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170 lines
3.6 KiB
Plaintext
170 lines
3.6 KiB
Plaintext
─────────────
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1.3
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(includes handout regarding ultraviolet catastrophe and the development of the planck function)
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Properties of probability distributions:
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b
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Pₐᵦ = ∫ ρ(x) dx.
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a
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∞
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1 = ∫ ρ(x) dx.
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-∞
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∞
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<x> = ∫ x ρ(x) dx.
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-∞
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∞
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<f(x)> = ∫ f(x) ρ(x) dx.
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-∞
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σ² ≝ <Δx²> = <x²> - <x>² ≥ 0. (proof in Chapter 1, pg 9)
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Example 1.1,
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for example when looking for time-average distance of a time-dependent path, find x(t), then consider the average time interval
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dt/T.
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find dt from x(t) and T from total time of travel, which gives
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ρ(x) dx.
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looking at problems...
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problem 1.1 is clear
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problem 1.2 is clear
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problem 1.3,
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ρ(x) = A exp(-λ(x-a)²). ← Gaussian distribution.
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(a) trans. Normalize to find A.
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A, λ, a are all real constants.
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∞
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1 = A ∫ exp(-λ(x-a)²) dx.
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-∞
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for λ ≥ 0,
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∞
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∫ exp(-λ(x-a)²) dx = √π/√λ, so
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-∞
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A = √λ/√π.
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∴ ρ(x) = A exp(-λ(x-a)²).
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∞
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<x> = ∫ √λ/√π exp(-λ(x-a)²) = √λ/√π √πa/√λ = πa.
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-∞
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∞
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<x²> = ∫ √λ/√π exp(-λ(x-a)²) = √λ/√π √πa/√λ = πa.
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-∞
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─────────────
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1.4
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proof that normalized solutions to the schrodinger equation stay normalized with time, pg 13
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problems...
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1.4 computing probabilty amplitude and probability of a triangle wavefunction
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1.5
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Ψ(x,t) = A exp(-λ│x│) exp(-ιωt)
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∞
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∫ Ψ(x,t)* Ψ(x,t) = 1.
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-∞
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Ψ(x,t)* Ψ(x,t) = A² exp(-λ│x│) exp(-ιωt + ιωt) exp(-λ│x│) = A² exp(-2λ│x│).
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∞ ∞
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∫ Ψ(x,t)* Ψ(x,t) = ∫ A² exp(-2λ│x│) = 1.
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-∞ -∞
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A² ∫ dx exp(-2λx) = A² -1/2λ exp(-2λx).
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A² ∫ dx exp(2λx) = A² 1/2λ exp(2λx).
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∞
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A² ∫ dx exp(-2λ│x│) = 1
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-∞
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0 ∞
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= A² ∫ dx exp(2λx) + A² ∫ exp(-2λx)
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-∞ 0
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= A² 1/2λ [exp(0) - exp(-∞)] + A² -1/2λ [exp(-∞) - exp(0)]
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= A² (1/2λ [1 - 0] - 1/2λ [0 - 1])
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= A² (1/2λ + 1/2λ)
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= A² λ = 1.
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∴ A = 1/√λ.
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expectations values come from
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∫ dx x Ψ*Ψ and ∫ dx x² Ψ*Ψ, now with the constant A known
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standard deviation of x comes from σ² = <x>² - <x²>. Compute those above and then subtract.
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─────────────
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1.5
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equation 1.29 computes the change in expectation value of x with time
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the expectation values of the velocity is equal to the time derivative of the expectation value of position
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<v> = d/dt <x>
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customary to work with momentum instead,
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<p> = m d/dt <x> = -ιħ ∫ (Ψ* ∂/∂x Ψ) dx.
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integration by parts twice to show above
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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.
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Ahh, these theorems can be summed up as "expectation values obey classical laws", itneresting.
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d/dt <p> = < -∂/∂x V >.
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d/dt <p> = m d²/dt² <x> = -ιħ d/dt ∫ (Ψ* ∂/∂x Ψ) dx.
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to find this consider ∂/∂x V = ∂/∂x (T - E), using the schrodinger equation to solve
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─────────────
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1.6
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qualitative argument using rope on a string to demonstrate that the momentum is less defined as the position becomes more defined.
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σ𝓍σₚ ≥ ħ/2
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problem 1.9 pretty straight-forward, moving on
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