Ladder and number operators of the quantum harmonic oscillator

[@BYUImageAI]

I have taken 3 semesters of QM and these videos explain the subject so well. I would even say they are better than both of the free MIT courses.

[@ProfessorMdoesScience]

Great to hear you find the videos useful :)

[@WhiskeyTobin]

Wow, discovered these videos just in time for a test on quantum harmonic oscillators and these are just so clear and concise. The speaker is well articulated, the handwriting is legible, the colors in the writing help to keep everything from jumbling up. These videos are great.

[@WhiskeyTobin]

Also the syntax is very close to JJ Sakurai, so it's really easy for me to follow as that's the text I'm using.

[@ProfessorMdoesScience]

Glad you like them and find them useful! :)

[@phonon1]

I am trained in math (no physics). I love that these video can be understood for a broader audience.

[@ProfessorMdoesScience]

Glad you find them useful! :)

[@ziadyoussef9157]

one of the most underrated channels thank you!!!!

[@ProfessorMdoesScience]

I appreciate that, thanks!!

[@yashwath5056]

The pronounciation though. Never heard english with this much clarity

[@ProfessorMdoesScience]

Glad you find it clear! :)

[@redaabakhti768]

Best course on Qm online BY FAR

[@ProfessorMdoesScience]

Glad you like it! :)

[@binabedin9823]

Why this channel is remained under rated ?!!!!!!!!!?

[@ProfessorMdoesScience]

Thanks for your support, slowly growing! :)

[@FrostmanJack]

I understood 0% of three videos so far, I love them 💯

[@ayushdhingra855]

One of best videos on ladder operators

[@ProfessorMdoesScience]

Thanks for watching and glad you liked it! :)

[@NommerDonzwaaa]

Just sent a link to this video to all my friends doing this! Amazing stuff

[@ProfessorMdoesScience]

Thanks for your support! :)

[@AmanGovindSoni5-YearIDDChemist]

How did you get N^a^|Lambda> = a^N^|Lambda> - a^|Lambda> timestamp : 8:38

[@ProfessorMdoesScience]

I'm using the expression for the commutator of N and a: [N,a]= -a By the definition of a commutator, this can be expanded as: Na - aN = -a Re-arranging, I get: Na = aN - a I hope this helps!

[@debrajbanerjee2428]

Please continue the great work....

[@ProfessorMdoesScience]

Thanks for your support! :)

[@guidomarcucci1893]

Phenomenally clear

[@ProfessorMdoesScience]

Glad you like it!

[@yugeshbhoge4950]

Very beautifully explained 🙌

[@ProfessorMdoesScience]

Glad you like it! :)

[@manujsharma1432]

Thank you for such a nice explanation 👍

[@ProfessorMdoesScience]

Glad you like it!

[@whilewecan]

Thank you very much!! I understood well.

[@ProfessorMdoesScience]

This is great to hear! :)

[@enricolucarelli816]

👏👏👏👏👏It’s amazing how you manage to make these complex concepts to look almost trivial! There is just one doubt for me to get everything strait: In classical mechanics omega is the square root of k/m, where k is the constant in Hooke’s law, but I doubt that in quantum mechanics we can make use of Hooke’s law just like that. So how do we calculate this omnipresent parameter omega in the quantum mechanics environment?

[@ProfessorMdoesScience]

Glad you like the videos! One way to think about the harmonic oscillator is to think about the form of the potential energy, which is quadratic. In this context, the square of omega is simply the curvature of the potential. The reason why harmonic potentials are so useful is because we can use them to approximate more general potentials near minima, and again in all these cases, omega is related to the curvature of these potentials around local minima. We briefly cover this concept in this video: kzbin.info/www/bejne/hZXMq4WLmp1nmMk I hope this helps!

[@enricolucarelli816]

@@ProfessorMdoesScience Omega is simply the square root of the curvature of the approximation of the potential to a parabola! OFF COURSE! 🤦‍♂️I was so stuck trying to give this parameter a concrete physical origin, whereas it just comes from the mathematical approximation of a local mínima to whatever potential! THANK YOU VERY MUCH!! ❤️

[@alfiashaikh99]

Great explanation!! I have a question. For 2 particle states such as |00>, |01>, |10>, |11> how will the number operator act on these states and what will be its eigenvalue?

[@ProfessorMdoesScience]

We cover multi-particle systems in these two series of videos: * Basics of identical particles in quantum mechancis: kzbin.info/aero/PL8W2boV7eVfnJ6X1ifa_JuOZ-Nq1BjaWf * Second quantization: kzbin.info/aero/PL8W2boV7eVfnSqy1fs3CCNALSvnDDd-tb I hope this helps!

[@danielisenberg2360]

I'm not sure if this is an issue of dialect (European English vs American English) or just a concept I am not familiar with, but what do you mean by "at joint" (Could be "ad joint"?) mentioned at 3:19 ?

[@ProfessorMdoesScience]

The word is "adjoint", referring to an operator. For a given operator A, its adjoint is written with a little "dagger", and usually also called A-dagger. It represents the action of the operator in the dual space. I hope this helps!

[@Andre-hp9hy]

In 9:24 i don't understand why you wrote a|lambda>=C_|lambda-1>. Before that you wrote that N a|lambda>= (lambda-1)a|lambda> so the eigenvalue is lambda -1. But after that you wrote lambda -1 as eigenstate. Can you explain the reason?

[@ProfessorMdoesScience]

We use standard notation in which an eigenstate label is written as the corresponding eigenvalue. I hope this helps!

[@DrMarcoArmenta]

Is there a good reason to denote the ladder operators as the creation and annihilation operators for bosons? I've seen some papers where they call the ladder operator of the harmonic oscillator like this. Are they the same? in which sense? thanks in advance!

[@ProfessorMdoesScience]

The bosonic creation and annihilation operators obey the commutation relation [a,a^dagger]=1. In this sense, the ladder operators we've defined for the quantum harmonic oscillator can be interpreted as bosonic creation/annihilation operators as they obey the same commutation relation. From a physical point of view, in the standard approach we view the ladder operators as adding or removing quanta of energy hbar*omega. If we instead interpret these as creation/annihilation operators, then you can think of them as adding/removing particles each of energy hbar*omega. For a general overview of creation/annihilation operators, I recommend our series on second quantization: kzbin.info/aero/PL8W2boV7eVfnSqy1fs3CCNALSvnDDd-tb I hope this helps!

[@oraange]

Why do you write â|lamda> = |lambda-1>, I'm confused by your notation. If you have any examples to explain me, that would be super!

[@ProfessorMdoesScience]

The label inside the ket can in general be anything we want that allows us to identify the ket. In this particular case, we are working with the kets associated with the eigenvalues lambda of the number operator N, so the label we use to distinguish the various kets is directly the eigenvalue they are associated with. If we now compare the kets |lambda> and |lambda-1>, the first ket is the eigenstate of the operator N with eigenvalue lambda, and the second ket is the eigenstate of the operator N with eigenvalue lambda-1. The equation you write, a|lambda>=|lambda-1> then tells us that, by acting on |lambda> with the operator a, we get another state |lambda-1>. More generally, we explain this notation of using the corresponding eigenvalue to label kets that correspond to eigenstates in this video: kzbin.info/www/bejne/pmLdmGCZZtOprbM I hope this helps!

[@irshadahmadi6497]

good explanation

[@ProfessorMdoesScience]

Thanks for watching! :)

[@knalltuete7311]

at 6:46 you switched the order. so if i would stay at the same order it would be N*=(a*a)*=(a*)*a*=aa* /unequal a*a (with *=dagger). Why am i allowed to change the order?

[@ProfessorMdoesScience]

In this step we are using a general property of the adjoint of a product of two operators. In general, for two operators A and B, we have: (AB)*=B*A* (where *=dagger). We go over this in detail in this video: kzbin.info/www/bejne/mJCnlKaMeNmDa6s I hope this helps!

[@knalltuete7311]

@@ProfessorMdoesScience thank you very much

[@almisratiali6375]

Great work ... could please explain what is the physical meaning of ladder operators or just it is quantum mathematics method

[@ProfessorMdoesScience]

The ladder operators are the operators that, when starting with an energy eigenstate of the system, raise it by one quantum of energy (a-dagger) or lower it by one quantum of energy (a). We explore this in detail in the video on the eigenvalues of the quantum harmonic oscillator: kzbin.info/www/bejne/fZy4iaaZmbGEh5I I hope this helps!

[@rimon9697]

i had a question from one of my assignment. It asked me to derive the equtions of motions for the creation and annihilation operators. I had been stuck in this since 2 days and have no idea on how to start. Would be great if you could provide me with a lead or a hint of some kind.

[@ProfessorMdoesScience]

In which context are you doing this? Are you studying the Heisenberg picture of time evolution in quantum mechanics?

[@rimon9697]

@@ProfessorMdoesScience yeah

[@ProfessorMdoesScience]

So here are a few hints to get you started. In your studies of the Heisenberg picture, you've probably come across the equation of motion for operators in the Heisenberg picture. It essentially involves a time derivative of the operator in question, and then a commutator of the operator with the Hamiltonian of the system. I imagine that what you are interest in here is to use this equation, where the operator of interest is one of the ladder operators, and the Hamiltonian is that of the harmonic oscillator. Once you have set this up, I suspect that it will be easiest to work with the Hamiltonian of the quantum harmonic oscillator written in terms of ladder operators, as then you'll be able to use the properties of ladder operators, such as their commutation relations. I hope this helps!

[@rimon9697]

@@ProfessorMdoesScience I just forgot to update on whether i was able to do it or not😅😅well i was able to do it by plugging the ladder operators in the heisenberg picture and using their associated commutator relations. Thanks for the hint. Also, i have been meaning to ask if covering spatial rotation operator and scattering theories was a possibility.

[@ProfessorMdoesScience]

@@rimon9697 Glad it worked out! And thanks for the suggestions! Scattering theory is something we do already have on our list. As to spatial rotation operators, we hope to cover them when we consider the wider topic of group theory applied to quantum mechanics.

[@eveaaron2885]

hello, thank you for this video, it is so useful ! Could you share this lecture notes ?

[@ProfessorMdoesScience]

Glad you found it useful! Unfortunately we don't have shareable lecture notes at present, but we are working on additional material (lecture notes, problems+solutions, etc) that will hopefully become available soon!

[@LifeIzBeautiful10]

Awesome!

[@ProfessorMdoesScience]

Glad you like it!

[@toufiksalhioui913]

thanks for the video very clear could you explain to me why c+=sqrt(lambda +1) and not c+=sqrt(lambda)? my calculation below: |c+|^2=||a+lambda||^2====lambda

[@ProfessorMdoesScience]

In your derivation, you set aa+=N+, but this is not quite correct. Note that the adjoint of a product of operators is equal to the reverse product of adjoints: (AB)+=B+A+, so actually N+=N. Instead, what you need to do is to use the commutation relation for a and a+ to re-write aa+ in terms of a+a, so that you can then introduce N. Specifically, you get: [a,a+]=1 --> aa+ = a+a +1, where the +1 is just a "plus 1". This means you introduce this extra +1 comming from the commutator. I hope this helps!

[@toufiksalhioui913]

@@ProfessorMdoesScience Got it! Thank you very much!

[@ProfessorMdoesScience]

@@toufiksalhioui913 Incidentally, we've just launched two problem sets (+solutions) on the quantum harmonic oscillator, including several problems dealing with ladder operators, just in case you want to practice: professorm.learnworlds.com/

[@ziadyoussef9157]

@@ProfessorMdoesScience thanks for the question and for the reply

[@gokhilgs3320]

can you explain how expression for raising operator is obtained

[@ProfessorMdoesScience]

The ladder operators are simply defined as described in the video. You can obtain the raising operator as the adjoint of the lowering operator, and to calculate the adjoint you need to remember that scalars become their complex conjugates (hence the change in sign for the i term) and operators become their adjoint. The last point means that we should, in principle, have x^dagger and p^dagger for the raising operator, but we also know that x and p are Hermitian operators, so that x^dagger = x and p^dagger = p, and we use this property to write the final expression. I hope this helps!

[@ronaldjorgensen6839]

thank ytou

[@ProfessorMdoesScience]

Thanks for watching!

[@sebastiangudino9377]

Where do these operator come from?

[@canyadigit6274]

The ladder operators seem unnatural to me. Like you said it looks like they were taken out of a hat. How did someone just go “hey let’s define this ladder operator in this specific way” and it turns out that they work? In other words, can you motivate the operators?

[@ProfessorMdoesScience]

Adding to the great response by Bavithra, I will provide yet another point of view. We can actually also define analogous quantities for the *classical* harmonic oscillator. These are a=sqrt(m*omega/2)x+i/(sqrt(2m*omega))p and its complex conjugate. You then find that classically, the equation of motion for a is given by (d/dt)a=-i omega*a, which means that a(t)=a(0)*exp(-i omega*t). The complex number "a" rotates in the complex plane as time progresses, and the projection on the real axis gives the displacement x(t), and the projection on the imaginary axis gives the momentum p(t). This approach is rather useful, for example, the fact that the magnitude of a(t) is fixed leads to energy conservation. The ladder operators are then simply the quantized version of this classical description.

[@canyadigit6274]

Thanks to both of you!

[@quantum4everyone]

They were originally introduced in matrix mechanics as a way of moving up or down rows of the matrix, and equivalently up and down the energy eigenvalues; they already appear in the paper by Born and Jordan in 1925. But a better way to think of them is the way @Bavithra describes them. Here, they are a way to factorize the Hamiltonian into the form A*A+E. It turns out all of the exactly solvable problems can also be factorized this way (the methodology for this was first worked out by Schroedinger in 1940). Using this factorization, one can obtain eigenvalues without computing eigenvectors. This is pretty neat. Too bad it isn’t taught in nearly all textbooks. It does not look like it will be discussed further here either, seeing the titles for the other videos.

[@mp3lwgm]

Overly complicated. The wave-mechanical solution is more satisfying to me.

[@mp3lwgm]

Overly complicated. The wave-mechanical solution is more satisfying to me.