Sarahisme Posted April 30, 2006 Share Posted April 30, 2006 Hi all I was wondering how you go about normalising the even bound state wavefunction for the finite square well? eg. so i have to find F and D. this what i got so far: i can't seem to find the answer to this 'normalising' problem anywhere! any ideas guys? Thanks Sarah Link to comment Share on other sites More sharing options...
Severian Posted April 30, 2006 Share Posted April 30, 2006 [math]|\phi|^2(x)[/math] is the probability of finding the particle in position x. So [math]\int dx |\phi|^2[/math] is the probability of finding the particle anywhere. What should this probability be? Link to comment Share on other sites More sharing options...
Sarahisme Posted May 1, 2006 Author Share Posted May 1, 2006 should be 1? Link to comment Share on other sites More sharing options...
5614 Posted May 1, 2006 Share Posted May 1, 2006 Sounds right and it's what I thought... but don't take my word for it, I really don't know for sure. Link to comment Share on other sites More sharing options...
Sarahisme Posted May 1, 2006 Author Share Posted May 1, 2006 so can i go: [math] \int^a_0D^2cos^2(lx)dx+\int^\infty_aF^2e^{-2kx}dx=1/2 [/math] ?? Link to comment Share on other sites More sharing options...
swansont Posted May 1, 2006 Share Posted May 1, 2006 Since the square of the wave function is symmetric about x = 0, that should be fine. Link to comment Share on other sites More sharing options...
Severian Posted May 1, 2006 Share Posted May 1, 2006 Yes. That is what I meant. You need a further condition though. This just gives you a relation between D and F. You need another relation to solve for them. So, what should happen to the funtional form of [math]\psi(x)[/math] as you pass [math]x=a[/math]? Link to comment Share on other sites More sharing options...
Sarahisme Posted May 2, 2006 Author Share Posted May 2, 2006 ok so could also use the continuity of psi at x = a [math] Fe^{-ka} = Dcos(la) [/math] ? Link to comment Share on other sites More sharing options...
Sarahisme Posted May 2, 2006 Author Share Posted May 2, 2006 then i get this [math] D = \sqrt{\frac{2kl}{\frac{1}{2}k(sin(2la)+la/k)+lcos^2(la)}} [/math] and [math] F = \sqrt{\frac{2kle^{2ka}cos^2(la)}{\frac{1}{2}k(sin(2la)+la/k)+lcos^2(la)}} [/math] Link to comment Share on other sites More sharing options...
Sarahisme Posted May 5, 2006 Author Share Posted May 5, 2006 k thanks, i got it now! Link to comment Share on other sites More sharing options...
Severian Posted May 5, 2006 Share Posted May 5, 2006 Incidentally, there was a slight subtlety in this problem. Normally one would expect the second derivative to be continuous also - so the gradiaent of the wavefunction would also be smooth. But in this case, the potential itself is not continuous in its gradient (it jumps from -V0 to 0 at x=a) so the gradient of psi is not continuous either. Link to comment Share on other sites More sharing options...
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