Finite square well scattering states

[@fr0styy202]

its 2 am and my final at 11am. As a frequent college all-nighter, I thank you for the clarity, good sir.

[@sasasatatata602]

solving for T=1 we require sin ( 2a/h * sqrt(2m(E + V0))) = 0 so 2a/h * sqrt(2m(E+V0)) = pi * n so E + V0 = (pi * n * h / 2a)^2 / 2m this looks like E = p^2 / 2m and de brolie says p = h/lambda so lambda = 2a / (pi*n) so we have solutions 1,2,3,4.... so when the half wavelength is an integer multiple of the well width we have a momentum ~ energy which will transmit fully so a standing wave in the well lets a scatter wave "coast by"

[@vorsankoy3548]

thank you very much..I have used a lot from Brant Carlson, from first video till this, its great

[@valeriaitzelarteagamuniz5286]

good video! where did you make the transmission coefficient graph?

[@howwedoMichaelmath]

You're the man! Thanks for making these videos.

[@MiguelGarcia-zx1qj]

15:25 onwards: Say it is an intuition; I think that those dips in transmission are the result of interference when the QWF has a wavelength related to the well width. As Brant points out, akin to the reflective coating in optic systems. If my intuition is right, I'm missing an explicit statement about the issue; more over, considering we are studying wave functions (and phenomena).

[@lancehenning4187]

For anyone struggling with the algebra of resolving the boundary conditions, heres a useful read. quantummechanics.ucsd.edu/ph130a/130_notes/node160.html

[@besidehimself9813]

Very informative, thank you.

[@meeranagarajarao6887]

Most treatment of this physics problem becomes so mathematical and one loses the physical perspective. But this is a very neat approach. Thanks.

[@meeranagarajarao6887]

A very neat approach.

[@enchis1269]

Why you use trigonometrical function as the general solution for region 2 instead of exponential funtion?

[@ljrahn5619]

Yes wondering the same

[@captainhd9741]

@@ljrahn5619 As long as it is the same then it doesn’t matter. Trigonometric functions are related to the exponential function e^ix = cosx + isinx

[@Maymz-uf6bc]

GREAT VIDEO, UNFORTUNATELY MY ASSIGNMENT IS TO UNDERTAKE ALL THAT MESSY ALGEBRA AND END UP WITH the expressions for F and T :( sorry, only realised my caps lock was on half way through writing this...

[@andrewstallard6927]

We have to go through the algebra hell on our own for this one.

[@OM-el6oy]

Should have used .lower() method

[@TiagoBatista-uc8nm]

Same :(

[@captainhd9741]

@@TiagoBatista-uc8nm Actually you can just go through all that mess for T or R and then use a sneaky trick to get the other one without repeating all that hard work. The trick is to remember that T+R=1 so if you have the formula for T then just do 1-T and all of that will be the formula for R

[@TiagoBatista-uc8nm]

@@captainhd9741 oh thank you! I found some PDF on google with the algebric manipulation to get T, I was struggling with that part. But thanks for the response, I appreciate it!!

[@xXxBladeStormxXx]

Why did you skip the formalism chapter? Also, thanks for the videos.

[@arupbiswas8288]

extremely helpful videos!

[@cafe-tomate]

For about two hours now Ive tried to resolve the BC, the two results Ive found are either one nor the other good

[@SampleroftheMultiverse]

Very well done!

[@kartikeypandey499]

why we didn't apply that G=0 approximation on surface x=-a too???

[@Mathletelady]

Thank you for explaining :)

[@davidhand9721]

Here's how to calculate things with the wavefunction: Don't calculate things with the wavefunction.