚 𝐸 𝐸 − 𝑉!
4
ℏ!
𝑇=
2𝑚 𝐸 𝐸 − 𝑉!
2𝑚 𝐸 − 𝑉!
2𝑚𝐸
+2
+ !
ℏ!
ℏ!
ℏ
4 𝐸 𝐸 − 𝑉!
∴𝑇=
𝐸 − 𝑉! + 𝐸 𝐸 − 𝑉! + 𝐸
Note that classically a particle would always reflect, but here there is a finite probability
of transmission.
Plotting R (red) and T (blue) versus E.
6
3.024 Electrical, Optical, and Magnetic Properties of Materials
Notes by Adam Hannon
Spring 2013
For the case E < V0, iρ becomes real, so let iρ = α.
𝐵 𝛼 − 𝑖𝑘
=
𝐴 𝛼 + 𝑖𝑘
∗
𝐵 𝐵
𝛼 + 𝑖𝑘 𝛼 − 𝑖𝑘
𝛼! + 𝑘!
𝑅= ∗ =
= !
=1
𝐴𝐴
𝛼 − 𝑖𝑘 𝛼 + 𝑖𝑘
𝛼 + 𝑘!
For this case, the entire wave is reflected, analogous to the classical case. The wave
function in region II is a decaying exponential, which is not classical. This implies even
though electrons are reflected if their energy is lower than the barrier potential, they have
a finite probability of penetrating the step barrier before being reflected.
7
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3.024 Electronic, Optical and Magnetic Properties of Materials
Spring 2013
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