Consider the time-dependent Schrodiner equation

[Jamin2112]

"Consider the time-dependent Schrodiner equation ..."

Homework Statement

Consider the time-dependent Schrodinger equation

ih²ψ_t = [-h²/(2m)]ψ_xx + V(x)ψ​

which is the underlying equation of quantum mechanics. Here V(x) is a given potential, h is the Planck's constant, and m is the mass of the particle. ψ(x,t) is the amplitude of the wave that the particle traces out. i=√(-1) is the imaginary unit.

(a) Use the separation of variable ψ(x,t)=u(x)exp(-iEt/h) to derive the time-independent Schrodiner equation which governs u(x). Show that the resulting equation is a Sturm-Louiville eigenvalue problem with

p(x) = h²/(2m), q(x) = V(x), r(x) = 1, λ=E.​

The eigenvalue E represents the energy of the particle.

(b) Solve the Sturm-Louiville eigenvalue problem in the domain -1 < x < 1 with zero potential and homogenous Dirichlet boundary conditions. Sketch the ground state (the lowest non-zero energy state). What is the energy?

(c) Without further calculation, explain what would happen to the eigenfunctions and eigenvalues if the domain is cut in half, i.e. 0 < x < 1, with new boundary conditions u(0)=u(1)=0.

Homework Equations

A Sturm-Louville eigenvalue problem has the form

-p(x)u''(x)-p'(x)u'(x)+q(x)=λr(x)u(x)​

The Attempt at a Solution

Part (a) is trivial. For part (b), doesn't "zero potential" mean V(x)=0, in which case I seem to get a trivial solution?

 

[Pengwuino]

The Schrodinger's Equation with V(x,t) = 0 is the free particle situation. The solution is not trivial, though, but very easy to determine.

 

[Jamin2112]

The Schrodinger's Equation with V(x,t) = 0 is the free particle situation. The solution is not trivial, though, but very easy to determine.

How is that not trivial?

 

[Pengwuino]

Well, what does the Schrodinger Equation look like with V(x,t) = 0 and the time dependence accounted for by the e^{{{-i\hbar E}\over{t}}}?

 

[Jamin2112]

Well, what does the Schrodinger Equation look like with V(x,t) = 0 and the time dependence accounted for by the e^{{{-i\hbar E}\over{t}}}?

-h²/(2m) * u''(x) = E * u(x).

If V(x) = 0, then which implies that u(x) = ke^{i(√(2mE)/h)x}

 

[Jamin2112]

Hang on a moment! I'm not sure if this is correct, but I figured out that we'll only have a ψ(x,t)≠0 if √(2mE)/h = (2n+1)/2 for some integer n≥0. Is that correct?