Scattering state solutions to the delta function potential TISE

[@mlungost]

Just know, in 2018 these slides are still the holy grail for Quantum students

[@jasonyao3753]

Just know, in 2021 these slides are still the holy grail for Quantum students

[@dreamworld2584]

jst now in 2022 still holy grail

[@ya_a_qov2000]

2023

[@rakhuramai]

@@ya_a_qov2000 2024

[@tentrot4420]

2024 too

[@LydellAaron]

Beautiful. 10 years later. Scattering parameters and scattering transmission parameters are fundamental to performing quantum computing. I think it's becoming more understood or accepted, how these same scattering phenomenon can model energy reaction and propagation problems, including human behavior.

[@andrewstallard6927]

Q. Can any initial conditions be expressed as a sum of scattering state solutions? A. Given an infinite number of terms the answer is yes. Q. Explain why the bound state is not possible for the delta function barrier? I think is fairly obvious that if the energy barrier exists only on one side the wave has infinite freedom on the other side. If you have two delta-function barriers than we are back to the particle-in-the-box.

[@yasirkhanniazi4462]

i follow ur lectures nd they r most benificial to me, thanks fr all

[@rmenchoachupicachu]

Best explanation by far! Great thank you

[@mlungost]

I know right

[@JH-if5rv]

Thank you, excellent explanations

[@solsticetwo3476]

I don't get it. Why is not just a free particle when E >0? The analysis looks for me as the one for a delta barrier rather than for a delta well.

[@hershyfishman2929]

This is indeed an unintuative result of quantum mechanics. He discusses this at 19:27

[@1ashad1]

Can anyone help me with just solving for the reflection and transmission coefficients? I am unable to do the basic maths to get the answer (my only concern is i am not able to get rid of 'i' in denominator). Can anyone help? thanks

[@timduncankobebryant]

professor, how come for the general solution, neither of the terms blow up?. One of the coeffs should be zero right?

[@rmenchoachupicachu]

+timduncankobebryant These terms are representations of Euler's identity. You basically get sin and cos terms instead of exponentials.

[@hershyfishman2929]

For E0 the solutions have i in the exponents, so they just rotate forever in the complex plane.

[@jonelreidmanuel]

thank you so much for this.

[@EmmanuelMlambani]

That's excellent

[@abhasoodan7982]

why do we have iota in the solution of scattering states but not in bound states?

[@hershyfishman2929]

When E0 (scattering), -2mE/ℏ is negative, so the differential equation is ψ'' = -k^2ψ. The solution to such differential equations has i (taking the derivative of e^ikx twice will give you -k^2(e^ikx) and similarly for e^-ikx)

[@timetraveller1237]

why is it that in this lecture brant did not cancel the exponential powers which blow up to infinity like in the last lecture

[@juhokupiainen5515]

These solutions are free particle solutions (the slight difference of having imaginary numbers in the exponent ). Look back a few lectures. You normalize them by creating a wave packet.

[@timetraveller1237]

ooh so are you saying that because of the term "i" it is essentially just a complex rotation but in the previous lectures the missing "i" meant the solutions blew up so it is not necessary to cancel it for free particle. and for your second part i don't really understand the notion of wave packet and how that relates to Fourier transforms and how that helps us to normalize them but thanks anyways if i got what you said correctly if not please do feel free to correct me thanks anyways it has been bugging me for a long time!!!

[@arunkancharla4455]

if we expand e^ikx=cos(kx)+i*sin(kx) , the sine and cosine are finite for any value of x.

[@ENetArch]

How did you reduce f - g = a (1+ 2iq) - b (1 - 2iq) ???

[@ENetArch]

If you substitute the values that F and B equal into the equation, while setting g = 0, you get A = A. I'm trying to walk forwards now to the answer.

[@ENetArch]

set g = 0 Recall that F + B = A .. where A is the incident amplitude, and F and B are the reflection and transmission amplitudes. F is some percentage of A, or ( F = A (%)) while B is the remainder of the percentage (B = A (1 - %)). A = 100% B = A % F = A (1 - %) F / (1 -%) = A B / % = A F / (1 - %) + B / % = A F (%) / (1 - %) % + B (1 - %) / % (1 - %) = A F (1 - %) + B % = A % (1 - %) === B = iq / (1 - iq) F = 1 / (1 - iq) B + F = (1 + iq) / (1 - iq) = 1 or A === F = A (1 + 2iq) - B (1 -2iq) subtract b from the left and add b to the right F + B (1 -2iq) = A (1 + 2iq) divide both sides by (1 - 2iq) [ F + B (1 -2iq) ] / (1 + 2iq) = A times the left side by (1 + iq) / (1 - iq) [ F + B (1 -2iq) ] (1 + iq) / [ (1 - iq) (1 + 2iq) ] = A Distribute (1 + iq) [ F (1 + iq) + B (1 + iq) (1 -2iq) ] / [ (1 - iq) (1 + 2iq) ] = A === % = (1 + iq) (1 - %) = (1 - 2 iq) 1 / % = (1 - iq) 1 / (1 - %) = 1 / (1 + 2 iq) === [ F % + B % (1 - %) ] / [ % (1 - %) ] = A B = A % F = A (1 - %) B = A (1 + iq) F = A (1 - 2 iq)

[@taibilimunduan]

quantically, well = wall!

[@xinzeng-iq7zv]

is this guy a dentist on some naruto shit