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Feb 16, 2024 377 likes β€’534 views

EE201/MSE207 Lecture 6 Finite square well: scattering states > 0 = 0 , < < 0 , > width 2 , depth 0 0 2 2 2 +

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SLIDE 1

EE201/MSE207 Lecture 6 Finite square well: scattering states

width 2𝑏, depth π‘Š 𝑏 βˆ’π‘ βˆ’π‘Š

π‘Š 𝑦 = βˆ’π‘Š

0 ,

βˆ’π‘ < 𝑦 < 𝑏 0 , 𝑦 > 𝑏

𝐹 > 0 TISE

βˆ’ ℏ2 2𝑛 𝑒2πœ” 𝑒𝑦2 + π‘Š(𝑦)πœ” = πΉπœ”

Now 𝐹 is given (any energy is possible) Again 3 regions:

𝑦 < βˆ’π‘, 𝑙 = 2𝑛𝐹 ℏ

(definition of 𝑙 as for free particle)

𝑦 < 𝑏, πœ” 𝑦 = 𝐷 sin(π‘šπ‘¦) + 𝐸 cos(π‘šπ‘¦), π‘š = 2𝑛 𝐹 + π‘Š

0 /ℏ (as before)

𝑦 > 𝑏,

4 boundary conditions, 6 variables (𝐹 is given, no normalization) No hope to find unique solution. But there should not be a unique solution! Let us focus on physical meaning (important to find a proper question).

πœ” 𝑦 = 𝐡 𝑓𝑗𝑙𝑦 + πΆπ‘“βˆ’π‘—π‘™π‘¦, πœ” 𝑦 = 𝐺 𝑓𝑗𝑙𝑦 + 𝐻 π‘“βˆ’π‘—π‘™π‘¦

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SLIDE 2

Add time dependence

𝑦 < βˆ’π‘, Ξ¨ 𝑦, 𝑒 = 𝐡 𝑓

𝑗𝑙 𝑦 βˆ’ ℏ𝑙 2𝑛 𝑒 + 𝐢 𝑓 βˆ’π‘—π‘™ 𝑦 + ℏ𝑙 2𝑛 𝑒

incident wave

  • utgoing wave

(as for free particle; different phase and group velocities, but the same direction)

𝐡 𝐢 Similarly for 𝑦 > 𝑏 𝐡 𝐢 𝐺 𝐻

(if necessary, wave packets can be constructed later; in reality nobody usually does it because it is too complicated; instead, people work with unnormalized states)

Assume that the wave is incident from the left, then 𝐻 = 0 𝐡 𝐢 𝐺 𝐡 is incident wave amplitude 𝐢 is reflected wave amplitude 𝐺 is transmitted wave amplitude We have 5 variables (𝐡, 𝐢, 𝐷, 𝐸, 𝐺) and 4 equations. Equations are linear. Can express 𝐢, 𝐷, 𝐸, 𝐺 as functions of 𝐡 (incident amplitude).

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SLIDE 3

𝐡 𝐢 𝐺 𝐡 is incident wave amplitude 𝐢 is reflected wave amplitude 𝐺 is transmitted wave amplitude

Proper questions

Goal: find ratios 𝑠 = 𝐢

𝐡

and 𝑒 = 𝐺

𝐡

(these ratios are called reflection and transmission amplitudes)

(assume a wave incident from the left)

Reflection coefficient (probability of reflection)

𝑆 = 𝑠 2 = 𝐢 2 𝐡 2

Transmission coefficient (probability of transmission)

π‘ˆ = 𝑒 2 = 𝐺 2 𝐡 2

From physical meaning

π‘ˆ + 𝑆 = 1

Remark 1. Definition of π‘ˆ is sometimes different (discuss later, Γ— 𝑀𝑠/π‘€π‘š) Remark 2. Terminology: Reflection/transmission amplitudes (𝑠, 𝑒) and coefficients (𝑆, π‘ˆ) Remark 3. We defined 𝑆 and π‘ˆ as ratios; they become probabilities for wave packets (possible to show). Quadratic because probability ∝ Ξ¨ 2.

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SLIDE 4

𝐡 𝐢 𝐺

Finding π‘ˆ and 𝑆

Boundary conditions: Simple to exclude 𝐷 and 𝐸 (similar combinations), then 2 equations with 𝐡, 𝐢, 𝐺

𝑦 < βˆ’π‘, πœ” 𝑦 = 𝐡 𝑓𝑗𝑙𝑦 + 𝐢 π‘“βˆ’π‘—π‘™π‘¦ 𝑦 < 𝑏, πœ” 𝑦 = 𝐷 sin(π‘šπ‘¦) + 𝐸 cos(π‘šπ‘¦) 𝑦 > 𝑏, πœ” 𝑦 = 𝐺 𝑓𝑗𝑙𝑦

𝑗𝑙 𝐡 π‘“βˆ’π‘—π‘™π‘ βˆ’ 𝐢𝑓𝑗𝑙𝑏 = π‘š [𝐷 cos π‘šπ‘ + 𝐸 sin π‘šπ‘ ] 𝐷 sin π‘šπ‘ + 𝐸 cos π‘šπ‘ = 𝐺 𝑓𝑗𝑙𝑏 𝐡 π‘“βˆ’π‘—π‘™π‘ + 𝐢𝑓𝑗𝑙𝑏 = βˆ’π· sin π‘šπ‘ + 𝐸 cos π‘šπ‘ π‘š [𝐷 cos π‘šπ‘ βˆ’ 𝐸 sin π‘šπ‘ = 𝑗𝑙 𝐺 𝑓𝑗𝑙𝑏

𝑦 = βˆ’π‘ 𝑦 = 𝑏

Finally 𝐺 = π‘“βˆ’2𝑗𝑙𝑏 cos 2π‘šπ‘ βˆ’ 𝑗 sin 2π‘šπ‘ 2π‘™π‘š (𝑙2 + π‘š2) 𝐡 𝐢 = 𝑗 sin 2π‘šπ‘ 2π‘™π‘š (π‘š2 βˆ’ 𝑙2) 𝐺 𝑙 = 2𝑛𝐹 ℏ π‘š = 2𝑛 𝐹 + π‘Š ℏ

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SLIDE 5

𝐡 𝐢 𝐺

Finding π‘ˆ and 𝑆

𝑦 < βˆ’π‘, πœ” 𝑦 = 𝐡 𝑓𝑗𝑙𝑦 + 𝐢 π‘“βˆ’π‘—π‘™π‘¦ 𝑦 > 𝑏, πœ” 𝑦 = 𝐺 𝑓𝑗𝑙𝑦

π‘ˆ = 𝐺 2 𝐡 2 = 1 1 + π‘Š

2

4𝐹(𝐹 + π‘Š

0) sin2 2𝑏

ℏ 2𝑛 𝐹 + π‘Š Transmission probability

βˆ’π‘Š 𝐹

Reflection probability 𝑆 = 1 βˆ’ π‘ˆ (too long from 𝐢 2/ 𝐡 2)

  • Remark. π‘ˆ = 1 if

2𝑏 ℏ 2𝑛(𝐹 + π‘Š

0) = π‘œπœŒ

⟺ 𝐹 + π‘Š

0 = π‘œ2𝜌2ℏ2

2𝑛 2𝑏 2 This is exactly the β€œsimple” energy quantization (in infinite well). Explanation: destructive interference of reflected waves (similar to anti-reflective coating with quarter-wavelength films).

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SLIDE 6

𝐢 𝐻 𝐺

Now wave incident from the right

𝑦 < βˆ’π‘, πœ” 𝑦 = 𝐢 π‘“βˆ’π‘—π‘™π‘¦ 𝑦 > 𝑏, πœ” 𝑦 = 𝐺 𝑓𝑗𝑙𝑦 + 𝐻 π‘“βˆ’π‘—π‘™π‘¦

βˆ’π‘Š 𝐹

π‘ˆ

r = 𝑒r 2 = 𝐢 2

𝐻 2 Similarly, we can find transmission and reflection coefficients In our case because of symmetry However, this is always true (for any potential π‘Š 𝑦 and possibly different masses) 𝑆r = 𝑠

r 2 = 𝐺 2

𝐻 2 π‘ˆ

r + 𝑆r = 1

π‘ˆ

r = π‘ˆ l = π‘ˆ

𝑆r = π‘ˆ

l = 𝑆

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SLIDE 7

π‘ˆ and 𝑆 in general case

Why? Probability current If πœ” 𝑦 = 𝐡 𝑓𝑗𝑙𝑦, then 𝑆 = 𝐢 2 𝐡 2 = 𝑠 2 π‘Š βˆ’βˆž π‘Š +∞ 𝐹 π‘Š 𝑦 π‘Š βˆ’βˆž β‰  π‘Š ∞ and/or 𝑛1 β‰  𝑛2

𝑛1 𝑛2 𝐡 𝐢 𝐺

π‘ˆ = 𝐺 2 𝐡 2 𝐹 βˆ’ π‘Š(+∞) 𝐹 βˆ’ π‘Š(βˆ’βˆž) 𝑛1 𝑛2 = 𝐺 2 𝐡 2 𝑙2/𝑛2 𝑙1/𝑛1 = 𝐺 2 𝐡 2 𝑀2 𝑀1 = 𝑒 2 𝑀2 𝑀1 π‘ˆ + 𝑆 = 1 𝐾 = 𝑗ℏ 2𝑛 πœ” π‘’πœ”βˆ— 𝑒𝑦 βˆ’ πœ”βˆ— π‘’πœ” 𝑒𝑦 𝐾 = 𝐡 2 ℏ𝑙 𝑛 = 𝐡 2𝑀

(remember that π‘›βˆ’1 π‘’πœ”

𝑒𝑦 is

continuous, not just

π‘’πœ” 𝑒𝑦)

The same velocity for reflection, but may be different for transmission

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SLIDE 8

Transmission/reflection for -potential

𝐡 𝑓𝑗𝑙𝑦 𝐺 𝑓𝑗𝑙𝑦 𝐢 π‘“βˆ’π‘—π‘™π‘¦ 𝐻 π‘“βˆ’π‘—π‘™π‘¦ π‘Š 𝑦 = 𝛽 πœ€ 𝑦

TISE

βˆ’ ℏ2 2𝑛 𝑒2πœ” 𝑒𝑦2 + π‘Š(𝑦)πœ” = πΉπœ”

𝐹 > 0 Integrate TISE near zero,

βˆ’ ℏ2 2𝑛 πœ”β€² 𝜁 βˆ’ πœ”β€² βˆ’πœ + 𝛽 πœ”(0) = 0

βˆ’πœ 𝜁

… β‡’

𝜁 β†’ 0

π‘’πœ”(+0) 𝑒𝑦 βˆ’ π‘’πœ” βˆ’0 𝑒𝑦 = 2𝑛𝛽 ℏ2 πœ”(0)

With -potential, π‘’πœ”/𝑒𝑦 has a step (not continuous). Boundary conditions 𝐡 + 𝐢 = 𝐺 + 𝐻 𝑗𝑙 𝐺 βˆ’ 𝐻 βˆ’ 𝑗𝑙(𝐡 βˆ’ 𝐢) = 2𝑛𝛽 ℏ2 (𝐡 + 𝐢) If 𝐻 = 0 (incident from the left), then 𝐢 𝐡 = βˆ’π‘— 𝑛𝛽 ℏ2𝑙 1 + 𝑗 𝑛𝛽 ℏ2𝑙 , 𝐺 𝐡 = 1 1 + 𝑗 𝑛𝛽 ℏ2𝑙

𝑙 = 2𝑛𝐹/ℏ

𝛽 πœ€ 𝑦

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SLIDE 9

Scattering matrix (now waves incident from both sides)

Suppose we found transmission/reflection amplitudes (π‘’π‘š, 𝑠

π‘š) for the wave

incident from the left and also from the right (𝑒𝑠, 𝑠

𝑠).

It is convenient to write these 4 complex numbers as a 2 Γ— 2 matrix. For simplicity assume π‘Š βˆ’βˆž = π‘Š ∞ , 𝑛1 = 𝑛2 𝑦 β†’ βˆ’βˆž Out of 4 wave amplitudes (𝐡, 𝐢, 𝐺, 𝐻), 2 free parameters, and the other 2 can be calculated (linear relations) π‘Š 𝑦

𝐡 𝐢 𝐺 𝐻

𝑦 β†’ ∞ πœ” 𝑦 = 𝐡𝑓𝑗𝑙𝑦 + πΆπ‘“βˆ’π‘—π‘™π‘¦ πœ” 𝑦 = 𝐺𝑓𝑗𝑙𝑦 + π»π‘“βˆ’π‘—π‘™π‘¦ Why 2 free parameters? 1) it was 2 in rectangular well 2) TISE is a second-order dif. eq. οƒž 2 boundary conditions

𝑇 = 𝑠

π‘š

𝑒𝑠 π‘’π‘š 𝑠

𝑠

= 𝑇11 𝑇12 𝑇21 𝑇22

What is the meaning? (scattering matrix)

𝐢 𝐺 = 𝑇 𝐡 𝐻

(outgoing via incoming)

[not included into this course]

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SLIDE 10

Scattering matrix (S-matrix)

For simplicity assume π‘Š βˆ’βˆž = π‘Š ∞ , 𝑛1 = 𝑛2 𝑦 β†’ βˆ’βˆž π‘Š 𝑦

𝐡 𝐢 𝐺 𝐻

𝑦 β†’ ∞ πœ” 𝑦 = 𝐡𝑓𝑗𝑙𝑦 + πΆπ‘“βˆ’π‘—π‘™π‘¦ πœ” 𝑦 = 𝐺𝑓𝑗𝑙𝑦 + π»π‘“βˆ’π‘—π‘™π‘¦

𝑇 = 𝑠

π‘š

𝑒𝑠 π‘’π‘š 𝑠

𝑠

= 𝑇11 𝑇12 𝑇21 𝑇22

What is the meaning?

𝐢 𝐺 = 𝑇 𝐡 𝐻

(outgoing via incoming) Suppose 𝐻 = 0, then

𝐢 𝐺 = 𝑠

π‘š 𝐡

π‘’π‘šπ΅

Suppose A = 0, then

𝐢 𝐺 = 𝑒𝑠 𝐻 𝑠

𝑠𝐻

π‘ˆ

π‘š = π‘’π‘š 2 = 𝑇21 2, π‘†π‘š = 𝑠 π‘š 2 = 𝑇11 2

π‘ˆ

𝑠 = 𝑒𝑠 2 = 𝑇12 2, 𝑆𝑠 = 𝑠 𝑠 2 = 𝑇22 2

(remember that formulas for π‘ˆ in general case are different)

Let us prove symmetry:

π‘†π‘š = 𝑆𝑠 π‘ˆπ‘š = π‘ˆ

𝑠

(for brevity will use notation: π‘’π‘š = 𝑒, 𝑠

π‘š = 𝑠)

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SLIDE 11

Symmetry of the scattering matrix

Our proof will use β€œgraphical operations” with solutions of TISE

𝐡 = 1 𝑠 𝑒

TISE

Conjugate solution of TISE π‘ βˆ— 1 π‘’βˆ— Γ— βˆ’1 π‘ βˆ—

+

οƒž

𝑠 βˆ’ 1 π‘ βˆ— 𝑒 βˆ’ π‘’βˆ— π‘ βˆ— Now multiply by

βˆ’π‘ βˆ— π‘’βˆ—

οƒŸ

βˆ’π‘ βˆ— π‘’βˆ— 𝑠 βˆ’ 1 π‘ βˆ— =

βˆ’π‘ βˆ— 𝑒 π‘’βˆ— 1

= 1 βˆ’ |𝑠|2 π‘’βˆ— = 𝑒 2 π‘’βˆ— = 𝑒 therefore

𝑒𝑠 = 𝑒 𝑠

𝑠 = βˆ’π‘ βˆ— 𝑒

π‘’βˆ— 𝑇 = 𝑠 𝑒 𝑒 βˆ’π‘ βˆ— 𝑒 π‘’βˆ— οƒž π‘ˆ

𝑠 = π‘ˆπ‘š

𝑆𝑠 = π‘†π‘š

(conjugation = β€œtime reversal”)

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SLIDE 12

Symmetry of the S-matrix in general case

Without derivation, just a result

𝑇 = 𝑠 𝑒 𝑀2 𝑀1 𝑒 βˆ’π‘ βˆ— 𝑒 π‘’βˆ— οƒž π‘ˆ

𝑠 = π‘ˆπ‘š

𝑆𝑠 = π‘†π‘š

π‘Š βˆ’βˆž β‰  π‘Š ∞ , 𝑛1 β‰  𝑛2

𝑠

𝑠 = βˆ’π‘ βˆ— 𝑒

π‘’βˆ—

Still But

𝑒𝑠 = 1 βˆ’ 𝑠 2 π‘’βˆ— = 𝑒 2 𝑀2 𝑀1 π‘’βˆ— = 𝑒 𝑀2 𝑀1

(velocity 𝑀1 at the left, 𝑀2 at the right)

𝑀2 𝑀1 = 𝑙2/𝑛2 𝑙1/𝑛1

Then

π‘ˆ

𝑠 = 𝑒𝑠 2 𝑀1

𝑀2 = 𝑒 2 𝑀2 𝑀1 = π‘ˆπ‘š 𝑆𝑠 = π‘†π‘š

still

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SLIDE 13

Transfer matrix (M-matrix or T-matrix)

[also not included into this course]

Sometimes instead of

𝐢 𝐺 = 𝑇 𝐡 𝐻

𝐡 𝐢 𝐺 𝐻 it is more convenient to use

𝐺 𝐻 = 𝑁 𝐡 𝐢

right left (sometimes notation π‘ˆ instead of 𝑁)

If we know 𝑇, then it is easy to calculate 𝑁, and vice versa. Why 𝑁 is convenient? 𝑁1 𝑁2

𝑁total = 𝑁2 𝑁1

𝑁1 𝑁2 π‘π‘œ

𝑁total = π‘π‘œ π‘π‘œβˆ’1. . . 𝑁1

For a multi-barrier structure, all 𝑁𝑗 are similar, therefore it is simple to calculate 𝑁total. (Actually, each 𝑁𝑗 also contains a phase factor, depending on x-position.)

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SLIDE 14

M-matrix: symmetries and relation to S-matrix

Symmetries of M-matrix

𝐢 𝐺 = 𝑇11 𝑇12 𝑇21 𝑇22 𝐡 𝐻

1) 𝑁22 = 𝑁11

βˆ—

(simple case, 𝑀1 = 𝑀2) 2) 𝑁12 = 𝑁21

βˆ—

3) det

𝑁 = 1

(can be derived from symmetries of S-matrix)

𝐺 𝐻 = 𝑁11 𝑁12 𝑁21 𝑁22 𝐡 𝐢 𝑇11 = βˆ’ 𝑁21 𝑁22 = βˆ’ 𝑁12

βˆ—

𝑁22

Conversion

𝑇22 = 𝑁12 𝑁22 𝑇12 = 𝑇21 = 1 𝑁22 𝑁11 = 1 𝑇12

βˆ— = 1

𝑇21

βˆ—

𝑁22 = 1 𝑇12 𝑁12 = βˆ’ 𝑇11

βˆ—

𝑇12

βˆ— = 𝑇22

𝑇12 𝑁21 = βˆ’ 𝑇11 𝑇12