phy-520/journal/chap2.motes

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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



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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. 
    


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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.

    
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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.

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)