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. 𝑦 = 𝐺 𝑛 = − 𝑙 Classical physics 𝐺 = −𝑙𝑦 𝑛 𝑦 𝑦 𝑢 = 𝐵 cos(𝜕𝑢 + 𝜒) 𝜕 = 𝑙 𝑛 spring constant k 𝑊 𝑦 = 𝑙𝑦 2 𝐺 = − 𝑒𝑊 = 1 2 𝑛𝜕 2 𝑦 2 Potential energy 𝑒𝑦 2 𝑊 𝑦 (by the way, 𝑊 = 𝑈 = 𝐹 2)  𝑦  

Harmonic oscillator 𝑦 = 𝐺 𝑛 = − 𝑙 Classical physics 𝐺 = −𝑙𝑦 𝑛 𝑦 𝑦 𝑢 = 𝐵 cos(𝜕𝑢 + 𝜒) 𝜕 = 𝑙 𝑛 spring constant k 𝑊 𝑦 = 𝑙𝑦 2 = 1 2 𝑛𝜕 2 𝑦 2 Potential energy 2 𝑊 𝑦 Quantum mechanics  − ℏ 2 𝑒 2 𝜔 𝑒𝑦 2 + 1 2 𝑛𝜕 2 𝑦 2 𝜔 = 𝐹𝜔 TISE 𝑦 2𝑛 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) 1 1 ∓ℏ 𝜖 𝑏 ± = ∓𝑗 𝑞 + 𝑛𝜕 𝑦 = 𝜖𝑦 + 𝑛𝜕𝑦 Define 2ℏ𝑛𝜕 2ℏ𝑛𝜕 (hat means operator) 𝐼 = 1 𝐼 1 𝑞 2 + 𝑛𝜕 𝑞 2 + 𝑛𝜕 𝑦 2 𝑦 2 ℏ𝜕 = Idea why: 2𝑛 2ℏ𝑛𝜕 𝑏 2 +𝑐 2 = (𝑏 + 𝑗𝑐)(𝑏 − 𝑗𝑐) However, operators usually do not commute: 𝐵 𝐶 ≠ 𝐶 𝐵 1 𝑏 − 𝑏 + = 2ℏ𝑛𝜕 𝑗 𝑞 + 𝑛𝜕 𝑦 −𝑗 𝑞 + 𝑛𝜕 𝑦 = 1 ℏ𝜕 − 𝑗 𝐼 𝑦 2 − 𝑗𝑛𝜕( 𝑞 2 + 𝑛𝜕 = 2ℏ𝑛𝜕 [ 𝑦 𝑞 − 𝑞 𝑦)] = 2ℏ ( 𝑦 𝑞 − 𝑞 𝑦) 𝑦 ? This is called commutator, 𝐵, 𝐶 ≡ 𝐵 𝐶 − 𝐶 What is 𝑦 𝑞 − 𝑞 𝐵

Commutator [ 𝑦, 𝑞] 𝑦 𝑔 𝑦 = 𝑦 −𝑗ℏ 𝜖𝑔 𝜖 𝑦 𝑞 − 𝑞 𝜖𝑦 − −𝑗ℏ 𝜖𝑦 𝑦 𝑔 𝑦 = 𝑗ℏ 𝑔(𝑦) 𝑦, 𝑞 = 𝑗ℏ Therefore Now back to calculation of 𝑏 − 𝑏 + ℏ𝜕 − 𝑗 𝐼 ℏ𝜕 + 1 𝐼 𝑏 − 𝑏 + = 2ℏ 𝑦 𝑞 − 𝑞 𝑦 = 2 [ 𝑏 − , 𝑏 + ] = 1 ℏ𝜕 − 1 𝐼 𝑏 + 𝑏 − = Similarly 2 𝑏 − + ℏ𝜕 ( 𝑏 + 1 2)𝜔 = 𝐹 𝜔 TISE can be written as ℏ𝜕 ( 𝑏 − 𝑏 + − 1 2)𝜔 = 𝐹 𝜔 or as

If 𝜔 𝑦 satisfies TISE with energy 𝐹 , then 𝑏 + 𝜔 Lemma also satisfies TISE, with energy 𝐹 + ℏ𝜕 . (This is why 𝑏 + is called raising operator) Proof 1 2 1 2 𝑏 − + 𝑏 + + 𝐼( 𝑏 + 𝜔) = ℏ𝜕 𝑏 + 𝑏 + 𝜔 = ℏ𝜕 𝑏 + 𝑏 − 𝑏 + 𝜔 𝑏 + 1 2 𝜔 = = ℏ𝜕 𝑏 + 𝑏 − 𝑏 + + 𝐼 + ℏ𝜕 𝜔 = 𝑏 + 𝐹 + ℏ𝜕 𝜔 = ℏ𝜕 + 1 𝐼 2 + 1 = 𝐹 + ℏ𝜕 ( 𝑏 + 𝜔) 2 Q.E.D. If 𝜔 𝑦 satisfies TISE with energy 𝐹 , then 𝑏 − 𝜔 Lemma 2 also satisfies TISE, with energy 𝐹 − ℏ𝜕 . (similar proof) Now main trick ℏ𝜕 If we have one solution 𝜔 𝑦 , we can construct ℏ𝜕 many other solutions (ladder of solutions ) ℏ𝜕 𝐹 Process of going down should stop somewhere ( 𝐹 ≥ 0 ) ℏ𝜕 Then 𝑏 − 𝜔 0 = 0 (further derivation is simple) ℏ𝜕

Ground state of harmonic oscillator 1 ℏ 𝑒 𝑒𝜔 0 𝜖𝑦 = − 𝑛𝜕 𝑏 − 𝜔 0 = 0 ⇒ 𝑒𝑦 + 𝑛𝜕𝑦 𝜔 0 𝑦 = 0 ⇒ ℏ 𝑦𝜔 0 2ℏ𝑛𝜕 𝜔 0 𝑦 = 𝐵 exp − 𝑛𝜕 2ℏ 𝑦 2 Solution 1 4 𝜔 0 𝑦 = 𝑛𝜕 exp − 𝑛𝜕 Find 𝐵 from normalization 2ℏ 𝑦 2 𝜌ℏ ground state 𝐼𝜔 0 = ℏ𝜕 𝑏 − + 𝐼 = ℏ𝜕( 𝑏 + 1 2) What is its energy? 2 𝜔 0 𝐹 0 = 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 𝑏 + 𝑜 exp − 𝑛𝜕 1 2ℏ 𝑦 2 𝜔 𝑜 𝑦 = 𝐵 𝑜 𝑜 = 0, 1, 2, … 𝐹 𝑜 = 𝑜 + 2 ℏ𝜕 𝐵 𝑜 is normalization constant Useful relations: 𝑏 + 𝜔 𝑜 𝑦 = 𝑜 + 1 𝜔 𝑜+1 (𝑦) therefore 𝑏 − 𝜔 𝑜 𝑦 = 𝑜 𝜔 𝑜−1 (𝑦) 𝜔 𝑜 = 1 𝑏 + 𝑜 𝜔 0 𝑜! A few lowest levels: 1 4 𝜔 0 𝑦 = 𝑛𝜕 exp − 𝑛𝜕 2ℏ 𝑦 2 𝜌ℏ 1 4 𝜔 1 𝑦 = 𝑛𝜕 2𝑛𝜕 𝑦 exp − 𝑛𝜕 2ℏ 𝑦 2 𝜌ℏ ℏ 1 4 𝜔 2 𝑦 = 𝑛𝜕 2 𝑛𝜕 2 exp − 𝑛𝜕 ℏ 𝑦 2 − 2ℏ 𝑦 2 𝜌ℏ 2

Stationary states of harmonic oscillator Fig. 2.7 from Griffiths’ book

General stationary states for harmonic oscillator 𝜔 𝑜 = 1 𝑜 + 1 𝑏 + 𝑜 𝜔 0 𝐹 𝑜 = 2 ℏ𝜕 𝑜! Explicitly 𝐼 𝑜 (𝜊) exp − 𝜊 2 𝜔 𝑜 𝑦 = 𝑛𝜕 1 4 1 𝜌ℏ 2 2 𝑜 𝑜! 𝑛𝜕 𝜊 = 𝜌ℏ 𝑦 𝐼 𝑜 (𝜊) – Hermite polynomials General solution of SE ∞ Ψ 𝑦, 𝑢 = 𝑜=0 𝑑 𝑜 𝜔 𝑜 𝑦 exp −𝑗 𝑜 + 1 2 𝜕𝑢 𝑑 𝑜 2 = 1 ∞ 𝑜=0

Free particle − ℏ 2 d 2 𝜔 𝑊 𝑦 = 0 TISE 𝑒𝑦 2 = 𝐹𝜔 2𝑛 2𝑛𝐹 𝜔 𝑦 = 𝐵𝑓 𝑗𝑙𝑦 + 𝐶𝑓 −𝑗𝑙𝑦 𝑙 = solution: ℏ 𝑙 – “wave vector”, 𝑙 = 2𝜌 𝜇 with time dependence: Ψ 𝑦, 𝑢 = 𝐵 exp 𝑗𝑙 𝑦 − ℏ𝑙 + 𝐶 exp −𝑗𝑙 𝑦 + ℏ𝑙 2𝑛 𝑢 2𝑛 𝑢 propagates to the right propagates to the left with velocity 𝑤 𝑞ℎ = ℏ𝑙 with the same velocity 2𝑛 2𝐹 𝑛 = ℏ𝑙 𝑤 classical = interesting that classically (twice larger) 𝑛 (will discuss later, difference between phase and group velocities) Not normalizable! What to do? Construct a wave packet.

Wave packet ∞ 𝜚(𝑙) exp 𝑗 𝑙𝑦 − ℏ𝑙 2 1 𝑙 > 0 to the right Ψ 𝑦, 𝑢 = 2𝑛 𝑢 𝑒𝑙 𝑙 < 0 to the left 2𝜌 −∞ energy is not well-defined ∞ 1 At 𝑢 = 0 𝜚(𝑙) 𝑓 𝑗𝑙𝑦 𝑒𝑙 Ψ 𝑦, 0 = just a Fourier transform 2𝜌 −∞ Remind Fourier transform: ∞ ∞ 1 1 𝐺(𝑙) 𝑓 𝑗𝑙𝑦 𝑒𝑙 𝑔(𝑦) = 𝑔(𝑦) 𝑓 −𝑗𝑙𝑦 𝑒𝑦 𝐺(𝑙) = 2𝜌 2𝜌 −∞ −∞ So if we know Ψ(𝑦, 0) , then can find 𝜚 𝑙 , and then know Ψ(𝑦, 𝑢) ∞ 1 Ψ(𝑦, 0) 𝑓 −𝑗𝑙𝑦 𝑒𝑦 𝜚(𝑙) = 2𝜌 −∞ Normalization: ∞ ∞ 2 𝑒𝑦 = 1 𝜚(𝑙) 2 𝑒𝑙 = 1 Ψ 𝑦, 0 is equivalent to −∞ −∞ 𝜚 𝑙 is like wavefunction in k -space

Phase and group velocities 𝜕 = ℏ𝑙 2 ∞ 1 Ψ 𝑦, 𝑢 = 𝜚(𝑙) exp 𝑗 𝑙𝑦 − 𝜕𝑢 𝑒𝑙 2𝑛 2𝜌 −∞ 𝜚(𝑙) Assume that 𝜚(𝑙) is concentrated around 𝑙 0 , then         𝜕 𝑙 ≈ 𝜕 0 + 𝑒𝜕 𝑒𝑙 𝑙 − 𝑙 0 = 𝜕 0 + 𝜕 ′ Δ𝑙 𝑙  𝑙 0 0 ∞ 1 𝜚(𝑙 0 + Δ𝑙) exp 𝑗 𝑙 0 + Δ𝑙 𝑦 − 𝑗 𝜕 0 + 𝜕 ′ Δ𝑙 𝑢 𝑒Δ𝑙 Ψ 𝑦, 𝑢 ≈  2𝜌 −∞  ∞ 1 exp 𝑗𝑙 0 𝑦 − 𝜕 0 𝜚(𝑙 0 + Δ𝑙) exp 𝑗Δ𝑙 𝑦 − 𝜕 ′ 𝑢 = 𝑢 𝑒Δ𝑙 𝑙 0 2𝜌 −∞ group velocity: 𝑤 𝑕𝑠 = 𝜕 ′ = 𝑒𝜕 𝑒𝑙 phase velocity: 𝑤 𝑞ℎ = 𝜕 0 𝑙 0 𝑤 𝑕𝑠 = 𝑒𝜕 𝑒𝑙 = ℏ𝑙 So, in quantum case 𝑛 = 𝑤 classical In our case 𝑤 𝑕𝑠 = 2𝑤 𝑞ℎ ℏ𝑙 2 (2𝑛) 𝜕 = 𝑤 𝑞ℎ = 𝜕 𝑙 = ℏ𝑙 (shape faster than ripples) (For waves on water 𝑤 𝑞ℎ = 2𝑤 𝑕𝑠 ) 2𝑛

Another normalization Actually, it is difficult to work with wavepackets, so in practice people usually work with extended waves 1 2𝜌 e 𝑗𝑙𝑦 𝜔 𝑙 𝑦 = This function satisfies “normalization” ∞ ∗ 𝑦 𝜔 𝑙 ′ 𝑦 𝑒𝑦 = 𝜀(𝑙 − 𝑙 ′ ) 𝜔 𝑙 −∞ Some properties of delta-function 𝜀(𝑦) ∞ ∞ 𝑓 𝑗𝑏𝑦 𝑒𝑦 = 2𝜌 𝜀(𝑏) 𝑔 𝑦 𝜀 𝑦 𝑒𝑦 = 𝑔(0) −∞ −∞ ∞ (from Fourier transform theorem) 𝑔 𝑦 𝜀 𝑦 − 𝑏 𝑒𝑦 = 𝑔(𝑏) −∞