Eigenvalues of the quantum harmonic oscillator

[@garvitmakkar]

you can't believe me every lecture on this channel is a jam. you deserve millions of likes from myside. Thankyou so much for making us fall in love with physics😅

[@ProfessorMdoesScience]

Thank you for the encouraging words, and glad you like the videos! :)

[@tommysullivan]

Today I clicked into subscriptions and while it was loading I was like cmon cmon cmon let there be a new professor m does science today!!! 😁🙏

[@ProfessorMdoesScience]

Aw, you made our day!! Thank you :-)

[@paulbk2322]

Certainly, one of the most exciting and amazing learning experiences in a long time

[@ProfessorMdoesScience]

It is great you found this helpful! :)

[@senti3035]

The fact that all other allowed states differ by 1hw but the zero-point energy is only 1/2 of it feels really unsatisfying but its true. Awesome video, I wish my college classes were these concise and easy to understand

[@ProfessorMdoesScience]

Glad you like the videos!

[@maxwindiff]

Thank you so much for making these videos! I only read about energy quanta, zero point energy, etc in pop sci books and it’s wonderful to see them derived in such a clear way.

[@ProfessorMdoesScience]

Really glad to hear you like the videos! :)

[@kazinahid5166]

I am from bangladesh ,enjoying your video so much ,,,thanks for making such video

[@ProfessorMdoesScience]

Glad you like the video!

[@thephenomenalone3391]

Hello, can you please suggest any problems book to solve on quantum harmonic oscillator

[@ProfessorMdoesScience]

Most quantum mechanics textbooks have problems, which would include some on the QHO. Books we like include Cohen-Tannoudji, Sakurai, Merzbacher, Shankar. We also hope to launch a website in the near future with more content to supplement our videos, and that will include problems (and solutions!).

[@ekhibaathaiyaar1610]

Thankyou so much for these awesome videos, it would be really helpful if you could link some good assignments or workout problems along with these.

[@ProfessorMdoesScience]

Thanks for the suggestion! We have been working for a while on material complementing the videos, including problems+solutions. But getting things ready always takes longer than expected... we hope to get there soon!

[@passionisinspiration1912]

@@ProfessorMdoesScience hi is it available now?

[@shashwatbhatnagar659]

I don't know how can I thank you ,I was really struggling this topic .😇😇😇

[@ProfessorMdoesScience]

Glad to be helpful!

[@manujsharma1432]

Thank you for such nice explanations. 😊

[@ProfessorMdoesScience]

Glad you like them!

[@subhajitsadhukhan8521]

Professor.... In the video of ladder and number operators we saw |lamda> is eigenstates of number operator. But here you're using |lamda> as the eigenstates of Hamiltonian as well. Why? Is this because number operator commutes with Hamiltonian ( I calculated the commutator and it is 0)and we can use a common set of eigenstates of this two operators?

[@ProfessorMdoesScience]

Yes, the number operator and the Hamiltonian share the same eigenstates as they commute. I hope this helps!

[@versatile5307]

Awesome explanation ..... My professor also shows your videos in classes .. Keep up the good work

[@ProfessorMdoesScience]

Glad you like this! Out of curiosity, what university are you at?

[@versatile5307]

@@ProfessorMdoesScience SSSIHL ,India

[@marit903]

You explain it really clear and have a really clean way of working and clear voice!! It helps me a lot, thank you :)!!! Greetings from the Netherlands :)

[@ProfessorMdoesScience]

Glad you like it and find it useful! :)

[@senkum80]

Best explanation !!!

[@ProfessorMdoesScience]

Glad you like it! :)

[@narfwhals7843]

I'm enjoying these a lot. What exactly does "killing the state" mean? If λ=1 we get â|λ>=1|0> But the wavefunction corresponding to 0 is not "nothing"? Is this the vacuum state with zero point energy? And â acting on _this_ yields the zero vector, which I suppose means the field does not exist? (perhaps I should just watch the next video)

[@ProfessorMdoesScience]

Your description is correct: |0> is the "zero-point energy state", and acting on it with a "kills the state", ie. you end up with the zero vector.

[@tomkubalik4878]

Each energy eigenvalue is separated by ℏωk from neighboring eigenvalues (3/2ℏωk-1/2ℏωk). Is this energy gap between two neighboring eigenvalues ℏωk different for each mode k? If omega is different for each k, doesn't it mean that omega is different for each mode? Doesn´t it also mean that ℏωk is different for each mode? Thanks a lot

[@ProfessorMdoesScience]

In this video we are considering the 1D harmonic oscillator which is characterized by a single omega. Its value determines the curvature of the harmonic potential, and indeed the spacing between the energy levels. I imagine that the situation you are referring to corresponds to a system in which you have several harmonic oscillators, each characterized by a different omega_k. In this case, as you suggest each oscillator will have its own energy level spacing, given by hbar*omega_k. An example of a physical system in which this happens are the vibrational properties of ions in solids, characterized by quasiparticles called phonons, each of which behaving like a harmonic oscillator, but each with a potentially different frequency omega_k. I hope this helps!

[@matthewpenner2348]

Am I ever watching all these the night before my midterm on 2x speed

[@ProfessorMdoesScience]

Good luck with the exam! Where are you studying?

[@sayanjitb]

Dear ma'am, I have one doubt. In the case of QHO, the position basis is continuous and spans the Hilbert space. But after solving the eigenvalue equation for the harmonic oscillator, we get another infinite discrete set of basis vectors that also span the Hilbert space. But these two kinds of basis sets are completely different. One is a continuous basis set, and the latter is discrete. Why is there such inconsistency? Or am I missing something? TIA

[@ProfessorMdoesScience]

Great question! This is a rather subtle point that physicists typically gloss over, and a full answer would require much more than a comment here. However, let me give you some thoughts to get you started in thinking about this. Mathematically, you are absolutely correct, these two bases are fundamentally different in that there is no one-to-one map from one to the other because of their different cardinality. However, when we use them to study quantum mechanics, this difference becomes irrelevant. Roughly speaking, this is because you can change a continuous function at a countable (or even uncountable) number of places and still get the same L^2 norm, which is what matters in quantum mechanics. I hope this helps to get you started!

[@sayanjitb]

@@ProfessorMdoesScience Thank you so much indeed for your constant support. Could you please be a little bit explicit on this "... because you can change a continuous function at a countable (or even uncountable) number of places and still get the same L^2 norm," Thank you

[@ProfessorMdoesScience]

First a disclaimer: we are not experts in the mathematics behind some of this, so I recommend you do your own reading. Having said that, in the position representation the relevant quantity is the L^2 inner product (mathworld.wolfram.com/L2-InnerProduct.html) between two wave functions. One can prove that this inner product cannot distinguish between two functions f(x) and g(x) such that their norm square vanishes: ||f-g||^2=integral(dx |f(x)-g(x)|^2)=0. It should be somewhat intuitive that if f and g only differ at a countable number of points, then this norm will vanish (as the functions are continuous). It is admittedly less intuitive that this norm will also vanish in some cases when they differ by an uncountable set, but this also turns out to be true. Given this, any physical prediction arising from f or g would be the same, hence the reason why the difference is irrelevant in quantum mechanics. I hope this helps to at least get you started learning more about this!

[@sayanjitb]

@@ProfessorMdoesScience Thank you so much for your effort to help me. Thank you indeed.

[@materiasacra]

There is more to this matter of continuous and discrete bases than has been discussed here. Physically the presence of an energy continuum means that the system is broken with its parts flying around having kinetic energy that can take on any value. In physics we tend to be a little cavalier with the mathematical treatment here. In developing QM we postulate that a plane wave proportional to e^{i(kx-\omega t)} is the wave function of a free particle with momentum p=\hbar k and energy E=\hbar \omega. Then we proceed to interpret this wave function as a probability amplitude, and we run headlong into a problem with normalization, because the 2-norm of a plane wave over all of real space is infinite. If we are somewhat careful, we fix this by assuming the particle is confined to a finite box of size L which is required to be much larger than any physically relevant size in the system. This introduction of a finite box turns the energy spectrum from continuous to discrete, albeit with very fine energy spacing. The corresponding set of eigenstates of the the Hamiltonian is thinned out from continuously infinite to countably infinite. All of this carries over to other physical systems, including interactions. In QM spectra and bases are ALWAYS DISCRETE in principle, but sometimes the discrete nature is physically irrelevant as it merely describes confinement of an experiment to a large 'laboratory'. Whenever we say in QM that a system has a continuous part in its spectrum, we mean that it can break and its parts then correspond to waves in a very large box. We mean it, but we don't say it. Hence students sometimes worry about the uncountably infinite collection of states. They need to keep the box in mind always while working with unnormalizable wave functions, and be very afraid of a 'true' continuum, as it breaks the probability interpretation. As they progress through the material, and reach the level where they are expected to do actual computations, they are automatically led to consider all the correct normalization factors in relation to actual experimental parameters such as the size and distance of detectors, and all will be well. Mathematicians frown on all of this. Their minds are not set up to 'keep in mind' implied conditions that are not written down and strictly adhered to. The proper physicist reaction is to smile and describe the difference in mindset and practice, and point out that we do in fact worry about the conditions, but only when they are important for our purpose. We're not fools, we're efficient! :+)

[@123Handbuch]

So quantization comes from the system being bounded, right? Is there a video about it?

[@ProfessorMdoesScience]

You are correct. We are planning a video on bound vs unbound states, but it is not yet available. Will let you know when we do!

[@afborro]

@@ProfessorMdoesScience This is one of the fundamental criteria in building QM course one should introduce sooner than later (properties of valid QM wave functions and their properyies,). it should come before any of this IMO. While I enjoy the videos, they are all over the place. It would be better to build a QM course in the right order, otherwise people are jumping all over the place cross referencing other videos, which makes it hard to follow. I've been doing that already. For me it mostly okay since I studied it before (but in the right order). Just my O2 :) Often too many things are thrown in at once, as is often stated in a video, "watch this video for x" It breaks the flow. 1) Build QM framework and postulates, properties of wave functions. then 2) free motion unbound, bound, barriers, particle in box, then the harmonic oscillator, angular momentum etc. There is excellent material here, but in the wrong order. Some applications of some of these things would also help in future using real data. eg. Harmonic oscillator is used in diatomic molecules and spectroscopy.

[@ProfessorMdoesScience]

@@afborro Thanks for all the suggestions! We are fully aware that our videos at present correspond roughly to a second undergraduate course in QM (with some more advanced topics), so there may be some gaps for someone who is only starting. The reason for this is based on our actual teaching needs as we use these videos with our students. Moving forward we hope to set up a more complete course, building everything from the beginning and with an index to guide self-study. We also hope to include problems and solutions. However, KZbin is something we are currently doing in our free time (our main work being research and teaching at the university), so it all takes much longer than we would like... In the meantime, we hope people can still gain something from our videos! :)

[@afborro]

@@ProfessorMdoesScience Sure. I understand. FWIW, I think there are so many bits 'n' pieces of QM on youtujbe already, high quality too. Clearly the knowledge is there with your staff to bring it to that next level with the potential I suggested, a more structured approach . I just went through revising QM after many years and used a book(s) building it up step by step, in that sense these videos are complementary and do help, that is to see the material from different angles, but not to learn QM from the ground up. Cheers.

[@quantum4everyone]

Here, you can see they arise simply from the fact that the commutator of the raising and lowering operators are 1. This commutator determines the energy difference of the bound state energies. This is why the energies are discrete for this problem.

[@nthumara6288]

thank you madem this is really helpful

[@ProfessorMdoesScience]

Glad you find it helpful!

[@oraange]

What is exactly the zero point energy ? I haven't well understood

[@ProfessorMdoesScience]

Let's consider a harmonic potential V(x)=0.5*m*omega^2*x^2. A classical particle moving in this potential can have any energy between 0 and infinity, and in particular, the lowest energy it can have is 0, which corresponds to the particle being at rest in the potential. By contrast, the lowest energy that a quantum particle can have is not 0, but 0.5*hbar*omega. This minimum energy is called the zero point energy (or sometimes also the quantum zero point energy). The fact that the lowest possible energy is not zero is related to some of the "strange" properties of quantum particles, for example in this case we can associate it with the uncertainty principle which says that the quantum particle cannot be fully at rest in the potential.

[@MrFilipebento]

Very good! I am learning a lot and I am trying to keep up with your classes. However, I suggest you use the same video names in description's fields "Background" and "What next?". For example, this video is called "Eigenvalues of the quantum harmonic oscillator", but it's reported in description of the other videos as "Harmonic oscillator eigenvalues". This difference occurs in several videos of your channel. The meanning is the same, but it may cause doubts in beginners.

[@ProfessorMdoesScience]

Glad you are learning and many thanks for the suggestion! We have corrected it in this video, and will do so whenever we identify similar issues in other ones.

[@quantum4everyone]

The main issue is that the position eigenstates are not well defined. They are not in L2 and hence not part of the Hilbert space. So to work with them, you need to use some regularization procedure. One is to broaden the sharp delta functions, and then you think of the position eigenstates as being defined only for a discrete subset of the continuous line. Then you have countable states for both position and the harmonic oscillator. Only when you make the gaussian’s infinitesimally thin do you get the continuum. This limit must be handled carefully and requires advanced math. When you study squeezed states, you can also take those to the limit where they are squeezed to delta functions as well, and one can see that there is a connection from position eigenstates to momentum eigenstates through the squeezed states.

[@ProfessorMdoesScience]

Thanks for your comments! We do have a few videos discussing some of these ideas in the pipeline (starting with bound vs unbound states, and so on), hopefully we'll get there soon!

[@quantum4everyone]

You should be sure to look at the relationship between squeezed states and position and momentum eigenstates. It is not very well known, but makes for a quite nice illustration of how this works.

[@asmitghosh2888]

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