chapter 6 300 fall 19... · 2019-10-21 · fall 2018 prof. sergio b. mendes 9 𝑖ℏ , =− ℏ2 2...

Quantum Mechanics II

Fall 2018 Prof. Sergio B. Mendes 1

CHAPTER 6

Topics


6.1 The Schrödinger Wave Equation

6.2 Expectation Values

6.4 Finite Square-Well Potential

6.3 Infinite Square-Well Potential

6.5 Three-Dimensional Infinite-Potential Well

6.6 Simple Harmonic Oscillator

6.7 Barriers and Tunneling

6.1 Schrödinger Equation


Erwin Schrödinger (1887–1961) was an Austrian who worked at several European universities before fleeing Nazism in 1938 and accepting a position at the University of Dublin, where he remained until his retirement in 1956.

His primary work on the wave equation was performed during the period he was in Zurich from 1920 to 1927.

Schrödinger worked in many fields including philosophy, biology, history, literature, and language.

Free Particle: 𝑉 = 0


𝐸 = 𝐾 + 𝑉

𝜓 𝑥, 𝑡 = 𝐴 𝑒𝑖 𝑘 𝑥 − 𝜔 𝑡 + 𝜀 𝜓 𝑥, 𝑡 2 = 𝐴 2

𝐸 = ℎ 𝑓 = ℏ 𝜔 𝑝 =ℎ

𝜆= ℏ 𝑘

=𝑚 𝑣2

2+ 0 =

𝑝2

2 𝑚

The Total Energy:


𝜕𝜓 𝑥, 𝑡

𝜕𝑡

𝜓 𝑥, 𝑡 = 𝐴 𝑒𝑖 𝑘 𝑥 − 𝜔 𝑡 + 𝜀

= − 𝑖 𝜔 𝜓 𝑥, 𝑡= − 𝑖 𝜔 𝐴 𝑒𝑖 𝑘 𝑥 − 𝜔 𝑡 + 𝜀

𝐸 = ℏ 𝜔 =ℏ

− 𝑖

1

𝜓 𝑥, 𝑡

𝜕𝜓 𝑥, 𝑡

𝜕𝑡

The Kinetic Energy:


𝜓 𝑥, 𝑡 = 𝐴 𝑒𝑖 𝑘 𝑥 − 𝜔 𝑡 + 𝜀

𝜕𝜓 𝑥, 𝑡

𝜕𝑥= 𝑖 𝑘 𝐴 𝑒𝑖 𝑘 𝑥 − 𝜔 𝑡 + 𝜀 = 𝑖 𝑘 𝜓 𝑥, 𝑡

𝜕2𝜓 𝑥, 𝑡

𝜕𝑥2=𝜕

𝜕𝑥

𝜕𝜓 𝑥, 𝑡

𝜕𝑥=𝜕

𝜕𝑥𝑖 𝑘 𝜓 𝑥, 𝑡 = − 𝑘2𝜓 𝑥, 𝑡

𝐾 =𝑝2

2 𝑚=

ℏ2

2 𝑚𝑘2 = −

ℏ2

2 𝑚

1

𝜓 𝑥, 𝑡

𝜕2𝜓 𝑥, 𝑡

𝜕𝑥2

Bringing the pieces together for a free particle:


ℏ 𝜔 =ℏ2

2 𝑚𝑘2

ℏ

− 𝑖

1

𝜓 𝑥, 𝑡

𝜕𝜓 𝑥, 𝑡

𝜕𝑡= −

ℏ2

2 𝑚

1

𝜓 𝑥, 𝑡

𝜕2𝜓 𝑥, 𝑡

𝜕𝑥2

𝐸 = 𝐾

In the presence of a potential 𝑉:


ℏ 𝜔 =ℏ2

2 𝑚𝑘2 + 𝑉

ℏ

− 𝑖

1

𝜓 𝑥, 𝑡

𝜕𝜓 𝑥, 𝑡

𝜕𝑡= −

ℏ2

2 𝑚

1

𝜓 𝑥, 𝑡

𝜕2𝜓 𝑥, 𝑡

𝜕𝑥2+ 𝑉 𝑥, 𝑡

𝑖 ℏ𝜕𝜓 𝑥, 𝑡

𝜕𝑡= −

ℏ2

2 𝑚

𝜕2𝜓 𝑥, 𝑡

𝜕𝑥2+ 𝑉 𝑥, 𝑡 𝜓 𝑥, 𝑡

𝐸 = 𝐾 + 𝑉

Schrödinger Equation



𝜕𝑡= −

ℏ2

2 𝑚

𝜕2𝜓 𝑥, 𝑡

𝜕𝑥2+ 𝑉 𝑥, 𝑡 𝜓 𝑥, 𝑡

• Not valid for relativistic particles (relativistic particles require Dirac Equation)

• Not valid for photons (light requires quantization of the electromagnetic field, which is called Second Quantization)

• What is really the wave function 𝜓 𝑥, 𝑡 or what is its physical meaning ?


• Actually, we just know about the physical meaning of 𝜓 𝑥, 𝑡 2.

• The Schrödinger Equation is for 𝜓 𝑥, 𝑡 .

• 𝜓 𝑥, 𝑡 2 describes the probability density of finding the wave-particle at point 𝑥 at time 𝑡.

• However, we don’t have an equation for 𝜓 𝑥, 𝑡 2.

A Few Remarks

Probability Density and Probability


Because 𝜓 𝑥, 𝑡 2 describes the probability density, then:

𝜓 𝑥, 𝑡 2 𝑑𝑥 gives us the probability of finding the wave-particle between (𝑥) and (𝑥 + 𝑑𝑥) at time (𝑡).

𝑥1𝑥2 𝜓 𝑥, 𝑡 2𝑑𝑥 gives us the probability of

finding the wave-particle between (𝑥1) and (𝑥2) at time (𝑡).

The probability of finding the wave-particle somewhere between −∞,∞will always be 1 at any time:


න−∞

+∞

𝜓 𝑥, 𝑡 2𝑑𝑥 = 1

Normalization Condition

Example 6.4


𝜓 𝑥, 𝑡 = 𝐴 𝑒−𝛼 𝑥 𝑒 ൗ− 𝑖 𝐸 𝑡ℏ

𝜓 𝑥, 𝑡 2 = 𝐴 2 𝑒−2𝛼 𝑥

1 = න−∞

+∞

𝜓 𝑥, 𝑡 2𝑑𝑥

= 2 𝐴 2න0

+∞

𝑒−2 𝛼 𝑥 𝑑𝑥 = 2 𝐴 2𝑒−2 𝛼 𝑥

−2 𝛼𝑥=0

𝑥=+∞

=𝐴 2

𝛼

𝐴 = 𝛼 𝑒𝑖 𝜙

= න−∞

+∞

𝐴 2𝑒−2𝛼 𝑥 𝑑𝑥

How can we get predictions from the wave function for physical quantities (position, energy, linear momentum,

angular momentum, …) ??


6.2 Physical Observables and Expectation Values

We need to confront those predictions against experimental measurements !!

Expectation Value for 𝑥 :


𝑎𝑔𝑒 =𝑁1 𝑎𝑔𝑒1 + 𝑁2 𝑎𝑔𝑒2 + 𝑁3 𝑎𝑔𝑒3 + 𝑁4 𝑎𝑔𝑒4 +⋯

𝑁1 + 𝑁2 +𝑁3 + 𝑁4 +⋯

=σ𝑖 𝑁𝑖 𝑎𝑔𝑒𝑖

σ𝑖 𝑁𝑖

𝑥 =∞−+∞

𝑃 𝑥 𝑥 𝑑𝑥

∞−+∞

𝑃 𝑥 𝑑𝑥=∞−+∞

𝜓 𝑥, 𝑡 2 𝑥 𝑑𝑥

∞−+∞

𝜓 𝑥, 𝑡 2 𝑑𝑥

Expectation Value for 𝑔 𝑥 :


𝑔 𝑥 =∞−+∞

𝜓 𝑥, 𝑡 2 𝑔 𝑥 𝑑𝑥

∞−+∞


Expectation Value for 𝑥2:

𝑥2 =∞−+∞

𝜓 𝑥, 𝑡 2 𝑥2𝑑𝑥

∞−+∞


Expectation Value for 𝑝:


𝑝 =∞−+∞

𝜓 𝑥, 𝑡 2 𝑝 𝑑𝑥

∞−+∞

𝜓 𝑥, 𝑡 2 𝑑𝑥=∞−+∞

𝜓∗ 𝑥, 𝑡 𝑝 𝜓 𝑥, 𝑡 𝑑𝑥

∞−+∞


= 𝑝 𝜓 𝑥, 𝑡

𝜕

𝜕𝑥𝜓 𝑥, 𝑡 = 𝑖 𝑘 𝜓 𝑥, 𝑡−𝑖 ℏ

−𝑖 ℏ𝜕

𝜕𝑥𝜓 𝑥, 𝑡 = ℏ 𝑘 𝜓 𝑥, 𝑡

=∞−+∞

𝜓∗ 𝑥, 𝑡 −𝑖 ℏ𝜕𝜕𝑥

𝜓 𝑥, 𝑡 𝑑𝑥

∞−+∞


Operator for Linear Momentum:


𝑝 =∞−+∞



∞−+∞


− 𝑖 ℏ𝜕

𝜕𝑥≡ ො𝑝

=∞−+∞

𝜓∗ 𝑥, 𝑡 Ƹ𝑝 𝜓 𝑥, 𝑡 𝑑𝑥

∞−+∞



𝑝2

=∞−+∞

𝜓∗ 𝑥, 𝑡 − ℏ2𝜕2

𝜕𝑥2𝜓 𝑥, 𝑡 𝑑𝑥

∞−+∞


Expectation Value for 𝑝2:

=∞−+∞


−𝑖 ℏ𝜕𝜕𝑥


∞−+∞


=∞−+∞

𝜓 𝑥, 𝑡 2 𝑝2𝑑𝑥

∞−+∞


Expectation Value for 𝐸:


𝐸 =∞−+∞

𝜓 𝑥, 𝑡 2 𝐸 𝑑𝑥

∞−+∞

𝜓 𝑥, 𝑡 2 𝑑𝑥=∞−+∞

𝜓∗ 𝑥, 𝑡 𝐸 𝜓 𝑥, 𝑡 𝑑𝑥

∞−+∞


= 𝐸 𝜓 𝑥, 𝑡

𝜕

𝜕𝑡𝜓 𝑥, 𝑡 = − 𝑖 𝜔 𝜓 𝑥, 𝑡𝑖 ℏ

𝑖 ℏ𝜕

𝜕𝑡𝜓 𝑥, 𝑡 = ℏ𝜔 𝜓 𝑥, 𝑡

=∞−+∞

𝜓∗ 𝑥, 𝑡 𝑖 ℏ𝜕𝜕𝑡


∞−+∞


Operator for Energy:


𝐸 =∞−+∞

𝜓∗ 𝑥, 𝑡 𝑖 ℏ𝜕𝜕𝑡


∞−+∞


=∞−+∞


∞−+∞


𝑖 ℏ𝜕

𝜕𝑡≡ 𝐸

Predicting Outcomes from the Theory:


𝑔 𝑥 =∞−+∞

𝜓∗ 𝑥, 𝑡 𝑔 𝑥 𝜓 𝑥, 𝑡 𝑑𝑥

∞−+∞


𝑝 =∞−+∞


∞−+∞

𝜓 𝑥, 𝑡 2 𝑑𝑥ො𝑝 ≡ −𝑖 ℏ

𝜕

𝜕𝑥

𝐸 =∞−+∞


∞−+∞

𝜓 𝑥, 𝑡 2 𝑑𝑥𝐸 ≡ 𝑖 ℏ

𝜕

𝜕𝑡

An useful result to rememberfrom Statistical Theory


𝜎𝑥2 ≡ 𝑥 − 𝑥 2 = 𝑥2 − 2 𝑥 𝑥 + 𝑥 2

= 𝑥2 − 2 𝑥 𝑥 + 𝑥 2

= 𝑥2 − 𝑥 2

= 𝑥2 − 2 𝑥 𝑥 + 𝑥 2

= 𝑥2 − 2 𝑥 2 + 𝑥 2


Schrödinger Equation becomes simpler when the potential energy 𝑉

does not depend on time 𝑡

Time Independent: 𝑉 𝑥, 𝑡 = 𝑉 𝑥



𝜕𝑡= −

ℏ2

2 𝑚

𝜕2𝜓 𝑥, 𝑡

𝜕𝑥2+ 𝜓 𝑥, 𝑡𝑉 𝑥, 𝑡𝑉 𝑥

𝜓 𝑥, 𝑡 ≡ 𝜓 𝑥 𝑓 𝑡

= −ℏ2

2 𝑚𝑓 𝑡

𝑑2𝜓 𝑥

𝑑𝑥2+ 𝑉 𝑥 𝜓 𝑥 𝑓 𝑡

𝑖 ℏ1

𝑓 𝑡

𝑑𝑓 𝑡

𝑑𝑡= −

ℏ2

2 𝑚

1

𝜓 𝑥

𝑑2𝜓 𝑥

𝑑𝑥2+ 𝑉 𝑥 = 𝐸

𝑖 ℏ 𝜓 𝑥𝑑𝑓 𝑡

𝑑𝑡

1

𝜓 𝑥 𝑓 𝑡×

Time Dependence of Wave Function


𝑓 𝑡 = 𝑒 ൗ−𝑖 𝐸 𝑡ℏ

𝑖 ℏ1

𝑓 𝑡

𝑑𝑓 𝑡

𝑑𝑡= 𝐸

𝜓 𝑥, 𝑡 ≡ 𝜓 𝑥 𝑓 𝑡 = 𝜓 𝑥 𝑒 ൗ−𝑖 𝐸 𝑡ℏ

𝜓 𝑥, 𝑡 2 = 𝜓 𝑥 𝑒 ൗ−𝑖 𝐸 𝑡ℏ2= 𝜓 𝑥 2

= 𝜓 𝑥 𝑒− 𝑖 𝜔 𝑡


−ℏ2

2 𝑚

1

𝜓 𝑥

𝑑2𝜓 𝑥

𝑑𝑥2+ 𝑉 𝑥 = 𝐸

−ℏ2

2 𝑚

𝑑2𝜓 𝑥

𝑑𝑥2+ 𝑉 𝑥 𝜓 𝑥 = 𝐸 𝜓 𝑥

Time-Independent Schrödinger Equation:

𝜓 𝑥 ×

𝜓 𝑥, 𝑡 = 𝜓 𝑥 𝑒 ൗ−𝑖 𝐸 𝑡ℏ

Requirements on the Wave Function:


• 𝜓 𝑥 and 𝑑𝜓 𝑥

𝑑𝑥must be finite

everywhere.


𝑑𝑥must be single-valued

everywhere.


𝑑𝑥must be continuous (no

jumps) everywhere, at least where 𝑉 𝑥is finite.

• 𝜓 𝑥 → 0 when 𝑥 → ±∞ .

Predicting Expected Values for Physical Observables :


𝑥 =∞−+∞

𝜓∗ 𝑥, 𝑡 𝑥 𝜓 𝑥, 𝑡 𝑑𝑥

∞−+∞


𝑝 =∞−+∞


∞−+∞

𝜓 𝑥, 𝑡 2 𝑑𝑥ො𝑝 ≡ −𝑖 ℏ

𝜕

𝜕𝑥

𝐸 =∞−+∞


∞−+∞

𝜓 𝑥, 𝑡 2 𝑑𝑥𝐸 ≡ 𝑖 ℏ

𝜕

𝜕𝑡

ො𝑥 ≡ 𝑥


6.3 Infinite Square-Well Potential

−ℏ2

2 𝑚

𝑑2𝜓 𝑥

𝑑𝑥2= 𝐸 𝜓 𝑥+ 𝑉 𝑥 𝜓 𝑥

−𝑑2𝜓 𝑥

𝑑𝑥2=2 𝑚 𝐸

ℏ2𝜓 𝑥 = 𝑘2 𝜓 𝑥

𝜓 𝑥 = 𝐴 𝑠𝑖𝑛 𝑘 𝑥 + 𝐵 𝑐𝑜𝑠 𝑘 𝑥

0 ≤ 𝑥 ≤ 𝐿

𝑘 ≡2 𝑚 𝐸

ℏ2


Infinite Square-Well Potential, cont.

𝜓 𝑥 = 𝐴 𝑠𝑖𝑛 𝑘 𝑥 + 𝐵 𝑐𝑜𝑠 𝑘 𝑥 0 ≤ 𝑥 ≤ 𝐿

𝜓 𝑥 = 0 𝑥 ≥ 𝐿𝑥 ≤ 0 and

𝜓 𝑥 = 0 𝜓 𝑥 = 0

𝜓 𝑥 = 𝐴 𝑠𝑖𝑛 𝑘 𝑥 + 𝐵 𝑐𝑜𝑠 𝑘 𝑥

𝜓 𝑥 = 0 = 𝐵 = 0


Infinite Square-Well Potential, cont.

𝜓 𝑥 = 𝐴 𝑠𝑖𝑛 𝑘 𝑥 0 ≤ 𝑥 ≤ 𝐿𝜓 𝑥 = 0 𝜓 𝑥 = 0

𝜓 𝑥 = 𝐴 𝑠𝑖𝑛 𝑘 𝑥

𝜓 𝑥 = 𝐿 = 𝐴 𝑠𝑖𝑛 𝑘 𝐿 = 0

𝑘𝑛 = 𝑛𝜋

𝐿

𝜓𝑛 𝑥 = 𝐴 𝑠𝑖𝑛 𝑘𝑛 𝑥

𝑛 = 1, 2, 3, …

= 𝐴 𝑠𝑖𝑛 𝑛𝜋

𝐿𝑥

Discrete Energy Levels:


=2 𝑚 𝐸𝑛ℏ2

𝑛𝜋

𝐿= 𝑘𝑛

𝑛 = 1, 2, 3, …

𝜓𝑛 𝑥 = 𝐴 𝑠𝑖𝑛 𝑛𝜋

𝐿𝑥

𝐸𝑛 = 𝑛2ℏ2 𝜋2

2 𝑚 𝐿2

𝜓𝑛 𝑥, 𝑡 2 = 𝐴 2 𝑠𝑖𝑛2 𝑛𝜋

𝐿𝑥

Normalizing the Wave Function:


𝜓 𝑥, 𝑡 2 = 𝜓𝑛 𝑥, 𝑡 2 = 𝐴 2 𝑠𝑖𝑛2 𝑛𝜋

𝐿𝑥

න−∞

+∞

𝜓 𝑥, 𝑡 2𝑑𝑥 = 1

න0

𝐿

𝐴 2 𝑠𝑖𝑛2 𝑛𝜋

𝐿𝑥 𝑑𝑥 = 1

𝐴 2𝐿

2= 1 𝐴 =

2

𝐿

𝑠𝑖𝑛2 𝜃 =1

21 − 𝑐𝑜𝑠 2𝜃

0 ≤ 𝑥 ≤ 𝐿

Bringing All Together:


𝜓 𝑥, 𝑡 = 𝜓𝑛 𝑥 𝑒− 𝑖 𝜔 𝑡

= 𝐴 𝑠𝑖𝑛 𝑘𝑛 𝑥 𝑒− 𝑖 𝜔 𝑡

=𝐴

2 𝑖𝑒+𝑖 𝑘𝑛 𝑥 − 𝑒−𝑖 𝑘𝑛 𝑥 𝑒− 𝑖 𝜔 𝑡

=𝐴

2 𝑖𝑒+𝑖 𝑘𝑛 𝑥 − 𝜔 𝑡 − 𝑒−𝑖 𝑘𝑛 𝑥 + 𝜔 𝑡

𝜔 ≡𝐸

ℏ

Two Counter-Propagating Waves

A More Realistic Potential Energy:


6.4 Finite Square-Well Potential


−ℏ2

2 𝑚

𝑑2𝜓 𝑥


Region II: 0 ≤ 𝑥 ≤ 𝐿


−ℏ2

2 𝑚

𝑑2𝜓 𝑥


−𝑑2𝜓 𝑥



𝜓𝐼𝐼 𝑥 = 𝐴𝐼𝐼 𝑠𝑖𝑛 𝑘 𝑥 + 𝐵𝐼𝐼 𝑐𝑜𝑠 𝑘 𝑥

𝑘 ≡2 𝑚 𝐸

ℏ2

Region I: 𝑥 ≤ 0


−ℏ2

2 𝑚

𝑑2𝜓 𝑥

𝑑𝑥2+ 𝜓 𝑥 = 𝐸 𝜓 𝑥𝑉 𝑥

𝑑2𝜓 𝑥

𝑑𝑥2=2𝑚 𝑉𝑜 − 𝐸

ℏ2𝜓 𝑥 = 𝛼2 𝜓 𝑥

𝜓𝐼 𝑥 = 𝐴𝐼𝑒−𝛼 𝑥 + 𝐵𝐼 𝑒

+𝛼 𝑥

𝛼 ≡2 𝑚 𝑉𝑜 − 𝐸

ℏ2

𝑉𝑜

= 𝐵𝐼 𝑒+𝛼 𝑥

𝐴𝐼 = 0to prevent the wave function to diverge when 𝑥 → −∞

Region III: 𝑥 ≥ 𝐿


−ℏ2

2 𝑚

𝑑2𝜓 𝑥


𝑑2𝜓 𝑥

𝑑𝑥2=2𝑚 𝑉𝑜 − 𝐸

ℏ2𝜓 𝑥 = 𝛼2 𝜓 𝑥

𝜓𝐼𝐼𝐼 𝑥 = 𝐴𝐼𝐼𝐼 𝑒−𝛼 𝑥 + 𝐵𝐼𝐼𝐼 𝑒

+𝛼 𝑥

𝛼 ≡2 𝑚 𝑉𝑜 − 𝐸

ℏ2

𝑉𝑜

= 𝐴𝐼𝐼𝐼 𝑒−𝛼 𝑥

𝐵𝐼𝐼𝐼 = 0to prevent the wave function to diverge when 𝑥 → +∞

Bringing All the Pieces Together:



𝜓𝐼𝐼𝐼 𝑥 = 𝐴𝐼𝐼𝐼 𝑒−𝛼 𝑥

𝜓𝐼 𝑥 = 𝐵𝐼 𝑒+𝛼 𝑥

Boundary Conditions for Wave Function:


𝑥 = 0 𝐵𝐼 = 𝐵𝐼𝐼

𝑥 = 𝐿 𝐴𝐼𝐼 𝑠𝑖𝑛 𝑘 𝐿 + 𝐵𝐼𝐼 𝑐𝑜𝑠 𝑘 𝐿 = 𝐴𝐼𝐼𝐼 𝑒−𝛼 𝐿




Boundary Conditions for the Derivative:


𝑑𝜓𝐼𝐼 𝑥

𝑑𝑥= 𝐴𝐼𝐼 𝑘 𝑐𝑜𝑠 𝑘 𝑥 − 𝐵𝐼𝐼 𝑘 𝑠𝑖𝑛 𝑘 𝑥

𝑑𝜓𝐼𝐼𝐼 𝑥

𝑑𝑥= −𝐴𝐼𝐼𝐼 𝛼 𝑒

−𝛼 𝑥

𝑑𝜓𝐼 𝑥

𝑑𝑥= 𝐵𝐼 𝛼 𝑒

+𝛼 𝑥

𝑥 = 0 𝐵𝐼 𝛼 = 𝐴𝐼𝐼 𝑘

𝑥 = 𝐿 𝐴𝐼𝐼 𝑘 𝑐𝑜𝑠 𝑘 𝐿 − 𝐵𝐼𝐼 𝑘 𝑠𝑖𝑛 𝑘 𝐿 = −𝐴𝐼𝐼𝐼 𝛼 𝑒−𝛼 𝐿

After Some Algebra (Appendix 1):


2 𝑡𝑎𝑛−1𝑉𝑜 − 𝐸

𝐸+ 𝑛 − 1 𝜋 =

2 𝑚 𝐸

ℏ2𝐿

𝑘𝑛 ≡2𝑚 𝐸𝑛ℏ2

𝛼𝑛 ≡2𝑚 𝑉𝑜 − 𝐸𝑛

ℏ2𝑛 = 1, 2, 3, …

𝜓𝐼,𝑛 𝑥 = 𝐵𝐼𝐼,𝑛 𝑒+𝛼𝑛 𝑥

𝜓𝐼𝐼,𝑛 𝑥 = 𝐵𝐼𝐼,𝑛𝛼𝑛𝑘𝑛

𝑠𝑖𝑛 𝑘𝑛 𝑥 + 𝑐𝑜𝑠 𝑘𝑛 𝑥

𝜓𝐼𝐼𝐼,𝑛 𝑥 = 𝐵𝐼𝐼,𝑛 𝑒+𝛼𝑛 𝐿

𝛼𝑛𝑘𝑛

𝑠𝑖𝑛 𝑘𝑛 𝐿 + 𝑐𝑜𝑠 𝑘𝑛 𝐿 𝑒−𝛼𝑛 𝑥


𝑎 ≡2 𝑚 𝐿2 𝑉𝑜

ℏ22 𝑡𝑎𝑛−1

1

𝑥− 1 + 𝑛 − 1 𝜋 = 𝑎 𝑥 𝑥 ≡

𝐸

𝑉𝑜

𝑎 = 10

𝐸1𝑉𝑜

𝐸2𝑉𝑜

𝐸3𝑉𝑜

𝐸4𝑉𝑜

Discrete Energy Levelsand Associated Wave Functions


PhET


https://phet.colorado.edu/en/simulation/legacy/bound-states

Number of Confined Wave Functions


𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐶𝑜𝑛𝑓𝑖𝑛𝑒𝑑 𝑊𝑎𝑣𝑒 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 = 𝐼𝑛𝑡𝑒𝑔𝑒𝑟𝑎

𝜋+ 1

= 𝐼𝑛𝑡𝑒𝑔𝑒𝑟2 𝑚 𝐿2 𝑉𝑜ℏ2 𝜋2

+ 1

= 𝐼𝑛𝑡𝑒𝑔𝑒𝑟8 𝑚 𝐿2 𝑉𝑜

ℎ2+ 1

Classical Physics


6.7 Potential Barrier


𝐸 ≥ 𝑉𝑜

Total energy higher than the barrier

Region I: 𝑥 ≤ 0


−ℏ2

2 𝑚

𝑑2𝜓 𝑥


−𝑑2𝜓 𝑥



𝜓𝐼 𝑥 = 𝐴𝐼 𝑒+𝑖 𝑘 𝑥 + 𝐵𝐼 𝑒

−𝑖 𝑘 𝑥

𝑘 ≡2 𝑚 𝐸

ℏ2

Region III: 𝑥 ≥ 𝐿


−ℏ2

2 𝑚

𝑑2𝜓 𝑥


−𝑑2𝜓 𝑥



𝜓𝐼𝐼𝐼 𝑥 = 𝐴𝐼𝐼𝐼 𝑒+𝑖 𝑘 𝑥 + 𝐵𝐼𝐼𝐼 𝑒

−𝑖 𝑘 𝑥

𝑘 ≡2 𝑚 𝐸

ℏ2

= 𝐴𝐼𝐼𝐼 𝑒+𝑖 𝑘 𝑥 𝐵𝐼𝐼𝐼 = 0

Region II: 0 ≤ 𝑥 ≤ 𝐿


−ℏ2

2 𝑚

𝑑2𝜓 𝑥


−𝑑2𝜓 𝑥

𝑑𝑥2=2 𝑚 𝐸 − 𝑉𝑜

ℏ2𝜓 𝑥 = 𝑘′2 𝜓 𝑥

𝑘′ ≡2 𝑚 𝐸 − 𝑉𝑜

ℏ2

𝜓𝐼𝐼 𝑥 = 𝐴𝐼𝐼 𝑒+𝑖 𝑘′ 𝑥 + 𝐵𝐼𝐼 𝑒

−𝑖 𝑘′ 𝑥

𝑉𝑜




−𝑖 𝑘 𝑥

𝜓𝐼𝐼𝐼 𝑥 = 𝐴𝐼𝐼𝐼 𝑒+𝑖 𝑘 𝑥



Boundary Conditions for Wave Function:



−𝑖 𝑘 𝑥

𝜓𝐼𝐼𝐼 𝑥 = 𝐴𝐼𝐼𝐼 𝑒+𝑖 𝑘 𝑥



𝑥 = 0

𝑥 = 𝐿

𝐴𝐼 + 𝐵𝐼 = 𝐴𝐼𝐼 + 𝐵𝐼𝐼

𝐴𝐼𝐼 𝑒+𝑖 𝑘′ 𝐿 + 𝐵𝐼𝐼 𝑒

−𝑖 𝑘′ 𝐿 = 𝐴𝐼𝐼𝐼 𝑒+𝑖 𝑘 𝐿

Boundary Conditions for the Derivative:



𝑑𝑥= 𝑖 𝑘′ 𝐴𝐼𝐼 𝑒

+𝑖 𝑘′ 𝑥 − 𝐵𝐼𝐼 𝑒−𝑖 𝑘′ 𝑥


𝑑𝑥= 𝑖 𝑘 𝐴𝐼𝐼𝐼 𝑒

+𝑖 𝑘 𝑥

𝑑𝜓𝐼 𝑥

𝑑𝑥= 𝑖 𝑘 𝐴𝐼 𝑒

+𝑖 𝑘 𝑥 − 𝐵𝐼 𝑒−𝑖 𝑘 𝑥

𝑥 = 0

𝑥 = 𝐿

𝑘 𝐴𝐼 − 𝐵𝐼 = 𝑘′ 𝐴𝐼𝐼 − 𝐵𝐼𝐼

𝑘′ 𝐴𝐼𝐼 𝑒+𝑖 𝑘′ 𝐿 − 𝐵𝐼𝐼 𝑒

−𝑖 𝑘′ 𝐿 = 𝑘 𝐴𝐼𝐼𝐼 𝑒+𝑖 𝑘 𝐿

Reflectance and Transmittance


𝑇 ≡𝐴𝐼𝐼𝐼

2

𝐴𝐼2

𝑅 ≡𝐵𝐼

2

𝐴𝐼2

𝜓𝐼 𝑥= 𝐴𝐼 𝑒

+𝑖 𝑘 𝑥

+ 𝐵𝐼 𝑒−𝑖 𝑘 𝑥

𝜓𝐼𝐼𝐼 𝑥= 𝐴𝐼𝐼𝐼 𝑒

+𝑖 𝑘 𝑥

𝑇 = 1 +𝑉0

2 𝑠𝑖𝑛2 𝑘′𝐿

4 𝐸 𝐸 − 𝑉0

−1

𝑘′𝐿 = 𝑛 𝜋 𝑇 = 1

see Appendix 2


𝐸 ≤ 𝑉𝑜Potential Barrier:

Region II: 𝑥 ≤ 0


−𝑑2𝜓 𝑥

𝑑𝑥2=2 𝑚 𝐸 − 𝑉𝑜

ℏ2𝜓 𝑥 = − 𝜅2 𝜓 𝑥

𝜅 ≡2 𝑚 𝑉𝑜 − 𝐸

ℏ2

𝜓𝐼𝐼 𝑥 = 𝐴𝐼𝐼 𝑒+ 𝜅 𝑥 + 𝐵𝐼𝐼 𝑒

− 𝜅 𝑥

−ℏ2

2 𝑚

𝑑2𝜓 𝑥

𝑑𝑥2+ 𝜓 𝑥 = 𝐸 𝜓 𝑥𝑉 𝑥𝑉𝑜

= 𝑖 𝑘′

Tunneling Probability


𝑇 = 1 +𝑉0

2 𝑠𝑖𝑛ℎ2 𝜅 𝐿

4 𝐸 𝑉0 − 𝐸

−1

𝐸

𝑉0

≅16 𝐸 𝑉0 − 𝐸

𝑉02 𝑒− 2 𝜅 𝐿

𝜅 ≡2 𝑚 𝑉𝑜 − 𝐸

ℏ2

Tunneling in Classical Optics


Quantum Tunneling


PhET


https://phet.colorado.edu/en/simulation/legacy/quantum-tunneling

Scanning Tunneling Microscope


Heinrich Rohrer (right, 1933– ) and GerdBinnig (1947– ) received the Nobel Prize for Physics in 1986 for their design of the scanning tunneling micro-scope.

The Swiss Rohrer was educated at the Swiss Federal Institute of Technology in Zurich and joined the IBM Research Laboratory in Zurich in 1963.

The German Binnig received his doctorate from the University of Frankfurt (Germany) in 1978 and then joined the same IBM Research Laboratory. He moved to the IBM Physics Group in Munich in 1984

Xe Atoms on Ni Surface


Photo taken with an STM, show xenon atoms placed on a nickel surface. The xenon atoms are 0.16 nm high and adjacent xenon atoms are 0.5 nm apart (the vertical scale has been exaggerated). The small force between the STM tip and an atom is enough to drag one xenon atom at a time across the nickel. The nickel atoms are represented by the black-and-white stripes on the horizontal surface. The image is magnified about 5 million times.

Ammonia Inversion


Field Effect Transistor


6.6 Harmonic Oscillator


Potential Energy and Wave Function


Lowest Order Solutions:


𝐸𝑛 = 𝑛 +1

2ℏ 𝜔

N = 10


6.5 Three-Dimensional, Time-Independent, Schrödinger Equation


−ℏ2

2 𝑚

𝑑2𝜓 𝑥


−ℏ2

2 𝑚

𝜕2

𝜕𝑥2+

𝜕2

𝜕𝑦2+

𝜕2

𝜕𝑧2𝜓 𝑥, 𝑦, 𝑧 + 𝑉 𝑥, 𝑦, 𝑧 𝜓 𝑥, 𝑦, 𝑧 = 𝐸 𝜓 𝑥, 𝑦, 𝑧

1D

3D

Appendix 1


Matching Boundary Conditions on Finite Square-Well Potential

Matching Boundary Conditions:


𝑥 = 0 𝐵𝐼 = 𝐵𝐼𝐼

𝑥 = 𝐿 𝐴𝐼𝐼 𝑠𝑖𝑛 𝑘 𝐿 + 𝐵𝐼𝐼 𝑐𝑜𝑠 𝑘 𝐿 = 𝐴𝐼𝐼𝐼 𝑒−𝛼 𝐿




Bound. Cond. for the Derivatives:



𝑑𝑥= 𝐴𝐼𝐼 𝑘 𝑐𝑜𝑠 𝑘 𝑥 − 𝐵𝐼𝐼 𝑘 𝑠𝑖𝑛 𝑘 𝑥


𝑑𝑥= −𝐴𝐼𝐼𝐼 𝛼 𝑒

−𝛼 𝑥

𝑑𝜓𝐼 𝑥

𝑑𝑥= 𝐵𝐼 𝛼 𝑒

+𝛼 𝑥

𝑥 = 0 𝐵𝐼 𝛼 = 𝐴𝐼𝐼 𝑘

𝑥 = 𝐿 𝐴𝐼𝐼 𝑘 𝑐𝑜𝑠 𝑘 𝐿 − 𝐵𝐼𝐼 𝑘 𝑠𝑖𝑛 𝑘 𝐿 = −𝐴𝐼𝐼𝐼 𝛼 𝑒−𝛼 𝐿


𝐵𝐼 = 𝐵𝐼𝐼

𝐴𝐼𝐼 𝑠𝑖𝑛 𝑘 𝐿 + 𝐵𝐼𝐼 𝑐𝑜𝑠 𝑘 𝐿 = 𝐴𝐼𝐼𝐼 𝑒−𝛼 𝐿

𝐵𝐼 𝛼 = 𝐴𝐼𝐼 𝑘

𝐴𝐼𝐼 𝑘 𝑐𝑜𝑠 𝑘 𝐿 − 𝐵𝐼𝐼 𝑘 𝑠𝑖𝑛 𝑘 𝐿 = −𝐴𝐼𝐼𝐼 𝛼 𝑒−𝛼 𝐿

𝐵𝐼𝐼 𝛼 = 𝐴𝐼𝐼 𝑘

𝐴𝐼𝐼 𝑘 𝑐𝑜𝑠 𝑘 𝐿 − 𝐵𝐼𝐼 𝑘 𝑠𝑖𝑛 𝑘 𝐿 = − 𝛼 𝐴𝐼𝐼 𝑠𝑖𝑛 𝑘 𝐿 + 𝐵𝐼𝐼 𝑐𝑜𝑠 𝑘 𝐿

1

2

3

4

1 3&

2 4&


𝐵𝐼𝐼 𝛼 = 𝐴𝐼𝐼 𝑘

𝐴𝐼𝐼 𝑘 𝑐𝑜𝑠 𝑘 𝐿 − 𝐵𝐼𝐼 𝑘 𝑠𝑖𝑛 𝑘 𝐿 = − 𝛼 𝐴𝐼𝐼 𝑠𝑖𝑛 𝑘 𝐿 + 𝐵𝐼𝐼 𝑐𝑜𝑠 𝑘 𝐿

𝐵𝐼𝐼 𝛼 𝑘 𝑐𝑜𝑠 𝑘 𝐿 − 𝐵𝐼𝐼 𝑘2 𝑠𝑖𝑛 𝑘 𝐿 = − 𝛼 𝐵𝐼𝐼 𝛼 𝑠𝑖𝑛 𝑘 𝐿 + 𝐵𝐼𝐼 𝑘 𝑐𝑜𝑠 𝑘 𝐿

𝛼 𝑘 𝑐𝑜𝑠 𝑘 𝐿 − 𝑘2 𝑠𝑖𝑛 𝑘 𝐿 = −𝛼2 𝑠𝑖𝑛 𝑘 𝐿 − 𝛼 𝑘 𝑐𝑜𝑠 𝑘 𝐿

2 𝛼 𝑘 𝑐𝑜𝑠 𝑘 𝐿 = 𝑘2 − 𝛼2 𝑠𝑖𝑛 𝑘 𝐿

2 𝛼 𝑘

𝑘2 − 𝛼2= 𝑡𝑎𝑛 𝑘 𝐿

𝑡𝑎𝑛 2 𝛿 =2 𝑡𝑎𝑛 𝛿

1 − 𝑡𝑎𝑛2 𝛿= 𝑡𝑎𝑛 𝑘 𝐿

𝑡𝑎𝑛 𝛿 ≡𝛼

𝑘


𝑡𝑎𝑛 2 𝛿 = 𝑡𝑎𝑛 𝑘 𝐿

2 𝛿 + 𝑛 − 1 𝜋 = 𝑘 𝐿

2 𝑡𝑎𝑛−1𝛼

𝑘+ 𝑛 − 1 𝜋 = 𝑘 𝐿

2 𝑡𝑎𝑛−1𝑉𝑜 − 𝐸

𝐸+ 𝑛 − 1 𝜋 =

2 𝑚 𝐸

ℏ2𝐿

𝑛 = 1, 2, 3, …


𝑎 ≡2 𝑚 𝐿2 𝑉𝑜

ℏ22 𝑡𝑎𝑛−1

1

𝑥− 1 + 𝑛 − 1 𝜋 = 𝑎 𝑥 𝑥 ≡

𝐸

𝑉𝑜

𝑎 = 10

𝐸1𝑉𝑜

𝐸2𝑉𝑜

𝐸3𝑉𝑜

𝐸4𝑉𝑜

Number of Confined Wave Functions


𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝐶𝑜𝑛𝑓𝑖𝑛𝑒𝑑 𝑊𝑎𝑣𝑒 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛𝑠 = 𝐼𝑛𝑡𝑒𝑔𝑒𝑟𝑎

𝜋+ 1

= 𝐼𝑛𝑡𝑒𝑔𝑒𝑟2 𝑚 𝐿2 𝑉𝑜ℏ2 𝜋2

+ 1

= 𝐼𝑛𝑡𝑒𝑔𝑒𝑟8 𝑚 𝐿2 𝑉𝑜

ℎ2+ 1

Region I:


𝜓𝐼,𝑛 𝑥 = 𝐵𝐼,𝑛 𝑒+𝛼𝑛 𝑥


= 𝐵𝐼𝐼,𝑛 𝑒+𝛼𝑛 𝑥

Region II:


𝜓𝐼𝐼,𝑛 𝑥 = 𝐴𝐼𝐼,𝑛 𝑠𝑖𝑛 𝑘𝑛 𝑥 + 𝐵𝐼𝐼,𝑛 𝑐𝑜𝑠 𝑘𝑛 𝑥

= 𝐵𝐼𝐼,𝑛𝛼𝑛𝑘𝑛




Region III:


𝜓𝐼𝐼𝐼,𝑛 𝑥 = 𝐴𝐼𝐼𝐼,𝑛 𝑒−𝛼𝑛 𝑥

= 𝐵𝐼𝐼,𝑛 𝑒+𝛼𝑛 𝐿

𝛼𝑛𝑘𝑛



𝐴𝐼𝐼 𝑠𝑖𝑛 𝑘 𝐿 + 𝐵𝐼𝐼 𝑐𝑜𝑠 𝑘 𝐿 = 𝐴𝐼𝐼𝐼 𝑒−𝛼 𝐿


All Regions:


𝜓𝐼,𝑛 𝑥 = 𝐵𝐼𝐼,𝑛 𝑒+𝛼𝑛 𝑥

𝜓𝐼𝐼,𝑛 𝑥 = 𝐵𝐼𝐼,𝑛𝛼𝑛𝑘𝑛


𝜓𝐼𝐼𝐼,𝑛 𝑥 = 𝐵𝐼𝐼,𝑛 𝑒+𝛼𝑛 𝐿

𝛼𝑛𝑘𝑛


Verification:


𝑘𝑛 𝐵𝐼𝐼,𝑛𝛼𝑛𝑘𝑛

𝑐𝑜𝑠 𝑘𝑛 𝐿 − 𝑠𝑖𝑛 𝑘𝑛 𝐿 = −𝛼𝑛 𝐵𝐼𝐼,𝑛 𝑒+𝛼𝑛 𝐿

𝛼𝑛𝑘𝑛

𝑠𝑖𝑛 𝑘𝑛 𝐿 + 𝑐𝑜𝑠 𝑘𝑛 𝐿 𝑒−𝛼𝑛 𝐿

𝑘𝑛𝛼𝑛𝑘𝑛

𝑐𝑜𝑠 𝑘𝑛 𝐿 − 𝑠𝑖𝑛 𝑘𝑛 𝐿 = −𝛼𝑛𝛼𝑛𝑘𝑛

𝑠𝑖𝑛 𝑘𝑛 𝐿 + 𝑐𝑜𝑠 𝑘𝑛 𝐿

𝛼𝑛𝛼𝑛𝑘𝑛

− 𝑘𝑛 𝑠𝑖𝑛 𝑘𝑛 𝐿 = −𝑘𝑛𝛼𝑛𝑘𝑛

− 𝛼𝑛 𝑐𝑜𝑠 𝑘𝑛 𝐿

𝑡𝑎𝑛 𝑘𝑛 𝐿 =−𝑘𝑛

𝛼𝑛𝑘𝑛

− 𝛼𝑛

𝛼𝑛𝛼𝑛𝑘𝑛

− 𝑘𝑛

=2 𝑘𝑛 𝛼𝑛

𝑘𝑛2 − 𝛼𝑛

2

Appendix 2


Matching Boundary Conditions on Barrier Potential

𝐸 ≥ 𝑉𝑜


𝐴𝐼 + 𝐵𝐼 = 𝐴𝐼𝐼 + 𝐵𝐼𝐼

𝐴𝐼𝐼 𝑒+𝑖 𝑘′ 𝐿 + 𝐵𝐼𝐼 𝑒

−𝑖 𝑘′ 𝐿 = 𝐴𝐼𝐼𝐼 𝑒+𝑖 𝑘 𝐿

𝑘 𝐴𝐼 − 𝐵𝐼 = 𝑘′ 𝐴𝐼𝐼 − 𝐵𝐼𝐼


−𝑖 𝑘′ 𝐿 = 𝑘 𝐴𝐼𝐼𝐼 𝑒+𝑖 𝑘 𝐿


−𝑖 𝑘′ 𝐿 = 𝑘 𝐴𝐼𝐼 𝑒+𝑖 𝑘′ 𝐿 + 𝐵𝐼𝐼 𝑒

−𝑖 𝑘′ 𝐿

𝑘 2 𝐴𝐼 − 𝐴𝐼𝐼 − 𝐵𝐼𝐼 = 𝑘′ 𝐴𝐼𝐼 − 𝐵𝐼𝐼

1

2

3

4

1 3

2 4

&

&



−𝑖 𝑘′ 𝐿 = 𝑘 𝐴𝐼𝐼 𝑒+𝑖 𝑘′ 𝐿 + 𝐵𝐼𝐼 𝑒


𝐴𝐼 =𝑘 𝐴𝐼𝐼 + 𝐵𝐼𝐼 + 𝑘′ 𝐴𝐼𝐼 − 𝐵𝐼𝐼

2 𝑘=𝐴𝐼𝐼 𝑘 + 𝑘′ + 𝐵𝐼𝐼 𝑘 − 𝑘′

2 𝑘

𝐴𝐼𝐼 𝑒+𝑖 𝑘′ 𝐿 𝑘′ − 𝑘 = 𝐵𝐼𝐼 𝑒

−𝑖 𝑘′ 𝐿 𝑘′ + 𝑘

𝐵𝐼𝐼 = 𝐴𝐼𝐼 𝑒𝑖 2 𝑘′𝐿

𝑘′ − 𝑘

𝑘′ + 𝑘

𝑘 2 𝐴𝐼 − 𝐴𝐼𝐼 − 𝐵𝐼𝐼 = 𝑘′ 𝐴𝐼𝐼 − 𝐵𝐼𝐼


𝐴𝐼 =𝐴𝐼𝐼 𝑘 + 𝑘′ + 𝐵𝐼𝐼 𝑘 − 𝑘′

2 𝑘= 𝐴𝐼𝐼

𝑘′ + 𝑘 2 − 𝑒𝑖 2 𝑘′𝐿 𝑘′ − 𝑘 2

2 𝑘 𝑘′ + 𝑘

𝐴𝐼𝐼𝐼 = 𝑒− 𝑖 𝑘 𝐿 𝐴𝐼𝐼 𝑒+𝑖 𝑘′ 𝐿 + 𝐵𝐼𝐼 𝑒


= 𝐴𝐼𝐼 𝑒− 𝑖 𝑘 𝐿 𝑒+𝑖 𝑘

′ 𝐿 + 𝑒𝑖 2 𝑘′𝐿

𝑘′ − 𝑘

𝑘′ + 𝑘𝑒−𝑖 𝑘

′ 𝐿

= 𝐴𝐼𝐼 𝑒− 𝑖 𝑘 𝐿 𝑒+𝑖 𝑘

′ 𝐿2 𝑘′

𝑘′ + 𝑘

𝐵𝐼𝐼 = 𝐴𝐼𝐼 𝑒𝑖 2 𝑘′𝐿

𝑘′ − 𝑘

𝑘′ + 𝑘


=𝐴𝐼𝐼 𝑒

𝑖 𝑘′𝐿

𝑘′ + 𝑘

𝑒−𝑖 𝑘′𝐿 𝑘′ + 𝑘 2 − 𝑒𝑖 𝑘

′𝐿 𝑘′ − 𝑘 2

2 𝑘

𝐴𝐼 = 𝐴𝐼𝐼𝑘′ + 𝑘 2 − 𝑒𝑖 2 𝑘

′𝐿 𝑘′ − 𝑘 2

2 𝑘 𝑘′ + 𝑘

𝐴𝐼𝐼𝐼 =𝐴𝐼𝐼 𝑒

− 𝑖 𝑘 𝐿 𝑒+𝑖 𝑘′ 𝐿

𝑘′ + 𝑘2 𝑘′


𝑇 =𝐴𝐼𝐼𝐼

2

𝐴𝐼2

𝐴𝐼 =𝐴𝐼𝐼 𝑒

𝑖 𝑘′𝐿

𝑘′ + 𝑘

𝑒−𝑖 𝑘′𝐿 𝑘′ + 𝑘 2 − 𝑒𝑖 𝑘

′𝐿 𝑘′ − 𝑘 2

2 𝑘



𝑘′ + 𝑘2 𝑘′


2

𝐴𝐼2=

16 𝑘′2𝑘2

𝑘′ + 𝑘 4 + 𝑘′ − 𝑘 4 − 2 𝑘′ + 𝑘 2 𝑘′ − 𝑘 2 𝑐𝑜𝑠 2 𝑘′𝐿



2

𝐴𝐼2=

16 𝑘′2𝑘2

𝑘′ + 𝑘 4 + 𝑘′ − 𝑘 4 − 2 𝑘′ + 𝑘 2 𝑘′ − 𝑘 2 𝑐𝑜𝑠 2 𝑘′𝐿

=16 𝑘′

2𝑘2

𝑘′ + 𝑘 4 + 𝑘′ − 𝑘 4 − 2 𝑘′ + 𝑘 2 𝑘′ − 𝑘 2 1 − 2 𝑠𝑖𝑛2 𝑘′𝐿

1

𝑇=

𝑘′ + 𝑘 4 + 𝑘′ − 𝑘 4 − 2 𝑘′ + 𝑘 2 𝑘′ − 𝑘 2 +4 𝑘′ + 𝑘 2 𝑘′ − 𝑘 2𝑠𝑖𝑛2 𝑘′𝐿

16 𝑘′2 𝑘2

=𝑘′ + 𝑘 2 − 𝑘′ − 𝑘 2 2 +4 𝑘′ + 𝑘 2 𝑘′ − 𝑘 2𝑠𝑖𝑛2 𝑘′𝐿

16 𝑘′2 𝑘2

= 1 +𝑘′ + 𝑘 2 𝑘′ − 𝑘 2 𝑠𝑖𝑛2 𝑘′𝐿

4 𝑘′2 𝑘2


𝑘′ ≡2 𝑚 𝐸 − 𝑉𝑜

ℏ2

1

𝑇= 1 +

𝑘′ + 𝑘 2 𝑘′ − 𝑘 2 𝑠𝑖𝑛2 𝑘′𝐿

4 𝑘′2 𝑘2

𝑘 ≡2 𝑚 𝐸

ℏ2

1

𝑇= 1 +

𝐸 − 𝑉𝑜 − 𝐸 2 𝑠𝑖𝑛2 𝑘′𝐿

4 𝐸 − 𝑉𝑜2 𝐸2

= 1 +𝑉𝑜

2 𝑠𝑖𝑛2 𝑘′𝐿

4 𝐸 − 𝑉𝑜 𝐸

= 1 +𝑘′ 2 − 𝑘 2 2 𝑠𝑖𝑛2 𝑘′𝐿

4 𝑘′2 𝑘2

= 1 +𝑠𝑖𝑛2 𝑘′𝐿

4𝐸𝑉𝑜− 1

𝐸𝑉𝑜

Appendix 3


Matching Boundary Conditions on Barrier Potential

𝐸 ≤ 𝑉𝑜


𝜅 ≡ 𝑖 𝑘′𝑘′ ≡2 𝑚 𝐸 − 𝑉𝑜

ℏ2𝜅 ≡

2 𝑚 𝑉𝑜 − 𝐸

ℏ2

𝐴𝐼 =𝐴𝐼𝐼 𝑒

𝑖 𝑘′𝐿

𝑘′ + 𝑘

𝑒−𝑖 𝑘′𝐿 𝑘′ + 𝑘 2 − 𝑒𝑖 𝑘

′𝐿 𝑘′ − 𝑘 2

2 𝑘



𝑘′ + 𝑘2 𝑘′



2

𝐴𝐼2

chapter 6 300 fall 19... · 2019-10-21 · fall 2018 prof. sergio b. mendes 9 𝑖ℏ , =− ℏ2 2...

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