The Symmetry at the Heart of the Canonical Commutation Relation
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- čas přidán 10. 01. 2022
- The canonical commutator is one of the most fundamental equations of quantum mechanics. But where does it come from? I'll show you how symmetry leads straight to it! Get the notes for free here: courses.physicswithelliot.com...
The canonical commutation relation between the position operator and momentum operator is one of the defining equations of quantum mechanics. It's what leads to the Heisenberg uncertainty principle, that even if you know the quantum state (or wavefunction) of a particle, you can't precisely determine its position and momentum at the same time. In this video, I'll show how the CCR follows by defining the momentum as the generator of spatial translation symmetry. Along the way, we'll take a crash course in the fundamentals of quantum mechanics.
Get all the links here: www.physicswithelliot.com/ccr...
Related earlier videos:
- Poisson brackets demystified: • Before You Start On Qu...
- Hamiltonian Noether theorem: • The Most Beautiful Res...
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About physics mini lessons:
In these intermediate-level physics lessons, I'll try to give you a self-contained introduction to some fascinating physics topics. If you're just getting started on your physics journey, you might not understand every single detail in every video---that's totally fine! What I'm really hoping is that you'll be inspired to go off and keep learning more on your own.
About me:
I’m Dr. Elliot Schneider. I love physics, and I want to help others learn (and learn to love) physics, too. Whether you’re a beginner just starting out with your physics studies, a more advanced student, or a lifelong learner, I hope you’ll find resources here that enable you to deepen your understanding of the laws of nature. For more cool physics stuff, visit me at www.physicswithelliot.com. - Věda a technologie
The niche you’ve chosen is unique on CZcams, I believe. You delve into the math deeper than most dare, without apologizing too much. Which is fantastic for those of us who have exhausted the videos that just skim the surface, and who crave a more mathematically rigorous explanation of these concepts. Thank you.
Glad it's been helpful Darren!
See Sean Carrol's Biggest Ideas series as well
these are also really nice as refreshers
This exactly!
I'm in my 3rd year of PhD, and you've explained this better than anyone else I've heard/seen! Thank you! Great as a refresher especially
Elliot, you are becoming the new GOAT of youtube physics! Please keep the great vids coming bro :)
Thanks Nick!
these videos are so valuable and are beginning to become real resources to use along with textbooks and lectures as i’m learning about all this. thank you!
Thanks Kashu! Love to hear it!
All we need is a mathematically rigorous treatment of physics with a nice physical explanation of things. Most people just explain the theory (very well though), but math takes a back seat. I like your way of balancing both the essential components of physics. I know that your community would be a niche, but whatever be it , you are giving a "complete" explanation as far as CZcams videos are concerned
You don't really understand something unless you can explain it to your grandmother. This quote is attributed to Albert Einstein, but I must say that you have understood these things very well, that you explain them so beautifully, simply and fluently. Thank you very much.
Wow… thank you for explanation of symmetry operators. I couldn’t understand that exact meaning when I learned about it in college. It’s very helpful for me. I’m interested in physics but it’s very hard for me to study alone.
Thank you for making this video and making it available to us. I hope many students will find it. This simple origin of the canonical commutation relation is often glossed over, maybe because it removes a lot of it's mystical nature. You already discussed it a little in relation to vecort field with the poisson bracket but when you look at the ccr as shown in this video, you can easily then see the commutator as a derivative of operators and also the spatial derivative for the momentum in position space become obvious.
Thanks Louis!
Another great video. I'm curious: what software do you use to make the animations in your videos?
Thanks! For the last few videos I've done most of the animation in Keynote, and then brought it over to Final Cut to finish
Glad to see u here! Big fan of ur GR videos :) and awesome channel too Elliot!!
@@PhysicswithElliot Seriously you are reigniting my passion for physics.
hey, i watched your series on tensor calculus and Riemannian geometry! great series btw, I wish there was more actual math in youtube physics videos (like this channel, and your channel). keep up the great work!!
i always look for your comments on videos...
amazing to go back to the quantum mechanics fundamentals, exactly what I needed before going deeper in quantum field theory and quantum loop gravity
Actually you have assumed that the position operator is continuous and that the translation operator can be expanded. This approach is very nice as it shows how our intuition from classical mechanics should lead to the commutation relation. I would add that the commutation relation is somehow not just a consequence though, but at least as fundamental and things go both ways. By defining it the way we do, we can then derive the eigenspectrums of the position and momentum operators to be continuous as we physically expect (that is the relation ^x|x>=x|x> and ^p|p>=p|p>) . The commutation relation is actually somehow even more central as it is what defines the algebra of those observables and also naturally leads to uncertainty principles. It is also a concept easy to generalize to other pair of observables too as you mentioned in your previous video I think.
You make amazing videos. Please never stop ✌🏻
Thanks Alaa!
This is another excellent video Elliot! I hope you will continue to produce this kinda gems and take the Grant's (from 3blue1brown) place for Physics. All the best!
Thanks Abdullah!
I’m loving these videos. They’re an excellent supplement to the physics courses I’ve been taking at university c
Love to hear it Dahlen!
@@PhysicswithElliot They really help with a conceptual understanding of things. It’s one thing to be able to do the math, it’s another thing to interpret things, which you do an excellent job of communicating. Keep up the good work!
Really nice videos! There is a deep connection between classical mechanics and quantum mechanics and something that cannot be easily explained. Your lectures are by far the most successful ones in making this connection, which in turn demystifies the structure of quantum mechanics as a machinery. Excellent indeed! Thanks for making these!
Thanks Zain!
We need more videos like this bruv
Wow. The video was great. Thanks a lot for your efforts :)
Thanks Anvesh!
Excellent video. Thank you for sharing these ideas.
Thank you for these videos and the notes! They help so much in understanding these topics.
So glad to hear it!
I applaud your approach, you go way deeper into calculations than I would dare in a youtube video :)
In this video, I would say there is a potential snag in that at one point you simply define the translation generator to be -i*p/hbar and this gives the result. Maybe proving or at least motivating this connection would be helpful?
Thanks Chris! Make sure you watch the previous video about how momentum is the classical generator of translations for some of the motivation
@@PhysicswithElliot Will check that out!
Amazing content! Keep it coming.
how can I NOT click on a title like this?
I understand the point of the commutation relations now... thank you
This is excellent. I took quantum, but I never saw this derivation.
This is very helpful
Thank you
Thank you! Your video are very good for me. I am a Mathematics student who want to learn more Physics. Will you make some video about Gauge theory in future?
Thanks! Yes I will surely start discussing gauge theories before too long
These videos are great. Conviniently, I recently bought "A Students Guide to Lagrangians and Hamiltonians" by Patrick Hamill to learn more about this very subject.
Thanks Justin!
Dude you are just awesome
Thanks!
Brilliant!!!!!!!!THANKS
At 8:34, I am confused as to why you put the Unitary Operator in between the bracket, Why did you do "U inverse, x, U"
Under the transformation, |psi> turned into U|psi> while
Hurray new upload....
what a freakin gold mine
Knowledge of Linear Algebra is necessary to understand the Relativistic QM . Is there any video for that ?
Thanks for the video!
Just a small question: what needs to be preserved is the probability ||^2 so the phase of isn't important, right? which means you can also have operators that satisfy U^daggerU = -1 (or any other phase), and not just the U^daggerU=1 operators, no?
The state is only defined up to rescaling, so changing the normalization isn't physically meaningful. But there is a second class of symmetry operators which are called anti-unitary---the classic example is time reversal symmetry
Is p the generator of spatial translation symmetry or just spatial translation? Wouldn't only conserved momentum yield translation symmetry?
Well it's a matter of terminology; sometimes we call any unitary operator in quantum mechanics a symmetry because it preserves the probabilities. But yes, the statement of translation invariance is that the translation operator U(\lambda) leaves the Hamiltonian invariant, U(\lambda)^{-1} H U(\lambda) = H, which requires that [p, H] = 0. That's the quantum version of the classical condition {p, H} from the last video.
Great video! When you are talking about ‘generators’ is that related to generators within algebra like for a cyclic group? Or are they unrelated and just share the name?
They're different because here we're talking about the infinitesimal generator of a continuous symmetry, so that there is a continuous parameter like \lambda or t that you can smoothly dial down to zero. Cyclic groups are discrete, so each transformation is either turned on or off.
The idea is similar though in that you're writing a group element as a power of the generator. The difference is roughly whether you can take a continuous power or only discrete powers
Interpretation of the english word 'generator' in some way implies a motivation twixt one state and another. Its actual meaning depends on the context within which it is being used. Its use in Group Theory is entirely independent of its usage in Quantum Mechanics (I think).
Is there an intuitive way to justify the choice of the rescaling factor used in the video?
In i p/hbar? The hbar is there so that p has the usual units for momentum, and the i is there so that p comes out real in an appropriate sense (called Hermitian). It's analogous to writing a complex phase e^{i \theta} = 1 + i \theta + ..., where \theta is a real angle.
If x is to be conserved in time dx/dt=0, then also [x, H]=0?
What is the physical significance of commutators?
I still don't get what it means by "Momentum is the generator of spatial translations" classically before quantization and probabilities.
Thank you so much for your videos . You are the best . 😉😉😉😉😉
I want to be a theoretical physicist in future . Please can you tell me what books should I read to understand quantum mechanics . I have " Quantum Mechanics Concepts and Applications by Nouredine Zettili " Is this book , good ?
Thanks Emiliano! I don't know the book, but the main thing is to find one that's at the right level for you based on what you've already learned. So if that one feels good then go for it!
@@PhysicswithElliot
In terms of math I think I went to a considerable level to understand things at the micro and macro level (when I was in the 11th grade I worked with national and international Mathematics and Physics Olympiads) ( Last year I ranked as First in my country for Physics )
Being a fan of Elon Musk ,I want to not be influenced by school because education today ( at high school)does not measure intelligence but memory ( Of course you can say that I need a degree , but my principles are two : 1 - education is not the learning of many facts but the training of the mind to think
2 - when something is important enough , you will do it even if the odds are not in you favour
In Short ( one person must have a master degree but still be an idiot if he has paid money to get that degree )
In my country, believe it or not, we get cosine and sine theorem in the last year of high school ( is shameful for education today )
Since no Physics professor in my city has reached this level, then I started working on my own.
So I want to thank you professor for every videos that you have posted on youtube ,,, have helped me a lot .
Now I am learning theories of these topics ( Quantum Meachanics and Astronomy ) and certainly associated with relevant problems,
can you help me with a list of books( 2 or 3) that you think I should read on both of the topics I mentioned above to get the best education ( certainly with your videos )
🤯
greats concepts mate! a little bit too fast in my opinion... not only for understanding the maths (if you dont have the background you will never get) but to digest what you've just said... it's hard to follow.
Definitely a ton in there to digest! I suggest watching again ;)
You introduce the i/hbar factor to get the right units and to ensure that the result is real. Do we get a different commutation relation if we don't enforce that requirement? Could we have a version of quantum mechanics with complex momenta and [x,p]=1?
Real momentum has a direct interpretation in experiments. How would you relate the results from experiments with the eigenvalues of an anti-Hermitian operator?
@@zray2937 I haven't the faintest idea. Perhaps a subset of a theory from which observable reality emerges.
But I'm not actually necessarily looking for a theory that describes reality. Just wondering if it makes sense mathematically. if it could describe _some_ reality.
The i/hbar factor came straight out from the generator. We just conveniently extract it because we want the generator to be real, but it’s always there. Let’s suppose we don’t do that, that means to make it generic, any operator might not include i, but it’s adjoint will have i, due to the requirement that it needs to be hermitian. so it will still include i at the end.
A different choice would just amount to rescaling the definition of p, but we want it to coincide with the usual definition of momentum in the classical limit
@@PhysicswithElliot I think the point is about 'i', h bar is OK since we need to have the right unit for momentum. However, you did not tell us why do we need an 'i' here. Why it would ensure the generator to be real. It would be great if you could explain it. Otherwise it's like a magic trick to introduce the 'i' to the commutation relation. Anyway this is still a nice video.
You said "we introduce this factor of i/h as a matter of convention" but then it ended up being integral to the solution :|
The hbar is there to get the units right and the i makes p real in an appropriate sense (called a Hermitian operator). You could pick a different overall scaling, but then p wouldn't correspond to what we usually call momentum in the classical limit
Hate to be a hater, but I find the explanation in this video to be lacking and circular. Okay, so we're going to pull out this factor (1/ih_bar) and this here is a quantum generator(because it just is), and now that we've pulled out of our hat that it's the quantum generator, it should be momentum. For me, Dirac's approach of getting the quantum Poisson bracket from the classical one is much more satisfying and then using the canonical commutation relations you got from that you find the translation operator and all the other stuff you want.
As always great video. Just a correction at 7:19, It's not adjoint of U. It's U dagger. adjoint is different.
Let me see if I can summarize the steps here, in order to test my understanding:
We want U(\lambda) to be a one-parameter unitary family of operators (or, more specifically, a continuous group homomorphism from (\br,+) to the unitary group) such that, for any state |\psi> , U(\lambda) |\psi> is such that the expected value of the position in this state, is the same as \lambda plus the expected value of the position in the state |\psi>
i.e. () = + \lambda =
and to first order, U(\lambda) = 1 + \lambda * (-i /\hbar) * U'(0) , where we name U'(0) to be called \hat{p} ,
and where the factor of (-i /\hbar) makes it so that \hat{p} is hermitian, and the minus sign part of it is there to... make it have the sign we would expect for momentum?
and so then,
well, there's the \hat{x} + (i /\hbar) \lambda \hat{p} \hat{x} - (i/ \hbar) \lambda \hat{x} \hat{p} + \lambda^2 \hbar^{-1} \hat{p}^2
ok, yeah, and, uh, this being the same (up to first order in \lambda) as \hat{x} + \lambda, implies that (i/ \hbar) \lambda [\hat{p},\hat{x}] = \lambda
and so [\hat{p},\hat{x}] = - i \hbar
and so [\hat{x},\hat{p}] = i\hbar
but I glossed over the "why can we go from it looking at it through expectations, to it being like this" step...
Uh, something about, quadratic forms and bilinear forms being in correspondence under some conditions, maybe, gives us that?
Unsure.
Can we phrase all these "to first order in \lambda" as just, taking the derivative wrt \lambda? Seems like should be able to.
If U(\lambda)^\dagger \hat{x} U(\lambda) = \hat{x} + \lambda
and then we take the derivative of this with respect to \lambda,
then we get U'(\lambda)^\dagger \hat{x} U(\lambda) + U(\lambda)^\dagger \hat{x} U'(\lambda) = 1
then, evaluating this at \lambda = 0, the U(0) = 1
and so we get the U'(0)^\dagger \hat{x} + \hat{x} U'(0) = 1
Which is, basically the same thing, yeah,
just need to get that U'(0)^\dagger = - U'(0)
And, that comes from, uh, the generators of the unitary group being, err..
the skew-hermitian operators?
Yeah. That's exactly what is needed.
Ok, cool!
That gives that [\hat{x},U'(0)] = 1
(and the i \hbar comes from just, pulling that out of the U'(0) )
And so, if you have some operator A,
and there's an analogy of translation but for the values of this operator,
then, a similar thing happens, yeah. Ok.
What if, like, we try to do this with "angle"?
Like, suppose the operator A gives us the direction something is facing,
and we are trying to get the generator of the rotation?
Well, a problem with that is that the angle doesn't really, uh,
if it has real number values that we are adding (say, they are from 0 to 2 \pi)
then, at the point where they wrap around, they don't add correctly across that.
Can still like, do stuff for little intervals of angles I guess,
and definitely we can define the rotations,
ok yeah, we can define the U(\lambda) which rotates things around by \lambda radians,
and we can talk about the generator of that,
but, I guess there's just no appropriate "angle" operator, for which it would have the corresponding commutation relation?
But I expect that the operators that do exist and are kind of like one, would be in some sense close to having the corresponding commutation relation? idk.
Yes you don't need to restrict to infinitesimal transformations; taking the derivative of the finite transformation e^{i \lambda p/\hbar} x e^{-i \lambda p} = x + \lambda at \lambda = 0 will give the same result
A perfect example of the dual usage of the term 'generator' in a quantum thesis.
Sir can you help me on writing synopsis for MPhil degry?
You should add some cuts in your video showing your face. I feel I understand the material better when someone explains it to me with a serious face while wearing glasses.