The others have a point; the solutions you 'should' assume had the form
\psi(x)=Ce^{zx}+De^{-zx}
For some z that may end up being real, imaginary or complex; this is how you get the original ('particle in a box') solution in the first place.
So remember how you get k in the first place in the region where the potential is 0:
\frac{-\hbar^2}{2m}\frac{∂^2\psi}{∂x^2}=E\psi
The solution comes from 'guessing' ψ = A\sin(kx) + B\cos(kx), and plugging this into the differential equation and algebraically solving for k.
Now consider...
To answer the simplest question, \psi^* is just the complex conjugate of ψ:
For some complex number c=a+bi, c^*=a-bi.
For a real number a, a=a^*
In the case of complex exponential functions:
c=Ae^{ix}, c^*=A^*e^{-ix}
Finally, cc^*=c^*c=|c|^2, which is a positive real number.
The...
I finally found the information I was looking for here:
http://vizier.u-strasbg.fr/viz-bin/VizieR
This is by far the most comprehensive supernova catalog I have ever seen, by the way.
I need a list of type Ia supernova with redshifts, Distance Modulus, Distance modulus error, and position in galactic coordinates, but I havn't been able to find one that has all of that information (I have found several with some of the information, but not all of it). Does anyone know where I...
You see the distance between the two light beams to be decreasing at a speed c: If the distance between the photons if x, then you see dx/dt=c.
In special relativity the velocity addition formula isn't just v1+v2=v3