Finite square well bound states

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This video describes the finite square well general solutions, boundary condition matching, even and odd structure of the solutions, and a graphical representation of the solution to the equation for allowed energies. (This lecture is part of a series for a course based on Griffiths' Introduction to Quantum Mechanics. The Full playlist is at ru-vid.com?list=...)

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31 июл 2024

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Комментарии : 52

@leetingfung 7 лет назад

video less than 30min still way better than lectures in whole semester

@changbadinesh 4 года назад

I need exact analytical solution to the energies ..can YOu do it for me?

@annakiseleva7418 4 года назад

THANK YOU SO MUCH YOU'RE LITERALLY SAVING MY INTEREST IN QUANTUM HERE

@monicapym8948 2 года назад

This is exactly what I was looking for, amazing!!

@SampleroftheMultiverse 9 лет назад

Your video is well done!

@himangshuchakraborty1760 3 года назад

Thank you so much.. On 31st may'2021 I've presentation on "Finite square well potential". Nd I found this☺

@vivekpanchal3338 5 лет назад

Thanks for explainning Better than everyone. 👌

@aniruddhavichare5425 8 лет назад

best explanation

@davidhand9721 3 года назад

Do you have a course on QFT, too? Can anyone recommend one that is as clear and good as this?

@sharairamirez9274 3 года назад

hi, great video c: I was wondering, what happens if the potential is odd?

@user-rg1nt9lf4s 5 лет назад

very good content sir. thank you .. sir kindly make a video on Bound States for Potential Wells with no rigid walls.

@emillytabara9410 3 года назад

Hi! great explanation . Which software do you use to write?

@Dr.kcMishra 6 лет назад

Nicely Explained

@ashwith 10 лет назад

Would be incorrect to not use the fact that \Psi can be odd or even and do the problem the hard way?

@onlineearning8323 3 года назад

I need a pdf solution of chapter 11 problems. quantum scattering.. can I get it from you?

@edithtea9477 7 лет назад

Thanks for the explanation! Where did you graph the functions?

@the-fantabulous-g 4 года назад

Program's called SAGE, you should be able to find it by searching sage graphing or something similar

@BLVGamingY 5 месяцев назад

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

@arupmarik 5 лет назад

What about last slide (no9)

@eliaskaroui5665 4 года назад

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 ?

@puikihung5882 4 года назад

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.

@Chemidan92 10 лет назад

At 9:30 "if we want to have an even function for psi we cannot have a sin term". Can someone explain please

@LimosRock1 9 лет назад

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

@rukwoo4418 5 лет назад

What about cos( la)=0?

@datsmydab-minecraft-and-mo5666 3 года назад

Shouldnt it be the second derivative of psi for when you wrote the time independant schrodinger equation?

@ifrazali3052 Месяц назад

Yes

@ifrazali3052 Месяц назад

Yes

@niamphmotley Год назад

where does the 4 come from ?

@pranavgeorge992 3 года назад

In the 2nd region E>V, does this mean the particle is free in that region?

@account1307 3 года назад

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 :)

@BLVGamingY 5 месяцев назад

@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

@shivanandashekhar6580 9 лет назад

Easily understood

@kokori100 10 лет назад

SOS!I have a question @ 04:30 . for x

@kartikaloria8256 5 лет назад

|E|

@rubenlauwaert6673 5 лет назад

I think it is just an assumption that (E

@puikihung5882 4 года назад

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.

@xichen9674 8 лет назад

Why is E negative? Is it always negative?

@monku1521 5 лет назад

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.

@YourAverageHater 9 лет назад

What happens if the potential is positive? Solution without imaginary roots?

@zeenaligog 8 лет назад

+Дејан Гујић yes if the V is +, inside the well ,the solutions are exponential without imaginary

@beedeelovesyouall 2 года назад

Thank you so muccchhhhhh. Ur GODDDDDDDD

@JohnDavid-iq9rz 6 лет назад

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 4 года назад

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

@frede1905 4 года назад

@@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.).

@kemalaziz9696 5 лет назад

Look at the bottom of 3:26, there is an error.

@NiflheimMists 4 года назад

Yeah, leftmost term should have d2ψ/dx2 instead of just ψ

@Paradox586 4 года назад

Why only even solution boundary conditions? I don’t get why sine is not considered for even solutions

@NiflheimMists 4 года назад

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.