11:20 the equation is for x not =0 , but at around 14:00 we are considering the node for the first excited state (psi=0 at x=0)for the same equation and then try to figure out a solution. isn't that a contradiction in itself? that maybe why we don't get any solution for the 1st excited state and so on.
Hi, I have the same question. I thought maybe it just a kind of analytic continuation. Consider that the left and right sides must be connected somehow at x=0. One possible way is mixing the wave function of both sides, namely, 1/2(exp(-Kx)±exp(Kx)) which gives the guessing solutions, sinh(x) and cosh(x). How do you think about it now?
The delta function has been seen as a finite well which becomes more and more deep and narrow. So, however small the width of the potential is there will be a node at x=0, for the first excited state. So, I think in the limiting case also, we can expect the node for the first excited state.
because delta function is even, so its bound states must be even or odd, and having increasing nodes by the increment of 1. since the ground state is even and has no node, the 1st excited state must be odd and have 1 node. And since the 1st excited state is also odd, the node has to be at x=0
@@bhaskararaosuravarapu9260 that's a different problem. If your potential is possitive, it acts as a opaque barrier. First of all, there's no bound strate since your potential is possitive. Then, as a repulsive potential, it makes the probability of finding the particle at x=0 go down so it alters the waveform of your solution. I think it's a scattering/tunnelling problem.
Its is attractive dirac delta potential & therefore here, the energy is negative for V tends to + & - infinity & here the system is treated as bound state. Also for repulsive dirac delta, Potential is +ve & it is treated as scattered states.
We don't take the well upper side because we know that if energy level is lesser than the minimum value of potential then any physical solution doesn't exist! For your case we have to choose +ve energy levels only! But if you make delta function -ve then you have the chance to vary your energy levels from -ve to +ve any value! You would get more fascinating result and a complete analysis for the discrete energy levels! That's why we usually do so!
Using units to figure out E? Smart!!! I'm so happy that I found this video!
This professor is excellent.....I enjoyed his books too.
please tell me the book's name
@@edufijiks3705 mastering quantum physics
11:20 the equation is for x not =0 , but at around 14:00 we are considering the node for the first excited state (psi=0 at x=0)for the same equation and then try to figure out a solution. isn't that a contradiction in itself? that maybe why we don't get any solution for the 1st excited state and so on.
Hi, I have the same question. I thought maybe it just a kind of analytic continuation. Consider that the left and right sides must be connected somehow at x=0. One possible way is mixing the wave function of both sides, namely, 1/2(exp(-Kx)±exp(Kx)) which gives the guessing solutions, sinh(x) and cosh(x). How do you think about it now?
The delta function has been seen as a finite well which becomes more and more deep and narrow. So, however small the width of the potential is there will be a node at x=0, for the first excited state. So, I think in the limiting case also, we can expect the node for the first excited state.
because delta function is even, so its bound states must be even or odd, and having increasing nodes by the increment of 1. since the ground state is even and has no node, the 1st excited state must be odd and have 1 node. And since the 1st excited state is also odd, the node has to be at x=0
Absolutely well done and definitely keep it up!!! 👍👍👍👍👍
i realize it is quite randomly asking but do anybody know a good site to stream new tv shows online ?
@Bjorn Douglas try flixzone. You can find it by googling :)
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@Bjorn Douglas i watch on flixzone. You can find it by googling :)
What will be changed if strenth alpha is doubled ...???
great
Can anyboday explain why that negative sign for Dirac delta potential
Its because its a well. So the potential is negative
@@jakeobrien809 then why don't we take it as upper side
@@bhaskararaosuravarapu9260 that's a different problem. If your potential is possitive, it acts as a opaque barrier. First of all, there's no bound strate since your potential is possitive. Then, as a repulsive potential, it makes the probability of finding the particle at x=0 go down so it alters the waveform of your solution. I think it's a scattering/tunnelling problem.
Its is attractive dirac delta potential & therefore here, the energy is negative for V tends to + & - infinity & here the system is treated as bound state. Also for repulsive dirac delta, Potential is +ve & it is treated as scattered states.
We don't take the well upper side because we know that if energy level is lesser than the minimum value of potential then any physical solution doesn't exist! For your case we have to choose +ve energy levels only! But if you make delta function -ve then you have the chance to vary your energy levels from -ve to +ve any value! You would get more fascinating result and a complete analysis for the discrete energy levels! That's why we usually do so!
Lol 15:21
What will be changed if strenth alpha is doubled ...???
Height will be short..