Infinite square well example computations and simulation

[@matrixate]

Absolutely the most comprehensive tutorial I've seen. This smokes everyone trying to teach QM. I wish someone like you could have been my professor Dr. Carlson. I will pay this forward one day in honor of you.

[@AnthonyDavid59]

Loved seeing Sage in your video. I used it for my ODE modelling over a decade ago.

[@aprodriguesoficial]

Wouldn't the sign in the exponentials (in blue) at about 15:56 be a negative sign instead of a plus?

[@jimdogma1537]

I've been following this series so far and it has been truly fantastic. Sadly, though, this episode has left me a bit confused. I am uncertain as to how the time part of the SE is applied to the stationary states to affect their evolution. You've been saying all along that the time component of the equation is trivial, but if it affects the evolution of the states, it doesn't seem like it would be trivial. The simulation at the end seemed important and fascinating, but I didn't understand what was going on because all of a sudden everything was relating to the "phase" of something. I don't remember you mentioning the word phase at all up until then in this video or the entire series so far. Darn! So, I'm left a bit confused. On a smaller note, at 14:25, you're dividing the denominator by h-bar? Is that right? The form you have it in earlier, say at 1:50, simply has the h-bar in the denominator. Which one is right? Other than that, great series, looking forward to the rest!

[@sphericalchicken]

The main idea is that the solutions to the full time-dependent Schrodinger equation can be expressed as a sum of the time-dependent form of the solutions to the time-independent Schrodinger equation. The "time-dependent form" of the solutions to the time-independent Schrodinger equation are simply the "stationary states" multiplied by exp(-i E t / hbar). That exp(...) factor is what I'm referring to as a "phase". It's a complex number with magnitude 1 that rotates around in the complex plane at a rate given by E/hbar. That evolution is "trivial" in the sense that it's easy to write down and compute, and in that it doesn't affect any measurement outcome (hence the term "stationary state"). The time evolution of the full time-dependent Schrodinger equation is more complicated, though, since you have a superposition of stationary states. Each stationary state evolves in the "trivial" way (the phase in the complex plane changes with time), but since each stationary state evolves at a different rate (they have different energies, so the exp(-i E t / hbar) factors will be different for each state), the overall superposition behaves in a rather complicated way -- at one time, the exp(-i E t / hbar) factors may all be positive and real, at another time one might be positive while the rest are negative, etc., and there's the difficulty associated with complex numbers... The simulation shows the results of doing such analysis.

[@jimdogma1537]

Brant Carlson Thank you so much for responding. I really like your teaching style and approach to the subject and want to understand as much as I can. Hopefully, you'll do the same for the other classes you may teach as you did for this one, so we can all enjoy them.

[@SampleroftheMultiverse]

Interesting, personal i like the U shape wave point before the transition to the next energy levels. I seen where using F transformer using all wave frequencies you can make U shape waves. 31:20

[@manaoharsam4211]

Excellent teacher once again.

[@MrEzystreet]

The Falstad applet is amazing. Thank you!

[@m1ke50]

sir, at 13:35, why did C2 = 0 and how did you find the C3 and C1 ?

[@airscience1597]

I think it was with Sage

[@nisarqasim9965]

with all due respect sir which software you used in lecture please tell me how to access this computational software

[@κπυα]

30:19 This psi is on fireeeee 🎶🎶🔥🔥

[@imppie3754]

At 2:55 how did we draw that graph? Sorry for such a lame question but i'm really bad with graphs.

[@hershyfishman2929]

just plug in different values of x and mark the corresponding point on the graph. At some point, you'll have enough points on the graph to get the picture of the entire graph

[@mohamadmartini8559]

But the initial conditions are the V(x) and the boundaries not a new psi(x)?! where did that new psi(x) come from? Shouldn't the initial psi be the first n of the sine wave function as a solution of Schrodinger in the particle-in-a-box problem?

[@MiguelGarcia-zx1qj]

I was about to ask something similar. I'm well versed on Fourier Analysis, and on the PDE of the (say, non Quantum) onedimensional wave propagation. So all the mathematical trickery is a piece of cake (even the use of a symbolic math computer tool). But my very big "but" is which physical (or experimental) setup leads to this "initial conditions" Psi(x,0)? I don't see how this inicitial conditions are related to the infinite square well setup ...

[@fireblizzard2287]

30:20 wave functions are on fire.

[@fpucol554]

Great example! Thank you! I tried to do the example on my computer using Mathematica but I changed the initial conditions to ONE from zero to a and ZERO otherwise. The Fourier series looks great with 25 to 30 terms and the sum-of-c[n]^2 terms add up nicely to one as they should. But the c[n] now depends on 1/n instead of 1/n^2 as in the example, and when c[n]^2 is multiplied by the n^2 dependence of E[n], the n's cancel and the series to calculate the total energy diverges. My question: Why would some perfectly innocent looking initial conditions make the total energy go to infinity? Is there some additional restriction on the initial conditions besides going to zero at plus-minus infinity?

[@turboleggy]

You know as soon as he put the graph up there of the initial conditions I noticed the kink. The kink would make the first derivative undefined and the second derivative and even worst undefined thing. I'm not exactly sure how having the undefined second derivative makes the energy infinite other than you'd need infinite fourier terms to make the kink but this link helps: physics.stackexchange.com/questions/262671/can-a-physical-wavefunction-be-non-smooth-its-first-derivative-is-discontinuous#:~:text=You%20cannot%20add%20a%20finite,get%20a%20non%2Dsmooth%20function.

[@Dekoherence-ii8pw]

2:50 When you drew that tent... MASSIVE FLASHBACK. Scout camp. Putting up tents. 34 years ago!

[@xXxBladeStormxXx]

How exactly do we even give a quantum system initial conditions like the "tent"? When I studied Vibrations and Waves this was all very intuitive since you could actually pluck a string. Here it's a particle! represented by a Wave equation!! which is like a "tent"!!! at t=0? How is that even possible. I may be completely missing the point here, but this is a difficult subject so go easy please...

[@MiguelGarcia-zx1qj]

I asked almost the same question as a response to Mohamad Martini :) After thinking about it a bit, I think the "tent" is not a realistic initial conditions type. I think that what it is physically sound is the case of Psi (x, 0) equal to a Dirac delta, which would be obtained after observing the position of the particle. The math part is slightly more complicated (but you can't study Fourier Analysis seriously without getting familiar with the Dirac delta)

[@Dekoherence-ii8pw]

30:20 Looks like it's on fire!

[@BPHSadayappanAlagappan]

How the two facts are related!?

[@lifeofphyraprun7601]

Time variability arises due to superposition of stationary states.Let's say 2 of them psi_a(x,t),with energy Ea and psi_b(x,t) with energy Eb.We take their superposition to get psi(x,t) = C1 psi_a(x,t) + C2 psi_b(x,t). The probability density function is psi×psi* which can be computed to be made up of terms whose time factors (exponentials) either cancel out or are of the form e^((Ea-Eb)×constant).This makes sense,because the frequency of time variation of the probability density function is a physical thing, while the energies of particular states,are as always, defined with respect to an arbitrary reference.The dependence of any expression on that arbitrary reference automatically cancels out once we take the difference of two energies,and hence the time variability doesn't depend upon it.This is clearly analogous to how we are only concerned with differences of potential energy in classical mechanics,and what the exact value of potential energy is,in a particular is immaterial,and has no physical significance.The value of an expression for anything that has physical significance should not depend upon that arbitrary reference.

[@imppie3754]

I'm so damn analog....