EE201/MSE207 Lecture 3 Time-independent Schrdinger equation = 2 - - PowerPoint PPT Presentation

β–Ά
ee201 mse207 lecture 3 time independent schr dinger
SMART_READER_LITE
LIVE PREVIEW

EE201/MSE207 Lecture 3 Time-independent Schrdinger equation = 2 - - PowerPoint PPT Presentation

Nov 15, 2022 112 likes β€’235 views

EE201/MSE207 Lecture 3 Time-independent Schrdinger equation = 2 2 , (, ) SE (time- + , (, ) dependent) 2 2 If , = () (i.e.,

slide-1
SLIDE 1

will be energy, now just a constant

EE201/MSE207 Lecture 3

SE (time- dependent)

𝑗ℏ πœ–Ξ¨(𝑦, 𝑒) πœ–π‘’ = βˆ’ ℏ2 2𝑛 πœ–2Ξ¨ 𝑦, 𝑒 πœ–π‘¦2 + π‘Š 𝑦, 𝑒 Ξ¨(𝑦, 𝑒)

Time-independent SchrΓΆdinger equation

If π‘Š 𝑦, 𝑒 = π‘Š(𝑦) (i.e., potential energy does not change with time), then simplification: SE can be solved by separation of variables Assume

Ξ¨ 𝑦, 𝑒 = πœ” 𝑦 𝑔(𝑒)

πœ–Ξ¨ πœ–π‘’ = πœ” 𝑦 𝑒𝑔 𝑒𝑒 , πœ–2Ξ¨ πœ–π‘¦2 = 𝑒2πœ” 𝑒𝑦2 𝑔 𝑒 ,

𝑗ℏ πœ” 𝑒𝑔 𝑒𝑒 = βˆ’ ℏ2 2𝑛 d2πœ” 𝑒𝑦2 𝑔 + π‘Š 𝑦 πœ”π‘”

Γ— 1 πœ” 𝑔 ⟹

𝑗ℏ 1 𝑔 𝑒𝑔 𝑒𝑒 = βˆ’ ℏ2 2𝑛 1 πœ” d2πœ” 𝑒𝑦2 + π‘Š 𝑦 = const

Now two equations (in time and in space)

= 𝐹

slide-2
SLIDE 2

𝑒𝑔 𝑒𝑒 = βˆ’π‘— 𝐹 ℏ 𝑔

⟹

𝑗ℏ 1 𝑔 𝑒𝑔 𝑒𝑒 = βˆ’ ℏ2 2𝑛 1 πœ” d2πœ” 𝑒𝑦2 + π‘Š 𝑦 = 𝐹

First equation:

𝑔 𝑒 = 𝐷 𝑓

βˆ’π‘—πΉ ℏ 𝑒

Can choose 𝐷 = 1 𝐹 must be real (for normalization)

Ξ¨ 𝑦, 𝑒 = πœ” 𝑦 𝑔(𝑒) 𝑔 𝑒 = exp βˆ’π‘— 𝐹 ℏ 𝑒

Second equation:

βˆ’ ℏ2 2𝑛 d2πœ” 𝑦 𝑒𝑦2 + π‘Š 𝑦 πœ” 𝑦 = 𝐹 πœ” 𝑦

Time-independent SchrΓΆdinger equation (TISE)

πΌπœ” = 𝐹 πœ”

(this hints that E is energy)

Ξ¨ 𝑦, 𝑒 = πœ” 𝑦 exp βˆ’π‘— 𝐹 ℏ 𝑒

Normalization

βˆ’βˆž ∞

Ξ¨ 2𝑒𝑦 = 1,

while exp βˆ’π‘— 𝐹 ℏ 𝑒 = 1 therefore

βˆ’βˆž ∞

πœ”(𝑦) 2𝑒𝑦 = 1 πœ” normalized in the same way as Ξ¨

slide-3
SLIDE 3

𝑅 =

βˆ’βˆž ∞ πœ” 𝑦 𝑓 βˆ’π‘—πΉπ‘’ ℏ βˆ—π‘… 𝑦, βˆ’π‘—β„ πœ– πœ–π‘¦ [πœ” 𝑦 𝑓 βˆ’π‘—πΉπ‘’ ℏ] 𝑒𝑦

For any observable 𝑅(𝑦, π‘ž) , the average 𝑅(𝑦, βˆ’π‘—β„ πœ– πœ–π‘¦) does not depend on time β‡’ stationary state time-dependence cancels

Ξ¨ 𝑦, 𝑒 = πœ” 𝑦 exp βˆ’π‘— 𝐹 ℏ 𝑒

Stationary state

Let us show that E is energy

πΌπœ” = πΉπœ”

Average energy:

𝐼 =

βˆ’βˆž ∞ πœ”βˆ—

πΌπœ” 𝑒𝑦 =

βˆ’βˆž ∞ πœ”βˆ—πΉπœ” 𝑒𝑦 =

= 𝐹

βˆ’βˆž ∞ πœ” 2 𝑒𝑦 = 𝐹

Similarly

𝐼2 =

βˆ’βˆž ∞ πœ”βˆ—

𝐼 πΌπœ” 𝑒𝑦 =

βˆ’βˆž ∞ πœ”βˆ—πΉπΉπœ” 𝑒𝑦 = 𝐹2

No variance ⟹ energy is E (definite energy, β€œwell-defined” energy, rare case in QM)

(H is operator of total energy)

slide-4
SLIDE 4

TISE solutions (infinitely many): SE is linear ⟹ any linear combination of solutions is a solution

βˆ’ ℏ2 2𝑛 d2πœ”π‘œ 𝑒𝑦2 + π‘Š(𝑦) πœ”π‘œ = πΉπ‘œ πœ”π‘œ Ξ¨ 𝑦, 𝑒 =

π‘œ=1 ∞

π‘‘π‘œ πœ”π‘œ 𝑦 exp βˆ’π‘— πΉπ‘œ ℏ 𝑒 π‘œ = 1, 2, 3, …

General solution of SE (for time-independent π‘Š(𝑦))

Any complex coefficients π‘‘π‘œ ; however, Ξ¨ should be normalized. As we will see, with normalized πœ”π‘œ, this requires normalization for π‘‘π‘œ as

π‘œ=1 ∞

π‘‘π‘œ 2 = 1

slide-5
SLIDE 5

Theorem

Stationary states πœ”π‘œ(𝑦) and πœ”π‘› 𝑦 with different energies,

πΉπ‘œ β‰  𝐹𝑛 , are orthogonal:

βˆ’βˆž ∞ πœ”π‘› βˆ— 𝑦 πœ”π‘œ 𝑦 𝑒𝑦 = 0.

Orthogonality of stationary states

Remarks: - orthonormal (since normalized)

  • we treat functions as vectors, inner product (generalized dot-product)

𝑔 𝑕 ≑

βˆ’βˆž ∞ π‘”βˆ— 𝑦 𝑕 𝑦 𝑒𝑦, orthogonal: 𝑔 𝑕 = 0

Proof Start with βˆ’ ℏ2

2𝑛 d2πœ”π‘œ 𝑒𝑦2 + π‘Š πœ”π‘œ = πΉπ‘œ πœ”π‘œ

Γ— πœ”π‘›

βˆ— , integrate: βˆ’ ℏ2

2𝑛 πœ”π‘› βˆ— d2πœ”π‘œ 𝑒𝑦2 𝑒𝑦 + πœ”π‘› βˆ— π‘Šπœ”π‘œπ‘’π‘¦ = πΉπ‘œ πœ”π‘› βˆ— πœ”π‘œπ‘’π‘¦

exchange π‘œ ↔ 𝑛,

conjugate:

βˆ’ ℏ2

2𝑛 πœ”π‘œ d2πœ”π‘›

βˆ—

𝑒𝑦2 𝑒𝑦 + πœ”π‘œπ‘Šπœ”π‘› βˆ— 𝑒𝑦 = 𝐹𝑛 πœ”π‘œπœ”π‘› βˆ— 𝑒𝑦

equal (by parts) equal

β‡’ also equal

(πΉπ‘œβˆ’πΉπ‘›) πœ”π‘›

βˆ— πœ”π‘œπ‘’π‘¦ = 0

If πΉπ‘œ β‰  𝐹𝑛, then πœ”π‘›

βˆ— πœ”π‘œπ‘’π‘¦ = 0

Q.E.D. If energies are the same, usually also choose orthogonal functions πœ”π‘œ(𝑦)

slide-6
SLIDE 6

Normalization for coefficients π‘‘π‘œ

Use orthogonality of πœ”π‘œ

Ξ¨ 𝑦, 𝑒 =

π‘œ=1 ∞

π‘‘π‘œ πœ”π‘œ 𝑦 exp βˆ’π‘— πΉπ‘œ ℏ 𝑒 1 =

βˆ’βˆž ∞

Ξ¨(𝑦, 𝑒) 2𝑒𝑦 =

π‘œ,𝑛

𝑑𝑛

βˆ— π‘‘π‘œ exp βˆ’π‘— πΉπ‘œ βˆ’ 𝐹𝑛

ℏ 𝑒

βˆ’βˆž ∞

πœ”π‘›

βˆ— 𝑦 πœ”π‘œ 𝑦 𝑒𝑦

  • nly n=m

=

π‘œ

π‘‘π‘œ 2 Γ— 1 Γ—

βˆ’βˆž ∞

πœ”π‘œ 𝑦

2𝑒𝑦 = π‘œ

π‘‘π‘œ 2

This is how we obtain normalization for π‘‘π‘œ:

π‘œ=1 ∞

π‘‘π‘œ 2 = 1

slide-7
SLIDE 7

General solution for evolution of Ξ¨(𝑦, 𝑒)

Ξ¨ 𝑦, 𝑒 =

π‘œ=1 ∞

π‘‘π‘œ πœ”π‘œ 𝑦 exp βˆ’π‘— πΉπ‘œ ℏ 𝑒 πœ€π‘œπ‘›

Suppose we know wave function at 𝑒 = 0: Ξ¨(𝑦, 0) Then we can find coefficients π‘‘π‘œ in Ξ¨ 𝑦, 0 = π‘œ π‘‘π‘œπœ”π‘œ 𝑦 Again use orthogonality: multiply by πœ”π‘›

βˆ— (𝑦) and integrate

πœ”π‘›

βˆ— 𝑦 Ξ¨ 𝑦, 0 𝑒𝑦 = π‘œ π‘‘π‘œ πœ”π‘› βˆ— 𝑦 πœ”π‘œ 𝑦 𝑒𝑦 = 𝑑𝑛

Therefore

π‘‘π‘œ =

βˆ’βˆž ∞ πœ”π‘œ βˆ— 𝑦 Ξ¨ 𝑦, 0 𝑒𝑦

Then we can find Ξ¨(𝑦, 𝑒) at any time 𝑒 :

slide-8
SLIDE 8

Example: Infinite square well

𝑙 = 2𝑛𝐹 ℏ

(one of the most important systems in this course) Boundary conditions: x

a π‘Š 𝑦 = 0, 0 ≀ 𝑦 ≀ 𝑏 ∞,

  • therwise

(similar to quantum well in semiconductors, except different effective mass and finite depth) Solve TISE inside

βˆ’ ℏ2 2𝑛 d2πœ” 𝑦 𝑒𝑦2 = 𝐹 πœ” 𝑦 β‡’ πœ” 𝑦 = 𝐡 sin 𝑙𝑦 + 𝐢 cos(𝑙𝑦) πœ” 0 = 0 πœ” 𝑏 = 0

(because πœ” 𝑦 = 0 outside and πœ”(𝑦) is always continuous)

πœ” 𝑦 = 𝐡 sin π‘™π‘œπ‘¦

therefore

π‘™π‘œ = πœŒπ‘œ 𝑏 , π‘œ = 1, 2, 3, … πΉπ‘œ = ℏ2π‘™π‘œ

2

2𝑛 = π‘œ2𝜌2ℏ2 2𝑛𝑏2 π‘Š(𝑦)

slide-9
SLIDE 9

Infinite square well

x

a πœ”π‘œ 𝑦 = 𝐡 sin π‘™π‘œπ‘¦ , π‘™π‘œ = π‘œπœŒ 𝑏 πΉπ‘œ = ℏ2π‘™π‘œ

2

2𝑛 = π‘œ2𝜌2ℏ2 2𝑛𝑏2 π‘œ = 1, 2, 3, …

Normalization

1 =

𝑏 πœ” 𝑦 2𝑒𝑦 = 𝐡 2 1 2 𝑏 β‡’ 𝐡 =

2 𝑏

πœ”π‘œ 𝑦 = 2 𝑏 sin π‘œπœŒ 𝑏 𝑦

(because sin2(𝑦) = 1/2)

n=1 – lowest energy,

β€œground state”

nο‚³2 – β€œexcited states”

ο‚· Similar to atom states ο‚· Even lowest state has positive energy (understanding

from uncertainty principle)

π‘Š(𝑦)

1 ο€­   οƒ— οƒ—  οƒ— οƒ— οƒ—

n=1 n=2 n=3 a πœ”π‘œ(𝑦)

slide-10
SLIDE 10

Infinite square well: Summary

x

a

πΉπ‘œ = π‘œ2𝜌2ℏ2 2𝑛𝑏2

π‘œ = 1, 2, 3, …

πœ”π‘œ 𝑦 = 2 𝑏 sin π‘œπœŒ 𝑏 𝑦

n=1 – lowest energy,

β€œground state”

nο‚³2 – β€œexcited states” π‘Š(𝑦) π‘™π‘œ = π‘œπœŒ 𝑏

ο€­    οƒ— οƒ— 4   οƒ— οƒ— 9  οƒ— οƒ—  οƒ—  οƒ— 

n=1 n=2 n=3 𝐹1 𝐹2 𝐹3

1 ο€­   οƒ— οƒ—  οƒ— οƒ— οƒ—

n=1 n=2 n=3 a πœ”π‘œ(𝑦)

slide-11
SLIDE 11

Infinite square well: properties of stationary states

x

a π‘œ = 1, 2, 3, … πœ”π‘œ 𝑦 = 2 𝑏 sin π‘œπœŒ 𝑏 𝑦 π‘Š(𝑦) ο‚· orthogonal to each other,

𝑏 πœ”π‘› βˆ— 𝑦 πœ”π‘œ(𝑦) 𝑒𝑦 = πœ€π‘›π‘œ

ο‚· complete set, any 𝑔 𝑦 can be represented as 𝑔 𝑦 = π‘œ=1

∞

π‘‘π‘œ πœ”π‘œ 𝑦

if 𝑔 0 = 𝑔 𝑏 = 0 How to find π‘‘π‘œ? Already know the trick:

πœ”π‘›

βˆ— 𝑦 𝑔(𝑦) 𝑒𝑦 = π‘œ π‘‘π‘œ πœ”π‘› βˆ— 𝑦 πœ”π‘œ 𝑦 𝑒𝑦 = 𝑑𝑛

Therefore

π‘‘π‘œ =

𝑏

πœ”π‘œ

βˆ— 𝑦 𝑔(𝑦) 𝑒𝑦

(i.e., orthonormal basis)

1 ο€­   οƒ— οƒ—  οƒ— οƒ— οƒ—

n=1 n=2 n=3 a πœ”π‘œ(𝑦)

slide-12
SLIDE 12

General solution

x

a π‘Š(𝑦)

ο€­    οƒ— οƒ— 4   οƒ— οƒ— 9  οƒ— οƒ—  οƒ—  οƒ— 

n=1 n=2 n=3 𝐹1 𝐹2 𝐹3 Ξ¨ 𝑦, 𝑒 =

π‘œ=1 ∞

π‘‘π‘œ 2 𝑏 sin π‘œπœŒ 𝑏 𝑦 exp βˆ’π‘— π‘œ2𝜌2ℏ 2𝑛𝑏2 𝑒 π‘œ=1

∞

π‘‘π‘œ 2 = 1

If only one coefficient π‘‘π‘œ is non-zero, then stationary state If more than one level occupied (β€œsuperposition”), then non-stationary, state evolves in time (interference) Classical motion can be represented as combination

  • f many high-numbered states (transition to classical

picture when energies are ≫ 𝐹1,2,3) Is a general (superposition) state localized in space? – No Is it localized in energy? – Also no! However, if we measure energy, then we will get some result: some πΉπ‘œ (nothing in between!) If measure again – will get the same result (collapse)