at around 6:00 you assume the function only contains sines and cosines instead of letting it be complex exponentials, it ends up the same because once you assume those complex exponentials form an even function, then you arrive at sine anyways
No a particle is only free if its energy is greater than the potential everywhere, in the second region the energy of the particle is greater than V but it is still less than the maximum value of the potential globally :)
If E = Potential Energy(V) + Kinetic Energy(K), then, how E could be lesser than V ? For E< V, K will have to be negative, which is impossible. Please comment.
@@NiflheimMists Yes, E CAN be less than V(x) for several points (so for several values of x). It's just that E can't be less than V(x) for ALL points (so for all values of x), because then the wave function can't be normalized (it will either be 0 everywhere, or it will blow up at x=-inf. and/or x=inf.).
The outside of the well is 0 joules. It's when the potential is zero. Since we can't escape the potential (bound states), the Energy is less than zero.
Sine is an odd function, because it is antisymmetric about the origin. Cosine is an even function because it is symmetric about the origin. Both symmetric (even) and antisymmetric (odd) wavefunctions are both, in general, solutions for symmetric potentials, because the magnitude squared of either a symmetric or antisymmetric function is symmetric.
video less than 30min still way better than lectures in whole semester
I need exact analytical solution to the energies ..can YOu do it for me?
THANK YOU SO MUCH YOU'RE LITERALLY SAVING MY INTEREST IN QUANTUM HERE
Thank you so much.. On 31st may'2021 I've presentation on "Finite square well potential". Nd I found this☺
at around 6:00 you assume the function only contains sines and cosines instead of letting it be complex exponentials, it ends up the same because once you assume those complex exponentials form an even function, then you arrive at sine anyways
Thanks for explainning Better than everyone.
👌
This is exactly what I was looking for, amazing!!
hi, great video c: I was wondering, what happens if the potential is odd?
best explanation
Your video is well done!
Look at the bottom of 3:26, there is an error.
Yeah, leftmost term should have d2ψ/dx2 instead of just ψ
Shouldnt it be the second derivative of psi for when you wrote the time independant schrodinger equation?
Yes
Yes
very good content sir. thank you .. sir kindly make a video on Bound
States for Potential Wells with no rigid walls.
Hi! great explanation . Which software do you use to write?
please help
why E is greater then the potential V in Region 2 ?2:35
why is V greter then E in Region 1 and 3 ?
see the graph at 1:57. E(x) is always in between o and -V0. Hence E(x) is a negative number. In region 1 and 3, by definition, V(x) is 0, so EV.
In the 2nd region E>V, does this mean the particle is free in that region?
No a particle is only free if its energy is greater than the potential everywhere, in the second region the energy of the particle is greater than V but it is still less than the maximum value of the potential globally :)
@account1307 bruh he meant in the region and, in the region, the answer is yes, but since the region is finite there are some limitations
21:52 n should only take odd integer values no?
SOS!I have a question @ 04:30 . for x
|E|
I think it is just an assumption that (E
see the graph at 1:57. E(x) is always in between o and -V0. Hence E(x) is a negative number. In region 1 and 3, by definition, V(x) is 0, so EV.
The S.E. is missing the second derivative with respect to x in minute 3:00
Do you have a course on QFT, too? Can anyone recommend one that is as clear and good as this?
Thanks for the explanation!
Where did you graph the functions?
Program's called SAGE, you should be able to find it by searching sage graphing or something similar
If E = Potential Energy(V) + Kinetic Energy(K), then, how E could be lesser than V ? For E< V, K will have to be negative, which is impossible. Please comment.
E cannot be less than V. But it can be negative, if V is.
V < 0
E < 0
V < E < 0 is an allowed energy
@@NiflheimMists Yes, E CAN be less than V(x) for several points (so for several values of x). It's just that E can't be less than V(x) for ALL points (so for all values of x), because then the wave function can't be normalized (it will either be 0 everywhere, or it will blow up at x=-inf. and/or x=inf.).
Would be incorrect to not use the fact that \Psi can be odd or even and do the problem the hard way?
Why is E negative? Is it always negative?
The outside of the well is 0 joules. It's when the potential is zero. Since we can't escape the potential (bound states), the Energy is less than zero.
Thank you so muccchhhhhh. Ur GODDDDDDDD
E is negative because its the binding energy
I need a pdf solution of chapter 11 problems. quantum scattering.. can I get it from you?
where does the 4 come from ?
Why only even solution boundary conditions? I don’t get why sine is not considered for even solutions
Sine is an odd function, because it is antisymmetric about the origin. Cosine is an even function because it is symmetric about the origin.
Both symmetric (even) and antisymmetric (odd) wavefunctions are both, in general, solutions for symmetric potentials, because the magnitude squared of either a symmetric or antisymmetric function is symmetric.
Just plot a graph of sin function from -pi/2 to +pi/2.
You will see sin function is not symmetric about vertical axis.
Easily understood
What happens if the potential is positive? Solution without imaginary roots?
+Дејан Гујић yes if the V is +, inside the well ,the solutions are exponential without imaginary
What about last slide (no9)
What about cos( la)=0?
At 9:30 "if we want to have an even function for psi we cannot have a sin term". Can someone explain please
sin(x) is an odd function so even if you had any even terms, the function for psi will always be odd if there's a sin
cos()/sin() = cot()