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covered:
deBroglie wavelengths
double-slit diffraction
Rutherford planetary model
Bohr Model
Correspondence Principle
This seems to be primarily historical motivation.
The correspondence principle, I'm not sure I've ever attempted to compute for even a simple system. Hmm... I see the note here
"What is the Correspondence principle? Derive the quantum condition from it." I don't know what is meant by the "quantum condition". I guess just that the states of a system are quantized. I have no clue how I would show that using the correspondence principle. Should ask about this.
planck determined that light transferred energy in "packets". De Broglie suggested a certain nature to the packets, that the wave nature of light was also applicable to matter. These ideas suggested that matter and light both exhibit properties of particles and waves.
the Kapitz-Dirac effect demonstrates wave-particle duality by showing that matter waves are diffracted by an incident electromagnetic field as predicted by the wave properties of the particles, i.e., by the diffraction equation nλ = 2dsin(Θ).

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basic properties of waves
λ - wavelength
ν - temporal freq
v - wave speed
A - amplitude
k the spatial frequency, i.e., 1/λ, or the wave number, the nubmer of crests in a unit length, so to speak.
ω the (angular) temporal frequency
not taking the time to solve things out in this run, unless I run into something I'm not familiar with.

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double-slit revisited, motivating why we look now at matter waves...
plane waves
general form
relation of k to ω, not generalized to a formula but always has some relation with the wave motion, e.g, ω = kv with v the speed of the plane wave.
group velocity is something I understood much less about.
From B&S 2.3.2, I see it is bound up in the relationship between k and ω, e.g.,
∂²/∂k² ω = β where k = k₀.
the fourier transformation gives a k-space function representation for an x-space wave function at a given time.
for a Gaussian,

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THe Born Rule, in English, states that the probability density of finding a particle at a given point is proportional to the magnitude of the particle's wavefunction at that point, squared.
A classical wave has an amplitude A. In a mechanical medium, the displacement of particles follows a path along the wave at that amplitude. A quantum particle does not directly follow the path of the wave function, but rather the wave function, or the "probability amplitude", provides a spread that, when squared, gives the probability density of finding the particle at a given point.
The probability current is essentially a flux of probability. The wave function spreads with time across space. We have
𝗷 = ħ/2mι (Ψ* ∇Ψ - Ψ ∇Ψ*).
recognize p̂ = -ιħ∇.
𝗷 = 1/2m (Ψ* p̂ Ψ - Ψ p̂ Ψ*).
this is in the position basis
∂/∂t ρ + ∇⋅𝗷 = 0 is the continuity equation, a statement that the density of a system can only lose quantity equal to its divergence.
the probability density is Ψ*Ψ.
in one dimension,
∂/∂t Ψ*Ψ + d/dx j = 0.
(∂/∂t Ψ*) Ψ + (Ψ* ∂/∂t Ψ) + d/dx j = 0.
...

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Ehrenfest Theorems
<p> = m d/dt <x> and <-∂/∂x V> = d/dt <p>
─────────────

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─────────────
1.3
(includes handout regarding ultraviolet catastrophe and the development of the planck function)
Properties of probability distributions:
b
Pₐᵦ = ∫ ρ(x) dx.
a
∞
1 = ∫ ρ(x) dx.
-∞
∞
<x> = ∫ x ρ(x) dx.
-∞
∞
<f(x)> = ∫ f(x) ρ(x) dx.
-∞
σ² ≝ <Δx²> = <x²> - <x>² ≥ 0. (proof in Chapter 1, pg 9)
Example 1.1,
for example when looking for time-average distance of a time-dependent path, find x(t), then consider the average time interval
dt/T.
find dt from x(t) and T from total time of travel, which gives
ρ(x) dx.
looking at problems...
problem 1.1 is clear
problem 1.2 is clear
problem 1.3,
ρ(x) = A exp(-λ(x-a)²). ← Gaussian distribution.
(a) trans. Normalize to find A.
A, λ, a are all real constants.
∞
1 = A ∫ exp(-λ(x-a)²) dx.
-∞
for λ ≥ 0,
∞
∫ exp(-λ(x-a)²) dx = √π/√λ, so
-∞
A = √λ/√π.
∴ ρ(x) = A exp(-λ(x-a)²).
∞
<x> = ∫ √λ/√π exp(-λ(x-a)²) = √λ/√π √πa/√λ = πa.
-∞
∞
<x²> = ∫ √λ/√π exp(-λ(x-a)²) = √λ/√π √πa/√λ = πa.
-∞
─────────────
1.4
proof that normalized solutions to the schrodinger equation stay normalized with time, pg 13
problems...
1.4 computing probabilty amplitude and probability of a triangle wavefunction
1.5
Ψ(x,t) = A exp(-λ│x│) exp(-ιωt)
∞
∫ Ψ(x,t)* Ψ(x,t) = 1.
-∞
Ψ(x,t)* Ψ(x,t) = A² exp(-λ│x│) exp(-ιωt + ιωt) exp(-λ│x│) = A² exp(-2λ│x│).
∞ ∞
∫ Ψ(x,t)* Ψ(x,t) = ∫ A² exp(-2λ│x│) = 1.
-∞ -∞
A² ∫ dx exp(-2λx) = A² -1/2λ exp(-2λx).
A² ∫ dx exp(2λx) = A² 1/2λ exp(2λx).
∞
A² ∫ dx exp(-2λ│x│) = 1
-∞
0 ∞
= A² ∫ dx exp(2λx) + A² ∫ exp(-2λx)
-∞ 0
= A² 1/2λ [exp(0) - exp(-∞)] + A² -1/2λ [exp(-∞) - exp(0)]
= A² (1/2λ [1 - 0] - 1/2λ [0 - 1])
= A² (1/2λ + 1/2λ)
= A² λ = 1.
∴ A = 1/√λ.
expectations values come from
∫ dx x Ψ*Ψ and ∫ dx x² Ψ*Ψ, now with the constant A known
standard deviation of x comes from σ² = <x>² - <x²>. Compute those above and then subtract.
─────────────
1.5
equation 1.29 computes the change in expectation value of x with time
the expectation values of the velocity is equal to the time derivative of the expectation value of position
<v> = d/dt <x>
customary to work with momentum instead,
<p> = m d/dt <x> = -ιħ ∫ (Ψ* ∂/∂x Ψ) dx.
integration by parts twice to show above
So, from the relationship we see from velocity, the momentum operater emerges, which takes the partial derivative of the wavefunction with respect to space, returning the momentum, harkening back to classical formulations of mechanics. This is all just for the expectation value, however, not for a momentum of a single particle.
Ahh, these theorems can be summed up as "expectation values obey classical laws", itneresting.
d/dt <p> = < -∂/∂x V >.
d/dt <p> = m d²/dt² <x> = -ιħ d/dt ∫ (Ψ* ∂/∂x Ψ) dx.
to find this consider ∂/∂x V = ∂/∂x (T - E), using the schrodinger equation to solve
─────────────
1.6
qualitative argument using rope on a string to demonstrate that the momentum is less defined as the position becomes more defined.
σ𝓍σₚ ≥ ħ/2
problem 1.9 pretty straight-forward, moving on

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─────────────
2.1
if the potential V is independent of time, Ψ is separable, Ψ = ψ(x) f(t)
∂/∂t Ψ = ψ d/dt f.
∂²/∂x² Ψ = d²/dx² ψ f.
ιħψ d/dt f = [ħ²/2m d²/dx² ψ f] + [V ψ f].
ιħ 1/f d/dt f = [ħ²/2m 1/ψ d²/dx² ψ] + [V].
critical: this can only be true if both sides are constant, otherwise variations could occur on either side, creating a contradiction
THis constant is called E, and reorganize to two ordinary differential equations
⎧-ħ²/2m d²/dx² ψ + V ψ = E ψ
⎨
⎩d/dt f = -ιE/ħ f
one function of time, one of space
??????????????????????????
would generalize directly to multiple dimensions? e.g.:
⎧ -ħ²/2m d²/dx² ψ + V ψ = E ψ
⎪
⎨ -ħ²/2m d²/dx² ψ + V ψ = E ψ
⎪
⎩ d/dt f = -ιE/ħ f
??????????????????????????
f(t) = exp(-ιEt/ħ)
separable solutions
1. are stationary states -- cover all stationary states?
2. have definite total energy, proof on pg 22
3. general solution is a linear combination of separable solutions
pg 23 shows there is a different wave function which is such a linear combination for each allowed energy, but how to know what the allowed energies are?
one general solutions:
∞
Ψ(x,t) = ∑ cₙ ψₙ(x) exp(-ιEₙt/ħ)
n
problems...
prove:
E must be constant for the solutions to be normalizable
ψ can always be chosen to be real
if V is even, ψ can be taken to be even or odd
these are very much worth doing -- should come back to this
problem two seems like it's saying "prove the conservation of energy", except with a bit of kinetic energy always needing to exist -- do this to explore
─────────────
2.2
infinite square well
V=0 potential gives simple harmonic oscillator solutions with boundary conditions ψ(0) = ψ(a) = 0.
These boundary conditions restrict the allowed wavelengths thus providing the answer to the earlier question "how do we find the allowed states" in this particular case.
← no general approach?
In this case we have kₙ = nπ/a with n=1,2,3,... .
The constant A must be found from normalization of ψ, i.e.,
a
∫ dx │A│² sin²(kx) = │A│² a/2 = 1, and pick positive real A.
0
ψₙ(x) = √2/√a sin(nπ/a x).
Dirichlet's thm -- orthogonal and complete set of functions
Fourier's trick is used to find cₙ coefficients, pg 28 -- GRiffiths calls it this in his E&M book, too.
In this case,
a
cₙ = √s/√a ∫ dx sin(nπ/a x) Ψ(x,0)
0
problems...
problem 2.3 should just come out to where the only viable solution is that the wave function is 0
2.4 actually solve the square well problem -- done this several times before
2.5 calculate <x>, <x²>, <p>, <p²>, σ𝓍, σₚ for the nth stationary state of the square well -- have not done this yet, think it's clear but should do it
2.6 state is superposition of two states
a) normalize
b) compute probabilty and amplitude
c) compute <x> -- other parts straightforward but do this keeping in mind seeing how <x> changes with time
d) compute <p> -- there is a quick way
e) Expecation value of H -- hmm, have to think about this
f) compare the frequency of the quantum particle with that of a classical particle bouncing off the walls
2.7 relative phase matters, this is obvious
2.8 compute state properties with another initial state, quadraitc in x
2.9 find the fourier series equivalent for the above state function
2.10 oh, this is interesting, the hamiltonian <H> is constant in time because of the sum ∑ │cₙ│² = 1.
─────────────
2.3
Quantum Harmonic Oscillator problem is to solve for
V(x) = m/2 ω² x²
THe ladder operators are an alternative to solving this problem using a power series expansion, but the power series expansion is a more important operation to understand for attacking systems in general.
Develops the ladder operators starting pg 32
a₊₋ ≝ √(1/2m) ( ħ/ι d/dx ±ιmωx)
[a₋,a₊] = ħω
pg 34, proof if ψ satisfies schrodinger eq with E, then a₊ψ satisfies it with E+ħω.
pg 36 concludes
ψₙ(x) = Aₙ(a₊)ⁿ exp(-mω/2ħ x²), Eₙ = (n + 1/2) ħω.
prob 2.12 works out Aₙ.
problem 2.11 seems like it's basically showing the lowering operator has to be zero when operating on the lowest state.
problem 2.12 deteremines Aₙ from above
problem 2.13
---
now he approaches the problem with a power series expansion, the Frobenius method
step 2.64, "follows from the uniqueness of te power series expansion" -- what's up with this?
recursion formula -- don't understand this, need to follow these through
ahh, OK, the recursion formula connections a₀ and a₁ to all of the proceeding aₙ values, thus determining the function by two constants.
allowed functions are still further limited by normalizability, which leaves us with K = 2n + 1, giving the same Eₙ as the first method.
─────────────
2.4 – Free particle
-ħ²/2m d²/dx² ψ = E ψ
k ≝ √(2mE)/ħ, so d²/dx² ψ = -k² ψ.
Solutions are obvious: ψ(x) = A exp(ιkx) + B exp(-ιkx).
Time dependence: exp(-ιEt/ħ) = exp(-ιk²ħ/2m t)
ψ(x,t) = A exp(ιkx - ιEt/ħ) + B exp(-ιkx - ιEt/ħ)
= A exp(ιkx - ιk²ħ/2m t) + B exp(-ιkx - ιk²ħ/2m t)
= A exp(ιk(x - kħ/2m t)) + B exp(-ιk(x + kħ/2m t))
These are travelling waves with a fixed profile. In fact, perhaps better to use
ψ(x,t) = A exp(ιk(x - kħ/2m t)),
where k < 0 means travelling to the left, and k > 0 means travelling to the right.
This appears to harken to the idea of group vs. particle velocity, as the wave propagates at v[quantum] = √(E/2m), but a particle moves at v[classical] = √(2E/m) = 2 v[quantum].
Would need to normalize to find A. According to Griffiths, this wave function is not normalizable .. why?
∫[-∞,∞]dx Ψ⃰ₖ Ψₖ = │A²│ ∫[-∞,∞]dx 1 = │A²│ ∞
The conclusion from this is that these are not allowed states... as GRiffiths puts it, "there is no such thing as a free particle with definite energy". OK, but then what is allowed?
Instead we look at
Ψ(x,t) = 1/√(2m) ∫[-∞,∞]dk ϕ(k) exp(ιk(x - kħ/2m t)).
Basically this means that for certain ϕ(k), I can normalize the wave function. It carries a range of ks.
Need to study problem 2.22
**** The dispersion relation is the formula for ω as a function of k. How does this fit with σ², the variance?
Regarding group and phase velocities,
Consider the general form
Ψ(x,t) = 1/√(2m) ∫[-∞,∞]dk ϕ(k) exp(ι(kx - ωt))
Here, ω represents the phase velocity, and ϕ is narrowly peaked about k₀.
<<<<<<< HEAD
*** dispersion notes: derive group velocity from 2 frequencies
*** dispersion comes from curvature of ω(k)
=======
A Taylor expansion helps elucidate the situation
ω(k) = ω₀ + ω′₀(k-k₀)
set s≝k - k₀
ψ(x,t) ≈ 1/√(2π) ∫[-∞,∞]ds ϕ(k₀ + s) exp(ι[(kₒ+s)x - (ω₀ + ω′₀ s)t])
At t=0, ψ(x,0) = 1/√(π) ∫[-∞,∞]ds ϕ(k₀ + s) exp(ι(kₒ+s)x),
and later,
ψ(x,t) = 1/√(π) exp(ι(-ω₀ + kₒ ω′₀)t∫[-∞,∞]ds ϕ(k₀ + s) exp(ι(kₒ+s)x)
>>>>>>> 9d60318cfe5d81c018be9b1fdbf72e03e8733ea0

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10-23
──────────────────────────────────────────────────────────────────────────
graphical review of the tangent/cotangent bundle to show phases
Linear Operators
─────────────
rank 0 scalar ℝ,ℂ
rank 1 vector basis(ê¹,ê²,ê³), components [v1, v2, v3]
rank 2 𝔸:𝘃→𝘄 Linear
???rank 3 εᵢⱼᵏ - cross product, gᵢ - dot product
graphically show the operation of operators
graphically show the operation of rotation operators
graphically distinguish active and passive transformations
active - includes evolution operator U⊹
passive - matrix change of basis
𝘄 = 𝔸 𝘃.
ℝ𝘄 = ℝ𝔸ℝᵀ ℝ𝘃.
𝘄′ = 𝔸′ 𝘃′.
──────────────────────────────────────────────────────────────────────────
2017-10-25
Linear transformations
⎛1 a⎞ ⎛0⎞ = ⎛a⎞
⎝0 1⎠ ⎝1⎠ ⎝1⎠
⎛λ₁ 0⎞ ⎛0⎞ = ⎛0 ⎞
⎝0 λ₂⎠ ⎝1⎠ ⎝1/2⎠
M = U(ϕ)⎛λ₁ 0⎞ U⁻¹(ϕ) R(θ)
⎝0 λ₂⎠ ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟
⏟⏟⏟⏟ Vᵀ
W
Sᵀ = S, symmetric → Hermitian
Rᵀ = R⁻¹ orthogonal → unitary
M = U W Vᵀ singular decomposition
= S R polar decomposition
normal matrix analogue
x ± ιn = r exp(ιϕ)
Theorem
─────────────
if H⁺ = H and H v⃗ᵢ = v⃗ᵢ λᵢ
then λᵢ real and v⃗ᵢ⋅v⃗ⱼ = 0 if λᵢ ≠ λⱼ
take ⊹: Vⱼ⊹ H = λ⃰ⱼ vⱼ⊹
λ⃰ⱼ v⃑ⱼ⊹ v⃑ᵢ = v⃑ⱼ⊹ H v⃑ᵢ = v⃑ⱼ⊹ v⃑ᵢ λᵢ
I i=j, ‖vᵢ‖ ≠ 0, λᵢ⃰ = λᵢ, ℝ ∋ λ
i ≠ j, λᵢ ≠ λⱼ, v⃗ⱼ⊹ v⃗ᵢ = 0
───────────────────────────────────────────────────────────────────────
10-30
"The importance of being Earnest"
0) Linear
linear combinations
separable diff. eqs.
superposition
basis
1) Unitary
Rotations
U⊹ U = I
change of coordinates
preserves angles and length
2) Hermitian
Stretches
H⊹ = H
real eigenvalues → observables
orthogonal eigenvectors
diagonalizable
H = U D U⊹ → similarity transform, rotate, stretch, rotate back
H U = U D ; H⊹ = H; U⊹ U = I; U⊹ = U⁻¹
H = U D U⁻¹ = U D U⊹
U⊹ H U = D
b
3) diagonal
element-wise multiplication
───────────────────────────────────────────────────────────────────────
11-01
From last time:
The importance of
1) linear
2) Unitary
3) Hermitian
Polar Decomposition -- A = P*U
H = U D U⊹
spectral decomposition
H = λ₁ P₁ + λ₂ P₂
4) Diagonal
Now:
5) Importance of Commuting
tied up with diagonal matrices, because diagonal matrices commute
Normal matrix: [N⊹,N] = 0.
similarity transform → change of basis
A = U D U⁻¹
A² = U D U⁻¹ U D U⁻¹ = U D² U⁻¹
f(A) = f₀ + f₁ A + 1/2! f₂ A²
= U U⁻¹ + U f₁ D U⁻¹ + U 1/2! f₂ D² U⁻¹
= U(1 + f₁ D + 1/2! f₂ D² + ...) U⁻¹
= U(diagonal matrix) U⁻¹
Obvious that diagonal matrices commute
commuting --
physical measurements represented by Hermitian operators H⊹ = H
λᵢ = what you can measure
𝘃ᵢ = definte states of 𝒪.
two measurements are compatible if they have the same eigenvectors, simulataneously diagonalizable
every physical state is reprsented as a dot on a unit circle, brilliant
expansion and projection postulate
diagonal matrices are automatically diagonalizable
A = U Dₐ U⁻¹, B = U Dᵦ U⁻¹
if [A,B] = 0, are A,B simulataneously diagonalizable?
ex [x,p] = ιħ.
Thm: if [A,B] = 0 and U⁻¹ A U = D, then U⁻¹ B U is also "block diagonal".
recall ladder operators:
if N|n> = n|n>, then N(a±|n>) = (n+1)|n>.
let A 𝐮ᵢ = λᵢ 𝐮ᵢ
A(B 𝐮ᵢ) = B(A 𝐮ᵢ)z
11-06
─────────────
Postulates - lead-up
a) How does the state evolve (when not watched)? Ĥ ψ = Ê ψ.
b) what happens when we measure? e.g. x
i) pick random x weighted by prob │ψ(x)│²
ii) collapses the wave function to δ(x-a) if measured "a"
c) Observation is richer than classical mechanics due to complementarity and superposition
1) Observables are a Hermitian operator
Q̂(x,-ιħ∂/∂x) on ❙ψ(x)❭
2) Every observable has "definitey states"
eigenstates of Q̂: Q̂❙ϕₙ❭ qₙ ❙ϕₙ❭
3) Any other state is a superposition of ❙ϕₙ❭
❙ψ❭ = ∑ cₙ ❙ϕₙ❭ cₙ = Probability Amplitude
4) Observation is an irreversible process, "collapses ❙ψ❭"
5) Measurements with same ❙ϕₙ❭ are compatible.
[Q̂,R̂] = 0.
Practical Application
1) Solve Ĥ ❙ψₙ❭ = Eₙ ❙ψₙ❭ stationary states
2) Find components of ❙ψ₀❭ = cₙ❙ψₙ❭
3) Rotate State in time ❙ψ(t)❭ = exp(-ιEₙt/ħ)cₙ❙ψ₀❭
4) Solve eigenstates of Q̂❙ϕₙ❭ = qₙ❙ϕₙ❭
5) find components of ❙ψ(t)❭ = aₙ❙ϕₙ❭
6) project the state ❙ψ(t)❭ ⟶ ❙ϕₙ❭
TOOLS:
❬ϕₙ❙ϕₘ❭ = δₙₘ -- orthonormality
∑ₙ ❙ϕₙ❭❬ϕₙ❙ = 𝕀 -- closure
Postulates
1) superposition
state ❙ψ❭ collection of probability amplitues cₙ with ∑ₙ│cₙ│² = 1.
2) expansion/projection
Q̂❙ϕₙ❭ = qₙ❙ψₙ❭
❙ψ❭ = aₙ ❙ϕₙ❭ aₙ = probability amplitude of measurement "q" after measurement qₙthe system collapses to state ❙ϕₙ❭
3) evolution Ĥψ = Êψ.
4) Uncertainty [x,p] = ιħ.
5) wait until next semester -- exclusion principle

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#include <math.h>
#include <vector>
#include <stdio>
#include <iomanip>
double c0 = 1;
double e = 1;
double pi = 3.14;
double ep_0 = 1;
double hbar = 1;
double m = 1;
double a = 4*pi*ep_0*hbar*hbar/m/e/e;
// hydrogen atom model
// kappa = sqrt(-2*m*E/hbar)
// kappa = m * e^2 / (4 * pi * ep_0 * hbar^2)
double kappa(double n)
{
double result = m * e^2 / (4 * pi * ep_0 * hbar^2);
result /= n;
return result;
}
double c_j (double j, double n, double l)
{
double result = (2 *(j+l+1-n)/(j+1)/(j+2*l+2));
result *= c_j(j);
}
double v(double n, double l, double rho)
{
double jmax = n-l-1;
std::vector<double> c;
for (double power=0; power < jmax + 1; power++) {
c[power] = c0
}
}
double R(double n, double l,
double r)
{
double rho = r/a*n;
double result = 1/r
* pow(rho,l+1)
* exp(-rho)
* v(n,l,rho);
}
normalize()
double Y(double l, double m,
double theta, double phi);
double psi(double n, double l, double m,
double r, double theta, double phi)
= R(n,l,r) * Y(l,m,theta,phi);
int main(int argc, char const *argv[])
{
// nanometers
for (double r=0; r<100; r += 0.005)
std::cout
<< std::scientific << std::setprecision(6)
<< r << '\t'
<< R(1,0,r) << '\t'
<< R(2,0,r) << '\t'
<< R(3,0,r) << '\t'
<< R(4,0,r) << '\t'
<< R(5,0,r) << '\t'
<< R(6,0,r) << '\t'
<< R(7,0,r) << '\t'
<< R(8,0,r) << '\t'
<< R(9,0,r) << '\t'
<< R(10,0,r) << '\t'
<< R(11,0,r) << '\t'
<< R(12,0,r) << '\t'
<< R(13,0,r) << '\t'
<< R(14,0,r) << '\t'
<< R(15,0,r);
//for (double r=0; r<100; r += 0.005)
//std::cout
// << std::scientific << std::setprecision(6)
// << r << '\t'
// << psi(1,0,0,r,0,0) << '\t'
// << psi(2,0,0,r,0,0) << '\t'
// << psi(3,0,0,r,0,0) << '\t'
// << psi(4,0,0,r,0,0) << '\t'
// << psi(5,0,0,r,0,0) << '\t'
// << psi(6,0,0,r,0,0) << '\t'
// << psi(7,0,0,r,0,0) << '\t'
// << psi(8,0,0,r,0,0) << '\t'
// << psi(9,0,0,r,0,0) << '\t'
// << psi(10,0,0,r,0,0) << '\t'
// << psi(11,0,0,r,0,0) << '\t'
// << psi(12,0,0,r,0,0) << '\t'
// << psi(13,0,0,r,0,0) << '\t'
// << psi(14,0,0,r,0,0) << '\t'
// << psi(15,0,0,r,0,0);
return 0;
}

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#include <math.h>
#include <vector>
#include <stdio>
#include <iomanip>
double c0 = 1;
double e = 1;
double pi = 3.14;
double ep_0 = 1;
double hbar = 1;
double m = 1;
double a = 4*pi*ep_0*hbar*hbar/m/e/e;
// hydrogen atom model
// kappa = sqrt(-2*m*E/hbar)
// kappa = m * e^2 / (4 * pi * ep_0 * hbar^2)
double kappa(double n)
{
double result = m * e^2 / (4 * pi * ep_0 * hbar^2);
result /= n;
return result;
}
double c_j (double j, double n, double l)
{
double result = (2 *(j+l+1-n)/(j+1)/(j+2*l+2));
result *= c_j(j);
}
double v(double n, double l, double rho)
{
double jmax = n-l-1;
std::vector<double> c;
for (double power=0; power < jmax + 1; power++) {
c[power] = c0
}
}
double R(double n, double l,
double r)
{
double rho = r/a*n;
double result = 1/r
* pow(rho,l+1)
* exp(-rho)
* v(n,l,rho);
}
normalize()
double Y(double l, double m,
double theta, double phi);
double psi(double n, double l, double m,
double r, double theta, double phi)
= R(n,l,r) * Y(l,m,theta,phi);
int main(int argc, char const *argv[])
{
// nanometers
for (double r=0; r<100; r += 0.005)
std::cout
<< std::scientific << std::setprecision(6)
<< r << '\t'
<< R(1,0,r) << '\t'
<< R(2,0,r) << '\t'
<< R(3,0,r) << '\t'
<< R(4,0,r) << '\t'
<< R(5,0,r) << '\t'
<< R(6,0,r) << '\t'
<< R(7,0,r) << '\t'
<< R(8,0,r) << '\t'
<< R(9,0,r) << '\t'
<< R(10,0,r) << '\t'
<< R(11,0,r) << '\t'
<< R(12,0,r) << '\t'
<< R(13,0,r) << '\t'
<< R(14,0,r) << '\t'
<< R(15,0,r);
//for (double r=0; r<100; r += 0.005)
//std::cout
// << std::scientific << std::setprecision(6)
// << r << '\t'
// << psi(1,0,0,r,0,0) << '\t'
// << psi(2,0,0,r,0,0) << '\t'
// << psi(3,0,0,r,0,0) << '\t'
// << psi(4,0,0,r,0,0) << '\t'
// << psi(5,0,0,r,0,0) << '\t'
// << psi(6,0,0,r,0,0) << '\t'
// << psi(7,0,0,r,0,0) << '\t'
// << psi(8,0,0,r,0,0) << '\t'
// << psi(9,0,0,r,0,0) << '\t'
// << psi(10,0,0,r,0,0) << '\t'
// << psi(11,0,0,r,0,0) << '\t'
// << psi(12,0,0,r,0,0) << '\t'
// << psi(13,0,0,r,0,0) << '\t'
// << psi(14,0,0,r,0,0) << '\t'
// << psi(15,0,0,r,0,0);
return 0;
}

120
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To model a quantum particle in a radial potential, we addressed the derivation of the wave function parallel with the development in Griffiths, pg. 145, and in Dr. Crawford's course notes, and some various online sources.
The potential V(r) = 1/r in spherical coordinates describes the system as having constants of motion in the angular coordinates theta and phi, thus conserving angular momentum. Still, it admits an effective potential that includes a centrifugal factor,
V[eff](r) = V(r) + ħ²/2m l(l+1)/r².
The Schrodinger Equation with this potential may be solved using separation of variables. In this development, we focus on the radial wave function. The Schrodinger equation reads
-ħ²/2m d²/dr² ψ + [-1/r + ħ²/2m l(l+1)/r²] ψ = E ψ.
[-1/r + ħ²/2m l(l+1)/r²] ψ = E ψ + ħ²/2m d²/dr² ψ.
1/-E [-1/r + ħ²/2m l(l+1)/r²] ψ = -ψ + ħ²/-2mE d²/dr² ψ.
[1/rE + ħ²/-2mE l(l+1)/r²] ψ + ψ = ħ²/-2mE d²/dr² ψ.
We can introduce the variable κ = √(-2mE)/ħ, giving
[1 + 2m/ħ²κ 1/κr + l(l+1)/(κr)²] ψ = 1/κ² d²/dr² ψ.
Set ρ = κr, revealing ρ₀ = 2m/ħ²κ, which now gives
[1 + ρ₀/ρ + l(l+1)/ρ²] ψ = d²/dρ² ψ.
Note that by scaling ρ, we can translate this problem from the problem V(r) = 1/r to that of the hydrogen atom, with its specific radius. This is planned for later in the development.
Using an asymptotic breakdown, we can find the solution in three parts,
ρ → ∞: d²/dρ² ψ = ψ,
with solution ψ(ρ) = A exp(-ρ) + B exp(ρ),
and since the second factor is not normalizable,
ψ(ρ) = A exp(-ρ).
ρ → 0: d²/dρ² ψ = l(l+1)/ρ² ψ,
with solution ψ(ρ) = C ρˡ⁺¹ + D ρ⁻ˡ,
but Dρ⁻ˡ disobeys normalizability, so
ψ(ρ) = C ρˡ⁺¹.
The complete solution is then constructed assuming a power series v(ρ),
ψ(ρ) = A exp(-ρ) C ρˡ⁺¹ v(ρ),
∞
where v(ρ) = ∑ cⱼ ρʲ.
j=0
The recursion relationship for the cⱼ coefficients is given by
⎧ 2(j + l + 1) - ρ₀ ⎫
cⱼ₊₁ = ⎨─────────────────⎬ cⱼ,
⎩(j + 1)(j + 2l + 2)⎭
where c₀ is normalized by total probability equal to one.
In general, this power series does result in unnormalizable states. There must be a maximum value of j, then, above which the coefficients are all zero. This jmax is related to the quantum numbers n and l by
jmax = n - l + 1.
Looking back at the recursion relationship, it becomes clear that
2(j + l + 1) - ρ₀ = 2n - ρ₀, and ρ₀ = 2n.
This gives us a relationship between the principle quantum number n and the energy of a state n, recalling that the angular momentum is conserved and so the energy value does not rely on the quantum numbers l or m.
Eₙ = -ħ²κ²/2m.
κ = 2m/ħ²ρ₀.
Eₙ = -2m/ħ²ρ₀² = -m/2ħ²n² = E₁/n², n = 1,2,3... ,
where E₁ = -m/2ħ².
Actually, this is not an energy. The scaling factor bringing this to solutions of the hydrogen atom would make it an energy. It comes back to the units of V(r) = 1/r. This of course has units of energy, but what are the constants that make it that way? We've used no constants here. This should at least be V(r) = E₀/r, for some energy.
This reveals the dependence on the principle quantum number of the previous quanties,
κ = 2m/ħ²ρ₀ = a/n,
where a = m/ħ² is the Scrodinger constant factor relating the action and mass of this system (or if scaled to the correct units, the inverse of the Bohr radius of the Hydrogen atom);
ρ = κr = ar/n.
This is where it becomes obvious the scaling would be good, because here we would see that the form reveals r/(bohr radius).
The spatial wave function is defined by three quantum numbers n, l, and m, and as previously stated, can be solved by separation of various such that
Ψₙₗₘ(r,θ,ϕ) = Rₙₗ(r) Yₗᵐ(θ,ϕ),
where we have already found the radial function R,
Rₙₗ(r) = 1/r exp(-ρ) ρˡ⁺¹ v(ρ).
The model we've developed is nearly capable of handling the associated LaGuerre polynomials, but they become difficult to plot. So, for now, we focus on the radial function, only.
Substituting our new information back into the recursion relationship gives us the recursion in terms of the quantum numbers,
⎧ 2(j + l + 1 - n) ⎫
cⱼ₊₁ = ⎨────────────────⎬ cⱼ.
⎩(j + 1)(j + 2l + 2)⎭
This is sufficient information to model the radial component of the wave function, except for normalizing the coefficient c₀. This is done using the expression of probability conservation
∞
∫ │Rₙₗ│² r² dr = 1.
o
This is difficult to model numerically in the general case, and while interesting, the more practical approach is probably to normalize any states we're interested in by hand. The normalization of c₀ is different for each set of values {n,l}.
A model has (nearly) been written in c++ to compute the wavefunction resultant from this potential. The data is output to tab-separated data tables in the format
r R₁₀ R₂₀ R₃₀ ...
Still deciding how to output the different values of l.