segre/phys406/15S/lecture_02.pdf · Ideal gas example Treat an ideal gas as a 3D particle in an in...

Today’s Outline - January 15, 2015

• The ideal gas

• Black body radiation

• Perturbation theory

• First order PT

• First order PT examples

Reading Assignment: Chapter 6.2

Homework Assignment #01:Chapter 5: 3,4,7,11,16,34due Tuesday, January 20, 2015

C. Segre (IIT) PHYS 406 - Spring 2015 January 15, 2015 1 / 21

• The ideal gas

• First order PT

• The ideal gas

• First order PT

• The ideal gas

• First order PT

• The ideal gas

• First order PT

• The ideal gas

• First order PT

• The ideal gas

• First order PT

• The ideal gas

• First order PT

• The ideal gas

• First order PT

Ideal gas example

Treat an ideal gas as a 3D particle in an infinite square well

assuming non-interacting particles,the energy is

as before, we have a volume π3/Vper state in reciprocal space and thenumber of states in a shell of thick-ness dk is simply the degeneracy ofthe state with energy Ek

for distinguishable particles

Ek =~2

(πnxlx,πnyly,πnzlz

4πk2dk

(π3/V )

2π2k2dk

Nn= dne−(α+βEn)

and the constraint that N =∑

Nn becomes

2π2e−α

∫ ∞0

k2e−β~2k2/2mdk

= Ve−α(

2πβ~2

e−α =N

(2πβ~2

Ideal gas example

Treat an ideal gas as a 3D particle in an infinite square wellassuming non-interacting particles,the energy is

Ek =~2

4πk2dk

(π3/V )

2π2k2dk

Nn= dne−(α+βEn)

Nn becomes

2π2e−α

∫ ∞0

k2e−β~2k2/2mdk

= Ve−α(

2πβ~2

e−α =N

(2πβ~2

Ideal gas example

Ek =~2

4πk2dk

(π3/V )

2π2k2dk

Nn= dne−(α+βEn)

Nn becomes

2π2e−α

∫ ∞0

k2e−β~2k2/2mdk

= Ve−α(

2πβ~2

e−α =N

(2πβ~2

Ideal gas example

Ek =~2

4πk2dk

(π3/V )

2π2k2dk

Nn= dne−(α+βEn)

Nn becomes

2π2e−α

∫ ∞0

k2e−β~2k2/2mdk

= Ve−α(

2πβ~2

e−α =N

(2πβ~2

Ideal gas example

Ek =~2

4πk2dk

(π3/V )

2π2k2dk

Nn= dne−(α+βEn)

Nn becomes

2π2e−α

∫ ∞0

k2e−β~2k2/2mdk

= Ve−α(

2πβ~2

e−α =N

(2πβ~2

Ideal gas example

Ek =~2

4πk2dk

(π3/V )

2π2k2dk

Nn= dne−(α+βEn)

Nn becomes

2π2e−α

∫ ∞0

k2e−β~2k2/2mdk

= Ve−α(

2πβ~2

e−α =N

(2πβ~2

Ideal gas example

Ek =~2

4πk2dk

(π3/V )=

2π2k2dk

Nn= dne−(α+βEn)

Nn becomes

2π2e−α

∫ ∞0

k2e−β~2k2/2mdk

= Ve−α(

2πβ~2

e−α =N

(2πβ~2

Ideal gas example

Ek =~2

4πk2dk

(π3/V )=

2π2k2dk

Nn= dne−(α+βEn)

Nn becomes

2π2e−α

∫ ∞0

k2e−β~2k2/2mdk

= Ve−α(

2πβ~2

e−α =N

(2πβ~2

Ideal gas example

Ek =~2

4πk2dk

(π3/V )=

2π2k2dk

Nn= dne−(α+βEn)

Nn becomes

2π2e−α

∫ ∞0

k2e−β~2k2/2mdk

= Ve−α(

2πβ~2

e−α =N

(2πβ~2

Ideal gas example

Ek =~2

4πk2dk

(π3/V )=

2π2k2dk

Nn= dne−(α+βEn)

Nn becomes

2π2e−α

∫ ∞0

k2e−β~2k2/2mdk

= Ve−α(

2πβ~2

e−α =N

(2πβ~2

Ideal gas example

Ek =~2

4πk2dk

(π3/V )=

2π2k2dk

Nn= dne−(α+βEn)

Nn becomes

2π2e−α

∫ ∞0

k2e−β~2k2/2mdk

= Ve−α(

2πβ~2

e−α =N

(2πβ~2

Ideal gas example

Ek =~2

4πk2dk

(π3/V )=

2π2k2dk

Nn= dne−(α+βEn)

Nn becomes

2π2e−α

∫ ∞0

k2e−β~2k2/2mdk = Ve−α

2πβ~2

e−α =N

(2πβ~2

Ideal gas example

Ek =~2

4πk2dk

(π3/V )=

2π2k2dk

Nn= dne−(α+βEn)

Nn becomes

2π2e−α

∫ ∞0

k2e−β~2k2/2mdk = Ve−α

2πβ~2

e−α =N

(2πβ~2

Significance of α and β

The constraint on the total energy, E =∑

NnEn becomes, in this context

2π2e−α

∫ ∞0

k4e−β~2k2/2mdk

2βe−α

2πβ~2

substituting for e−α obtained pre-viously

this is similar to the classical ther-modynamic expression for energyper molecule

we postulate that β is related to thetemperature

while α is related to the chemicalpotential

e−α =N

(2πβ~2

µ(T ) ≡ −αkBT

2π2e−α

∫ ∞0

k4e−β~2k2/2mdk

2βe−α

2πβ~2

e−α =N

(2πβ~2

µ(T ) ≡ −αkBT

2π2e−α

∫ ∞0

k4e−β~2k2/2mdk =

2βe−α

2πβ~2

e−α =N

(2πβ~2

µ(T ) ≡ −αkBT

2π2e−α

∫ ∞0

k4e−β~2k2/2mdk =

2βe−α

2πβ~2

e−α =N

(2πβ~2

µ(T ) ≡ −αkBT

2π2e−α

∫ ∞0

k4e−β~2k2/2mdk =

2βe−α

2πβ~2

e−α =N

(2πβ~2

µ(T ) ≡ −αkBT

2π2e−α

∫ ∞0

k4e−β~2k2/2mdk =

2βe−α

2πβ~2

e−α =N

(2πβ~2

µ(T ) ≡ −αkBT

2π2e−α

∫ ∞0

k4e−β~2k2/2mdk =

2βe−α

2πβ~2

e−α =N

(2πβ~2

µ(T ) ≡ −αkBT

2π2e−α

∫ ∞0

k4e−β~2k2/2mdk =

2βe−α

2πβ~2

e−α =N

(2πβ~2

µ(T ) ≡ −αkBT

2π2e−α

∫ ∞0

k4e−β~2k2/2mdk =

2βe−α

2πβ~2

e−α =N

(2πβ~2

µ(T ) ≡ −αkBT

2π2e−α

∫ ∞0

k4e−β~2k2/2mdk =

2βe−α

2πβ~2

e−α =N

(2πβ~2

µ(T ) ≡ −αkBT

2π2e−α

∫ ∞0

k4e−β~2k2/2mdk =

2βe−α

2πβ~2

e−α =N

(2πβ~2

µ(T ) ≡ −αkBT

Particle distributions

It is conventional to divide the expressions for number of particles in astate by the degeneracy of the state to get an occupation fraction for eachstate and use the reduced energies ε = En/kBT

dn= n(ε) =

[e(ε−µ)/kBT ]−1, Maxwell− Boltzmann

[e(ε−µ)/kBT + 1]−1, Fermi− Dirac

[e(ε−µ)/kBT − 1]−1, Bose− Einstein

if we apply the same constraints as before to identical fermions or bosonsin the 3-D infinite well, we obtain

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

2π2~2

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

dn= n(ε) =

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

2π2~2

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

dn= n(ε) =

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

2π2~2

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

dn= n(ε) =

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

2π2~2

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

dn= n(ε) =

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

2π2~2

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

dn= n(ε) =

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

2π2~2

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

dn= n(ε) =

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

2π2~2

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

dn= n(ε) =

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

2π2~2

∫ ∞0

e [(~2k2/2m)−µ]/kBT ± 1dk

Comparison of statistics

The three statistics lead to very different behavior in occupation number

0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3

1 K50 K300 K

Maxwell-Boltzmann

0.9 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3

Fermi-Dirac

0 0.2 0.4 0.6 0.8 1 1.2 1.4

1 K50 K

Bose-Einstein

Maxwell-Boltzmann: n(ε)→∞ as ε < µ and exponential behavior withtemperature

Bose-Einstein: occupation number is always infinite for ε < µ

Fermi-Dirac: n(ε) ≡ 1 as T → 0

0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3

1 K50 K300 K

Maxwell-Boltzmann

0.9 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3

Fermi-Dirac

0 0.2 0.4 0.6 0.8 1 1.2 1.4

1 K50 K

Bose-Einstein

0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3

1 K50 K300 K

Maxwell-Boltzmann

0.9 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3

Fermi-Dirac

0 0.2 0.4 0.6 0.8 1 1.2 1.4

1 K50 K

Bose-Einstein

0.4 0.5 0.6 0.7 0.8 0.9 1 1.1 1.2 1.3

1 K50 K300 K

Maxwell-Boltzmann

0.9 0.95 1 1.05 1.1 1.15 1.2 1.25 1.3

Fermi-Dirac

0 0.2 0.4 0.6 0.8 1 1.2 1.4

1 K50 K

Bose-Einstein

Blackbody spectrum

Photons are bosons and so now we can use our understanding of quantumstatistics to compute the distribution of phonons in a rectangular box withthe following assumptions

1 The energy of a photon is E = hν = ~ω2 The wave number is k = 2π/λ = ω/c

3 Only two spin states occur, s = 1 but m = ±1

4 The number of photons is NOT conserved

since the number of photons is notconserved, we can set α → 0 andwe have

the degeneracy in terms of ω is

the energy density, Nω~ω/V in theinterval dω is

Nω =dk

e~ω/kBT − 1

π2c3ω2dω

ρ(ω)dω =~ω3dω

π2c3(e~ω/kBT − 1)

Blackbody spectrum

1 The energy of a photon is E = hν = ~ω

2 The wave number is k = 2π/λ = ω/c

Nω =dk

e~ω/kBT − 1

π2c3ω2dω

ρ(ω)dω =~ω3dω

Blackbody spectrum

Nω =dk

e~ω/kBT − 1

π2c3ω2dω

ρ(ω)dω =~ω3dω

Blackbody spectrum

Nω =dk

e~ω/kBT − 1

π2c3ω2dω

ρ(ω)dω =~ω3dω

Blackbody spectrum

Nω =dk

e~ω/kBT − 1

π2c3ω2dω

ρ(ω)dω =~ω3dω

Blackbody spectrum

Nω =dk

e~ω/kBT − 1

π2c3ω2dω

ρ(ω)dω =~ω3dω

Blackbody spectrum

Nω =dk

e~ω/kBT − 1

π2c3ω2dω

ρ(ω)dω =~ω3dω

Blackbody spectrum

Nω =dk

e~ω/kBT − 1

π2c3ω2dω

ρ(ω)dω =~ω3dω

Blackbody spectrum

Nω =dk

e~ω/kBT − 1

π2c3ω2dω

ρ(ω)dω =~ω3dω

Blackbody spectrum

Nω =dk

e~ω/kBT − 1

π2c3ω2dω

ρ(ω)dω =~ω3dω

Blackbody spectrum

Nω =dk

e~ω/kBT − 1

π2c3ω2dω

ρ(ω)dω =~ω3dω

Blackbody spectrum

0 0.5 1 1.5 2 2.5 3

nsity (

Wavelength (µm)

5000 K

4000 K

3000 K

ρ(ω) =~ω3

Universal curves with reducedtemperature

As ω → 0 (λ → ∞) dropsoff as λ−3 avoiding ultravioletcatastrophe

As ω → ∞ (λ → 0) dropsto zero exponentially, movingpeak position up in λ

Blackbody spectrum

0 0.5 1 1.5 2 2.5 3

nsity (

Wavelength (µm)

5000 K

4000 K

3000 K

ρ(ω) =~ω3

Blackbody spectrum

0 0.5 1 1.5 2 2.5 3

nsity (

Wavelength (µm)

5000 K

4000 K

3000 K

ρ(ω) =~ω3

Blackbody spectrum

0 0.5 1 1.5 2 2.5 3

nsity (

Wavelength (µm)

5000 K

4000 K

3000 K

ρ(ω) =~ω3

Fundamentals of perturbation theory

Most problems in quantum mechanics are not exactly soluble. It isimportant, therefore to develop approximation methods. One of the mostuseful is perturbation theory.

consider a Hamiltonian which is ex-actly soluble and has a complete setof orthonormal, non-degenerate so-lutions

suppose we have a “real” potentialwhich consists of a slight perturba-tion of the exactly soluble system

the solutions to this potential canbe determined using the solution ofthe unperturbed system (they are acomplete set)

H0ψ0n = E 0

〈ψ0n|ψ0

m〉 = δnm

H = H0 + H ′

Hψn = Enψn

ψn =∑m

c(n)m ψ0

H0ψ0n = E 0

〈ψ0n|ψ0

m〉 = δnm

H = H0 + H ′

Hψn = Enψn

ψn =∑m

c(n)m ψ0

H0ψ0n = E 0

〈ψ0n|ψ0

m〉 = δnm

H = H0 + H ′

Hψn = Enψn

ψn =∑m

c(n)m ψ0

H0ψ0n = E 0

〈ψ0n|ψ0

m〉 = δnm

H = H0 + H ′

Hψn = Enψn

ψn =∑m

c(n)m ψ0

H0ψ0n = E 0

〈ψ0n|ψ0

m〉 = δnm

H = H0 + H ′

Hψn = Enψn

ψn =∑m

c(n)m ψ0

H0ψ0n = E 0

〈ψ0n|ψ0

m〉 = δnm

H = H0 + H ′

Hψn = Enψn

ψn =∑m

c(n)m ψ0

H0ψ0n = E 0

〈ψ0n|ψ0

m〉 = δnm

H = H0 + H ′

Hψn = Enψn

ψn =∑m

c(n)m ψ0

H0ψ0n = E 0

〈ψ0n|ψ0

m〉 = δnm

H = H0 + H ′

Hψn = Enψn

ψn =∑m

c(n)m ψ0

First order perturbation theory

Begin by writing the full Hamil-tonian as a sum,

where λ issmall.

Use this same parameter to ex-pand the the energy and wave-function of the full Hamiltonian.

Using these three equations, wewrite the Schrodinger equationas.

H = H0 + λH ′

ψn = ψ0n + λψ1

n + λ2ψ2n + · · ·

En = E 0n + λE 1

n + λ2E 2n + · · ·

Hψn = Enψn

(H0 + λH ′

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]Expanding this equation and grouping the same powers of λ

Begin by writing the full Hamil-tonian as a sum, where λ issmall.

H = H0 + λH ′

ψn = ψ0n + λψ1

n + λ2ψ2n + · · ·

En = E 0n + λE 1

n + λ2E 2n + · · ·

Hψn = Enψn

(H0 + λH ′

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

H = H0 + λH ′

ψn = ψ0n + λψ1

n + λ2ψ2n + · · ·

En = E 0n + λE 1

n + λ2E 2n + · · ·

Hψn = Enψn

(H0 + λH ′

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

H = H0 + λH ′

ψn = ψ0n + λψ1

n + λ2ψ2n + · · ·

En = E 0n + λE 1

n + λ2E 2n + · · ·

Hψn = Enψn

(H0 + λH ′

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

H = H0 + λH ′

ψn = ψ0n + λψ1

n + λ2ψ2n + · · ·

En = E 0n + λE 1

n + λ2E 2n + · · ·

Hψn = Enψn

(H0 + λH ′

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

H = H0 + λH ′

ψn = ψ0n + λψ1

n + λ2ψ2n + · · ·

En = E 0n + λE 1

n + λ2E 2n + · · ·

Hψn = Enψn

(H0 + λH ′

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

H = H0 + λH ′

ψn = ψ0n + λψ1

n + λ2ψ2n + · · ·

En = E 0n + λE 1

n + λ2E 2n + · · ·

Hψn = Enψn

(H0 + λH ′

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

H = H0 + λH ′

ψn = ψ0n + λψ1

n + λ2ψ2n + · · ·

En = E 0n + λE 1

n + λ2E 2n + · · ·

Hψn = Enψn

(H0 + λH ′

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

H = H0 + λH ′

ψn = ψ0n + λψ1

n + λ2ψ2n + · · ·

En = E 0n + λE 1

n + λ2E 2n + · · ·

Hψn = Enψn

(H0 + λH ′

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

Expanding this equation and grouping the same powers of λ

H = H0 + λH ′

ψn = ψ0n + λψ1

n + λ2ψ2n + · · ·

En = E 0n + λE 1

n + λ2E 2n + · · ·

Hψn = Enψn

(H0 + λH ′

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

Isolating the first order correction

Each order of λ results in a separate equation

(H0 + λH ′

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]H0ψ0

n = E 0nψ

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

H0ψ2n + H ′ψ1

n = E 0nψ

2n + E 1

nψ1n + E 2

H0ψ3n + H ′ψ2

n = E 0nψ

3n + E 1

nψ2n + E 2

nψ1n + +E 4

Each order of λ results in a separate equation(H0 + λH ′

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

H0ψ0n = E 0

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

H0ψ2n + H ′ψ1

n = E 0nψ

2n + E 1

nψ1n + E 2

H0ψ3n + H ′ψ2

n = E 0nψ

3n + E 1

nψ2n + E 2

nψ1n + +E 4

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]H0ψ0

n = E 0nψ

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

H0ψ2n + H ′ψ1

n = E 0nψ

2n + E 1

nψ1n + E 2

H0ψ3n + H ′ψ2

n = E 0nψ

3n + E 1

nψ2n + E 2

nψ1n + +E 4

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]H0ψ0

n = E 0nψ

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

H0ψ2n + H ′ψ1

n = E 0nψ

2n + E 1

nψ1n + E 2

H0ψ3n + H ′ψ2

n = E 0nψ

3n + E 1

nψ2n + E 2

nψ1n + +E 4

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]H0ψ0

n = E 0nψ

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

H0ψ2n + H ′ψ1

n = E 0nψ

2n + E 1

nψ1n + E 2

H0ψ3n + H ′ψ2

n = E 0nψ

3n + E 1

nψ2n + E 2

nψ1n + +E 4

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]=(E 0n + λE 1

n + λ2E 2n + · · ·

) [ψ0n + λψ1

n + λ2ψ2n + · · ·

]H0ψ0

n = E 0nψ

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

H0ψ2n + H ′ψ1

n = E 0nψ

2n + E 1

nψ1n + E 2

H0ψ3n + H ′ψ2

n = E 0nψ

3n + E 1

nψ2n + E 2

nψ1n + +E 4

First order energy correction

Taking the inner product of the first order equation with⟨ψ0n

∣∣

⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= ��

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟩H0 is hermitian, so

which cancels the identicalterm on the right

and by definition of the un-perturbed basis set

this gives us an expression forthe first order perturbationtheory energy correction

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

⟩Thus, the first order energy correction is simply the expectation value ofthe perturbing Hamiltonian in the unperturbed eigenfunction.

∣∣⟨ψ0n

∣∣H0ψ1n

+⟨ψ0n

∣∣H ′ψ0n

⟩= ��

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

∣∣⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

= ��E 0n

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

∣∣⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= E 0

⟨ψ0n

∣∣ψ1n

+ E 1n

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

∣∣⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

H0 is hermitian, so

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

∣∣⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

∣∣⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

∣∣⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

= E 0n

⟨ψ0n

∣∣ψ1n

⟩⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

∣∣⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

∣∣��⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= ��

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

∣∣��⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= ��

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

∣∣��⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= ��

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩⟨ψ0n

∣∣ψ0n

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

∣∣��⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= ��

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

∣∣��⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= ��

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

∣∣��⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= ��

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

Thus, the first order energy correction is simply the expectation value ofthe perturbing Hamiltonian in the unperturbed eigenfunction.

∣∣��⟨ψ0n

∣∣H0ψ1n

⟩+⟨ψ0n

∣∣H ′ψ0n

⟩= ��

⟨ψ0n

∣∣ψ1n

⟩+ E 1

⟨ψ0n

∣∣ψ0n

⟨ψ0n

∣∣H0ψ1n

⟩=⟨H0ψ0

∣∣ψ1n

⟩= E 0

⟨ψ0n

∣∣ψ1n

⟩⟨ψ0n

∣∣ψ0n

⟩= 1

E 1n =

⟨ψ0n

∣∣H ′∣∣ψ0n

First order eigenfuction correction

We can use the same first or-der equation (in λ) to derivean expression for the first or-der correction to the eigen-function

Assuming a solution for ψ1n

in terms of the complete un-perturbed eigenfunction ba-sis set

Pulling the operators into thesum

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

(H0 − E 0

)ψ1n = −

(H ′ − E 1

ψ1n =

∑m 6=n

c(n)m ψ0

(H0 − E 0

)∑m 6=n

c(n)m ψ0

m = −(H ′ − E 1

)ψ0n∑

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

(H0 − E 0

)ψ1n = −

(H ′ − E 1

ψ1n =

∑m 6=n

c(n)m ψ0

(H0 − E 0

)∑m 6=n

c(n)m ψ0

m = −(H ′ − E 1

)ψ0n∑

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

(H0 − E 0

)ψ1n = −

(H ′ − E 1

ψ1n =

∑m 6=n

c(n)m ψ0

(H0 − E 0

)∑m 6=n

c(n)m ψ0

m = −(H ′ − E 1

)ψ0n∑

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

(H0 − E 0

)ψ1n = −

(H ′ − E 1

ψ1n =

∑m 6=n

c(n)m ψ0

(H0 − E 0

)∑m 6=n

c(n)m ψ0

m = −(H ′ − E 1

)ψ0n∑

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

(H0 − E 0

)ψ1n = −

(H ′ − E 1

ψ1n =

∑m 6=n

c(n)m ψ0

(H0 − E 0

)∑m 6=n

c(n)m ψ0

m = −(H ′ − E 1

)ψ0n∑

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

(H0 − E 0

)ψ1n = −

(H ′ − E 1

ψ1n =

∑m 6=n

c(n)m ψ0

(H0 − E 0

)∑m 6=n

c(n)m ψ0

m = −(H ′ − E 1

∑m 6=n

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

(H0 − E 0

)ψ1n = −

(H ′ − E 1

ψ1n =

∑m 6=n

c(n)m ψ0

(H0 − E 0

)∑m 6=n

c(n)m ψ0

m = −(H ′ − E 1

∑m 6=n

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

(H0 − E 0

)ψ1n = −

(H ′ − E 1

ψ1n =

∑m 6=n

c(n)m ψ0

(H0 − E 0

)∑m 6=n

c(n)m ψ0

m = −(H ′ − E 1

)ψ0n∑

(E 0m − E 0

)c(n)m ψ0

= −(H ′ − E 1

H0ψ1n + H ′ψ0

n = E 0nψ

1n + E 1

(H0 − E 0

)ψ1n = −

(H ′ − E 1

ψ1n =

∑m 6=n

c(n)m ψ0

(H0 − E 0

)∑m 6=n

c(n)m ψ0

m = −(H ′ − E 1

)ψ0n∑

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

First order eigenfuction correction∑m 6=n

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

Taking the inner product with⟨ψ0k

∣∣∑m 6=n

(E 0m − E 0

)c(n)m

⟨ψ0k |ψ0

⟩= −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩+ E 1

⟨ψ0k |ψ0

⟩If k = n, we simply get theenergy correction again, butif k 6= n, then only the m =k term survives in the sum

The first order correction tothe eigenfunction is thus

(E 0k − E 0

)c(n)k = −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩c(n)m =

⟨ψ0m |H ′|ψ0

⟩E 0n − E 0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

This holds as long the the unperturbed energy spectrum is non-degenerate

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

∣∣

∑m 6=n

(E 0m − E 0

)c(n)m

⟨ψ0k |ψ0

⟩= −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩+ E 1

⟨ψ0k |ψ0

(E 0k − E 0

)c(n)k = −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩c(n)m =

⟨ψ0m |H ′|ψ0

⟩E 0n − E 0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

∣∣∑m 6=n

(E 0m − E 0

)c(n)m

⟨ψ0k |ψ0

= −⟨ψ0k

∣∣H ′∣∣ψ0n

⟩+ E 1

⟨ψ0k |ψ0

(E 0k − E 0

)c(n)k = −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩c(n)m =

⟨ψ0m |H ′|ψ0

⟩E 0n − E 0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

∣∣∑m 6=n

(E 0m − E 0

)c(n)m

⟨ψ0k |ψ0

⟩= −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩+ E 1

⟨ψ0k |ψ0

If k = n, we simply get theenergy correction again, butif k 6= n, then only the m =k term survives in the sum

(E 0k − E 0

)c(n)k = −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩c(n)m =

⟨ψ0m |H ′|ψ0

⟩E 0n − E 0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

∣∣∑m 6=n

(E 0m − E 0

)c(n)m

⟨ψ0k |ψ0

⟩= −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩+ E 1

⟨ψ0k |ψ0

(E 0k − E 0

)c(n)k = −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩c(n)m =

⟨ψ0m |H ′|ψ0

⟩E 0n − E 0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

∣∣∑m 6=n

(E 0m − E 0

)c(n)m

⟨ψ0k |ψ0

⟩= −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩+ E 1

⟨ψ0k |ψ0

(E 0k − E 0

)c(n)k

= −⟨ψ0k

∣∣H ′∣∣ψ0n

⟩c(n)m =

⟨ψ0m |H ′|ψ0

⟩E 0n − E 0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

∣∣∑m 6=n

(E 0m − E 0

)c(n)m

⟨ψ0k |ψ0

⟩= −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩+ E 1

⟨ψ0k |ψ0

(E 0k − E 0

)c(n)k = −

⟨ψ0k

∣∣H ′∣∣ψ0n

c(n)m =

⟨ψ0m |H ′|ψ0

⟩E 0n − E 0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

∣∣∑m 6=n

(E 0m − E 0

)c(n)m

⟨ψ0k |ψ0

⟩= −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩+ E 1

⟨ψ0k |ψ0

(E 0k − E 0

)c(n)k = −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩c(n)k

⟨ψ0m |H ′|ψ0

⟩E 0n − E 0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

∣∣∑m 6=n

(E 0m − E 0

)c(n)m

⟨ψ0k |ψ0

⟩= −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩+ E 1

⟨ψ0k |ψ0

(E 0k − E 0

)c(n)k = −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩c(n)k =

⟨ψ0k |H ′|ψ0

⟩E 0n − E 0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

∣∣∑m 6=n

(E 0m − E 0

)c(n)m

⟨ψ0k |ψ0

⟩= −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩+ E 1

⟨ψ0k |ψ0

(E 0k − E 0

)c(n)k = −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩c(n)m =

⟨ψ0m |H ′|ψ0

⟩E 0n − E 0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

∣∣∑m 6=n

(E 0m − E 0

)c(n)m

⟨ψ0k |ψ0

⟩= −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩+ E 1

⟨ψ0k |ψ0

(E 0k − E 0

)c(n)k = −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩c(n)m =

⟨ψ0m |H ′|ψ0

⟩E 0n − E 0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

∣∣∑m 6=n

(E 0m − E 0

)c(n)m

⟨ψ0k |ψ0

⟩= −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩+ E 1

⟨ψ0k |ψ0

(E 0k − E 0

)c(n)k = −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩c(n)m =

⟨ψ0m |H ′|ψ0

⟩E 0n − E 0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

(E 0m − E 0

)c(n)m ψ0

m = −(H ′ − E 1

∣∣∑m 6=n

(E 0m − E 0

)c(n)m

⟨ψ0k |ψ0

⟩= −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩+ E 1

⟨ψ0k |ψ0

(E 0k − E 0

)c(n)k = −

⟨ψ0k

∣∣H ′∣∣ψ0n

⟩c(n)m =

⟨ψ0m |H ′|ψ0

⟩E 0n − E 0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

Example 6.1

Constant potential, V0 in aninfinite square well from x =0 to x = a

The unperturbed wavefunc-tions are

ψ0n(x) =

asin(nπ

the perturbing potential isH ′ = V0 and the first orderenergy correction is

all the terms in the sum van-ish and the eigenfunction so-lution is exact

E 1n = 〈ψ0

n |V0|ψ0n〉

= V0〈ψ0n | ψ0

n〉 = V0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

ψ0m = 0

ψn ≡ ψ0n

Example 6.1

ψ0n(x) =

asin(nπ

E 1n = 〈ψ0

n |V0|ψ0n〉

= V0〈ψ0n | ψ0

n〉 = V0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

ψ0m = 0

ψn ≡ ψ0n

Example 6.1

ψ0n(x) =

asin(nπ

E 1n = 〈ψ0

n |V0|ψ0n〉

= V0〈ψ0n | ψ0

n〉 = V0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

ψ0m = 0

ψn ≡ ψ0n

Example 6.1

ψ0n(x) =

asin(nπ

the perturbing potential isH ′ = V0

and the first orderenergy correction is

E 1n = 〈ψ0

n |V0|ψ0n〉

= V0〈ψ0n | ψ0

n〉 = V0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

ψ0m = 0

ψn ≡ ψ0n

Example 6.1

ψ0n(x) =

asin(nπ

E 1n = 〈ψ0

n |V0|ψ0n〉

= V0〈ψ0n | ψ0

n〉 = V0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

ψ0m = 0

ψn ≡ ψ0n

Example 6.1

ψ0n(x) =

asin(nπ

E 1n = 〈ψ0

n |V0|ψ0n〉

= V0〈ψ0n | ψ0

n〉 = V0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

ψ0m = 0

ψn ≡ ψ0n

Example 6.1

ψ0n(x) =

asin(nπ

E 1n = 〈ψ0

n |V0|ψ0n〉 = V0〈ψ0

n | ψ0n〉

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

ψ0m = 0

ψn ≡ ψ0n

Example 6.1

ψ0n(x) =

asin(nπ

E 1n = 〈ψ0

n |V0|ψ0n〉 = V0〈ψ0

n | ψ0n〉 = V0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

ψ0m = 0

ψn ≡ ψ0n

Example 6.1

ψ0n(x) =

asin(nπ

E 1n = 〈ψ0

n |V0|ψ0n〉 = V0〈ψ0

n | ψ0n〉 = V0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

ψn ≡ ψ0n

Example 6.1

ψ0n(x) =

asin(nπ

all the terms in the sum van-ish

and the eigenfunction so-lution is exact

E 1n = 〈ψ0

n |V0|ψ0n〉 = V0〈ψ0

n | ψ0n〉 = V0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

ψn ≡ ψ0n

Example 6.1

ψ0n(x) =

asin(nπ

all the terms in the sum van-ish

and the eigenfunction so-lution is exact

E 1n = 〈ψ0

n |V0|ψ0n〉 = V0〈ψ0

n | ψ0n〉 = V0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

ψ0m = 0

ψn ≡ ψ0n

Example 6.1

ψ0n(x) =

asin(nπ

E 1n = 〈ψ0

n |V0|ψ0n〉 = V0〈ψ0

n | ψ0n〉 = V0

ψ1n =

∑m 6=n

〈ψ0m |H ′|ψ0

n〉(E 0

n − E 0m)

ψ0m = 0

ψn ≡ ψ0n

Example 6.1 (cont.)

Suppose, instead, that theperturbing potential extendsonly from 0 < x < a/2?

E 1n =

∫ a/2

(nπax)dx

∫ a/2

[1− cos

[x − a

2nπsin

)∣∣∣∣a/20

This is not an exact solution but the first term in a series of energycorrection terms.

The first order correction to the eigenfunction is an infinite sum of termsbut the first few are the most important.

Example 6.1 (cont.)

E 1n =

∫ a/2

(nπax)dx

∫ a/2

[1− cos

[x − a

2nπsin

)∣∣∣∣a/20

Example 6.1 (cont.)

E 1n =

∫ a/2

(nπax)dx

∫ a/2

[1− cos

[x − a

2nπsin

)∣∣∣∣a/20

Example 6.1 (cont.)

E 1n =

∫ a/2

(nπax)dx =

∫ a/2

[1− cos

[x − a

2nπsin

)∣∣∣∣a/20

Example 6.1 (cont.)

E 1n =

∫ a/2

(nπax)dx =

∫ a/2

[1− cos

[x − a

2nπsin

)∣∣∣∣a/20

Example 6.1 (cont.)

E 1n =

∫ a/2

(nπax)dx =

∫ a/2

[1− cos

[x − a

2nπsin

)∣∣∣∣a/20

Example 6.1 (cont.)

E 1n =

∫ a/2

(nπax)dx =

∫ a/2

[1− cos

[x − a

2nπsin

)∣∣∣∣a/20

Example 6.1 (cont.)

E 1n =

∫ a/2

(nπax)dx =

∫ a/2

[1− cos

[x − a

2nπsin

)∣∣∣∣a/20

Example 6.1 (cont.)

E 1n =

∫ a/2

(nπax)dx =

∫ a/2

[1− cos

[x − a

2nπsin

)∣∣∣∣a/20

Problem 6.2

For the harmonic oscillator[V (x) = (1/2)kx2

], the al-

lowed energies are

)~ω, (n = 0, 1, 2, . . . )

where ω =√

k/m is the classical frequency. Supposethe spring constant increases slightly: k → (1 + ε)k .

a Find the exact new energies. Expand your answer asa power series to second order in ε.

b Calculate the first-order perturbation in the energy.Compare your result with part (a).

Problem 6.2 (cont.)

a. The modified spring constantleads to a modified frequency

and a modified energy which can beexpanded using the binomial theo-rem

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

)b. The perturbing Hamil-tonian can be computedas

the first order energy cor-rection is thus

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

since 〈T 〉 = 〈V 〉 = E 0n /2 for a harmonic oscillator

Problem 6.2 (cont.)

ω′ ≡√

k(1 + ε)

= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

Problem 6.2 (cont.)

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

Problem 6.2 (cont.)

and a modified energy

which can beexpanded using the binomial theo-rem

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

Problem 6.2 (cont.)

and a modified energy

which can beexpanded using the binomial theo-rem

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

Problem 6.2 (cont.)

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

Problem 6.2 (cont.)

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

b. The perturbing Hamil-tonian can be computedas

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

Problem 6.2 (cont.)

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

Problem 6.2 (cont.)

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

H ′ =1

2kx2(1 + ε)− 1

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

Problem 6.2 (cont.)

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

Problem 6.2 (cont.)

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

Problem 6.2 (cont.)

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

Problem 6.2 (cont.)

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

Problem 6.2 (cont.)

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

Problem 6.2 (cont.)

ω′ ≡√

k(1 + ε)

m= ω√

1 + ε

)~ω√

1 + ε

En ≈(n +

2ε− 1

8ε2 + · · ·

H ′ =1

2kx2(1 + ε)− 1

2kx2 = ε

E 1n =

⟨ψ0n |εV |ψ0

⟩=ε

Problem 6.3

Two identical bosons are placed in an infinite square well.They interact weakly with one another, via the potential

V (x1, x2) = −aV0δ(x1 − x2)

(where V0 is a constant with dimensions of energy, anda is the width of the well).

a. Ignoring the interaction between particles, find the groundstate and the first excited state – both the wave functionsand the associated energies.

b. Use first-order perturbation theory to estimate the effect ofthe particle-particle interaction on the energies of theground state and the first excited state.

Problem 6.3 - solution

a. For bosons, the unperturbed ground state is the symmetrized wavefunction

and the unperturbed energy is

ψ01(x1, x2) = ψ1(x1)ψ1(x2) =

asin(πx1

)sin(πx2

)E 01 = 2E1 =

For the first excited state one of the bosons is in a higher state with energy

ψ02(x1, x2) =

[ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)]

[sin(πx1

(2πx2a

)+ sin

(2πx1a

)sin(πx2

)]E 02 = E1 + E2 =

ψ01(x1, x2) = ψ1(x1)ψ1(x2)

asin(πx1

)sin(πx2

)E 01 = 2E1 =

ψ02(x1, x2) =

[ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)]

[sin(πx1

(2πx2a

)+ sin

(2πx1a

)sin(πx2

)]E 02 = E1 + E2 =

ψ01(x1, x2) = ψ1(x1)ψ1(x2) =

asin(πx1

)sin(πx2

E 01 = 2E1 =

ψ02(x1, x2) =

[ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)]

[sin(πx1

(2πx2a

)+ sin

(2πx1a

)sin(πx2

)]E 02 = E1 + E2 =

a. For bosons, the unperturbed ground state is the symmetrized wavefunction and the unperturbed energy is

ψ01(x1, x2) = ψ1(x1)ψ1(x2) =

asin(πx1

)sin(πx2

E 01 = 2E1 =

ψ02(x1, x2) =

[ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)]

[sin(πx1

(2πx2a

)+ sin

(2πx1a

)sin(πx2

)]E 02 = E1 + E2 =

ψ01(x1, x2) = ψ1(x1)ψ1(x2) =

asin(πx1

)sin(πx2

)E 01 = 2E1

=π2~2

ψ02(x1, x2) =

[ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)]

[sin(πx1

(2πx2a

)+ sin

(2πx1a

)sin(πx2

)]E 02 = E1 + E2 =

ψ01(x1, x2) = ψ1(x1)ψ1(x2) =

asin(πx1

)sin(πx2

)E 01 = 2E1 =

ψ02(x1, x2) =

[ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)]

[sin(πx1

(2πx2a

)+ sin

(2πx1a

)sin(πx2

)]E 02 = E1 + E2 =

ψ01(x1, x2) = ψ1(x1)ψ1(x2) =

asin(πx1

)sin(πx2

)E 01 = 2E1 =

For the first excited state one of the bosons is in a higher state

with energy

ψ02(x1, x2) =

[ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)]

[sin(πx1

(2πx2a

)+ sin

(2πx1a

)sin(πx2

)]E 02 = E1 + E2 =

ψ01(x1, x2) = ψ1(x1)ψ1(x2) =

asin(πx1

)sin(πx2

)E 01 = 2E1 =

with energy

ψ02(x1, x2) =

[ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)]

[sin(πx1

(2πx2a

)+ sin

(2πx1a

)sin(πx2

)]E 02 = E1 + E2 =

ψ01(x1, x2) = ψ1(x1)ψ1(x2) =

asin(πx1

)sin(πx2

)E 01 = 2E1 =

with energy

ψ02(x1, x2) =

[ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)]

[sin(πx1

(2πx2a

)+ sin

(2πx1a

)sin(πx2

E 02 = E1 + E2 =

ψ01(x1, x2) = ψ1(x1)ψ1(x2) =

asin(πx1

)sin(πx2

)E 01 = 2E1 =

ψ02(x1, x2) =

[ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)]

[sin(πx1

(2πx2a

)+ sin

(2πx1a

)sin(πx2

E 02 = E1 + E2 =

ψ01(x1, x2) = ψ1(x1)ψ1(x2) =

asin(πx1

)sin(πx2

)E 01 = 2E1 =

ψ02(x1, x2) =

[ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)]

[sin(πx1

(2πx2a

)+ sin

(2πx1a

)sin(πx2

)]E 02 = E1 + E2

ψ01(x1, x2) = ψ1(x1)ψ1(x2) =

asin(πx1

)sin(πx2

)E 01 = 2E1 =

ψ02(x1, x2) =

[ψ1(x1)ψ2(x2) + ψ2(x1)ψ1(x2)]

[sin(πx1

(2πx2a

)+ sin

(2πx1a

)sin(πx2

)]E 02 = E1 + E2 =

b. The energy correction to first order is given by

E 11 =

⟨ψ01

∣∣H ′∣∣ψ01

⟩= (−aV0)

)2 ∫ a

0sin2(πx1

)sin2(πx2

)δ(x1 − x2) dx1dx2

= −4V0

0sin4(πx2

)dx2 = −4V0

a· aπ

∫ π

0sin4y dy

= −4V0

8− sin 2y

sin 4y

∣∣∣∣π0

= −3V0

The correction to the first excited state is

E 12 =

⟨ψ02

∣∣H ′∣∣ψ02

⟩= (−aV0)

)2 ∫ a

[sin(πx1

(2πx2a

(2πx1a

)sin(πx2

)]2δ(x1 − x2) dx1dx2

E 11 =

⟨ψ01

∣∣H ′∣∣ψ01

= (−aV0)

)2 ∫ a

0sin2(πx1

)sin2(πx2

)δ(x1 − x2) dx1dx2

= −4V0

0sin4(πx2

)dx2 = −4V0

a· aπ

∫ π

0sin4y dy

= −4V0

8− sin 2y

sin 4y

∣∣∣∣π0

= −3V0

E 12 =

⟨ψ02

∣∣H ′∣∣ψ02

⟩= (−aV0)

)2 ∫ a

[sin(πx1

(2πx2a

(2πx1a

)sin(πx2

)]2δ(x1 − x2) dx1dx2

E 11 =

⟨ψ01

∣∣H ′∣∣ψ01

⟩= (−aV0)

)2 ∫ a

0sin2(πx1

)sin2(πx2

)δ(x1 − x2) dx1dx2

= −4V0

0sin4(πx2

)dx2 = −4V0

a· aπ

∫ π

0sin4y dy

= −4V0

8− sin 2y

sin 4y

∣∣∣∣π0

= −3V0

E 12 =

⟨ψ02

∣∣H ′∣∣ψ02

⟩= (−aV0)

)2 ∫ a

[sin(πx1

(2πx2a

(2πx1a

)sin(πx2

)]2δ(x1 − x2) dx1dx2

E 11 =

⟨ψ01

∣∣H ′∣∣ψ01

⟩= (−aV0)

)2 ∫ a

0sin2(πx1

)sin2(πx2

)δ(x1 − x2) dx1dx2

= −4V0

0sin4(πx2

= −4V0

a· aπ

∫ π

0sin4y dy

= −4V0

8− sin 2y

sin 4y

∣∣∣∣π0

= −3V0

E 12 =

⟨ψ02

∣∣H ′∣∣ψ02

⟩= (−aV0)

)2 ∫ a

[sin(πx1

(2πx2a

(2πx1a

)sin(πx2

)]2δ(x1 − x2) dx1dx2

E 11 =

⟨ψ01

∣∣H ′∣∣ψ01

⟩= (−aV0)

)2 ∫ a

0sin2(πx1

)sin2(πx2

)δ(x1 − x2) dx1dx2

= −4V0

0sin4(πx2

)dx2 = −4V0

a· aπ

∫ π

0sin4y dy

= −4V0

8− sin 2y

sin 4y

∣∣∣∣π0

= −3V0

E 12 =

⟨ψ02

∣∣H ′∣∣ψ02

⟩= (−aV0)

)2 ∫ a

[sin(πx1

(2πx2a

(2πx1a

)sin(πx2

)]2δ(x1 − x2) dx1dx2

E 11 =

⟨ψ01

∣∣H ′∣∣ψ01

⟩= (−aV0)

)2 ∫ a

0sin2(πx1

)sin2(πx2

)δ(x1 − x2) dx1dx2

= −4V0

0sin4(πx2

)dx2 = −4V0

a· aπ

∫ π

0sin4y dy

= −4V0

8− sin 2y

sin 4y

∣∣∣∣π0

= −3V0

E 12 =

⟨ψ02

∣∣H ′∣∣ψ02

⟩= (−aV0)

)2 ∫ a

[sin(πx1

(2πx2a

(2πx1a

)sin(πx2

)]2δ(x1 − x2) dx1dx2

E 11 =

⟨ψ01

∣∣H ′∣∣ψ01

⟩= (−aV0)

)2 ∫ a

0sin2(πx1

)sin2(πx2

)δ(x1 − x2) dx1dx2

= −4V0

0sin4(πx2

)dx2 = −4V0

a· aπ

∫ π

0sin4y dy

= −4V0

8− sin 2y

sin 4y

∣∣∣∣π0

= −3V0

E 12 =

⟨ψ02

∣∣H ′∣∣ψ02

⟩= (−aV0)

)2 ∫ a

[sin(πx1

(2πx2a

(2πx1a

)sin(πx2

)]2δ(x1 − x2) dx1dx2

E 11 =

⟨ψ01

∣∣H ′∣∣ψ01

⟩= (−aV0)

)2 ∫ a

0sin2(πx1

)sin2(πx2

)δ(x1 − x2) dx1dx2

= −4V0

0sin4(πx2

)dx2 = −4V0

a· aπ

∫ π

0sin4y dy

= −4V0

8− sin 2y

sin 4y

∣∣∣∣π0

= −3V0

E 12 =

⟨ψ02

∣∣H ′∣∣ψ02

⟩= (−aV0)

)2 ∫ a

[sin(πx1

(2πx2a

(2πx1a

)sin(πx2

)]2δ(x1 − x2) dx1dx2

E 11 =

⟨ψ01

∣∣H ′∣∣ψ01

⟩= (−aV0)

)2 ∫ a

0sin2(πx1

)sin2(πx2

)δ(x1 − x2) dx1dx2

= −4V0

0sin4(πx2

)dx2 = −4V0

a· aπ

∫ π

0sin4y dy

= −4V0

8− sin 2y

sin 4y

∣∣∣∣π0

= −3V0

E 12 =

⟨ψ02

∣∣H ′∣∣ψ02

= (−aV0)

)2 ∫ a

[sin(πx1

(2πx2a

(2πx1a

)sin(πx2

)]2δ(x1 − x2) dx1dx2

E 11 =

⟨ψ01

∣∣H ′∣∣ψ01

⟩= (−aV0)

)2 ∫ a

0sin2(πx1

)sin2(πx2

)δ(x1 − x2) dx1dx2

= −4V0

0sin4(πx2

)dx2 = −4V0

a· aπ

∫ π

0sin4y dy

= −4V0

8− sin 2y

sin 4y

∣∣∣∣π0

= −3V0

E 12 =

⟨ψ02

∣∣H ′∣∣ψ02

⟩= (−aV0)

)2 ∫ a

[sin(πx1

(2πx2a

(2πx1a

)sin(πx2

)]2δ(x1 − x2) dx1dx2

E 12 = −2V0

[sin(πx2

(2πx2a

)+ sin

(2πx2a

)sin(πx2

)]2dx2

= −8V0

0sin2(πx2

(2πx2a

= −8V0

a· aπ

∫ π

0sin2y sin22y dy = −8V0

π· 4∫ π

0sin2y sin2y cos2y dy

= −32V0

∫ π

0sin4y

(1− sin2y

)dy = −32V0

∫ π

(sin4y − sin6y

= −32V0

8− sin 2y

sin 4y

32− 5y

15 sin 2y

64− 3 sin 4y

sin 6y

∣∣∣∣π0

= −32V0

8− 5π

]= −2V0

E 11 = −3

2V0 E 1

2 = −2V0

E 12 = −2V0

[sin(πx2

(2πx2a

)+ sin

(2πx2a

)sin(πx2

)]2dx2

= −8V0

0sin2(πx2

(2πx2a

= −8V0

a· aπ

∫ π

π· 4∫ π

= −32V0

∫ π

0sin4y

(1− sin2y

)dy = −32V0

∫ π

(sin4y − sin6y

= −32V0

8− sin 2y

sin 4y

32− 5y

15 sin 2y

64− 3 sin 4y

sin 6y

∣∣∣∣π0

= −32V0

8− 5π

]= −2V0

E 11 = −3

2V0 E 1

2 = −2V0

E 12 = −2V0

[sin(πx2

(2πx2a

)+ sin

(2πx2a

)sin(πx2

)]2dx2

= −8V0

0sin2(πx2

(2πx2a

= −8V0

a· aπ

∫ π

0sin2y sin22y dy

= −8V0

π· 4∫ π

= −32V0

∫ π

0sin4y

(1− sin2y

)dy = −32V0

∫ π

(sin4y − sin6y

= −32V0

8− sin 2y

sin 4y

32− 5y

15 sin 2y

64− 3 sin 4y

sin 6y

∣∣∣∣π0

= −32V0

8− 5π

]= −2V0

E 11 = −3

2V0 E 1

2 = −2V0

E 12 = −2V0

[sin(πx2

(2πx2a

)+ sin

(2πx2a

)sin(πx2

)]2dx2

= −8V0

0sin2(πx2

(2πx2a

= −8V0

a· aπ

∫ π

π· 4∫ π

= −32V0

∫ π

0sin4y

(1− sin2y

)dy = −32V0

∫ π

(sin4y − sin6y

= −32V0

8− sin 2y

sin 4y

32− 5y

15 sin 2y

64− 3 sin 4y

sin 6y

∣∣∣∣π0

= −32V0

8− 5π

]= −2V0

E 11 = −3

2V0 E 1

2 = −2V0

E 12 = −2V0

[sin(πx2

(2πx2a

)+ sin

(2πx2a

)sin(πx2

)]2dx2

= −8V0

0sin2(πx2

(2πx2a

= −8V0

a· aπ

∫ π

π· 4∫ π

= −32V0

∫ π

0sin4y

(1− sin2y

= −32V0

∫ π

(sin4y − sin6y

= −32V0

8− sin 2y

sin 4y

32− 5y

15 sin 2y

64− 3 sin 4y

sin 6y

∣∣∣∣π0

= −32V0

8− 5π

]= −2V0

E 11 = −3

2V0 E 1

2 = −2V0

E 12 = −2V0

[sin(πx2

(2πx2a

)+ sin

(2πx2a

)sin(πx2

)]2dx2

= −8V0

0sin2(πx2

(2πx2a

= −8V0

a· aπ

∫ π

π· 4∫ π

= −32V0

∫ π

0sin4y

(1− sin2y

)dy = −32V0

∫ π

(sin4y − sin6y

= −32V0

8− sin 2y

sin 4y

32− 5y

15 sin 2y

64− 3 sin 4y

sin 6y

∣∣∣∣π0

= −32V0

8− 5π

]= −2V0

E 11 = −3

2V0 E 1

2 = −2V0

E 12 = −2V0

[sin(πx2

(2πx2a

)+ sin

(2πx2a

)sin(πx2

)]2dx2

= −8V0

0sin2(πx2

(2πx2a

= −8V0

a· aπ

∫ π

π· 4∫ π

= −32V0

∫ π

0sin4y

(1− sin2y

)dy = −32V0

∫ π

(sin4y − sin6y

= −32V0

8− sin 2y

sin 4y

32− 5y

15 sin 2y

64− 3 sin 4y

sin 6y

∣∣∣∣π0

= −32V0

8− 5π

]= −2V0

E 11 = −3

2V0 E 1

2 = −2V0

E 12 = −2V0

[sin(πx2

(2πx2a

)+ sin

(2πx2a

)sin(πx2

)]2dx2

= −8V0

0sin2(πx2

(2πx2a

= −8V0

a· aπ

∫ π

π· 4∫ π

= −32V0

∫ π

0sin4y

(1− sin2y

)dy = −32V0

∫ π

(sin4y − sin6y

= −32V0

8− sin 2y

sin 4y

32− 5y

15 sin 2y

64− 3 sin 4y

sin 6y

∣∣∣∣π0

= −32V0

8− 5π

= −2V0

E 11 = −3

2V0 E 1

2 = −2V0

E 12 = −2V0

[sin(πx2

(2πx2a

)+ sin

(2πx2a

)sin(πx2

)]2dx2

= −8V0

0sin2(πx2

(2πx2a

= −8V0

a· aπ

∫ π

π· 4∫ π

= −32V0

∫ π

0sin4y

(1− sin2y

)dy = −32V0

∫ π

(sin4y − sin6y

= −32V0

8− sin 2y

sin 4y

32− 5y

15 sin 2y

64− 3 sin 4y

sin 6y

∣∣∣∣π0

= −32V0

8− 5π

]= −2V0

E 11 = −3

2V0 E 1

2 = −2V0

E 12 = −2V0

[sin(πx2

(2πx2a

)+ sin

(2πx2a

)sin(πx2

)]2dx2

= −8V0

0sin2(πx2

(2πx2a

= −8V0

a· aπ

∫ π

π· 4∫ π

= −32V0

∫ π

0sin4y

(1− sin2y

)dy = −32V0

∫ π

(sin4y − sin6y

= −32V0

8− sin 2y

sin 4y

32− 5y

15 sin 2y

64− 3 sin 4y

sin 6y

∣∣∣∣π0

= −32V0

8− 5π

]= −2V0

E 11 = −3

2V0 E 1

2 = −2V0

segre/phys406/15S/lecture_02.pdf · Ideal gas example Treat an ideal gas as a 3D particle in an in...

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