1. Elements of Quantum Mechanics
Chapter 2
Elements of Quantum Mechanics
Classical (Newtonian) mechanics is inaccurate when applied to electrons in crystals or any systems with atomic dimensions.
Quantum mechanics is necessary background knowledge for understanding electrons in crystals.

2. THE QUANTUM CONCEPT Blackbody Radiation 𝐵 λ = 2𝑐𝑘𝑇 λ4 𝐸𝑛=𝐧ℎ𝜈=𝑛ℏ𝜔
An opaque non-reflective ideal body
classical model, assuming continuum of allowed energy
(valid for long wave length)
𝐵 λ = 2𝑐𝑘𝑇 λ4
real observation
explained by Max Planck, assuming discrete energy spectrum
𝐸𝑛=𝐧ℎ𝜈=𝑛ℏ𝜔
𝐧=0, 1, 2, 3,………
𝐵 λ = 2ℎ𝑐2 λ5 1 𝑒 ℎ𝑐/λ𝑘𝑇 −1
Vibrating atoms in a material can only radiate or
absorb energy in discrete packets (energy quantization)
λ (𝜇m)

3. The Bohr Atom 𝐿𝑛=𝑚0ʋ 𝑟 𝑛 =𝒏ℏ For the simple hydrogen atom with Z = 1,
Postulations:
Electrons exist in certain stable circular orbits.
Electrons can shift between orbitals by gaining or losing energy.
Angular momentum is quantized.
𝐿𝑛=𝑚0ʋ 𝑟 𝑛 =𝒏ℏ
𝐧=1, 2, 3,………
centripetal force −𝑞 +𝑞 𝑟𝑛
𝐧=1
𝐧=2
𝐧=3
= 𝑚0ʋ2 𝑟𝑛
Continuous energy radiation
electromagnetic radiation
Coulombl force
= 𝑞2 4𝜋𝜖0 𝑟 𝑛 2
Classically, accelerating charge radiates electromagnetic wave.

4. 𝑟𝑛= 4𝜋𝜖0 𝒏ℏ 2 𝑚0𝑞2 𝑚0ʋ2 𝑟𝑛 = 𝑞2 4𝜋𝜖0 𝑟 𝑛 2 P.E = - 𝑞2 4𝜋𝜖0𝑟𝑛
𝑚0ʋ2 𝑟𝑛 = 𝑞2 4𝜋𝜖0 𝑟 𝑛 2
K.E = 1 2 𝑚0ʋ2= 1 2 ( 𝑞2 4𝜋𝜖0𝑟𝑛 )
P.E = - 𝑞2 4𝜋𝜖0𝑟𝑛
E = K.E + P.E
𝐸𝑛=− 𝑚0𝑞4 2 4𝜋𝜖0𝒏ℏ 2 =− 𝒏2 𝑒𝑉

5. Wave-Particle Duality
Light: wave nature (diffraction, refraction, interference….)
particle nature (photoelectric effect, Compton effect)
called “photon”
Electron: particle nature (m0, q ….)
wave nature (SEM, TEM, diffraction, refraction, interference…)
De Broglie’s matter wave: All particles have the properties of wave.
λ= ℎ 𝑚ʋ
p= ℎ λ
de Broglie’s hypothesis
Low dimensional materials (2D, 1D) in terms of electrical properties: size of reduced dimensions approaches λ of electron

6. One-electron Schrödinger equation
General Formulation
There exist a wavefunction, Ψ(x, y, z, t), describing
the dynamic behavior of the system.
mathematically complex quantity
(2) Ψ and 𝛻Ψ must be continuous, finite, and single-valued
for all values of x, y, z, and t.
not single-valued
not continuous
not finite
(3) Ψ∗ΨdV = Ψ 2 dV : the probability that the particle will be
found in the spatial volume element dV.
Ψ∗ΨdV = 1:
normalization

7. (4) There is a unique mathematical operator with each dynamic variable
such as position or momentum. The expectation value can be obtained
by operating on the wavefunction.
<𝛼> = 𝑉 Ψ∗𝛼𝑜𝑝ΨdV
< 𝑝 𝑥 > = 𝑉 Ψ∗( ℏ 𝑖 𝜕 𝜕𝑥 Ψ)dV =ℏ𝑘 𝑉 Ψ∗ΨdV =ℏ𝑘
<𝑥> = 𝑉 Ψ∗𝑥 ΨdV
If Ψ = ej(kx - 𝜔𝑡)
<𝐸> = 𝑉 Ψ∗(− ℏ 𝑖 𝜕 𝜕𝑡 Ψ)dV =ℏ𝜔 𝑉 Ψ∗ΨdV =ℏ𝜔

8. Where does operator come from?
For plane wave solution( or harmonic solutions of the wave):
Ψ = Cei(kx – 𝜔t)
operator
𝜕Ψ 𝜕𝑥 =𝑖𝑘Ψ
ℏ 𝑖 𝜕Ψ 𝜕𝑥 =ℏ𝑘Ψ
expectation value
𝜕2Ψ 𝜕𝑥2 =− 𝑘2Ψ
− ℏ2 2𝑚 𝜕2Ψ 𝜕𝑥2 = ℏ2𝑘2 2𝑚 Ψ
𝜕Ψ 𝜕𝑡 =− 𝑖𝜔Ψ
− ℏ 𝑖 𝜕Ψ 𝜕𝑡 =ℏ𝜔Ψ
Time-independent Formulation
E = K.E + P.E
− 𝑃2 2 𝑚 0 +𝑈 𝑥, 𝑦, 𝑧 =𝐸:𝑐𝑙𝑎𝑠𝑠𝑖𝑐𝑎𝑙 𝑒𝑥𝑝𝑟𝑒𝑠𝑠𝑖𝑜𝑛
In operator form with Ψ
− ℏ2 2 𝑚 0 𝛻2+𝑈(𝑥, 𝑦, 𝑧, 𝑡)) Ψ=− ℏ 𝑖 𝜕Ψ 𝜕𝑡
time-dependent Schrödinger equation
𝐻 Ψ=− ℏ 𝑖 𝜕Ψ 𝜕𝑡 =𝐸Ψ
Hamiltonian operator

9. (3’) ψand 𝛻ψmust be continuous, finite, and single-valued
− ℏ 𝑖 𝜕Ψ 𝜕𝑡 =𝐸Ψ
𝑉 Ψ∗(− ℏ 𝑖 𝜕 𝜕𝑡 Ψ)dV =𝐸 𝑉 Ψ∗ΨdV =𝐸:𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
The equation has the solution of the form
Ψ 𝑥, 𝑦, 𝑧, 𝑡 = ψ 𝑥, 𝑦, 𝑧 𝑒 −𝑖𝐸𝑡/ℏ
Substituting this into time dependent equation and by using separation of variables.
𝛻2ψ+− 2 𝑚 0 ℏ2 [ 𝐸 −𝑈 𝑥, 𝑦, 𝑧 ]ψ=0
time-independent Schrödinger equation
(3’) ψand 𝛻ψmust be continuous, finite, and single-valued
for all values of x, y, and z.
(4’) Ψ∗Ψ=ψ∗ψdV = ψ 2 dV : the probability that the particle will be
found in the spatial volume element dV. Likewise
𝑉 ψ∗ψdV = 1
(5’) The expectation value of the system variable 𝛼 is given by
<𝛼> = 𝑉 ψ∗𝛼𝑜𝑝ψdV

10. q ℰ ( r , t) = 𝛻(-qV) = 𝛻U( r , t)
The Ψ for a given system and specified system constraint
is determined by solving the time dependent Schrödinger wave equation,
Classically, the position ( 𝑟 )=(x, y, z) of an electron in an electric field ℰ ( 𝑟 , t) = ℰ 𝑥, 𝑦, 𝑧, 𝑡 evolves with time (t) according to Newton’s second law.
𝑚 0 𝑑 2 𝑟 𝑑𝑡 = - q ℰ ( r , t)
Quantum mechanical descriptions are always in terms of potentials rather than electric field.
q ℰ ( r , t) = 𝛻(-qV) = 𝛻U( r , t)
Quantum mechanically, the dynamics of an individual electron is described by the one-electron Schrödinger equation.
− ℏ 2 2 𝑚 0 𝛻2 Ψ 0 (x, y, z, t) + U(x, y, z, t) Ψ 0 (x, y, z, t) = iℏ 𝜕 𝜕𝑡 Ψ 0 (x, y, z, t) or
− ℏ 2 2 𝑚 0 𝛻2 Ψ 0 ( r , t) + U( r , t) Ψ 0 ( r , t) = iℏ 𝜕 𝜕𝑡 Ψ 0 ( r , t)

11. U( r , t) = U 𝐿 ( r ) + U 𝑆 ( r , t) + U 𝐸 ( r , t)
The potential energy can be separated into a macroscopic part and a microscopic part.
U( r , t) = U 𝐿 ( r ) + U 𝑆 ( r , t) + U 𝐸 ( r , t)
U 𝐸 ( r , t)
: potential from macroscopic electric fields due to any externally
applied voltage and/or macroscopic space charge
U 𝐿 ( r , t)
: potential from microscopic electric fields due to a perfect static lattice
U 𝑆 ( r , t)
: called scattering potential
random fluctuating potential from microscopic electric fields from
defects and impurities (impurity scattering), or phonons (phonon
scattering), electron-electron scattering(electron-electron scattering)

12. Effective mass equation
If we consider only the periodic potential (neglecting US and UE), the solutions to the Schrödinger equation can be written in the form of Bloch waves.
U 𝐿 ( r )
The wavefunction of an electron in a band ν with wave vector 𝑘 is written as
Ψ 0 ( r ,𝑡) = 𝑢 ν,𝑘 r exp(𝑖 k ∙ r )exp(−𝑖 𝐸 ν ( k )𝑡/ℏ)
Ψ 0 ( r ) = 𝑢 ν,𝑘 r exp(𝑖 k ∙ r )
the form of (plane wave) ×
(function with the periodicity of the Bravais lattice)
where 𝑢 ν,𝑘 𝑟 is a periodic function with the same period as the lattice potential U 𝐿 ( r ) that is different for each ν and each 𝑘 .
Use the plane waves to expand the envelope function
Ψ( r ,𝑡).
Ψ r ,𝑡 = 𝑘 ψ 𝑘(t) exp(𝑖 k ∙ r )
Ψ 0 r ,𝑡 ≈ 𝑢 ν,𝑘 r Ψ
r ,𝑡
The true wavefunction is the approximately equal to the product of the periodic part of the Bloch wavefunction, u r and the envelope Ψ r ,𝑡 .

13. iℏ 𝜕 𝜕𝑡 Ψ r , 𝑡 = 𝐸 ν k Ψ r , 𝑡 + (U 𝑆 ( r , t) + U 𝐸 ( r , t))Ψ r , 𝑡
iℏ 𝜕 𝜕𝑡 Ψ 0 ( r , t) = iℏ 𝑢 ν,𝑘 r 𝜕 𝜕𝑡 Ψ r , 𝑡 = iℏ 𝑢 ν,𝑘 r (− i ℏ )Ψ r , 𝑡 = 𝑢 ν,𝑘 r 𝐸 ν k Ψ r , 𝑡
Including US and UE,
(U 𝑆 ( r , t) + U 𝐸 ( r , t))Ψ 0 ( r , t)= 𝑢 ν,𝑘 r (U 𝑆 ( r , t) + U 𝐸 ( r , t))Ψ r , 𝑡
Then, the Schrödinger equation is reduced to the equation called
effective mass equation.
iℏ 𝜕 𝜕𝑡 Ψ r , 𝑡 = 𝐸 ν k Ψ r , 𝑡 + (U 𝑆 ( r , t) + U 𝐸 ( r , t))Ψ r , 𝑡 or
iℏ 𝜕 𝜕𝑡 Ψ r , 𝑡 = 𝐸 ν − 𝑖𝛻 Ψ r , 𝑡 + (U 𝑆 ( r , t) + U 𝐸 ( r , t))Ψ r , 𝑡
For a conduction band with a parabolic dispersion law,
𝐸 𝑐 k = 𝐸 𝑐𝑜 + ℏ 2 𝑘 2 2 𝑚 ∗
iℏ 𝜕 𝜕𝑡 Ψ r , 𝑡 = - ℏ 2 2 𝑚 ∗ 𝛻 2 Ψ r , 𝑡 + ( 𝐸 𝑐𝑜 + U 𝑆 ( r , t) + U 𝐸 ( r , t))Ψ r , 𝑡
iℏ 𝜕 𝜕𝑡 Ψ r , 𝑡 = - ℏ 2 2 𝑚 ∗ 𝛻 2 Ψ r ,𝑡 + 𝐸 𝑐 ( r , t)Ψ r , 𝑡 + U 𝑆 ( r , t)Ψ r , 𝑡
Where Ec = Ec0 + UE is the band edge energy we usually draw when draw band diagrams for devices.

14. n( r , t) = <Ψ∗ r ,𝑡 Ψ r , 𝑡 >
The electron density n that is obtained by summing the probability densities Ψ∗Ψ, of all the electrons
n( r , t) = <Ψ∗ r ,𝑡 Ψ r , 𝑡 >
does not reflect the periodic variations at the atomic level (we have used < > to denote summation over all the electrons.
This is expected since the periodic lattice potential does nor appear in the effective mass equation that is used to calculate Ψ.
In contrast, the electron density n0 obtained from the true wavefunction Ψ0 will be peaked near the ion cores; n( r , t) is a smoothed out version of n0( r , t) in which the rapid variations over a unit cell are averaged out.
The current density,
𝐽 ( r , t) = −𝑖𝑞ℏ 2 𝑚 ∗ < (𝛻Ψ) ∗ Ψ - Ψ∗(𝛻Ψ)Ψ >
The current continuity equation,
𝛻∙ 𝐽 ( r , t) - q 𝜕𝑛 𝜕𝑡 = 0
Prove these three equations as homework.

15. In the effective mass equation,
iℏ 𝜕 𝜕𝑡 Ψ r , 𝑡 = - ℏ 2 2 𝑚 ∗ 𝛻 2 Ψ r , 𝑡 r ,𝑡 + 𝐸 𝑐 ( r , t)Ψ r , 𝑡 + U 𝑆 ( r , t)Ψ r , 𝑡
US : random microscopic potential
Ec : potential due to externally applied voltage and/or
macroscopic space charges
The macroscopic potential Ec has to be determined from the Poisson equation.
For electrons in the conduction band,
𝛻 2 𝑉 𝐸 = 𝑞 𝜖 (𝑁 𝐷 + - n)
𝑉 𝐸 is the electrostatic potential (= - UE/q)
𝑁 𝐷 + is the density of ionized donors
In quantum device analysis, instead of calculating the electron density and the current density from
the “drift-diffusion” equation, we have to calculate it from the equations n( r , t) and 𝐽 ( r , t) after solving
for the wavefunction Ψ r , 𝑡 from the effective mass equation.

16. iℏ 𝜕 𝜕𝑡 Ψ r , 𝑡 = - ℏ 2 2 𝑚 ∗ 𝛻 2 Ψ r , 𝑡 r ,𝑡 + 𝐸 𝑐 ( r , t)Ψ r , 𝑡
Assuming a perfect crystal with no scattering due to impurities and phonons (US = 0), the effective mass equation becomes
iℏ 𝜕 𝜕𝑡 Ψ r , 𝑡 = - ℏ 2 2 𝑚 ∗ 𝛻 2 Ψ r , 𝑡 r ,𝑡 + 𝐸 𝑐 ( r , t)Ψ r , 𝑡
The simplest case corresponds to a constant EC equal to EC0.
The solutions can be written as,
Ψ r , 𝑡 = Cexp(𝑖 k ∙ r )exp(−𝑖𝐸( k )𝑡/ℏ)
= Cexp(𝑖 k x x)exp(𝑖 k y y)exp(𝑖 k z z)exp(−𝑖𝐸( k )𝑡/ℏ)
Normalization constant C is a constant with the dimension of (volume)-1/2.
This is necessary in order that Ψ∗Ψ have the same units as the electron density.
The wavefunction satisfies the effective equation for any k , provided E is given by
E = E 𝑐𝑜 + ℏ 2 2 𝑚 ∗ ( k x k y k z 2 )
called the dispersion relation.
n( r , t) = <Ψ∗ r ,𝑡 Ψ r , 𝑡 > = < 𝐶 2 >
𝐽 ( r , t) = −𝑖𝑞ℏ 2 𝑚 ∗ < (𝛻Ψ) ∗ Ψ - Ψ∗(𝛻Ψ)Ψ > = 𝑞ℏ k 𝑚 ∗ < 𝐶 2 >
k = 𝑥 k x + 𝑦 k y + 𝑧 k z

17. E = E 𝑐𝑜 + ℏ 2 2 𝑚 ∗ ( k x 2 + k y 2 + k z 2 ) = p 2 2 𝑚 ∗
In classical relation,
𝐽 = - nq 𝑣
If we interpret ℏ k as the momentum p
p = m ∗ 𝑣 = ℏ k
Momentum is a “particle” concept while the wavevector k is a “wave” concept.
The dispersion relation, which is wave concept, can also be interpreted in particle-like terms:
E = E 𝑐𝑜 + ℏ 2 2 𝑚 ∗ ( k x k y k z 2 ) = p 2 2 𝑚 ∗
total energy
potential energy
kinetic energy
If EC is varying along the x-direction, but is constant in the other two dimensions y and z and in time t, then we can write.
Ψ r , 𝑡 = Cψ(x)(𝑖 k y y)exp(𝑖 k z z)exp(−𝑖𝐸𝑡/ℏ)
d 2 ψ(x) dx 𝑚 ∗ ℏ 2 E(x) ψ(x) = 0
E (x) =E − E 𝐶 (x) - ℏ 2 2 𝑚 ∗ ( k y k z 2 )
where

18. SIMPLE PROBLEM SOLUTIONS
Free Particle (Electron in Free Space)
U(x, y, z) = 0, more generally U(x, y, z) = constant
with total energy E and mass m
no force acting on the particle
In 1-D, time-independent Schrödinger equation
𝑑 2 ψ 𝑑 𝑥 𝑚𝐸 ℏ2 ψ=0
By introducing the constant, k ≡ 2 𝑚 0 E/ ℏ 2 or equivalently
E= ℏ 2 k 2 2 𝑚 0 = < p 𝑥 >2 2 𝑚 0
Solution becomes
ψ 𝑥 = 𝐴 + 𝑒 𝑖 𝑘 𝑥 𝑥 + 𝐴 − 𝑒 −𝑖 𝑘 𝑥 𝑥
Recalling the relationship,
Ψ 𝑥, 𝑦, 𝑧, 𝑡 = ψ 𝑥, 𝑦, 𝑧 𝑒 −𝑖𝐸𝑡/ℏ
Ψ 𝑥, 𝑡 = 𝐴 + 𝑒 𝑖( 𝑘 𝑥 𝑥− 𝐸𝑡 ℏ ) + 𝐴 − 𝑒 −𝑖( 𝑘 𝑥 𝑥+ 𝐸𝑡 ℏ )
Traveling wave moving (-) x-direction
Traveling wave moving (+) x-direction

19. Traveling wave (plane wave)
Electron in free space
Traveling wave (plane wave)
particle λ
𝑘 x = 2𝜋 λ :
Expectation value of momentum for free particle
wave number
The phase of the wave, 𝑘 𝑥 𝑥 –Et/ℏ = constant
< 𝑝 𝑥 > = −∞ ∞ ψ∗ ℏ 𝑖 𝑑 𝑑𝑥 ψ dx =ℏ𝑘 −∞ ∞ ψ∗ψdx = ℏ 𝑘 𝑥 = ℎ λ
𝑑 𝑑𝑡 𝑘 𝑥 𝑥 − 𝐸𝑡 ℏ =0
de Broglie’s hypothesis
Expectation value of the momentum for a free particle is exact , because ∆x = ∞ and hence ∆ p = 0.
𝑘 𝑥 𝑑𝑥 𝑑𝑡 = 𝑘 𝑥 ʋ= 2𝜋 λ 𝜈λ=2𝜋𝜈=𝜔= 𝐸 ℏ
From classical mechanics
From wave mechanics
E= 𝑝 2 2 𝑚 0 = 𝑚 0 𝑣2 2
E= ℏ 2 𝑘 2 2 𝑚 0 = < 𝑝 𝑥 >2 2 𝑚 0
continuous energy spectrum
Identical because ∆ p = 0 for free particle.

20. Potential Step
What is the fraction R of the incident current that is reflected and
what is the fraction T that is transmitted?
Neglecting band bending due to space-charge effects,
∆ E 𝑔 =∆ E 𝐶 + ∆ E 𝑉
J 𝑟 =R J 𝑖
J 𝑡 =T J 𝑖
A beam of light with frequency 𝜔 carrying Pi watts/m2 is incident from medium 1 on medium 2.
P 𝑟 =R P 𝑖
P 𝑡 =T P 𝑖
R +T = 1

21. T is function of function of electron energy.
Classical device analysis from Newton’s law, electrons are viewed as particles.
if E > ∆EC, R = 0 and T = 1.
The electrons are slowed down as they cross the interface since the up-step in potential energy ∆ EC is equivalent to a retarding field. But as long as E > ∆ EC the electrons are completely transmitted with no reflection.
if E < ∆EC, R = 1 and T = 0.
The electrons are completely turned around.

22. d 2 ψ(x) dx 2 + 2 𝑚 ∗ ℏ 2 E(x) ψ(x) = 0
Quantum mechanically, we have to view the electron as a wave obeying the effective mass equation. We have to solve the one-dimensional effective mass equation.
d 2 ψ(x) dx 𝑚 ∗ ℏ 2 E(x) ψ(x) = 0
E (x) =E − E 𝐶 (x) - ℏ 2 2 𝑚 ∗ ( k y k z 2 )
with E 𝐶 (x) = E 𝐶1 , x<0
= E 𝐶2 , x>0
E(x) = E 1 , x<0
where
E1=E − E 𝐶 ℏ 2 2 𝑚 ∗ ( k y k z 2 )
= E 2 , x>0
E2=E − E 𝐶 ℏ 2 2 𝑚 ∗ ( k y k z 2 )

23. ψ(x) = exp(i k 1 x) + rexp(−i k 1 x), x < 0 k 1 = 2 𝑚 ∗ E 1 /ℏ
On the left, we have an incident and a reflected wave
ψ(x) = exp(i k 1 x) + rexp(−i k 1 x), x < 0
where
k 1 = 2 𝑚 ∗ E 1 /ℏ
On the right, we have only a transmitted wave,
ψ(x) = texp(i k 2 x), x > 0
where
k 2 = 2 𝑚 ∗ E 2 /ℏ
Boundary conditions at x = 0,
1 + r = t
ψ(x) and ψ(x)/dx must be continuous.
k 1 (1 – r) = t k 2
Solving for r and t,
r = k 1 − k 2 k 1 + k 2
t = 2k 1 k 1 + k 2
From the current density equation,
𝐽 ( r , t) = −𝑖𝑞ℏ 2 𝑚 ∗ < (𝛻Ψ) ∗ Ψ - Ψ∗(𝛻Ψ)Ψ >
J(x < 0)= - 𝐶 2 𝑞ℏ 𝑘 1 𝑚 ∗ 1 − 𝑟 2 = J 𝑖 − J 𝑟
J(x > 0)= - 𝐶 2 𝑞ℏ 𝑘 2 𝑚 ∗ 𝑡 2 = J 𝑡
R≡ J r / J 𝑖 = r 2
J 𝑟 =R J 𝑖
T≡ J 𝑡 /J 𝑖 = t 2 𝑘 2 𝑘 1
J 𝑡 =T J 𝑖
T is function of function of electron energy.
R +T = 1

24. r’ = k 2 − k 1 k 1 + k 2 t’ = 2k 2 k 1 + k 2 R′ = r′ 2 = R R +T = 1
For the reverse problem, an incident wave from the right and a transmitted to the left
The coefficients,
r’ = k 2 − k 1 k 1 + k 2
t’ = 2k 2 k 1 + k 2
𝑘 1 and 𝑘 2 interchanged
The corresponding reflection and transmission coefficients,
R′ = r′ 2 = R
R +T = 1
T′ = t′ 2 𝑘 1 𝑘 2 = T
The results derived above are often written compactly in the form of a scattering matrix.
For a fixed energy E, we have a forward wave ψ 𝑓 (x) and a reverse wave ψ 𝑟 (x) at any point.
ψ(x) = ψ 𝑓 (x) + ψ 𝑟 (x)
If a wave is incident only from the left,

25. k = 𝑛𝜋 𝑎 , Particle in a 1-D Box ∴ ψ𝑛 𝑥 = 𝐴 𝑛 𝑠𝑖𝑛 𝑛𝜋𝑥 𝑎
i) Outside the box, U = ∞,
ψ 𝑥 =0
ii) Inside the box, U = 0.
𝑑 2 ψ 𝑑 𝑥 2 +− 2𝑚 ℏ2 (𝐸−𝑈)ψ=0
𝑑 2 ψ 𝑑 𝑥 2 + 𝑘 2 ψ=0
0 < x < a
where k ≡ 2𝑚𝐸/ ℏ 2 or
E= ℏ 2 𝑘 2 2𝑚
Solution becomes
ψ 0 =0 or
ψ 𝑥 =𝐴 𝑒 𝑖𝑘𝑥 + 𝐵 𝑒 −𝑖𝑘𝑥
Boundary conditions
ψ 𝑥 =𝐴𝑠𝑖𝑛𝑘𝑥 + 𝐵𝑐𝑜𝑠𝑘𝑥
ψ 𝑎 =0
k = 𝑛𝜋 𝑎 ,
ψ 0 =𝐵=0,
ψ 𝑎 =𝐴𝑠𝑖𝑛𝑘𝑎=0 ∴
n = ±1,±2, ±3,…….
Two waves with +, - direction form standing wave.
ψ𝑛 𝑥 = 𝐴 𝑛 𝑠𝑖𝑛 𝑛𝜋𝑥 𝑎
Particle was restricted to a finite range of coordinate value.

26. E= ℏ 2 𝑘 2 2𝑚 En= 𝑛 2 ℏ 2 𝜋 2 2𝑚 𝑎 2 discrete energy spectrum
standing wave
Particle’s momentum is zero for all energy states, since the particle periodically changes direction.
< 𝑝 𝑥 > = −∞ ∞ ψ∗ ℏ 𝑖 𝑑 𝑑𝑥 ψ dx =0

27. ψ∗ ψ= ψ 2
is called spatial density (probability density when it is normalized).
ψ 2
ψ∗ ψ𝑑𝑥= ψ 2 𝑑𝑥:
probability of finding the particle between x and x + dx x0 x
−𝑞ψ∗ ψ=−𝑞 ψ 2 :
spatial distribution of charge corresponding to a single electron
The electron is no longer considered to be identifiable as a point with particular position, the whole density distribution is the “particle”.
From normalization,
−∞ ∞ ψ∗ψdx = 0 𝑎 ψ∗ψdx = 0 𝑎 𝐴 𝑛 2 𝑠𝑖𝑛2 𝑛𝜋𝑥 𝑎 dx = 1 ∴
𝐴 𝑛 = ( 2 𝑎 )1/2
−∞ ∞ −𝑞ψ∗ψdx =−𝑞 0 𝑎 ψ∗ψdx =−𝑞

28. Finite Potential Well U(x) U0 U(x) = 0, 0 <x < a
𝑑 2 ψ 𝑑 𝑥 2 +− 2𝑚 ℏ2 (𝐸−𝑈)ψ=0,
U(x) = U0, otherwise x a
For 0 < x < a, U = 0,
𝑑 2 ψ0 𝑑 𝑥 2 + 𝑘 2 ψ0=0
where k ≡ 2𝑚𝐸/ ℏ 2
𝑑 2 ψ ± 𝑑 𝑥 2 + 𝛼 2 ψ ± =0
For x > a, x < 0, U = U0,
where α≡ 2𝑚(𝑈0 −𝐸)/ ℏ (0 < E < U0)
The general solutions,
ψ− −∞ =0
𝐵-, A+ = 0
ψ+ ∞ =0
ψ− 𝑥 =𝐴− 𝑒 𝛼𝑥 + 𝐵 −𝑒 −𝛼𝑥
…x < 0
ψ− 0 =ψ0 0
ψ0 𝑥 =𝐴0𝑠𝑖𝑛𝑘𝑥 + 𝐵0𝑐𝑜𝑠𝑘𝑥
…0 <x < a
continuity of ψ
B.C’s
ψ0 𝑎 =ψ+ 𝑎
ψ+ 𝑥 =𝐴+ 𝑒 𝛼𝑥 + 𝐵 +𝑒 −𝛼𝑥
…x > a
𝑑ψ−(0) 𝑑𝑥 = 𝑑ψ0(0) 𝑑𝑥
continuity of 𝑑ψ 𝑑𝑥
𝑑ψ0(𝑎) 𝑑𝑥 = 𝑑ψ+(𝑎) 𝑑𝑥
Four equations four unknowns

29. 𝐴0𝑠𝑖𝑛𝑘𝑎 + 𝐵0𝑐𝑜𝑠𝑘𝑎=𝐵 +𝑒 −𝛼𝑎 𝐴0 𝑘 2 − 𝛼 2 𝑠𝑖𝑛𝑘𝑎 −2𝛼𝑘𝑐𝑜𝑠𝑘𝑎 =0 𝛼𝐴−= 𝑘𝐴0
𝐴− = 𝐵0
𝐴0𝑠𝑖𝑛𝑘𝑎 + 𝐵0𝑐𝑜𝑠𝑘𝑎=𝐵 +𝑒 −𝛼𝑎
𝐴0 𝑘 2 − 𝛼 2 𝑠𝑖𝑛𝑘𝑎 −2𝛼𝑘𝑐𝑜𝑠𝑘𝑎 =0
𝛼𝐴−= 𝑘𝐴0
𝐴0 = 0; trivial solution
𝑘𝐴0𝑐𝑜𝑠𝑘𝑎 − 𝑘𝐵0𝑠𝑖𝑛𝑘𝑎=− 𝛼𝐵 +𝑒 −𝛼𝑎
Non-trivial solution,
𝑘 2 − 𝛼 2 𝑠𝑖𝑛𝑘𝑎 −2𝛼𝑘𝑐𝑜𝑠𝑘𝑎=0 or
𝑡𝑎𝑛𝑘𝑎 = 2𝛼𝑘 𝑘 2 − 𝛼 2
Introducing,
α0≡ 2𝑚𝑈0 /ℏ2 (α0 =constant) and ξ≡ 𝐸/𝑈 ( 0 <ξ < 1)
Then,
α=α0 1 −ξ and 𝑘=α0 ξ
and therefore
𝑡𝑎𝑛⁡(α0𝑎 ξ )= 2 ξ(1 −ξ) 2ξ −1 = 𝑓(ξ)
α0𝑎= 𝜋 4
assuming
ξ = 0.87
𝐸=0.87𝑈0

30. there is one and only one allowed energy level.
α0𝑎< 𝜋 𝑜𝑟 𝑈0< ℏ 2 𝜋 2 2𝑚 𝑎 2 ,
For very shallow wells,
there is one and only one allowed energy level.
When
𝜋< α0𝑎<2𝜋,
there is two allowed energy levels.
When
2𝜋< α0𝑎<3𝜋,
there is three allowed energy levels.
Visualization of quantum mechanical reflection
finite well
e-1 a
infinite well
penetration depth
Visualization of tunneling through a thin barrier