Potential functions in the Schrodinger equation

[@BongboBongbong]

My best guess: a) Bad wavefunction, has discontinuous first derivative (it's not defined in the borders), so the second derivative is also discontinuous so we can't say that the second derivative is 0 at the borders (as required by V(x) = E), b) Good wavefunction, second derivative = 0 at the borders, decreases exponentially outside of the box, tends to zero as x goes to infinity, it has definite parity (as required by symmetric potential), normalizable c) Bad wavefunction, not normalizable (the square of the modulus will not tend to zero as x goes to +- infinity), no definite parity while it should have definite parity, d) Bad wavefunction, first derivative is not smooth so second derivative is not continuous so no way to verify whether d^2/dx² ψ = 0 at the borders.

[@andreamelendez1423]

same same

[@matrixate]

B - smooth, continuous, and differentiable everywhere. It also follows the V>E=concave toward X-axis.

[@Justmetmt24]

I love that, "kisses the x-axis." Thought I could only get personification like that in literature. I'm gonna say that now.

[@MCube96]

Best explanation of the different potential scenarios that I found. Thank you for this! :)

[@Maxwell_Integral]

at 17:18 is he saying that the 2nd derivative of psi is continuous or discontinuous?

[@manaoharsam4211]

A very good teacher.

[@a1ang0r85]

For the region that E is greater than potential V, which makes second derivatives of psi negative, then psi should towards the axis, vice versa. Answer C matches what we have learnt. For the region on the right, the curve continuous pointing downward, which means that psi may be negative.

[@a1ang0r85]

However, when we consider the boundary condition for normalizing the wave function, the curve should towards the x-axis, so C is not the answer.

[@a1ang0r85]

For a and d, there is discontinuity is derivative which is not smooth. For b, the curve is smooth and continuous, and also towards the axis, so it is the answer.

[@neutropie6383]

B) is my best guess as it has continuous first derivative and also inside the potential region(where E>V(x)) the wave function is oscillatory and also have no sharpe edges i.e continuous at boundaries. According to boundary conditions If V(x) has finite discontinuity at x=x° the wave function must be continuous and differentiable at x=x°

[@kirthikaranantharam7380]

Dr. Carlson - you refer to "the text book" in your videos. Who is the author and what is the edition? Thank you!

[@FlintPet]

Griffith something

[@siddarthjawahar5452]

Umm.. What you are plotting here is the stationary state and not the actual wave function right? Coz the wave function has time dependence As well as well as space(assume 1-D) dependence which means we'd need a 3 dimensional plot

[@mike4ty4]

Another interesting and rather important system that can still be solved explicitly is the Morse potential. This one provides a more accurate model for the vibration of a chemical bond than the harmonic oscillator model does - in particular, while the latter is only good for the first few energy steps, the former is good enough all the way up to around bond rupture and in fact can model the rupture itself. The still more accurate Lennard-Jones potential, though, does _not_ have an explicit solution at least as a finite-term formula. (Expansion of the wave function as an infinite sum still is/should be possible.)

[@nailanoushin1442]

loved the video since it was explained in so detail and with basic step.Thank you!

[@bonbonpony]

01:30 So is V(x) the POTENTIAL or POTENTIAL ENERGY? Because, you know, these are not exactly the same things :q Make up your mind please. 09:48 What textbook you're talking about?

[@fireblizzard2287]

Its griffith's introduction to quantum mechanics

[@prathamhullamballi837]

It's the potential energy, but we normally just call it the potential even though they are not quite the same.

[@gerrynightingale9045]

"All of the energy and matter that existed still exists. Matter does not create energy of itself. The actions of matter enable energy to become manifest".

[@FlintPet]

my guess a) no istnt smooth, curves wrongon Left side b) smooth curves wrong on left side. looks good on right sidde c) All correct d) Has to curve When E

[@Oh4Chrissake]

18:00 "The first derivative of psi won't ever show a corner like this." Did you mean, that psi itself won't show a corner?

[@koenth2359]

Very helpful course, great teacher!

[@sujaysinha81]

Nicely explained

[@Thenewgreenmaju]

c and b are right-a and d have discontinous first derivaties at boundaries. am i right?

[@FlintPet]

b isnt correctly curved i think

[@ak49gun123]

c is not correct at the right hand side boundary.

[@ricomajestic]

Why is it that some teachers use potential and potential energy interchangeably? That is just sloppy teaching. In science, words matter and most things are precisely defined.

[@0Navin0]

good work sir..thank you.

[@richardwedlich1021]

great. thanks.

[@endaleabebekumsa7271]

Thanks

[@herman9737]

you never put my psi in the corner

[@yoavavinir6288]

thank you! !

[@Dragonshadowbob]

Hey howdy! I'm a student at Emory University down in Georgia, and I'm currently taking a Physical Chemistry course. I love my professor, but he is kind of a bad lecturer, but these videos are saving my bacon! It makes me sort of sad that most people won't ever go deep enough into chemistry to truly appreciate just how simple and understandable your explanations are. Just after this video alone, I feel like I've mad a serious breakthrough in my understanding of wavefunctions as they relate to energy and probability. I just want to take the time to thank you for putting the time and effort into something like this. There's not a huge audience for content like this, but I guarantee that my classmates and I will be binging your videos start to finish!

[@giannymarx4434]

james

[@Dragonshadowbob]

@@giannymarx4434 you have no proof

[@antongolles8896]

b