QM solutions to the Schrodinger Equation

Maybe_Memorie
Messages
346
Reaction score
0

Homework Statement



Here's something that's confusing me. Say we have a potential

V(x) = Vo if x < 0, x > a
and
V(x) = 0 if 0 < x < a

(yes I know the notation with greater than/equals etc isn't totally correct, but you know what I'm talking about.)

In the middle section, ψ'' + k2ψ = 0

Whenever I see the solutions it's always ψ = Asin(kx) + Bcos(kx) in the middle section where k2 = -2mE/h2

In the right/left sections, ψ'' + f2ψ = 0, with f2 = -2m(E - V0)/h2

The solutions here always seem to be complex exponentials.

Can someone please explain the difference to me in the solutions?
 
Physics news on Phys.org
You must know the identities:
exp(±iø) = cosø ± i*sinø
So the two sets of solutions you're referring to are basically equivalent once you take the real part, only the exponential form turns out to be more practical in most cases because easier to manipulate.
 
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
The value of H equals ## 10^{3}## in natural units, According to : https://en.wikipedia.org/wiki/Natural_units, ## t \sim 10^{-21} sec = 10^{21} Hz ##, and since ## \text{GeV} \sim 10^{24} \text{Hz } ##, ## GeV \sim 10^{24} \times 10^{-21} = 10^3 ## in natural units. So is this conversion correct? Also in the above formula, can I convert H to that natural units , since it’s a constant, while keeping k in Hz ?
Back
Top