Lecture 8: Quantum Harmonic Oscillator

[@lukelively4732]

I love this guy. I love when you can tell how much a professor loves a subject.

[@shibanisahani1033]

Yes it shows in his face. I love him too 💜

[@drummerboy91893]

ughhh gosh this is so good. my prof never goes through the mathematical steps like this one does. this one gives you mathematical insight and reasonable physical arguments. my prof treats us like we're physicists!

[@chandus2496]

AMEN!!! We should be thankful to MIT for doing this for free.

[@everlastinGamer]

When the lecture is so good that KZbin viewers clap alongside the live student audience. Awesome, thanks so much.

[@owen7666]

KZbin needs a clap button. Giving a like is too mainstream

[@saskiavanhoutert3190]

@@owen7666 Fine that you like it, kind regards

[@bonbonpony]

They only want to know if you watch their ads, not whether the content is actually useful to you. They'll rather remove the voting entirely than let you vote on videos.

[@chinonsougwumadu9340]

This is the only professor that makes quantum physics seem like some classical music... sometimes I just watch his lectures lying on my bed.

[@revanth36]

Right now I'm doing that😅😂

[@orsozapata]

I always watch science lectures lying in bed :)

[@charlesnathansmith]

I started watching this lecture series a few years ago and was so enamored with Adams' teaching style and personality that I wasn't giving the ones Zweibach filled in on as much attention. Rewatching I keep realizing just how unfair I was being. He has an insane passion for conveying information, and can immediately respond to any question no matter how tangential or based on a misunderstanding that ties it back into what he's covering in a way that gives some crazy new insight. That's how science communication is supposed to work and I'm fully here for it even if I'm a bit late to the party

[@CARLOSROMERO-MathFermat]

Solution of the Schrödinger Equation of the Quantum Harmonic Oscillator begins at: 25:30

[@brno322]

+CARLOS ROMERO thanks. And if you're already familiar with some of the relationships then it's at 34:04

[@dhoonygo]

You guys save my time. THX.

[@mehmoonarubab2064]

Thank you sir😂

[@saptarshidas4515]

And I am watching this comment at 25:27...ugh..

[@aldenphillip765]

dunno if anyone gives a damn but if you guys are bored like me during the covid times then you can stream all of the new movies on InstaFlixxer. Have been binge watching with my brother recently =)

[@anugrahmathewprasad172]

This is so good.. He explained quantisation so well! Prof. Adams set it up so well in the last lecture and here it just became so clear

[@robertsalazar2770]

Hey! I'm retired and watching these lectures for fun! I agree, the instructor's energy should be part of schodingers equation.😁

[@okotray4577]

I appreciate the online lectures from MIT, especially this professor.

[@Dr.Sortospino]

This professor speak so clear (nevertheless the spanish accent) that you can gain time putting the speed 1.5x :D

[@flatisland]

that was a fantastic suggestion! it's cool he doesn't speak like Mickey Mouse when you speed it up. And if you set it to 1.75 he almost sounds like Allan Adams :-)

[@PrabandhamSriniketan]

Watching this a day before the exam, wish me luck!!!

[@adrianlowenberg]

I am watching this content as a mathematics student, which is interested in physics. I guess this will be the topic of the everywhere announced 8.05, but the quantization follows generally as the Eigenvectors of a Hermitian operator, like E here, are orthogonal to each other, and the Hilbert Space where the Wavefunctions live in is of course separable, so there are only countably many different Eigenvalues, therefore possible energy measurements. This also explains why the Eigenfunctions are a class of orthogonal ploynomials. The physical argument given here is that the Wavefunction must be normalizable, but mathematically this only means that it is in the Hilbert Space, or that it is integrable. What I find most interesting here is the techniques used to gain intuition about the differential equation, and therefore solve it explicitly. I think however that a bigger mathematical background would help many students to know which "rough" solutions are okay and which cannot be. I am very grateful however to see this very different teaching style to the one I am used to for free!

[@DrDavidHerreraMath]

Eigenvectors do not necessary exist for general Hermitian operators on an infinite dimensional Hilbert space. For instance, the multiplication operator by x in L^2([0,1]) has spectrum [0,1] with no eigenvectors. Being Hermitian, just as in the case of matrices, implies that the eigenvectors are orthogonal. You calculate the inner product (f | g) of two eigenvectors f and g and use some algebra to show that E_1(f | g) = (E_1 f | g) = (P f | g) = ( f | P g) = E_2(f | g). So, (E_1 - E_2) (f | g) = 0. The existence is more subtle. On finite intervals, one can use en.wikipedia.org/wiki/Spectral_theory_of_ordinary_differential_equations#Spectral_theorem You can find a nice exposition of this in Haim Brezis' textbook "Functional Analysis, Sobolev Spaces and Partial Differential Equations" For unbounded intervals, you have to be more careful and many related topics is the subject of mathematical physics / functional analysis research. For instance, see this undergraduate summary for multiple dimensions: www.math.uchicago.edu/~may/VIGRE/VIGRE2009/REUPapers/Emberton.pdf Effectively, the second derivative (Laplacian in multiple dimensions) term is an unbounded self-adjoint operator. The potential there can be given conditions (like we saw in the previous lecture) so that the differential operator has a compact inverse. A compact operator is a limit of matrices where the "infinite part" gets smaller in contribution ("norm") as the matrices gets larger in dimension. Compact operators have complete spectrum made of discrete eigenvalues, the eigenvalues converging to 0. So, the inverse has discrete eigenvalues going to infinity.

[@comrade_marshal]

I hope that you didn't break your computer or worse still, had a heart attack after seeing the way we derive mathematical formulae 😅

[@bonbonpony]

I'm curious what would you say about the next lecture, @adrianlowenberg, from the mathematical point of view (I mean the operator approach to solving the same problem), because I also think that this technique should work for other mathematical problems without all those quantum shenanigans. I see that commutators for operators are not something unique to quantum physics, but a more general idea for handling the cross-terms in differential operator interactions, when usually there's a chain rule involved because variables are mutually dependent. Similarly, factoring operators should work in general. There's one thing though that I find kinda sus: the lowering operator acting on the ground state, because it uses a purely physical argument for why should there be a ground state that, when acted on by the lowering operator, becomes 0. There should be an equivalent mathematical argument that doesn't involve physical terms like energy levels, expectation values, wavefunction normalization etc.

[@pirate0bloodyskull]

Holy cow, this guy is so good.

[@kalj7]

Very nice and understandable derivation. Thank you!

[@DrBouwman]

Wow. Just wow. i wish I could give classes like that

[@sanketdeshpande345]

lecture was intense

[@ranjeetsingh-fw1ij]

Thanks mit . It really helps.

[@indiaview9414]

After so many approximation finally solution of harmonic oscillator land up on the Harmite Polynomial series...

[@darinpeev5820]

Very impressive! He's got a style!

[@bonbonpony]

05:40 Could this be also said that these nodes must be non-repeated roots? Because a minimum/maximum that touches the X axis seems to me like a *quadratic* repeated root, and the one that aligns with the X axis for a while and then sinks down again seems to me like a *cubic* repeated root. But those who cross the X axis at an angle seem to be just regular, non-repeated roots of the first order (linear). Is my observation correct?

[@flatisland]

36:01 1:05:44 1:13:26 he just swallows it I like this series! :-)

[@61gopalprabhulsm70]

With your observational skills, I hope you must have swallowed each and every bit of information he threw at us! XD

[@qpkwsxcb8277]

You should have more likes. I am here for more than a few times your comment makes me laugh every time.😂

[@oogaboogaman0046]

I woke up to this playing on my tv

[@flyingbirds6794]

This video is an excellent video!! I want to thank the lecturer a lot because he made the topic very easy and told in a very beutiful way. Especially the coefficients and exp(u^2) comparison was very very clever and an excellent explanation. Thank MIT ocw very much!!

[@productivelb]

Beautiful lecture!

[@piyushgalav6483]

20:56 only the slope is 0 at turning point not the wavefunction then why that is not possible.

[@SuperSvx]

i had the same doubt

[@richardneifeld7797]

No one said "that" was not possible. The reasoning relied only upon the conclusion that if both slope and value were zero then the only solution to the dif eq is the function is zero at all values of x.

[@bonbonpony]

He briefly hinted at the reason when he said that it's a second-order differential equation, so once you know the value of the function at some point, and its first derivative at some (possibly different) point, then your solution is pinned down (unique). And the only solution that has both the value of a function and its first derivative at some point equal to zero, is the trivial solution - the zero function ψ(x)=0.

[@meetghelani5222]

Love this series.

[@stefkakannenberg5154]

Yes. One has to take the solution for the ground state energy level at T=0K from Stochastic Elektrodynamics in order the differential equation to have reasonable solutions. Right. Why not to just calculate with Stochastic Elektrodynamics from the beginning.

[@ravitrivedi8654]

Really wonderful explanations......

[@ignaciomartinalliati6293]

23:42 as a physicist proof, can you just say that a wavefunction with more nodes has a shorter wavelength and thus more energy? Or the association of wavelength and energy is too sloppy in this context?

[@ey3796]

The wavefunction wont look like the infinite square well, rather itll be some complicated mess of wavelengths so we cant give this reason.

[@richardneifeld7797]

Qualitatively, yes.

[@bonbonpony]

"More nodes per the same unit of length" would be more precise. Because if there were more nodes, but spread further apart, that would still be less energy.

[@MuhammadAsif-nt7zv]

Professor Barton❤❤❤ form Pakistan for nice explainations.

[@sunnypala7098]

Explained very awesomely sr.veryyyyyyy helpful.thank you so much

[@MardukNHR]

A round of applause for the camera operator too.

[@abc99311]

58:18 -if he can start the dummy index from j=2 for second derivative of h, then,why can't he take the dummy index from j=1, for 1st derivative.further,on replacing j by j'+1 will make the coefficient different . Please someone explain this.

[@OcctySun]

S K I believe it is because there is no need to in the first place, for what we are aiming to do here is collect the common u^j factors so as to elucidate a recursive relationship between a_j and a_(j+2) for all integer values of j from 0 to infinity. Here, the -2u factor multiplied with dh/du gives the aforementioned u^j, hence not requiring any change of the dummy index j. If I remember correctly from my Mathematical Methods course, this is a standard way of solving second-order ordinary differential equations.

[@richardneifeld7797]

Shri, infinite series have the property that they are infinite. So there is no highest order term to account for as in a finite series. If that is your confusion then review Taylor series of 1/(1-x).

[@bonbonpony]

The index is just a dummy variable anyway, it doesn't matter how you name it, or where you start counting, as long as the counting is consistent and gives the same terms after expanding the sum. That's why this is legal.

[@nickanna8857]

what happens to the u^j on the end @1:04:04? Why must it be zero? I feel like the answer probably involves the phrase "linear independency" but it's not immediately apparent.

[@joeybf]

Nick Anna The equation is an equation of power series in u, so it simply means that the RHS is equal to the zero power series, i.e. the power series with all coefficients 0. Equality between power series only means that their coefficients are equal everywhere, the same way that two vectors (x_1, x_2) and (y_1, y_2) are equal if and only if x_i = y_i for all meaningful i.

[@LydellAaron]

My understanding of why it is set to zero is because these are types of "harmonic functions" because these functions help describe stable systems that oscillate back and forth (made up of vibrations).

[@richardneifeld7797]

Nick, no one said Uj is zero. Instead, the coefficient of Uj, which is a sub j, is zero for specific energy values. Only those specific energy values result in a cofficient being zero. From the coefficient relations all higher order coefficients also vanish.

[@pawelperkowski1971]

I wonder if the length parameter "a" introduced at 32:00 posses any name or any physical interpretation? In my opinion, the parameter "a" seems to be the width of the space classically available for a particle in a harmonic field, when the particle exists in the lowest possible state, with the energy E=1/2*h-dashed omega. Could anyone comment on it?

[@saskiavanhoutert3190]

Quantum physics is somehow good explained and interesting by this physician, thanks.

[@patrickwang671]

physician is a doctor, he's a physicist ;)

[@saskiavanhoutert3190]

@@patrickwang671 Somehow it's the same, perhaps he say's that because he doesn't like to be ill and has to go to a docter. And he does his job and you are present, don't go to a pub but make your homework please. kind regards and have a good life. And that is also a song, perhaps you know the singer. I Know him again James Brown.

[@Godakuri]

@@saskiavanhoutert3190 wtf did I just read

[@rubensverstappen3458]

wonderful discussion. thank you, MIT!

[@atithi8]

At 56:50 , how does (e-1)psi pop up , i am getting no -1 term

[@psharmacgk]

Be careful with your signs, simplifying your second derivative with the u^2 term should yield: -h''+uh'+h+uh'-eh=0 -> h''-2uh'-h+eh=0 -> h''-2uh'+(e-1)h=0 hope that helps :)

[@g1grace]

I am having the same result.

[@psharmacgk]

are you getting to the first equation in my previous comment?

[@myreneario7216]

We start with -ψ' ' + u^2 *ψ = ϵ * ψ which we can rewrite as: ψ' ' - u^2 *ψ + ϵ * ψ = 0 We calculate ψ' ': ψ' ' = (h*e^(-u^2 /2))' ' = (h' * e^(-u^2 /2) - h * u * e^(-u^2 /2)) ' = = h' ' * e^(-u^2/2) - h' * u * e^(-u^2 /2) - h' * u * e^(-u^2 /2) - h * 1 * e^(-u^2 / 2) + h * u^2 * e^(-u^2 /2) If we divide that by e^(-u^2/2) we get: ψ' ' / e^(-u^2 /2) = h' ' - 2h'*u - h + h*u^2 So if we divide our previous equation by e^(-u^2/2) we get: h' ' - 2h'*u - h + h*u^2 - h*u^2 + ϵ * h = 0 which can be simplified to: h' ' - 2h'*u + (ϵ-1) * h = 0 If you don´t get the -1 term, then you probably made a mistake when calculating ψ' '. Maybe you should first try to calculate (e^(-u^2 /2))' ' , because if you do it correctly you´ll see where the -1 term comes from.

[@thesingingtown]

I don't understand any of the math, but it's so soothing to listen to 😂❤️

[@bonbonpony]

This happens when you jump in in the middle of something without learning all that leads up to it first: you don't understand much, because you don't have the full context. In that case, you can either start with something simpler first and gradually move on to more complicated things, or (if you're skilled in learning), you might as well go backward, by making a list of all the things that you hear for the first time and didn't understand, and then going through that list one item by one, trying to find more information about them. This will often just make your list grow with some more unknown terms, but if you repeat this process long enough, you'll eventually make a connection with something you already know and the dots will start connecting.

[@itsanki]

just becuase a polynomial has infinite terms doesnt mean it will blow up at infinity. A simple examle is series expansion of 1/(1+x) for x>=0 has form a0 + a1*x + a2*x^2 + a3*x^3 + ...... but this doesnt blow up at infinity, rather it goes to zero. So it really depends upon the coefficients of the series obtained. We need further analysis to show that the series obtained in this lecture(hermite equation) actually blows up. The result of analysis would be that both of the series grow like e^(y^2) . And the overall solution psi = h*u grows like e^(y^2 / 2 ) . That is non normalisable . So the only way out is that the series terminate.

[@ArnoldSommerfeld]

Expecting rigor from Zwiebach is asking rather a lot. LOL.

[@bonbonpony]

Isn't it what he did though beginning from 01:07:50 ? He compared the orders of growth with the exponential. You're right though that this should be done in a rigorous derivation, and oftentimes lecturers just say "whoah, an infinite series! it surely must blow up to infinity!", which of course is wrong, as you said. Reminds me though when I was trying to explain the calculation of compound interest to my father. As soon as I said "Well, let's take the twelfth power of this interest rate…" he immediately said: "Are you outta your mind?! A twelfth power?! That would be a humongous number!" :) It took me the next couple of minutes to convince him that it won't, if the magnitude of the number being exponentiated is smaller than 1, because then its subsequent powers are getting smaller and smaller.

[@saskiavanhoutert3190]

The point sight images are correctly to be seen, just a guess of psi , thanks and kind regards. Very important for designing corners. Thanks and kind regards.

[@phillip76]

fine lecture

[@pedropuglia2650]

1:14:30 where did u^(j-1) go?

[@robinsonjunior7641]

1:10:05 he changed variables from n running on the naturals to j running on the evens. I thinks that’s why there is no j-1

[@DxPain]

shouldn't it be sin(n) piex/a instead of sin(n+1) piex/a??

[@dhoonygo]

Why does a(n) in h(u) can be written in 2^n?? a(n) could be anything?

[@buzzlightyear7238]

Because those constants are related by steps of two. i hope this reply isn't too late lol.

[@rafaelmera3865]

Wow, I love this chanel and this university. I am from Ecuador and I want to study in this university

[@bonbonpony]

Well, in a way, you do, just remotely. And you won't get any certificate from that, just knowledge (which is much more valuable).

[@IliaToli]

At 43:19, how did you ignore epsilon*psi?

[@BArdekani]

Because we are talking about very large u's, so the term u^2*psi dominates epsilon*psi since epsilon is expected to be a constant.

[@vedantdhawale4812]

The Epsilon*psi is fixed quantity So in diff eqn at infinity u^2 tends to infinity so (Epsilon -u^2) ≈ -u^2

[@athul_c1375]

1:01:02 I only wish if Griffith explaind this

[@bryanortiz2098]

Psi_n(x) has n nodes? Could someone explain to me? Thanks

[@hurtcolor]

9:42 11:00 nodes

[@standardcoder1184]

Dem This is good stuff Even Griffiths' did a bad bad job in explaining this

[@bonbonpony]

Try Pauling instead ;)

[@ilanle]

Gotta love this accent :-)

[@proexcel123]

can someone please explain to me or show me how did he get d/dx = (1/a)(d/du) from x=au?

[@psharmacgk]

u=x/a. Letting it act on a function y, dy/dx = (dy/du)(du/dx) by chain rule. As an operator this is just (d/du)(du/dx) = (1/a)(d/du) since du/dx is a constant 1/a.

[@proexcel123]

@P Sharma thanks for explaining it plainly to me! It's a great help!

[@psharmacgk]

:)

[@johnhobson2106]

Thanks very much for this...I would have been totally lost otherwise. Greatly appreciated!

[@mindfreak932]

So this goes to show that if the harmonic oscillator will have only of energy E=hw(n+0.5) only or have the energy stated to be in a quantum mechanical behavior?

[@blueman69-x2o]

+Uday Patel The solution to the differential equation for h(u) diverges if energy is not quantized. There was no assumption made.

[@timetraveler5128]

I don't understand why the whole wavefunction is zero if the wavefunction and the slope vanishes at a single point. Anyone please explain.

[@emmygaming6895]

The answer lies right there in the blackboard at 1:56. Differentiating the time-independent schrodinger eq. with respect to x allows you to find third derivative of psi in terms of psi and its first derivative. If both of these are zero at x0 then the third derivative is also zero at this point. You can keep on differentiating the equation and express the nth derivative of psi in terms of psi and its lower derivatives. This will show that all the derivatives of psi vanish at x0. This means that if you were to try to expand psi about x0 using Taylor series, you will find that psi is identically equal to zero. This is how I try to make sense out of it. I might be wrong though.

[@vjfperez]

The claim is that a solution for the stationary equation (energy eigen state) cannot have zero derivative at a node. If you notice the solution satisfies a linear second order ordinary differential equation. For this kind of equation if you specify its value and its derivative at one point you have a unique solution. So if both are zero at one point, you know a solution (zero everywhere), so the unique must be zero everywhere. But zero is not a valid wavefunction, since the norm must be 1 (postulate of QM). Hence you can't accept energy eigenstates with vanishing derivatives at nodes. For a general wave function (i.e. one that is not an energy eigenstate) this does not apply, as you can have any well behaved C2 function that is normalized as a valid state, it won't be a stationary one though.

[@not_amanullah]

This is helpful ❤️🤍

[@ricomajestic]

Anyone know of great lectures on special relativity or General Relativity? I am looking for something similar to these lectures on QM from Adams/Zweibach which I feel are the best on KZbin. Lectures where the instructor tries to develop the student's physical intuition, enhances his conceptual understanding, and also gives the student mathematical insight of the subject.

[@Gooselabs]

Leonard Susskind's lectures are a good introduction. Not as rigorous perhaps, but great for developing a physicist's intuition.

[@fjanoos]

+ricomajestic check out Prof. Shankar's lectures on special relativity on the yale-courses feed.

[@jetpaq]

+ricomajestic Leonard Susskind has a few great ones on youtube . search him my friend!

[@8cec]

Shankar's lectures are just amazing.

[@alkagupta4130]

Thanks @fugdus...

[@dhoonygo]

How is En related to temperature? I mean, are there any specific relationship between En and temperature of molecule with some equations?

[@durgeshgaikwad741]

Temperature is the average kinetic energy of molecules

[@FPSIreland2]

(probably) the energy eigenvalues En describe the energy of an oscillation and temperature is as mentioned the average energy of all of the oscillations within the lattice (I aasume youre talking about a partivle in a lattice structure)

[@richardneifeld7797]

No. Temp relates to distribution of energy in allowed states. Allowed states are not funtions of temperature.

[@neelmanichaturvedi7901]

Can someone please tell me what is the significance of 'u' here , why it is taken in this case?

[@not_amanullah]

Thanks 🤍❤️

[@bobjones5869]

he forgot a pi^2 at 10:02

[@LydellAaron]

He asks what we've done so far? 54:30 my interpretation is that he has cleverly used multiple substitution methods (as used in differential equations) in order to clean up the differential equation? 57:52 is a power series but there are many types of power series. I have found a prime series most helpful if you want to design and build a working quantum computer (e.g. series must terminate 1:13:38). 1:16:36 "Hermite" functions. Not sure that the representation at 1:20:37 is accurate pictoral representation.

[@TORMENTUMM]

i dont understand... why the edges of psi_0 at the infinite potential aren't nodes?

[@rubbeldiekatz85]

can you post the time stamp?

[@telelight]

+TORMENTUMM They aren't nodes by definition. The meaning of node here is slightly different then the meaning of nodes on a vibrating string. The "nodes" or points of zero amplitude on the boundary are not considered quantum mechanical nodes, its just simply the way the meaning was chosen for the word. Of course the reason its done is this way to is correspond nicely to the nth energy eigenstate e.g. the nth energy eigenstate has n nodes.

[@thenateman27]

+TORMENTUMM If you're referring to the part at 9:40 when he says it's not technically a node, notice that a node crosses the axis and has the property that it has a defined, non-negative derivative. At the edges, it is actually not differentiable, and it doesn't cross the x-axis, so it's not technically a node. True, it touches the x-axis, but it doesn't continue as it's derivative (from the side within the well) would indicate, and the derivative from outside the well is zero since Psi is just zero outside of the well. As a side note, Psi is almost always differentiable EXCEPT for infinite potentials.

[@richardneifeld7797]

Tormentum, another way to look at the problem is that the "infinite" square well is really not physical. It is instead the limit of a square well potential is on the height of the potential outside the V=0 region continuously increases to infinite. In the corresponding physical system having a finite height potential walls, the stationary state solutions (of the wave functions for each allowed energy value) have tails (exponentially decaying values) that extend beyond the walls and therefore the wave function does not in fact a "zero" value at the potential energy wall. Physics has to do with reality and not the mathematical absolutes, such as the mathematical solutions to the non physical "infinite" square well potential.

[@jamesgarcia-in2vm]

infiniti equals infinity pi pi negative square infiniti equal positive infiniti square square square infiniti pi square.-N.A.S.A.

[@1_adityasingh]

What do you mean?

[@bonbonpony]

@@1_adityasingh Overloaded neurons most likely.

[@prachiargulewar2409]

What are smooth potentials ?

[@bonbonpony]

Something that physicists "just draw on a chalkboard" and mathematicians spend many sleepless nights upon and eventually go nuts :J ("Define 'smooth' …")

[@joansans9263]

I am a little confused by the way the professor calls the expression for the energy of the system. At 26:10 he calls the expression of energy "Energy operator" while in 29:10 he referes to is as the Hamiltonian. By energy operator I understand something like Eop = ih_d/dt (h_ refers to reduced Plank constant) that would operate on the wave function while the expression on the blackboard is what I would expect to be called the system Hamiltonian. Any clue why he does this? or it is because the energy operator, acting on a wave function brings you the Hamiltonian of the system? Thanks!

[@TheBob5260]

If you’re still wondering, that expression is still an operator in the same way what you wrote is. Apply each term in the E-hat operator on an arbitrary wavefn. The p-hat^2 term involves derivatives in x. The second term just gives you what you see multiplied by the wavefn. In the end, it’s still an operator. This energy operator is a Hamiltonian because it gives you the kinetic energy and potential energy terms for the system (essentially).

[@bonbonpony]

Yes, his notation might differ from the one usually used in this context (the H hat, i.e. the Hamiltonian operator, where he uses E hat). But the meaning behind both is the same, and they are equivalent. You might think of the Hamiltonian as a more detailed expression of the total energy operator, in which you actually specify the kinetic and potential terms separately, instead of just bunching them together and hiding under a single symbol.

[@shivamgakkhar9537]

Can anyone answer, why a node appear everytime when we jump to one energy level to another????

[@santiagoarce5672]

it's not that a node appears when we jump to another energy level, more like the fact that energy levels with more nodes have more energy. The reason they're all sine waves with a certain number of nodes is because that is what solutions to the schrödinger equation look like for the infinite square well.

[@shivamgakkhar9537]

@@santiagoarce5672 sir, this is good to explain by mathematical point of view. But just assume that there is a student who doesn't want to deal with the mathematical part (I know quantum mechanics cannot be imagined without mathematics...but still let's assume) how you are gonna explain to him that why there are some certain points in higher energy level which our particle can never access...

[@richardneifeld7797]

Shivam, because the wave function must vanish at infinity for any bound state. (To conserve probability and to recover the classical limit.) As explained in prior lectures, that only occurs for special wave functions, specifically wave function that have a value and a slope at the classical turning point that allows smooth and continuous match to exponentially decaying functions outside the classical turning point.

[@richardneifeld7797]

And that can only occur for one wave function within the classical turning point that has a specified number of node.

[@xinzeng-iq7zv]

no idea what equation represent that line

[@MeadowBrook2000]

I literally broke my headset unintentionally while trying to understand this LoL! But great video as always

[@thecaptainindia9790]

if 2j+1 =e for some j but it will not terminate the other half coefficients

[@rebekahshtayfman1967]

I'm a little confused. He went back to the postulate that says ψ(a) = 0, ψ'(a) = 0, then ψ(x) = 0... and he made the argument in the "physics" proof saying that you can't obtain a node because of this postulate. If you did, hypothetically, obtain a node out of nowhere (which obviously can't happen), you would have ψ(a) = 0, and ψ'(b) = 0. In other words, they would be at different points, so this doesn't say that the wave function has to be zero. He used this in the proof as if this would be occurring at the same point, and clearly it wouldn't.

[@smoothtriston6203]

As the screened potential is expanded, in order for the ground state to transform continuously and form a node, phi(a) and phi'(a) would both have to be zero at the same time somewhere during the transformation. Because of this, it can be said that any state for any potential (of that general shape) will have as many nodes as it did for the infinite well. Which for the ground state is zero.

[@richardneifeld7797]

Rebekah you confused his reasoning. The point he made is that if ψ(a) = 0, and ψ'(a) = 0, that is the value and first derivative *** at one point location in space*** is zero, then that requires ψ(x) = 0 everywhere in order to satisfy the differential equation. Node at x=a means the value of ψ(a) = 0. That says nothing about the first derivative ψ'(x) at x=b.

[@timetraveler5128]

I don't understand why the whole wavefunction is zero if the wavefunction and the slope vanishes at a single point.

[@eloymartinsuarez3946]

te amo

[@Breaker2361]

Does the term 'bound state' just mean a normalisable wavefunction?

[@ether5463]

I'm not sure if I'm correct on this, I'm just an undergrad student, but I think that answer to you're question is no. I'm pretty sure a free particle can have a wave-function that still decays to zero and is therefore normalizable. A bound state is one in which the energy is less than the potential energy *at both ends*, consequently the energy levels are quantized.

[@psharmacgk]

A completely free particle shouldn't have a wavefunction that decays to 0 though because there's no potential to make that happen, right? The only free particle wavefunctions I know of are delta functions in either position space or momentum space. These are normalizable of course, but I can't see a wavefunction in a system with no potential having any reason to decay at infinity without some outside influence.

[@libere1001]

P Sharma -- Those are the free particle eigenfunctions. But you can put together a bunch of free particle eigenfunctions into a fourier series and make a lot of different kinds of things. Importantly, you can make a little lump sort of thing that looks like a particle and that travels around with some velocity. Check this out: en.wikipedia.org/wiki/Free_particle#/media/File:Quantum_mechanics_travelling_wavefunctions.svg

[@psharmacgk]

I'm aware of this lol, I was just answering the free particle question. Thanks anyway though :)

[@smoothtriston6203]

A bound state is a quantum state where some potential in the system has bound a particle to a certain region or regions of space.

[@anmolsharma8002]

Sir, You have said that ¥1(psi 1) has one node. But in psi 1, there are only 2 end points, no in between. So, how did you say that? i didnt get. Please explain!

[@anmolsharma8002]

..

[@priyanksharma1124]

that is psi _0 with two end points and no nodes.psi_1 has one node

[@anabilkanungoe1728]

Is this an undergrad or a graduate course ??

[@mitocw]

This is an undergraduate course. See the course on MIT OpenCourseWare at: ocw.mit.edu/8-04S13 for more information.

[@pandiatonizm]

This is primary school stuff in China

[@TupacMakaveli1996]

@@pandiatonizm low key Einstein was Chinese so was Planck. They disguised as western to counter racism.

[@siulapwa]

Differential equations can be painful

[@bonbonpony]

This usually means that one uses a wrong tool for the job. But if you don't know any better tool (because it hasn't been invented yet), you don't have much of a choice.

[@kartikanand5374]

Art

[@ДенисМайфат-ш6ю]

10.47 there is no pi in energy

[@lachlangray8120]

Why am I paying my university thousands of dollars when I can watch this???

[@schmetterling4477]

Because they will give you a title that will allow you to get high paying jobs. Watching this won't.

[@bonbonpony]

@@schmetterling4477 It will give something much more valuable though: knowledge. Jobs are overrated.

[@schmetterling4477]

@ This won't give you much useful knowledge, either. The only people who use this stuff are physicists who are doing research. That's not even a high paying job. ;-)

[@meetghelani5222]

Barton ❤️👏🫡

[@sumitparida7993]

I don't understand why did he choose the complicated way to solve the differential equation

[@vedantdhawale4812]

Because this was the method through which the Harmonic osscialltor eqn was derived first time. It was difficult but it convinced the Physicists. Later Operator method confirmed the results

[@richardneifeld7797]

To provide physical insight and to have students believe the solution to be correct, and to connect Hermite polynomials and their properties to HO problem's. Solution. HO is the fundament on which field theories and string theories rely. It pervades physics, very important.

[@jamesgarcia-in2vm]

infiniti equals infinity pi pi negative square infiniti equal positive infiniti square square square infiniti pi square.

[@robertwest5746]

Nice lecture, but I'm not satisfied with the explanation in the beginning about the nodes. You expand the well and the energy becomes lower, why should we then assume the wavefn will morph into one that represents an excited state which has higher energy? It seems to be quite an assumption to wave your hands and state that the slope of the grnd fn decreases and eventually flips on one side to look like the 1st excited state. There seems to be no motivation for assuming that.

[@anmolsharma8002]

Robert West , well i didnt satisfied with the way he explained nodes

[@tobyhardcastle6830]

He states the opposite doesn’t he? He says it can’t create more nodes if you start with a ground state wave. He only talks about how it would be created to show that it’s impossible for it to do so

[@richardneifeld7797]

He did not state that "wavefn will morph into one that represents an excited state which has higher energy" and in fact stated the opposite. And in fact, as the walls of the potential well are move farther apart, each stationary state (allowed energy level with no time dependence) solution will retain its number of nodes and its spacial dependence will slowly spread its oscillations in the classically allowed region to longer oscillations. But the number of nodes for each stationary state will remain the same.

[@atamertaydn4703]

Awesome! Algebra and mathematicians are upset after this lesson :)

[@neotixx.]

Even me as a 13 year old can understand this! Love this guy:)

[@willemesterhuyse2547]

He set j =j' + 2, then in the next step he set j = j'. This is inconsistent. (Timestamp 1:01:13).

[@bonbonpony]

That's because it's a dummy variable after all. It doesn't matter how you name it, as long as you keep the numbering of terms in the sum consistent and it produces the same result when expanded. Try expanding each of them and you'll see that they are in fact consistent. What might have confound you is that he reused the same symbol as before (which previously had a different meaning). But think of it this way: First he moved on from j to j', and after that, he no longer needed the old j, because the new notation already expresses the same with a different symbol (which is arbitrary anyway). It wouldn't really matter if he used a different symbol in each of those sums, but then it would be hard to merge them into one sum, with all those different symbols. Hence he simply renamed the new symbol j' as j (which was no longer used in this particular sum), so that he had the same symbol (j) in all sums and didn't have to think too much on them when trying to merge them into one.

[@chanchalsharma2738]

Where is part 2

[@mitocw]

Part two is lecture 9 (but is not titled as such): kzbin.info/www/bejne/oHu7kGSwiZxmi5I. Best wishes on your studies!

[@chanchalsharma2738]

Thanku ❤️

[@bonbonpony]

@@mitocw Yeah, but it's a different lecturer.

[@txominpenasantacruz2472]

I invite you to consult (simple atomic oscillator) and (Atomos de Santa Cruz)

[@DevangLiya]

Watch it at 1.25x. You're welcom!

[@chymoney1]

Devang Liya *2.00

[@xiaoqilu1353]

With Chrome extension you can watch this with 2.50x speed.

[@saadibnasaadhusain]

Ladder operators are so much simpler!

[@AlexHandle355]

Amén haha Totaly agree

[@bonbonpony]

But they're also more obfuscated and some people might find them to be too much trickery. Therefore it's good to know the usual power-series approach first to ground your understanding before moving on to trickery.

[@xinzeng-iq7zv]

unfortnately, i need a physics degree to get a physics degree

[@jeremygong9410]

don't really think you need to solve it in such manner, and again, the manner of solving this equation wasn't even analytical, wasted 20 mins on the last part of this lecture :( I'm looking forward to the lecture 9, and hopefully there are some practical means to solve it. P.S: ALWAYS HATED INFINITE SUMS!!!!

[@xinzeng-iq7zv]

one slip up can cause u to drop the class

[@dimibe918]

Is the 1.25 speed the original one? xd

[@raphaelmonteiro8291]

26:19 he forgot the hat on X

[@richardneifeld7797]

No one's perfect.

[@Nick15632]

lmao he dots the t at 3:47

[@mikedelmonaco6193]

Are these precise results? These seem like insane approximations where we just throw terms away and stuff. This really doesn't seem ok to me, but I've only taken high school physics that didn't use calculus so I don't really know what I'm talking about. I'm used to calculating everything exactly so all of this estimation seems crazy. Are these kinds of approximations normal? Do they just happen to coincide with real, rigorous calculations, but there just isn't enough time to do things rigorously so they just estimate? I can't imagine a physicist saying "forget about these" and proceeding in an actual calculation.

[@galvinng1997]

I can see why Prof Barton chose to drop off the remaining terms. 1) it speeds up the computation because you don't have to worry about the not important terms 2)in a way, it's neater, the non-important terms need not worry you as you do the workings, allowing you to focus on the essential terms to get your result! :)

[@richardneifeld7797]

Yes. They are precise.

[@bonbonpony]

Physicists often drop terms that are too small to be noticeable anyway, because this saves calculations and can often simplify them without damaging the end result in a noticeable way. For example, there's no need to measure the length of a brick to a micrometer if the gaps between bricks and their overall sizes can differ in more than a millimeter. In this case though, the lecturer dropped the terms because he knew from experience with such equations that they won't play any role. So it still gives the correct and exact answer from mathematical standpoint, just saves you all the headache of grinding through complicated calculations that would cancel at the end anyway. Otherwise this lecture would have been much longer and everyone would get bored before reaching the solution. It's the type of calculations that you can check in more detail on your own if you're not convinced. There's also the procedure he used for finding the "asymptotic behavior". In this approach, you also drop some terms from the differential equation because for large values of x they contribute less and less to the function you're looking for. This allows you to focus on the most important parts of the equation first and *their* contributions to the unknown function so that you had the rough form of your trial function as something you could start with. Then you can eliminate this part from the problem and simplify the equation, which now focuses only on the part that you're still missing. If you compare the end result with this first rough estimate, you will see that these functions indeed behave in the predicted way for large values of x and approach them more and more closely as x gets bigger. Meanwhile, the Hermite polynomial part contributes to the deviation from this overall behavior near 0, for small values of x, where you need more precision.

[@LaurenceBrown-rx7hx]

brutal

[@tanelihautaniemi6771]

Fuck this is hc. Thank you!

[@psharmacgk]

Did you get to the point I mentioned in my comment?

[@pablo_brianese]

1:13 APPLAUSE