Splitting 2nd order DE into two 1st order DE's

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markov4
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Homework Statement


Text page giving the time-independent Schrödinger equation for an electron in a quantum well: -(hbar^2/2m) d/dz(dPsi/dz) + [U(z) - En]Psi(z) = 0, with m = 0.067m0, En and U(z) expressions.


1.a) Taking the Schrödinger second order differential equation given, split it into two first order differential equations for numerical solving.
We're given the relevant constants in a table, such as [tex]\hbar[/tex], [tex]\alpha[/tex], etc. 'z' is taken to be an indepedent variable which we incorporate into our initial conditions (which we must state).

The Attempt at a Solution


Handwritten derivation splitting the second-order equation -(h^2/2m) d/dz(dPsi/dz) + [U(z) - En]Psi(z) = 0 into two first-order equations by setting dPsi/dz = v, with dv/dz = -(2m/h^2)[U(z) - En]Psi(z) shown below.


Have i done this right? Is it right to say that [tex]\Psi(z)[/tex]=vz ? Since the [tex]\Psi(z)[/tex] that is in the equation, the one multiplied by ..+[U(z)-En][tex]\Psi(z)[/tex] has to be set to something too (?).
 
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  • Text page with the time-independent Schrödinger equation for a particle in a quantum well: -(ħ²/2m) d/dz(dΨ(z)/dz) + [U(z) − E_n]Ψ(z) = 0, with m = 0.067m₀, E_n = −U₀[1 − ħ/((2mU₀α²)^½)·(n + ½)]² and U(z) = U₀(e^(−2z/α) − 2e^(−z/α)).
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markov4 said:
Is it right to say that [tex]\Psi(z)[/tex]=vz ?

No, just leave [itex]\Psi(z)[/itex] as [itex]\Psi(z)[/itex]

It's true that [itex]d\Psi=vdz[/itex], but until you solve for [itex]v(z)[/itex] you have no way of integrating [itex]vdz[/itex].
 
^^But for the second part of the question, we're required to "Write a program that uses an inbuilt numerical solver (such as Matlab's ode45) to numerically solve for [tex]\Psi(z)[/tex] for z1<z<z2..." so we have to choose our own initial conditions to incorporate into our program, along with using ode45 in matlab.

So then does that mean [tex]\Psi(z)[/tex]=vz still? I don't understand why you would integrate it though? Unless I've completely missed something in the numerical solving part of this course.
 
v depends on z, so you can't integrate [tex]\frac{d\Psi}{dz}=v[/tex] to get [tex]\Psi=vz[/tex].
 
^^So when i put these coupled equations into matlab, what do i define [tex]\Psi(z)[/tex] to be?
 
markov4 said:
^^So when i put these coupled equations into matlab, what do i define [tex]\Psi(z)[/tex] to be?

It's one of the functions you are trying to solve your system of equations for...you simply leave it as [itex]\Psi(z)[/itex]
 
^^I see then. So I'm assuming that ODE45 in MATLAB will accommodate for leaving [tex]\Psi(z)[/tex] in the equation? I'm reading up on it now (ODE45) but i don't think i understand it.
 
Ok after doing some reading on ode45, i think i get it. So yes i'll leave Psi(z) as it is, and implementing it into my MATLAB code, it will be solved according to the initial conditions i set. Just need to try it out in MATLAB now!