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