Solution to time dependent wave equation

[jaejoon89]

Homework Statement

Show that Y(x,t) = cos(kx)exp(-iwt) is a solution to the time-dependent Schrödinger wave equation.

where k is the wavenumber and w is the angular frequency

Homework Equations

Hamiltonian of Y(x,t) = ihbar d/dt Y(x,t)

The Attempt at a Solution

When I plug everything in and do the calculus, I get

(hbar / 2m)*k^2 = w

I've been paying around with it but for the life of me cannot figure out how the left hand side is equal to w even though the unit check suggests it could. So my question is, how can you simplify the left hand side to w

(hbar / 2m)*k^2 = ?

 

[gabbagabbahey]

Isn't there some relation between wavenumber and angular frequency given in your textbook?

 

[jaejoon89]

I can't find it. I know angular frequency is 2pi*nu and wavenumber is one over lambda. And I know nu*lambda = c, if that's the equation you mean, but still cannot get it to work...!

(hbar / 2m)(1/lambda^2) = (h/(4pi*m*lambda^2)) =...?

 

[gnurf]

Alternatively, you can end your confusion by visiting these wikipedia pages:

http://en.wikipedia.org/wiki/Wavenumber"
http://en.wikipedia.org/wiki/Planck_constant#The_reduced_Planck_constant"

 

[jaejoon89]

Thanks! I got it now. A quick follow up: why is it assumed that the energy component for momentum (related to the wavenumber in this problem) is entirely kinetic?

i.e.,

p = mv => p^2 = m^2 v^2 = 2m (1/2) m v^2 = 2m E => p = sqrt(2mE)

 

[gabbagabbahey]

A free particle (I assume your problem is for a free particle, although you didn't actually say so in your problem statement) is subject to zero external forces/potentials, so its energy (in non-relativistic QM) is entirely kinetic.