How do I solve this integration problem in quantum mechanics?

sciboudy
Messages
89
Reaction score
0
in Quantum Mechanics Problem
it must be ∫ψ*d3ψ\dx3 dx =∫d2ψ\dx2 dx how ? ? i think it will be by inGration by parts i need steps
 
Physics news on Phys.org
d3\dx3 ∫ψ*ψdx = d2\dx ∫ψ*dψ\dx dx

d3\dx3 ∫ψ*ψdx = d2\dx d\dx ∫ ψ*ψ dx

d3\dx3 ∫ψ*ψdx = d3\dx3 ∫ ψ*ψ dx

and from normlization we found that

∫ψψ*=1

so right hand side = left hand side =d3\dx3 ∫ψ*ψdx = d2\dx ∫ψ*dψ\dx dx is this solution true ?
 
I can't tell what you're asking. Are you saying
\int \psi^* d^3 \psi dx^3 dx = \int d^2 \psi dx^2 dx
or
\int \psi^* \frac{d^3 \psi}{ dx^3} dx= \int \frac{d^2 \psi}{dx^2} dx?
I'm guessing it's the latter.
 
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
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...
Back
Top