Recent content by WrongMan

first off, the graf you called the "modulus" is actually the probability distrbution. wth that in mind, the lines "n=x" are (must be) your 0 line, they are just ploted over each other. was that your question? E=(n+1/2)ħw so with n increasing you have more energy therefore you can "travel"...

Homework Statement An electron coming from the left encounters/is trapped the following potential: -a<x<0; V=0 0<x<a; V=V0 infinity elsewhere the electron has energy V0 a)Write out the wave function b)normalize th wave function Homework EquationsThe Attempt at a Solution for -a<x<0...

Your problem is the substitution of bounds into the integral What happened to the "lower bound"?

Ok have redone this one last time and i think i got it finally... Vcm i got exactly 1/3 c And final energy i got 2.95*10^-13j (comparing to initial of 2.98 and accounting for rounding mistakes i think this is pretty good!) I resolved the problem like i described in my last post, difference was...

ok so i took a break from this, i was obviously tired of studying this, and moved on to other things. Today i came back to this. i quickly got the same expression for vcm and i got Vcm=0.045c and in the new frame got Ephoton=1.039*10-13J and, Ee-=Ee+=Ephoton, Pe+=-Pe-=1.87*10-22mkg/s ok i must...

thanks for yur answers sorry about the overlap thing it was a translation problem. and the problem i presentd was in an actual exam but all of them always have a problem like that. im going to take a carefull look at both books I've got

so I am finishing up my studies of intro to quantum mechanics, and this is not in my book and looking at previous exams i have to know this for single electron atoms/ions. one of the problems was somethin like "the wave function of an electron is the overlap of the orbitals: Ψ=aΨ1s+i/√3Ψ2p+¾Ψ3s...

so for photon 1 p1c= E1 ; (v/c)E= vp; p'1=γ[E1−(v)p1]; then p'2=-p'1; E1-vp1=-E2+vp2; E1+E2=v(p1+p2); v=c

ok so since p' is the same for both photons i get: p1c+p2c=v/c(-E2+E1) this doesn't work

ok but doesnt give |p2cm| = |p1cm|

yes so i used : to find pcm

momentum of photon 1 in CM frame

i don't get it photons have no mass how do i find "v" of the cm with photons? am i to use: γ(p1cmc−(vcm/c)E)= γ(p2cmc−(vcm/c)E) ?

doing this did not give correct results i got 0.9c and 0.72c this does not conserve energy... i got vcm =0.30c; i used the previously calculated energy and momentum, and used the formula you stated to calculate the new energy; γ[E−(v/c)pc]., with γ= 1/sqrt(1-v2/c2) E and p of the particle and v...

And the speed of the second frame is Pcm/2me?