EE201/MSE207 Lecture 4 Harmonic oscillator Important for optics - - PowerPoint PPT Presentation

EE201/MSE207 Lecture 4 Harmonic oscillator

Important for optics (photons) and phonons, though they are massless

• particles. Also rare examples like electrons in a very smooth Q.Well.

Recently became important for NEMS. Very important as a fundamental example, starting point for many problems.

Classical physics 𝑊 𝑦 = 𝑙𝑦2 2 = 1 2 𝑛𝜕2𝑦2

spring constant k

𝐺 = −𝑙𝑦 𝑦 = 𝐺 𝑛 = − 𝑙 𝑛 𝑦 𝑦 𝑢 = 𝐵 cos(𝜕𝑢 + 𝜒) 𝜕 = 𝑙 𝑛

Potential energy

𝐺 = − 𝑒𝑊 𝑒𝑦

(by the way, 𝑊 = 𝑈 = 𝐹 2)

  

𝑦 𝑊 𝑦

Harmonic oscillator

Classical physics 𝑊 𝑦 = 𝑙𝑦2 2 = 1 2 𝑛𝜕2𝑦2

spring constant k

𝐺 = −𝑙𝑦 𝑦 = 𝐺 𝑛 = − 𝑙 𝑛 𝑦 𝑦 𝑢 = 𝐵 cos(𝜕𝑢 + 𝜒) 𝜕 = 𝑙 𝑛

Potential energy

  

𝑦 𝑊 𝑦 Quantum mechanics TISE − ℏ2 2𝑛 𝑒2𝜔 𝑒𝑦2 + 1 2 𝑛𝜕2𝑦2 𝜔 = 𝐹𝜔

Need to find energies and corresponding wavefunctions Two ways to solve: 1) solve as a differential equation (Hermite polynomials, etc.) 2) solve using a trick: algebraic technique of ladder operators We will consider only the second way (important for second quantization, similar for angular momentum, spin, etc.)

Ladder operators

(raising/lowering or creation/annihilation) Define

𝑏± = 1 2ℏ𝑛𝜕 ∓𝑗 𝑞 + 𝑛𝜕 𝑦 = 1 2ℏ𝑛𝜕 ∓ℏ 𝜖 𝜖𝑦 + 𝑛𝜕𝑦

(hat means operator) Idea why:

𝐼 = 1 2𝑛 𝑞2 + 𝑛𝜕 𝑦 2 𝐼 ℏ𝜕 = 1 2ℏ𝑛𝜕 𝑞2 + 𝑛𝜕 𝑦 2 𝑏2+𝑐2= (𝑏 + 𝑗𝑐)(𝑏 − 𝑗𝑐)

However, operators usually do not commute: 𝐵 𝐶 ≠ 𝐶 𝐵

𝑏− 𝑏+ = 1 2ℏ𝑛𝜕 𝑗 𝑞 + 𝑛𝜕 𝑦 −𝑗 𝑞 + 𝑛𝜕 𝑦 = = 1 2ℏ𝑛𝜕 [ 𝑞2+ 𝑛𝜕 𝑦 2 − 𝑗𝑛𝜕( 𝑦 𝑞 − 𝑞 𝑦)] = 𝐼 ℏ𝜕 − 𝑗 2ℏ ( 𝑦 𝑞 − 𝑞 𝑦)

What is 𝑦 𝑞 − 𝑞 𝑦 ? This is called commutator, 𝐵, 𝐶 ≡ 𝐵 𝐶 − 𝐶 𝐵

Commutator [ 𝑦, 𝑞]

𝑦 𝑞 − 𝑞 𝑦 𝑔 𝑦 = 𝑦 −𝑗ℏ 𝜖𝑔 𝜖𝑦 − −𝑗ℏ 𝜖 𝜖𝑦 𝑦 𝑔 𝑦 = 𝑗ℏ 𝑔(𝑦)

Therefore

𝑦, 𝑞 = 𝑗ℏ

Now back to calculation of 𝑏− 𝑏+

𝑏− 𝑏+ = 𝐼 ℏ𝜕 − 𝑗 2ℏ 𝑦 𝑞 − 𝑞 𝑦 = 𝐼 ℏ𝜕 + 1 2

Similarly

𝑏+ 𝑏− = 𝐼 ℏ𝜕 − 1 2 [ 𝑏−, 𝑏+ ] = 1

TISE can be written as

ℏ𝜕 ( 𝑏+ 𝑏− + 1 2)𝜔 = 𝐹 𝜔

• r as

ℏ𝜕 ( 𝑏− 𝑏+ − 1 2)𝜔 = 𝐹 𝜔

Lemma If 𝜔 𝑦 satisfies TISE with energy 𝐹, then

𝑏+𝜔

also satisfies TISE, with energy 𝐹 + ℏ𝜕. (This is why 𝑏+ is called raising operator) Proof

𝐼( 𝑏+𝜔) = ℏ𝜕 𝑏+ 𝑏− +

1 2

𝑏+𝜔 = ℏ𝜕 𝑏+ 𝑏− 𝑏+ +

1 2

𝑏+ 𝜔 = ℏ𝜕 𝑏+ 𝑏− 𝑏+ +

1 2 𝜔 =

𝑏+ 𝐼 + ℏ𝜕 𝜔 = 𝑏+ 𝐹 + ℏ𝜕 𝜔 = = 𝐹 + ℏ𝜕 ( 𝑏+𝜔)

𝐼 ℏ𝜕 + 1 2 + 1 2

Q.E.D. Lemma 2 If 𝜔 𝑦 satisfies TISE with energy 𝐹, then

𝑏−𝜔

also satisfies TISE, with energy 𝐹 − ℏ𝜕. (similar proof) Now main trick If we have one solution 𝜔 𝑦 , we can construct many other solutions (ladder of solutions )

𝐹

ℏ𝜕 ℏ𝜕 ℏ𝜕 ℏ𝜕 ℏ𝜕 Process of going down should stop somewhere (𝐹 ≥ 0) Then

𝑏−𝜔0 = 0 (further derivation is simple)

Ground state of harmonic oscillator

𝑏−𝜔0 = 0 ⇒ 1 2ℏ𝑛𝜕 ℏ 𝑒 𝑒𝑦 + 𝑛𝜕𝑦 𝜔0 𝑦 = 0 ⇒ 𝑒𝜔0 𝜖𝑦 = − 𝑛𝜕 ℏ 𝑦𝜔0

Solution

𝜔0 𝑦 = 𝐵 exp − 𝑛𝜕 2ℏ 𝑦2

Find 𝐵 from normalization

𝜔0 𝑦 = 𝑛𝜕 𝜌ℏ

1 4

exp − 𝑛𝜕 2ℏ 𝑦2

ground state What is its energy?

𝐼 = ℏ𝜕( 𝑏+ 𝑏− + 1 2) 𝐼𝜔0 = ℏ𝜕 2 𝜔0 𝐹0 = 1 2 ℏ𝜕

(“zero-point” energy, vacuum energy) Now can find all stationary states by repeatedly applying 𝑏+ to the ground state

Stationary states of harmonic oscillator

𝐵𝑜 is normalization constant

𝑏+𝜔𝑜 𝑦 = 𝑜 + 1 𝜔𝑜+1(𝑦) 𝑏−𝜔𝑜 𝑦 = 𝑜 𝜔𝑜−1(𝑦) 𝜔𝑜 𝑦 = 𝐵𝑜 𝑏+ 𝑜 exp − 𝑛𝜕 2ℏ 𝑦2

𝐹𝑜 = 𝑜 +

1 2 ℏ𝜕

Useful relations: therefore

𝜔𝑜 = 1 𝑜! 𝑏+ 𝑜𝜔0

A few lowest levels: 𝜔0 𝑦 = 𝑛𝜕 𝜌ℏ

1 4

exp − 𝑛𝜕 2ℏ 𝑦2 𝜔1 𝑦 = 𝑛𝜕 𝜌ℏ

1 4

2𝑛𝜕 ℏ 𝑦 exp − 𝑛𝜕 2ℏ 𝑦2 𝜔2 𝑦 = 𝑛𝜕 𝜌ℏ

1 4

2 𝑛𝜕 ℏ 𝑦2 − 2 2 exp − 𝑛𝜕 2ℏ 𝑦2

𝑜 = 0, 1, 2, …

Stationary states of harmonic oscillator

• Fig. 2.7 from Griffiths’ book

General stationary states for harmonic oscillator

𝐹𝑜 = 𝑜 + 1 2 ℏ𝜕

General solution of SE Ψ 𝑦, 𝑢 = 𝑜=0

∞

𝑑𝑜 𝜔𝑜 𝑦 exp −𝑗 𝑜 + 1 2 𝜕𝑢

𝜔𝑜 = 1 𝑜! 𝑏+ 𝑜𝜔0

𝜊 = 𝑛𝜕 𝜌ℏ 𝑦 𝜔𝑜 𝑦 = 𝑛𝜕 𝜌ℏ

1 4

1 2𝑜𝑜! 𝐼𝑜(𝜊) exp − 𝜊2 2 Explicitly 𝐼𝑜(𝜊) – Hermite polynomials

𝑜=0

∞

𝑑𝑜 2 = 1

Free particle

TISE − ℏ2 2𝑛 d2𝜔 𝑒𝑦2 = 𝐹𝜔 𝑊 𝑦 = 0 𝜔 𝑦 = 𝐵𝑓𝑗𝑙𝑦 + 𝐶𝑓−𝑗𝑙𝑦

solution:

𝑙 = 2𝑛𝐹 ℏ 𝑙 – “wave vector”, 𝑙 = 2𝜌 𝜇

with time dependence:

Ψ 𝑦, 𝑢 = 𝐵 exp 𝑗𝑙 𝑦 − ℏ𝑙 2𝑛 𝑢 + 𝐶 exp −𝑗𝑙 𝑦 + ℏ𝑙 2𝑛 𝑢

propagates to the right with velocity 𝑤𝑞ℎ = ℏ𝑙

2𝑛

propagates to the left with the same velocity

interesting that classically 𝑤classical = 2𝐹 𝑛 = ℏ𝑙 𝑛 (twice larger) (will discuss later, difference between phase and group velocities)

Not normalizable! What to do? Construct a wave packet.

Wave packet

𝑙 > 0 to the right Ψ 𝑦, 𝑢 = 1 2𝜌

−∞ ∞

𝜚(𝑙) exp 𝑗 𝑙𝑦 − ℏ𝑙2 2𝑛 𝑢 𝑒𝑙

energy is not well-defined

𝑙 < 0 to the left

At 𝑢 = 0

Ψ 𝑦, 0 = 1 2𝜌

−∞ ∞

𝜚(𝑙) 𝑓𝑗𝑙𝑦𝑒𝑙

just a Fourier transform Remind Fourier transform:

𝑔(𝑦) = 1 2𝜌

−∞ ∞

𝐺(𝑙) 𝑓𝑗𝑙𝑦𝑒𝑙 𝐺(𝑙) = 1 2𝜌

−∞ ∞

𝑔(𝑦) 𝑓−𝑗𝑙𝑦𝑒𝑦

So if we know Ψ(𝑦, 0), then can find 𝜚 𝑙 , and then know Ψ(𝑦, 𝑢)

𝜚(𝑙) = 1 2𝜌

−∞ ∞

Ψ(𝑦, 0) 𝑓−𝑗𝑙𝑦𝑒𝑦

Normalization:

−∞ ∞

Ψ 𝑦, 0

2 𝑒𝑦 = 1

is equivalent to

−∞ ∞

𝜚(𝑙) 2 𝑒𝑙 = 1

𝜚 𝑙 is like wavefunction in k-space

Phase and group velocities

Ψ 𝑦, 𝑢 = 1 2𝜌

−∞ ∞

𝜚(𝑙) exp 𝑗 𝑙𝑦 − 𝜕𝑢 𝑒𝑙

Assume that 𝜚(𝑙) is concentrated around 𝑙0, then group velocity: 𝑤𝑕𝑠 = 𝜕′ =

𝑒𝜕 𝑒𝑙 𝜕 = ℏ𝑙2 2𝑛 𝜕 𝑙 ≈ 𝜕0 + 𝑒𝜕 𝑒𝑙 𝑙 − 𝑙0 = 𝜕0 + 𝜕′Δ𝑙

  

     

 

𝜚(𝑙) 𝑙 𝑙0

Ψ 𝑦, 𝑢 ≈ 1 2𝜌

−∞ ∞

𝜚(𝑙0 + Δ𝑙) exp 𝑗 𝑙0 + Δ𝑙 𝑦 − 𝑗 𝜕0 + 𝜕′Δ𝑙 𝑢 𝑒Δ𝑙 = 1 2𝜌 exp 𝑗𝑙0 𝑦 − 𝜕0 𝑙0 𝑢

−∞ ∞

𝜚(𝑙0 + Δ𝑙) exp 𝑗Δ𝑙 𝑦 − 𝜕′𝑢 𝑒Δ𝑙

phase velocity: 𝑤𝑞ℎ =

𝜕0 𝑙0

In our case 𝜕 = ℏ𝑙2 (2𝑛)

𝑤𝑕𝑠 = 𝑒𝜕 𝑒𝑙 = ℏ𝑙 𝑛 = 𝑤classical 𝑤𝑞ℎ = 𝜕 𝑙 = ℏ𝑙 2𝑛

So, in quantum case

𝑤𝑕𝑠 = 2𝑤𝑞ℎ

(shape faster than ripples) (For waves on water 𝑤𝑞ℎ = 2𝑤𝑕𝑠)

Another normalization 𝜔𝑙 𝑦 =

1 2𝜌 e𝑗𝑙𝑦

Actually, it is difficult to work with wavepackets, so in practice people usually work with extended waves

−∞ ∞

𝜔𝑙

∗ 𝑦 𝜔𝑙′ 𝑦 𝑒𝑦 = 𝜀(𝑙 − 𝑙′)

This function satisfies “normalization”

Some properties of delta-function 𝜀(𝑦)

−∞ ∞

𝑔 𝑦 𝜀 𝑦 𝑒𝑦 = 𝑔(0)

−∞ ∞

𝑔 𝑦 𝜀 𝑦 − 𝑏 𝑒𝑦 = 𝑔(𝑏)

−∞ ∞

𝑓𝑗𝑏𝑦 𝑒𝑦 = 2𝜌 𝜀(𝑏)

(from Fourier transform theorem)