Ch 7: How are observables operators? | Maths of Quantum Mechanics

[@sarveshpadav2881]

As you mentioned in the video, any physical quantlty (called observables) associated with a particle is represented as a linear operators in QM which implies that 'mass' being a physical quantity should also have an operator representation right?

[@quantumsensechannel]

Hello! Thank you for watching. Ah, this is a good question. I think I may have been a little too loose with my language, and that is my mistake. You’re correct that mass is something that can be measured and observed, but it has one difference that other observed quantities don’t: mass is an inherent property of our particle. What does this mean? Well it means we never have a spectrum of masses for a single particle*. The mass we measure for the particle is the mass it has, so even though we could consider it an observable, it would be a very uninteresting one: it would only have a single eigenvalue. *Now, things change when we move to relativistic quantum mechanics (ie quantum field theory). And a beautiful example of this are neutrinos. It turns out, the flavors of neutrinos aren’t the same eigenstates as the mass eigenstates of the neutrino! What does this mean? Well when, say, an electron neutrino is emitted, it’s actually in a superposition of three mass eigenstates. The mass eigenstates are the free particle eigenstates, so as the electron neutrino moves through space, these mass eigenstates are each time evolving with their own phase. The result of this is that our electron neutrino now oscillates into a superposition of both the muon and tau neutrinos! This is the phenomenon known as neutrino oscillations. So you can have an electron neutrino be emitted, but measure a tau neutrino when you detect it later. So when you hear people talking about “measuring the masses of the neutrinos”, they’re not talking about any flavor in particular. They’re talking about measuring the mass in the mass eigenstates, which are superpositions of all three neutrino flavors. Cool right? I think quarks also have some mass mixing, but it’s much much smaller. So in QFT, it makes much more sense to consider mass as a true observable, but the idea of observables themselves is different altogether in QFT, so we need to be careful. Hopefully this answered some of your question! -QuantumSense

[@sarveshpadav2881]

@@quantumsensechannel First of all, thannks for answering my question along with sharing some interesting facts related to neutrinos and quarks. So basically in case of non-relativistic QM where in we consider the mass of the particle to be fixed, its not very useful to write it as an operator as it always has oone value. This much is clear to me from your response. In case of neutrinos (if I understand correctly) the flavour state the neutrino is in is a superposition of three mass eigen states. So, does it mean that a neutrino known to be in a particular flavour state can have three possible values of mass and each time we measure it there is some probability of getting any one of those values? and if this is true then mass of neutrino needs to be represented as an operator that has three eigenvalues corresponding to those three values of masses. Right?

[@quantumsensechannel]

@Sarvesh Padav hello, Yes, as far as I understand it, the measured mass of a neutrino in a definite flavor state follows some probability distribution, and you could get either of the three. As for your point about mass being an operator: we need to be careful. The framework of relativistic quantum mechanics (quantum field theory) is inherently different than the quantum theory we are building in this series. In QFT the only operators are the fields themselves, and any operators you build from these fields. For example, there is no position operator in QFT, it is now simply a parameter of an underlying quantum field operator. That being said, we can build an operator P^2* out of these fields that does indeed give masses as eigenvalues. *if you know any special relativity, then P^2 is just the scalar invariant of the field’s total four momentum, which gives E^2-p^2=m^2, as claimed above. -QuantumSense

[@sarveshpadav2881]

@@quantumsensechannel Thank you very much for clarifying my doubts also I would like to mention that I will be taking up a QM course (8.05x) on 'MITx online' platform that heavily relies on the math and formalism of QM developed in this video series. I thank you for creating such an amazing series and also to youtube for recommending your channel!

[@vicar86]

This is by far the best video series in KZbin for 2023! Thank you!

[@imrematajz1624]

It goes for 2024 too!

[@Self-Duality]

I’m absolutely loving this series! This is a perfect precursor to tackling the older treatises (e.g., Dirac’s writings) and newer textbooks.

[@it6647]

0:00-Recap 0:49-Observables and linear operator definition 3:55-Eigenvalues, eigenstates, and superposition 5:22-Properties of physical observables, proof for eigenstates forming a basis

[@stephanecouvreur1377]

I find that “foundations” help me a lot to understand a subject, by explaining how concepts and laws come about from physical intuitions or observations, and how they relate to one another to ensure internal consistency. It doesn’t need to be historically accurate, it is still of great pedagogical value. QM textbooks often skip this part by slamming the reader on the head with 7 or 8 postulates which seem to fall out of the sky (do you know any exception, is your approach inspired by a particular textbook?) Congratulations for taking a different route, I’m impatient to see the rest of the series! 👏👏👏

[@henrycgs]

I think the biggest lesson I'm learning from this series is how physics is built. We observe things that don't follow our current understanding of nature. We try to come up with a model that better explains what we observed. We then extrapolate the model based on its mathematical properties. And then we check if the extrapolations also hold true. If they do, that's a good model. Otherwise, back to step 1. And sometimes, these models can get pretty unexpected and unintuitive. But if it works, it works.

[@pixlark4287]

Something that confuses me greatly is the nature of these quantum states. I think I'm misunderstanding something. If we represent a particle with a quantum state |ψ>, that state can be constructed from a basis of eigenvectors for some particular property like energy, momentum, etc. So if |ψ> = c_1 * E_1 + c_2 * E_2 + ..., where E_i are possible *energies* that the ψ particle could be measured at, then we've completely encapsulated ψ's quantum state just in terms of its possible energy values. But what I don't understand is that |ψ> can be expanded into *any* property, not just energy. Does this not imply that if two particles have the same exact quantum state in regards to one property, say energy, that they must then be identical with every other property? But surely that can't be the case - if two particles have the same exact energy, for example, that doesn't mean they have the same exact position, or the same exact momentum. Two electrons for example could be in the same energy state around two different hydrogen atoms. Obviously I'm confusing something, but I can't figure out what that is...

[@lintstudios3072]

Maybe I am wrong, but maybe it is because you are changing the basis of the quantum state? Meaning that you cannot move from your quantum state written in the basis of Energies to Momentum, as they both have different information. However this may also imply inequality. Maybe it has something to do with the different fields of these bases.

[@stevenschilizzi4104]

This was one of the key questions I had been asking myself for a long time (and that no textbook ever really explains: “This is the formalism, mate; now shut up and calculate”… and become a donkey-physicist!). This video is quite illuminating. And physics isn’t even my field. So, well done and million thanks. Looking forward to more.

[@divyaanshu123]

Best playlist on QM for students jumping to graduate level QM. I highly appreciated the content of this channel because it took me a complete reading of QM book to learn this.

[@physicsbutawesome]

I am really intrigued by this channel. Either there are a lot of physics students watching youtube or there is a general audience that is not scared off by maths. Either way, fascinating!

[@pseudolullus]

There is a general audience

[@sr3klsxcfviou]

8:00 Great intuition!

[@kesleta7697]

I am confused on how the eigenvectors of an observables associated operator can form a basis for our Hilbert space. For example, I do not understand how we could add several possible measurements for energy and end up with a vector (or ket) that includes information on a specific position. Additionally and relatedly, what do you mean by the notation at 3:00, "| 2.44 N*m*s >"? This seems like it is constructing a ket that contains information only about angular momentum, which seems like is incomplete. Am I incorrect that the kets in the Hilbert space contain _all_ the information about a given particle? Thanks for the really great series! Additionally, what do these operators actually "do?". If I have a particle a represented by the ket |a>, what does E^ |a> actually represent? Is it a different particle?

[@LeTtRrZ]

I think there was an overuse of fancy language in this video. It seemed like a really really really complex way of saying “each possible measurement is unique and irrelevant to all other possible measurements, except by their probabilities which sum up to 1. Also, there are no possible measurements which are missing and there are no categories/units of our measurements that will pop up midway through.” None of this strikes me as anything any reasonable person would even have to think about when taking actual measurements, so to me it seems a little bit like a waste of time, unless I have completely missed the point.

[@matusmoro955]

I am currently in undergrad and even though I haven't had QM yet, your videos are still a joy to watch!

[@shiroufubuki6172]

This explanation for the quantum formalism is the best I've seen teacher thank you so much.

[@EriiikaGuerra]

Awesome! It's saving my ass in my quantum mech class at the moment. Can't wait for the next video!

[@charbinger3803]

I absolutely cannot wait for the next video, this helps a lot

[@pjotrstraver]

Amazing series, looking forward to the next chapters!

[@timuralmabetov2213]

This is amazing! Keep doing what you do and you will get your million followers very soon!

[@SeattleMarc]

These are great videos. I'm already familiar with a lot of the material but I am still learning things seeing the concepts from a unique perspective. Can I request that you make a playlist with all the videos in this series? It makes it easier to play them all end-to-end that way. (And I can also add the playlist to my library, share it out, etc.) Thanks!

[@quantumsensechannel]

Hello, thank you for watching. There is already a playlist with all current videos on my channel. -QuantumSense

[@SeattleMarc]

@@quantumsensechannel Oops I guess I missed that. Thanks.

[@aramsarkisyan8061]

Great video and amazing series. I don't yet get why eigenvectors and eigenvalues are used though. If I undestand correctly, we need a linear operaor that takes the quantum state vector and gives us the particular number for one of the observable properties of the particle (say, angular momentum). If we use eigenvectors, then using them as operators to quantum states will give us the same eigenvector times the eigenvalue. What I don't undestand is why those correspond state of the observable and values of it? I feel the intuition that the set of eigenvectors gives some kind of building blocks of the quantum state, that are least affected by the changes and thus can be considered as defining for the state, but it still seems vague.

[@amitshoval7653]

I love this series, thank you I have a few questions (sorry if they are basic) I understand that an observable are a linear operator which takes a ket and returns another ket First, is the observable a function from the space to itself? Second, what does the returned ket represent, and how do you use it to get the value this observable needs to meassure, like position, momentum, etc.?

[@angelmendez-rivera351]

The video already answers these questions.

[@siarez]

Love this series. By far the best intro into the topic.

[@Qrudi234]

Bro just popped up and started dropping the hardest physics tutorials

[@JonasBroeBendtsen]

Thank you for this amazing series. I love the style of the videos, and you do such a good job explaining fairly complex subjects! I hope the rest of the series will come out in time for my quantum mechanics exam in two weeks, as they really help me get som of the concepts i didn't get during lectures or while reading in Griffiths Introduction to Quantum Mechanics.

[@tedsheridan8725]

Great videos! Looking forward to the next one!

[@michaelpotter3418]

Thank you for your wonderful teaching!

[@greeeeeeeen]

Amazing series

[@andywong2135]

@quantumsensechannel Great video series! One question, does the following mentioned 'vector space' actually mean 'Hilbert space' ? 5:50 Observables' eigenstates must span the entire _vector space_ 8:44 span the entire _vector space_

[@briansteven6318]

Gracias por toda la serie bro

[@typha]

I think you've kind of got the direction of this whole thing backwards in a sense: We can first notice that any physical observable is going to be associated with a spanning set of mutually orthogonal definite states. Each of those definite states is also associated with a value of our observable. We want a single object that ties these two things together, and we can realize that any basis that is associated with a list of scalars can be used uniquely define a linear operator where the vectors are the eigenvectors and the scalars are the eigenvalues. So why are observables operators? We made that _choice_ because linear operators are a _seemingly natural_ way to encode the relationship between the vectors and the values that we want to associate with each other.

[@quantumsensechannel]

Hello! Thank you for watching! Yes! This is a really great interpretation of *why* observables are operators. I worked the other way around for concreteness, but your approach works more for understanding the appearance of operators in the first place. And in fact, in a future chapter we’ll discuss Hermitian operators, and we’ll learn how to write operators as a sum of their eigenkets/bras and their eigenvalues, which is exactly how you propose to define these operators. We’ll then use this form to prove that observables must be hermitian operators. -QuantumSense

[@IonfraX]

The last line in your first paragraph is super useful. That's the result that ties the whole thing together. Thanks for sharing!

[@HilbertXVI]

Not every physical observable has an orthonormal basis of eigenvectors. For instance, take the simplest ones: position or momentum

[@quantumsensechannel]

@@HilbertXVI Hello! This is a bit of a gray area, and I’ve been somewhat intentional with that. The approach I’ve taken in this series is that orthonormality in the continuous case (like position and sometimes momentum) is expressed as *Dirac orthonormality* , I.e. through the Dirac delta. This works well in physics, and it keeps all the math in check. That being said, you are correct in that formally, that doesn’t really count as normalization, and we run into a bit of an issue. One way we fix this is by widening our space a bit, by considering what’s known as a *rigged Hilbert Space* . This is a bit too niche to cover in this series, but it essentially allows us to work in a vector space that includes the poorly behaved bases like position and sometimes momentum. All in all, physicists can consider all operators to have an orthonormal eigenbasis, so long as we remember that we have to use Dirac orthonormalization every now and then. -QuantumSense

[@HilbertXVI]

@@quantumsensechannel That's fair, and I know this is the popular approach, but it's easier to drop that formalism all together and just admit that certain self-adjoint operators in certain cases have an orthonormal bases of genuine eigenvectors. Imho it's much less confusing when you can build an actual basis of functions for L^2(R) (or whatever your Hilbert space is, all of them are isometric anyway) instead of thinking of abstract entities like the dirac delta. When you're starting out learning QM you'd only ever think of operators on a case by case basis and that's all you'd ever need. (Of course, more generally to put this "continuous basis" approach on a formal footing, there's the spectral theorem for self-adjoint operators which gives you a spectral decomposition of your operator in terms of projection-valued measures, but that's too much formalism)

[@narfwhals7843]

Again a very nice summary! Goes a long way towards demystifying superposition. I have two gripes, though. At 5:05 you say "linear combination of an observable's eigenvectors" but this should just be "operator's eigenvectors". Superposition is not limited to observables and the arbitrariness of the basis is something you even mentioned in passing earlier in the video. And maybe I am misunderstanding what it means to "span" the space, but not all basis are equal in information. Many don't even have the same dimensionality as the space. The energy basis may well be infinite but the spin basis only has two. It only spans a subspace.

[@quantumsensechannel]

Hello! Thanks for watching! In what we are considering, I’ve only used superposition to refer to being in a linear combination of possible measurement outcomes, which implies observables. Although there are operators we use in quantum mechanics that are not observables, the ones that generate superpositions of outcomes are indeed observables. And in reference to spanning, I do indeed think I mean that eigenstates of an observable span the whole space of the vector space they act on. This last bit is important. You bring up spin, but remember that spin operators( eg Pauli Matrices) only act on their spin vector spaces (which is two dimensional for the Pauli case). So the spin eigenstates do indeed span the entire spin vector space, it is not a subspace - this is a big point. What you may be confusing is when we combine the spin vector space with our particle motion quantum vector space, which we do with a tensor product. This creates a larger vector space that includes both, but in this new vector space, every state has a spin, so the spin eigenstates still span the whole space (they’re just wildly degenerate now). Let me know if this clears it up! -QuantumSense

[@narfwhals7843]

@@quantumsensechannel Thank you for your reply! I don't think that's right about superpositions. An eigenstate of an observable can be expressed a superposition of some non-observable eigenvectors. Unless you are using that term specifically to mean what you said in the video "linear combination of observable eigenstates". I am using superposition and linear combination interchangeably. Is that wrong? So by "the space" I meant the _entire_ Hilbert space of general quantum states. The spin space is a subspace of that, is it not? Do you mean that for each problem we should restrict ourselves to the particular space of the observable we are interested in? Isn't that just a projection into that subspace? Or do I also not know what a subspace is? :D I don't think I understand how the spin eigenstates can span a space that has a larger dimensionality. You can't get out of the subspace with just the spin information. How could we possibly guarantee orthogonality?

[@quantumsensechannel]

@@narfwhals7843 hello! I think we’re just using terms a bit differently, but we’re agreeing. I use superposition to specifically mean those with quantum state outcomes, ie observable eigenstates. But yes you are 100% correct, general linear combinations need no mention of observables eigenvectors. And re: eigenstate span. No worries, this is a subtle issue. Let’s take the Pauli spin operator. By definition, this operator acts on spin 1/2 states, which are two dimensional. This 2D space is the entire Hilbert space for spin states. Note that this is a *different* Hilbert space than the one that the position eigenstates etc. live in. So to get our “full” Hilbert space, we need to combine the spin states, |s>, and our general quantum states |psi>, which live in their own Hilbert space. So we do this with a tensor product of the two hilbert spaces, H_spin*H_general. So the states that occupy this tensor product now look like |s>|psi>, where we have an implied tensor product between them. So every quantum state now has a spin attached to it. So now note that spin eigenstates can look like |s>|psi> or |s>|phi> or |s>|zeta> etc. Since the spin operator only acts on the spin part, these are still eigenstates of the spin operator, but they’re now horribly degenerate with any |psi> we want to choose. But do you kind of see how the spin states still span the entire larger space? In the tensor product, every state has a spin attached to it (which makes sense, since the quantum states of a spin 1/2 particle need spin!) Let me know if this is still sort of shakey! -QuantumSense

[@quantumsensechannel]

@@narfwhals7843 One last note: I think to be more succinct - the “full” Hilbert space is the tensor product of the spin and general hilbert spaces. So the spin space is not a subspace of this larger space, but rather a part of the tensor product that makes it up. -QuantumSense

[@narfwhals7843]

@@quantumsensechannel Thank you very much. That clears up a lot. So the basis for the combined statespace is not just |+>, |-> but |+>|psi>,|+>|phi>,|+>|ki>..., |->|psi>,|->|phi>,|->|ki>...etc So the spin _alone_ only spans the spin states. But the tensor product of the bases spans the tensor product space. That makes sense. And the spaces that make up the tensor product are not called subspaces of the tensor product space? That seems to be my error, then.

[@juniorcyans2988]

My goodness! You saved me!

[@hjs6102]

as always: great, thx

[@denisch_1010]

Nice videos. Some questions on this one however: 1) I think the jump from "observing a real number" to claiming that this is a eigenvalue is still too big; same problem as in all the QM books; after all observation is not an abstract process, it goes via a physical particle interaction; e.g. emission of a photon if spectrum is observed or absorption of a field photon if particle changes direction which would reveal its spin; how do we best show it in matrix terms that an eigenvalue will be revealed by an interaction? 2) Should not a full state vector contain sum of all Hilbert spaces involved? All energies, all positions, all angular momenta, all momenta, all spin values are dimensions of the combined Hilbert space. How do we motivate that we can work with e.g. definite energy eigenstates disregarding all the other dimensions corresponding to position, momenta, spin, etc, etc (which may be not observed and therefore may degenerate those observed eigenstates)?

[@quantumsensechannel]

Hello! Thank you for watching! As for question 1) I don’t really have a better motivation to be totally honest, since at the end of the day, connecting physical values to eigenvalues is an axiom. I merely justify this axiom in the episode, but any approach to quantum mechanics is bound to have a logical leap here. There may be some other approaches that make this leap easier, but hopefully mine wasn’t too bad! For question 2) I think there may be some confusion as to what space our eigenstates span. The position, momentum, energy, angular momentum etc. eigenstates all span the same Hilbert space. So the position eigenstates span the same space as the energy ones and vice versa. So when we expand our quantum state in terms of the energy eigenbasis, we are not ignoring the “position dimensions”, they are still in this space, and they are covered by the energy eigenbasis. Working with different eigenstate expansions is just a change of basis, but it’s a basis for the same hilbert space as a whole. So we don’t ignore any dimensions - hopefully this makes sense. The only exception to this is spin, which does indeed exist in its own Hilbert space (eg a 2D space for the case of a spin1/2 particle). To combine this into a “full Hilbert space” we take the tensor product of the two hilbert spaces, and now our quantum states are of the form |spin> |psi>, where a tensor product is implied between the two. I don’t plan on talking about spin yet, so there was no need to introduce the tensor product or how we combine Hilbert spaces. So far, we only have one Hilbert space, and just a bunch of different bases (position, momentum, etc) that can cover this space. Hopefully this cleared it up a bit! -QuantumSense

[@denisch_1010]

@@quantumsensechannel Thanks for the answer! I think that even though that connection is an axiom, it is possible to motivate it far enough so that it will feel like an almost rigorously derived conclusion. In fact here on KZbin I found a very good attempt to do it, albeit in a special case of spin: kzbin.info/www/bejne/aZnMpo2Ilqt8m6M. (especially from 1:10 to the end). Basically, what Noah is trying to say there is that given that a state can be written out as spanned over basis of eigenstates, then given sufficiently large number of tensor-multiplications of it with states of outer world, it will get entangled very many times and eventually get distilled into those eigenstates, which will reveal the eigenvalues. Thus, since in order to make a classical observation we by necessity need to let our particle interact with very large external system, then a measurement in practice will pick out one state from the eigenstate basis, and not any linear combination. However, I am still struggling with how to formulate this explanation in a simple, concise and generalised manner without a 1-2 hours long deliberation! Maybe you have some good ideas?

[@quantumsensechannel]

@@denisch_1010 hello, Ah I better understand your point now. Yes, what he is referencing in that part of the video is how we go from quantum superposition to classical state observation via interactions with your system. This is a bit different than what I understood originally. This isn’t so much a question about getting an eigenvalue, but rather a question about the nature of quantum collapse, and how we get the classical world we see around us from that mechanism. I think he summed it up fairly well, and it wasn’t quite what I was trying to get at in this video. That being said, in the next chapter, we will show how we mathematically describe quantum collapse, but that is as far as I will go. I won’t interpret it (since there are many interpretations!) and I won’t quite show how it leads to our classical world. To do so would require a video on its own, since there is a lot of background needed to understand it (as you can see from Noah’s explanation). -QuantumSense

[@Jamie-my7lb]

Hello! I have a question. Do you discuss duplicate eigenvalues? If the quantum states have dimension D, then the eigenstate basis is D independent vectors. That could give rise to D possible values of the physical property. It seems counterintuitive that the quantum state dimensions equal the number of values the property can take on, since physical properties can be highly constrained, i.e., one of two values, while quantum state may be higher dimensional.

[@physicsnabo]

thank you so much....

[@GosaDoz]

I have a few questions and well these videos belong to about a year ago... Is it still possible to ask questions?

[@yashpurohit6988]

for a quantum state phi of a particle's let's say, *Energy*, why does the observed list of measurements make eigenbasis, and not just, basis? and what kind of *transformation* would affect this basis,. since linear independence does not require orthogonality, why do we require these observed energy measurements to be orthogonal, it could be a fine basis for the quantum state with just being independent, and also I'm super confused why energy observables need normality. help?

[@aslpuppy1026]

Hi Quantum Sense, I understand why the eigenstates form on orthogonal basis, but why would the eigenstates also be orthonormal?

[@quantumsensechannel]

Hello! Thank you for watching. This is a good question. If the states can be normalized, then we can just divide each vector by their length to normalize them. If they can’t (like the position eigenbasis), then we can normalize them to the Dirac delta (if this is unfamiliar to you, watch Chapter 5 on the Dirac delta). Let me know if this didn’t clear it up! -QuantumSense

[@aslpuppy1026]

@@quantumsensechannel Thanks so much

[@truthseeker8228]

Hi, thank you for this nice series. Could you please suggest any text book on Quantum mechanics which has its contents organised like this series.

[@شعرکوتاه-ع7ظ]

Thank so much

[@tedsheridan8725]

Just curious what software you're using for text and animation. Manim?

[@quantumsensechannel]

Hello! Thank you for watching. Yes, all animations were made with Manim. -QuantumSense

[@koenvandamme9409]

You show why the eigenstates must be orthogonal, but not why they must have a norm equal to 1. Is that because each eigenstate has a probability of 1 for having its exact value?

[@quantumsensechannel]

Hello! Thank you for watching. Yes! This was intentional. In the next episode, we’ll show how this normalization comes about. In fact, we’ll derive that we can have any normalization, so long as we accordingly change our probability scheme. However, it’s convention to choose 1, since that makes all the formulae much simpler. -QuantumSense

[@LL-mq7gj]

The issue so far that I am having is where do the actual operators come from and how do we know their values. And the same with the states and associated probabilities?

[@pawelwelsberg6800]

I really enjoy your videos. One question regarding eigenvectors: you seem to imply that there is a specific eigenvector corresponding to a specific eigenvalue. Why is that? Isn't it that there's infinite number of eigenvectors with specific eigenvalue? The vectors that will be scaled by the operator located on the same line? Thanks

[@angelmendez-rivera351]

The space of eigenvectors corresponding a particular (nondegenerate) eigenvalue of an observable is spanned by exactly one vector.

[@pawelwelsberg6800]

Maybe what you are talking about linearly independent eigenvectors? If yes it is fine. But which one of linearly dependent eigenvectors do you choose? Or it doesn't matter in your opinion?

[@pawelwelsberg6800]

For anyone else wondering why/how do we choose an eigenvector for a specific non-degenerate eigenvalue from all the possible linearly dependent eigenvectors search therm "normalised eigenstates".

[@JDY0303]

I am a little confused between the difference between a measurement operator and an observable operator. Are they the same thing? Or can observable operator seen as the operator that changes the basis for the quantum state before the measurement? For instance, if you want to measure a qubit in |+> basis, you would need to multiply the Pauli X operator and then measure it. Is this correct?

[@palepoint7092]

Oh yeah 😍😍😍

[@drewnoren8416]

How do we know that the set of definite states is countable?

[@quantumsensechannel]

Hello! Great question, we don’t! In fact, they are not countable in the case of position. If you watch episode two, you’ll see how we deal with continuous definite states. The formalism changes slightly, but everything still works* *it works for physicists. The formal mathematics get murky, but physicists can still use the math to get all the right answers. -QuantumSense

[@Doozy_Titter]

Thank you

[@NovaWarrior77]

Wait I'm struggling to comprehend: are making these like every single day???

[@quantumsensechannel]

Hello! Thank you for watching. I have made these videos in advance, and am releasing them at a regular schedule. This series is a project I’ve been working on for about a year and a half in my free time. -QuantumSense

[@NovaWarrior77]

@@quantumsensechannel oh I see that makes more sense 😅 amazing amazing series by the way!!!

[@elizabethmartell5835]

What should I do if I am taking quantum mechanics (graduation requirement) but not able to or required to take linear? How would one be able to learn quantum and understand without? Most of my class has been taken past calc three and we are so lost but unable to ask for help because our professor doesn't understand why we do not just get it. Sorry this sounds like a rant but I am honestly asking because I have been trying to learn this for weeks now and still cannot comprehend how to do the homework questions. Like I look at a question and cannot comprehend why math would need to be present.

[@quantum4everyone]

Two more quibbles on this. (1) There are many observables from classical mechanics that can be said to be complex. One example is the complex impedance of an ac circuit. It is measured all the time. While measuring instruments are not calibrated in complex numbers, and measurements of complex numbers may require measuring more than one quantity (Like an amplitude and a phase), I don’t think it is correct to assume all measurements are real. This is especially true with quantum computers, where we can routinely measure complex objects like Green’s functions or matrix elements by measuring different components of ancilla spins. Just look up the Hadamard test. (2) Unless you postulate that the measuring device always measures all particles, the flaw in your completeness argument for measurement is if the particle has none as an eigenvalue, then it goes through the apparatus unmeasured. And it is always dangerous to describe measuring devices in the abstract. For example, how is momentum measured? Hint, usually it requires measuring position and inferring the result. I don’t think there is any simple way to measure angular momentum of a particle unless it is neutral and can have a Stern-gerlach experiment done on it (and then it measures total angular momentum) and so forth. What you have given is a simple heuristic, as the spectral theorem is much richer than what you describe here. Dangerous to call it a proof.

[@angelmendez-rivera351]

He never called it a proof. This entire objection is a giant strawman. This video series has been explicitly stated multiple times to be a series that intuitively motivates the axioms of quantum theory. It never has claimed to prove those axioms, especially since, apparently you did not know, axioms cannot be proven at all. We motivate the axioms intuitively, and then we test them experimentally. Since they match experimental data, we keep them. That is all there is to it. There is no "proof" in science, only empirical verification. This is what happens when people decide to jump onto a video series and ignore everything that was explicitly stated in the first video series: they start typing nonsense like this. *sigh*

[@quantum4everyone]

@@angelmendez-rivera351 The spectral theorem is not an axiom, even if some textbooks state it as so, and the standard theory of quantum measurement has many flaws. This is what my comment is about. Not sure why caused a strong response from you. But to ease your concerns, I have watched every video, not jumped in at the middle. What I have stated is mot nonsense. Let’s be polite.

[@angelmendez-rivera351]

@@quantum4everyone I never stated the spectral theorem is an axiom. I was very specific in my comment when I mentioned the axioms of quantum mechanics, of which the spectral theorem is not one of them. Reading comprehension goes a long way, see? The standard theory of quantum measurement? What on Earth are you talking about? Are you talking about nonrelativistic quantum mechanics (what would be called the Dirac-Neumann quantum theory)? Or are you talking about quantum field theory? Because these are very different mathematical frameworks, with very different levels of experimental success. There is nothing in physics that goes by the name "standard theory of quantum measurement." That just sounds like some pseudoscience a crackpot made up.

[@quantum4everyone]

@@angelmendez-rivera351 I can see this conversation will not go anywhere. I am sorry you are having trouble understanding what I said. Maybe someday you will. But I am not engaging further with you as you refuse to maintain decorum.

[@angelmendez-rivera351]

@@quantum4everyone I am not the one with the problem understanding here, as my previous comment I think made clear, but I will not force you to have a conversation you do not want to have. Good bye.

[@technicallittlemaster8793]

Since a nonzero subspace is infinite, every eigenvalue has infinitely many eigenvectors. Then how is it possible for a physical observation say momentum to be an eigen value from only a particular eigen vector?

[@tomtomtomtom691]

If position is continuous but energy is discrete, it seems that the position operator should have more eigenvectors than the energy one, but the cardinality of both bases must be infinite since they must span an infinite vector space. This confuses me.

[@rileymcnamara5841]

the position is continuous and thus doesn’t have the “definite states.” Only observables with definite states are expressed by and eigenvalue/vector equation

[@angelmendez-rivera351]

Yes, both spaces are infinite, but they do not have the same dimension. One of the spaces has countably infinite dimension, while the other has uncountably infinite dimension. The basis are not of the same size at all.

[@angelmendez-rivera351]

@@rileymcnamara5841 The position operator still has definite states, you just have to use the spectral theorem to meaningfully talk about those.

[@aydencook03]

I can intuitively understand that the eigenstates must span the entire space and that they must be linearly independent, but why must they also be orthogonal?

[@jamesu8033]

Orthoginality is required to satisfy the born rule. Imagine you are in a specific eigenstate with a know observable. The probability of measuring a different eigenvalue is 0 so these eigenstates must have 0 overlap by the born rule.

[@misterlau5246]

Ok let's see they don't commute, I fancy L± to take measurables of m, max on z with magnetic field to align spin... 🤔 And there's the leeway of heisenberg in the x abd and y axis. Ok, let's see this video... But we have expectation, expected value, and stdevs. It's a long way isn't it? Then, we still have to get the ketbra and the Fermi Dirac distribution for massive particles... And the B. E. etc. What else here, the vectorial space, it's OK you said it is spreaded through the whole thing. Then the eigenstuff, something outside the famous wavefunction gives us random values around the distribution, which is the wavefunction. It's just that it's too little time for each clip🤓😬😟 Hernitian. Well, the explanation is good, so is your voice and the graphics. Lab.. Oh particle lab... There's where your observables don't work all the time and you start tweaking parameters to adjust, then see which gives you the expected results, then you reverse engineer the observable to rebuild it so it fits and the grades stay on A😬😈

[@RurczakKurczak]

There are several things I don't understand. What actually are those operators? You said the physical quantities are represented with operators, but you didn't say why and just jumped to eigenvectors and eigenvalues of the operator that we don't even know where did it come from. I also rewatched the first episode, but I didn't find the answer. The second thing is: I understand that the definite states form some orthonormal basis, but why is it an eigenbasis? I just don't see it. Maybe it will be clearer after I understand the first problem.

[@mastershooter64]

Perhaps we need to introduce non-linear operators for quantum gravity haha

[@fawzibriedj4441]

Hello, First, thank you for the video :) I admit I didn't understand much in this video... I don't understand how physical quantities are linear operators, and even though the title indicates that this video will explain it, I don't see the explanation... Linear maps are transformations, I don't see why a physical quantity like angular momentum would be a transformation ? It transforms what in to what ? It's worse when we get to eigen vectors... These have a specific definition. They have to verify f(v)=k*v, v being the eigenvector and k the eigenvalue. I didn't see where it was mentioned that definite states verify this definition?

[@angelmendez-rivera351]

It seems to me that you need to rewatch the video and take notes, because this was all addressed in the video. Being confused is perfectly fine, but saying "the video did not address this" when it did is not fine. If some part of an explanation in the video is confusing, then you need to ask specific questions that make direct reference to which part of the explanation was confusing, not simply deny that the explanation exists.

[@fawzibriedj4441]

@@angelmendez-rivera351 I watched the video 3 times. And still I was more cautious than that, I didn't say "the video didn't adress this" (even though that's what I think) I said "I didn't see it". Also I tried to be specific about my questions: 1 - If physical quantities are linear operators, they transform what into what? 2 - If quantum states are eigenvectors, how are they verifying the defining property of eigenvectors? If you have answers to these questions or you know when it was adressed in the video, I would be grateful.

[@eqwerewrqwerqre]

You are a god amongst men and if you ever need US citizenship i will marry you in an instant.

[@officiallyaninja]

How do we find these operators? I accept that if you know the energy operator then you can find the eigen basis and use it to find the probabilities of each eigenstate, but to do that we need the energy operator. And how do we find that?

[@angelmendez-rivera351]

The Schrödinger equation is what you use for that.

[@Lenajj]

Don't 2 non colinear vectors of R2 form a base in R2 ? The orthonormality is a sufficient but not a necesseray condition to have a base in a vector space. Especially, if you have two vectors of R2 linearly independant, those vectors necesseraly span the entire vector space and therfore, this is sufficient to say that those vectors form a base in R2. In that sens, I don't understand why operator vectors should be orthogonal and not only linearly independant which doesn't necesseraly mean orthogonal. Could you please tell me if something is wrong in my reasoning ?

[@narfwhals7843]

He is showing that observable eigenstates must be orthogonal because the existence of an orthogonal component contradicts the idea that an Eigenstate corresponds to 100% sure of the measurement. That orthogonal component is not another basis vector but just some other possible orthogonal vector. He is saying any orthogonal component must _always_ be zero, therefore the Eigenstates are mutually orthogonal.

[@FiniteJest]

So if something is physically unobservable, we could create a framework to give it complex eigenvalues. I wonder if there is anything interesting with such a theory.

[@laurenzwicke1120]

What does it mean for an operator to "have" an eigenbasis? What is the term "having" mathematically?

[@quantumsensechannel]

Hello, thank you for watching. By this, we mean that the eigenvectors of the operator form a basis for the Hilbert space we are working in. -QuantumSense

[@laurenzwicke1120]

@@quantumsensechannel Ah yes, i just figured that out reading the comments. At first i didnt get that it would span the entire hilbert space and i also agree with the user that suggested tackling the intuition from actively building the operator from eigenstates instead of stating that the observable values result from the operator. Besides that, wonderfull series. Thanks for the lightning fast answer too.

[@stapleman007]

If you are lost, go study Linear Algebra. Gilbert Strang wrote the Gold Standard textbook. Else, you can flop around in the KZbin Wikipedia Soup trying to figure out Linear Algebra.

[@Machu_channel]

👍👍👍👍👍👍👍👍👍👍

[@curtpiazza1688]

😊

[@jessevanderkooi8794]

What i do not get, is how for example the energy eigenstates, can span the whole state space. Say that these eigenstates are countable and finite. How could the whole space be spanned by them when the position eigenstates are continuous and infinite dimensional? I guess i dont understand how something finitely dimensional can span something infinitely dimensional. Hope someone can help me understand :)

[@jessevanderkooi8794]

i think i get it, the eigenvalues are finite, but the eigenstates are infinite. essentially getting multiple states with the same energie eigenvalue. please correct me if I'm wrong tho :)

[@bharatjoshi9889]

Why orthonormal ? Does it’s component not exist ?…if the key is that ‘we can’t get another value from one value’ why orthonormal ?

[@4984christian]

What is the difference between 2D real numbers and complex numbers? You got me questioning myself... why can obersavles not be a 2D real number? I guess I am just mixing definitions unnessecarily ^^

[@APaleDot]

Complex numbers have an additional structure in their mathematics, specifically i² = -1. So, there is a difference between a 2D vector and a complex number. However, I've never really understood why that means it's literally _impossible_ for physical quantities to be complex. It seems like rotational quantities _should_ be complex.

[@4984christian]

@@APaleDot I think the i2=-1 mechanic can be replicated by geometry but maybe not.

[@quantumsensechannel]

@@APaleDot hello, I just wanted to pitch in with something kind of interesting. Usually we say physical quantities are real because our detectors must give us real numbers: the physical world is real. That being said, you CAN actually use imaginary energy values to model something unique in quantum mechanics: particle decay. You might have seen that the time evolution operator in quantum mechanics is e^(-i*E*t/hbar). So how do we model a particle whose probability of existing decays over time? Imaginary energy! If we have E = -i * Energy, then you see that we get a decaying exponential, which models a quantum state whose probability of existence decays with time. Now, note that we’re breaking a lot of rules here. 1. We can’t explicitly measure this energy 2. This energy is not Hermitian any more 3. Probability is no longer conserved. So it’s really more of a hack than anything, and you need the full framework of quantum field theory to model particle decay, but it’s cool that you actually CAN use imaginary observables to model phenomena in the universe. Whether this clarified things or gave you more questions, I’m not sure, but it was something I found really cool when first learning QM! -QuantumSense

[@angelmendez-rivera351]

@@APaleDot Has an instrument of measurement ever read a value that was not a real number? If yes, then provide the source. Otherwise, that actually answers your question.

[@APaleDot]

@@angelmendez-rivera351 But that's begging the question. Our instruments give us real numbers because that's how we interpret them. Is it literally _impossible_ to interpret a reading on an instrument as an imaginary or complex number? We interpret measurements as vector quantities all the time, so why not imaginary?

[@Alex-dn7jq]

Rx is leaking into quantum mechanics

[@danceordrink]

Totally makes sense to me thank you I hate books communicate by stating everything as facts they are stupid

[@antoinebachmann6253]

please disallow ads! especially as Google allows super long, non-shippable ads. So cannot watch your content

[@rentristandelacruz]

E!

[@lolmanthecat]

All this smokey explanation with no concrete objects confuses me a lot. I understand that it is done to explain the intuition behind the choice of our mathematical objects, and it shows, but aside from that nothing is clear. You never define an explicit space and just state how a particle is at the same time a combination of all its Eigenstates. Going with the other statement that eigenstates span the whole space and are a basis, and knowing that we can always change basis in a vector space, we get that two particles can have the same representation in one basis, they have the same energy, but different representation in a different basis, they have different angular momentum, which is nonsense. You make an example about angular momentum and write it as a scalar inside a ket, which makes it even more confusing, if legit at all (a 2D example maybe?). On this trend you continue by using the same nameholder for both the eigenvalue and eigenvector to make things more confusing. Was the assumption that a particle is an element of a space wrong? Was a particle also an operator on the space? Therefore can I state that a particle is particular linear combination of tensor products with some eigenstates times their corresponding bra, for a given observable that has only 1's as eigevalues? Still the assumption that all the states span the whole space will still make it impossible to differentiate two particles with the same energy but different angular momentum. So what's the problem? I have other questions about the fact that you say that observable quantities have moltitude of values while a compact operator has only a countable amount of them, and therefore wanted to ask about the fact if we were considering only non compact and worse unbounded operators. Now I realize that since this is Quantum mechanichs, it is here that we will break this stuff.

[@quantumsensechannel]

Hello! Thank you for watching! I think I may be misunderstanding some aspects of your question. You say that two states can have the same energy eigenbasis expansion, but different angular momentum eigenbasis expansion. I don’t quite see where this is coming from. If two quantum states have the same energy eigenbasis expansions, then their angular momentum expansions (or any other eigenbasis expansions) must also be the same. It’s the same quantum state in every basis we choose, so they have to be the same. Maybe I’m misunderstanding? Once you write your quantum state as an expansion in one basis, you’ve fully defined your quantum state, and that quantum state exists outside of any basis you could choose. If they are not the same quantum state in the angular momentum basis, then they weren’t the same state in the energy basis. Let me know if this clears up that part, then I can do my best to address other questions you have. -QuantumSense

[@lolmanthecat]

@@quantumsensechannel The question here is what is a particle? If a particle is an element of a Hilbert space, then we incur in the problem outlined before, since a vector is uniquely determined by its representation in a fixed basis, and you claim that different basis are used to express this vector as a linear combination of one particular fixed physical quantity, what happens is that if one physical quantity is fixed so must all the others. This sound wrong, thus the question. Are you saying that a particle is not an element of the Hilbert space but instead it is a subset of operators over a Hilbert space? So that one observable for this particle, be it energy, angular momentum, or whatever, is an operator that commutes with a global object responsible to evaluate energy, angular momentum, or whatever? So that it must share its eigenvectors and therefore can be completely determined by a choice of eigenvalues?

[@gameofquantity96]

Sir can I have your instagram.

[@quantumsensechannel]

Hello, thanks for watching! I do not have any social media associated with this channel, sorry about that! -QuantumSense