One quick cute tale: if you read Dirac's "Principles of Quantum Mechanics," in which he widely introduced his notation, he actually makes no mention of the Riesz Representation Theorem. In fact, he says to assume the correspondence between a ket and the corresponding bra. We don't know if he had such mathematical prowess that he felt the Riesz Representation Theorem was obvious enough to assume, or if he simply was not aware of the theorem's importance. That being said, this is the same man who remained completely silent after a student said, "I don't understand the second equation," during a lecture. After being asked why Dirac didn't answer the student's question, Dirac said, "that was not a question, that was a statement." So who knows. He dances all around the Riesz Representation Theorem in his book, so he likely knew about it on some level. Anyway, hope to see you next episode where we learn how we represent observables as operators in quantum mechanics! -QuantumSense
Out of curiosity, are you using Ballentine’s “Quantum Mechanics: A Modern Development” as a reference, because I think the structure of the way you teach is very similar. In my opinion, Ballentine’s book is one of, if not, the best quantum book out there.
Sir RP asked PD to explain something about the Dirac approach to relativistics and realized dirac had no idea about the Reice theorem … since he asked RP directly… “what is that”.
I wish I took Intro to QM a semester later so I could watch your videos side-to-side with my lessons. But better late than never: your series will literally save my career now
Your dreams are someone's reality. Literally due to this series i am having a feel of the math behind quantum mechanics, all thank to Quantum sense making sense.
This really reminds me of how little i "got" QM before i took a functional analysis math course. QM is basically just that applied to the Schrödinger Equation :D
The Dirac notation makes a lot of connections look so elegant its insane. It took me a while to understand its usefulness. Why do we need a new symbol for a vector? But being coordinate free and able to basically just forget about the basis is so powerful! For me it's basically tensor algebra without the index juggling.
I've done vector math without using indices even without bra-ket notation. Indices are just a projection onto an arbitrary basis. Heck, I've technically done vector math without even knowing what dimension I'm working in, because coordinate free math is just that powerful. The example at the end is pretty obvious even without bra-ket notation since it's taking each coordinate and then multiplying them back with the basis that they came from. It's a pretty complicated way of saying 𝜓 = 𝜓, or maybe that 𝜓 can be broken into a linear combination of basis vectors, meaning that 𝜓 is a vector. In fact, I can see how the conclusion only holds if i iterates through every single basis vector, otherwise it's a projection matrix into a specific subspace rather than the identity. Of course this only comes from the fact that it's written using indices. An alternative could be to specify the projection onto a basis, and then everything else, such that v = Proj_x(v) + Rej_x(v) = (v·x/x·x)x + (v^x/x·x)x = (v·x)x¯¹ + (v^x)x¯1 = (v·x + v^x)x¯¹ = vxx¯¹ = v(x/x) = v. Basis? What's that? x is just some arbitrary vector. It doesn't even have to be unit length.
0:00-Recap 0:38-Introduction 1:35-Linear Functional Definition 2:48-Dual space 4:15-Necessity of linear functionals and bra vector 5:23-Riesz Representation Theorem 7:58-Example for bra-ket notation
Hey, i just want to let you know, thank you for this amazing series. Making it easy and simple to understand . Please keep posting more videos like this !
the equation at 9:06 makes me think about the "bi-dual" space , i.e: the space of linear form which takes a linear form as an input. When we think about it , this space would look exactly like the "original space" (more rigorously , they would be isomorphic) One example of an element of the bidual space would be Ev_x(f) : V*-> R | f-> f(x) , i.e , it takes a linear form and return it's evaluation at input x. if V=R² , and we "indentify" Ev_(e_1) (f):= e_1 and Ev_(e_2) (f):= e_2 , we can se that 9:02 as a "go and return" : The first "bra" goes from V to its dual , and the second does the opposite ! My point is that the "ket" notation makes a vector look likes an "operation" on "something" , and it's reminiscent of that bidual space stuff. PS: thanks for the videos !
Multiplying a vector by its transpose yields a matrix. The identity operator is the identity matrix. Each term of the sum has a distinct diagonal element equal to 1.
I have a question from the end of the video. I see that a bra followed by a ket is the inner product braket. But what is a ket followed by a bra? (What is a "ketbra"?)
Hello! Thank you for watching! And this is a great question! A ket followed by a bra is an operator. What does it do when acting on a vector like so: |A_i>? Well the bra first gives you the inner product, , which intuitively is the component along vector A_i (which makes sense, since remember that that inner product gives you the coefficient in the basis expansion!). So then you multiply the component in the direction of A_i by the vector |A_i>. So, the operator as a whole takes your quantum state, finds the component in that direction, then multiplies it by the vector in that direction. Thus, an operator of the form |A_i>. More generally, an operator of the form |psi>
@@quantumsensechannel Thank you very much! (Also for your videos, they are being amazing!) Yes, it answers perfectly. It is very easy to see what you say if you think in row vectors for "bras" and in column vectors for "kets".
@@quantumsensechannel I am having trouble getting intuition for this yet, I will have to rewatch a few times and think about it on my own. However, if I understand correctly, is that state itself. Through this understanding, I can intuitively understand what you said about how |A_i> would be like summing |A_i> for ever A_i, which is gonna give you the sum of all projection operators, which represent ever quantum state scaled by how much |phi> aligns with it. With that understanding, it makes absolute sense that this is just decomposing and resumming all the components of |phi>, which makes it obvious that this is just the identity operator. I wish you had shown this intuition in the video, rather than just showing how we can arrive at this conclusion through the math, because when I was first watching it I didn't have that concrete of an understanding of what are (and I still don't exactly, even though I had already watched nearly everything and I am now rewatching it), and I just accepted the conclusion because it's what the math said and I could understand every step
Not the video I thought I'd get when searching what are bras. and now i know why so few people understand them... You would have to know quantum mechanics.
I'm a bit confused. At one point is seemed as if the output of a linear functional is a scalar; but then its output was expressed as a row vector - which is not a scalar. It also seemed to me that a linear functional operated on a column vector and produced its transpose, but this isn't true - it selects one component of a column vector and produces a row with the same number of elements, but all the non-selected elements set to zero? Those replacements-by-zero are what prevents the output row vector as being a transpose of the original column, am I right?
7:24 The notation comes from the closure is valid if there is only one possible x state. Otherwise, there should be an integration over x. Though if you are not making a rigorous representation then its fine.
For me this was always one of the things that made QM for workable from a purely practical sense… Coming from GR, which is still a nightmare for me to this day, Trying to keep track of indices of your various square matrices of rank 3 tensors just makes it so much more legwork… so many component relationships to keep track of for metric tensor, et al…
The present paper consists of two parts. In the first part, we prove a noncommutative analogue of the Riesz(-Markov-Kakutani) theorem on representation of functionals on an algebra of continuous functions by regular measures on the underlying space.
Hello, thank you for watching! There are a ton of great textbooks for learning quantum mechanics. If you are learning it for the very first time, I don't think you can beat "Introduction to Quantum Mechanics" by Griffiths. He doesn't go too deep into the mathematical formalism (and some people despise the book for that), but I think that's good for your first pass at quantum mechanics -- I think you need to build intuition into quantum mechanics before diving deep into the math; walk before you run. After your first quantum mechanics pass, I'd suggest either Shankar or Sakurai. Both are classic graduate level quantum texts, with Sakurai maybe being a bit more advanced and modern than Shankar. That being said, my all time favorite quantum mechanics textbook is "Quantum Mechanics: A Modern Development" by Ballentine. This is the quantum textbook you read after you've read all the other ones. He really digs into every single nook and cranny of quantum mechanics, and doesn't leave any stone unturned. In the book, he has a habit of saying "Here's what other quantum mechanics textbooks say, and here's why they are wrong." It is a sufficiently advanced textbook, so I would only recommend it if you feel quite comfortable with quantum mechanics. This is also the textbook used when I took graduate quantum mechanics courses, so it has a good reputation. Hopefully this gave you a starting point of where to search! -QuantumSense
@@quantumsensechannel oh than you very much for your responding and giving this tips, I am gonna do as you have just told me... you are the best...and keep going ...God bless you...
Why do we always assume were working in a orthonormal basis? Is this something we can always assume or are we assuming this because it makes the math easier to understand? If the basis isn't orthonormal how would you figure other the quantum state?
At time 3:35 You mentioned that a vector space can be transformed into its dual vector space consisting of linear functionals. Are you considering each set of individual vector space in R2 example of vector space or combining the sets of the vector space into one dual vector space of linear functionals as showed in example?
In the initial example a single linear functional Lx was introduced which acts on 2-D vectors, v. Then the dual space was defined as the set of "all" linear functionals. Now what are the other functionals in this set?
Keep in mind If you multiply a matrix by any zero matrix, you get another zero matrix. Always a bunch of zeros, though potentially a different number of zeros.
Hello, thank you for watching. The prerequisites necessary are listed in the first episode. Essentially, just a basic knowledge of calculus and linear algebra is needed. -QuantumSense
This, in practice. Ket is our measurement, Bra is our DETECTOR. What cancels a ket, it's that bra, so our ket is the antibra. And ket bra, density matrix >< like X cross product? 🙁
Hello! Thank you for watching. In the first episode, I lay out all the chapters I have planned to release, with additional chapters added in the future. -QuantumSense
He's not exchanging the inputs of the inner product. He's just moving it to the other side of the vector. Since the inner product gives a scalar this makes no difference. Then he takes advantage of the Dirac notation to turn the KetBra into an operator.
Excellent! I’ve been struggling a bit with the notion of dual spaces, but this was clear and precise and helped a lot. Thanks! Great series! (I’ve been working through Dirac’s book _The Principles of Quantum Mechanics_ where he introduces the bra-ket notation. _Not_ an easy read!)
I don't see much benefit to bra-key notation to some alternatives I've seen, but at least it's better than Einstein notation in my opinion. Get those ugly indices out of here! What are they even indexing anyways? I thought relativity was all about not having a privileged reference frame, not about describing components in terms of some privileged reference frame.
@@angeldude101 Well not having a privileged reference frame does not mean that the tensors (which are those "reference frame free" thing that we are indexing) do not have a certain number of components(4 or more)
@@BlackHole-qw9qg But those components literally change depending on the angle you look at them from, so they're ultimately meaningless except in a context, and ideally we'd be doing math that doesn't care about or reference that kind of context. They're necessary for computation, but not for formulating the desired relationships and symmetries. And plus, math that doesn't even acknowledge the dimensionality of your space is easier to project down to something easier to visualize. As far as I'm concerned, the equations governing 1+2D spacetime are the same as the equations governing 1+3D spacetime.
@@angeldude101 Okay so first of all there is something called abstract index notation (maybe you didn't know about it) which make a use of indices only to specify the rank and type of tensor you're using. This notation doesn't care about reference frame and all operations done with them are right in every reference frame. And actually one of the motivation behind "replacing" Einstein notation by Abstract index notation was to recover a more reference frame-free notation. Plus, don't forget that sometimes it is necessary to see what happens in a specific reference frame (for example when we want to know what an observer falling in a black hole do see versus what is seen by an observer watching him falling). In those cases we use Einstein greek letters (by convention latin are for abstract index notation) and it is quite useful too. Another useful application to switching to a specific reference frame is that if a tensor equation (so a reference-free thing) simplify to T=0 or T=G we know that the tensor T=0 or T=G in every other reference frame cuz it does not depend to the reference frame. And for your 2nd paragraph, keep in mind that while 1+2D space is quite similar to 1+3D, at the end of the day we'll still have to do physics not maths, so we'll have to recover 1+3D. And as you said yourself and with all those notations it is actually not that much easier in 1+2 than in 1+3D. We're only just removing one index for naught. Hopefully it's more clear for you that we're not using all of those indices just for fun and because we failed in algebra class haha
@@BlackHole-qw9qg From quickly looking over the Wikipedia page, abstract index notation looks like it's just Einstein notation but with variables. I used 1+2D spacetime specifically because it's the lowest dimension that has any spacelike rotations. 1+1D spacetime only has timelike rotations, but again, the math shouldn't really care about what dimension we're in. A rotation or reflection is a rotation or reflection, whether spacelike, timelike, in 2D, 3D, 4D, or even 5D and beyond. There are differences in how 1+2D and 1+3D work due to formulae depending on the surface area of the unit N-sphere, but that can still be generalized, even if it makes Newtonian gravity freak out in 1+2D.
Sir Why you write it that Lx of column matrix give a row matrix . In matrices the first column represent all x element then why you pick up only the first element .2ndly i think, the resulted matrix will have some valve on first column element like [ 3 0] or [2 0] and the second element always be zero because Lx only only pick x element and in examples you also write some element in 2nd column like [ 2 7] . 3rdly why we use duel space (R²) why not R³ or R⁴ because in quantum mechanics vector space can be form from infinite bases . Sir please answer or give me your email address
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One quick cute tale: if you read Dirac's "Principles of Quantum Mechanics," in which he widely introduced his notation, he actually makes no mention of the Riesz Representation Theorem. In fact, he says to assume the correspondence between a ket and the corresponding bra. We don't know if he had such mathematical prowess that he felt the Riesz Representation Theorem was obvious enough to assume, or if he simply was not aware of the theorem's importance.
That being said, this is the same man who remained completely silent after a student said, "I don't understand the second equation," during a lecture. After being asked why Dirac didn't answer the student's question, Dirac said, "that was not a question, that was a statement."
So who knows. He dances all around the Riesz Representation Theorem in his book, so he likely knew about it on some level.
Anyway, hope to see you next episode where we learn how we represent observables as operators in quantum mechanics!
-QuantumSense
That's pretty interesting, dirac was one eccentric dude
2
Out of curiosity, are you using Ballentine’s “Quantum Mechanics: A Modern Development” as a reference, because I think the structure of the way you teach is very similar. In my opinion, Ballentine’s book is one of, if not, the best quantum book out there.
Sir RP asked PD to explain something about the Dirac approach to relativistics and realized dirac had no idea about the Reice theorem … since he asked RP directly… “what is that”.
I've been reading through this book by Dirac. I believe he simply pared down his presentation to be as simple as possible.
PSA: if anyone wants/needs more material about dual spaces etc, eigenchris's videos are also really good
Good to know, thanks!
Seconded, very good.
+1
Also Professor M does science, that's a real deal.
I wish I took Intro to QM a semester later so I could watch your videos side-to-side with my lessons. But better late than never: your series will literally save my career now
Your dreams are someone's reality. Literally due to this series i am having a feel of the math behind quantum mechanics, all thank to Quantum sense making sense.
This really reminds me of how little i "got" QM before i took a functional analysis math course. QM is basically just that applied to the Schrödinger Equation :D
I'm having my QM I exams in a month.... This channel is a miracle before my eyes🔥
The Dirac notation makes a lot of connections look so elegant its insane. It took me a while to understand its usefulness. Why do we need a new symbol for a vector? But being coordinate free and able to basically just forget about the basis is so powerful!
For me it's basically tensor algebra without the index juggling.
I've done vector math without using indices even without bra-ket notation. Indices are just a projection onto an arbitrary basis. Heck, I've technically done vector math without even knowing what dimension I'm working in, because coordinate free math is just that powerful.
The example at the end is pretty obvious even without bra-ket notation since it's taking each coordinate and then multiplying them back with the basis that they came from. It's a pretty complicated way of saying 𝜓 = 𝜓, or maybe that 𝜓 can be broken into a linear combination of basis vectors, meaning that 𝜓 is a vector. In fact, I can see how the conclusion only holds if i iterates through every single basis vector, otherwise it's a projection matrix into a specific subspace rather than the identity. Of course this only comes from the fact that it's written using indices.
An alternative could be to specify the projection onto a basis, and then everything else, such that v = Proj_x(v) + Rej_x(v) = (v·x/x·x)x + (v^x/x·x)x = (v·x)x¯¹ + (v^x)x¯1 = (v·x + v^x)x¯¹ = vxx¯¹ = v(x/x) = v. Basis? What's that? x is just some arbitrary vector. It doesn't even have to be unit length.
@@angeldude101 yes "coordinate free" Was the term I was looking for. That's the power of it.
@@angeldude101 ty for this comment
thank you for your handwork and inuitive explanations. have first QM test tomorrow
I'm taking quantum this semester! So lucky that I found your videos!
0:00-Recap
0:38-Introduction
1:35-Linear Functional Definition
2:48-Dual space
4:15-Necessity of linear functionals and bra vector
5:23-Riesz Representation Theorem
7:58-Example for bra-ket notation
In love with this series, it is lovely. Well made!
Dirac needed female liberation: bra burning
I am very happy this series exists now. It will help many people with quantum physics, Thank you!
Thank you from a graduate student going from an engineering undergraduate degree to a physics degree 🙏
im a math major and enjoying your videos ! who knows probably i will do master studies in physics
Bro this is simply awesome! Thank you so much for this! This video just cleared so many things for me.
Hey, i just want to let you know, thank you for this amazing series. Making it easy and simple to understand . Please keep posting more videos like this !
if possible, can you also make a video on quantum entanglement with mathematics perspective?
the equation at 9:06 makes me think about the "bi-dual" space , i.e: the space of linear form which takes a linear form as an input.
When we think about it , this space would look exactly like the "original space" (more rigorously , they would be isomorphic)
One example of an element of the bidual space would be Ev_x(f) : V*-> R | f-> f(x) , i.e , it takes a linear form and return it's evaluation at input x.
if V=R² , and we "indentify" Ev_(e_1) (f):= e_1 and Ev_(e_2) (f):= e_2 , we can se that 9:02 as a "go and return" :
The first "bra" goes from V to its dual , and the second does the opposite !
My point is that the "ket" notation makes a vector look likes an "operation" on "something" , and it's reminiscent of that bidual space stuff.
PS: thanks for the videos !
Incredibly clear presentation !
Multiplying a vector by its transpose yields a matrix. The identity operator is the identity matrix. Each term of the sum has a distinct diagonal element equal to 1.
I have a question from the end of the video. I see that a bra followed by a ket is the inner product braket. But what is a ket followed by a bra? (What is a "ketbra"?)
Hello! Thank you for watching!
And this is a great question! A ket followed by a bra is an operator. What does it do when acting on a vector like so: |A_i>? Well the bra first gives you the inner product, , which intuitively is the component along vector A_i (which makes sense, since remember that that inner product gives you the coefficient in the basis expansion!). So then you multiply the component in the direction of A_i by the vector |A_i>. So, the operator as a whole takes your quantum state, finds the component in that direction, then multiplies it by the vector in that direction. Thus, an operator of the form |A_i>.
More generally, an operator of the form |psi>
@@quantumsensechannel Thank you very much! (Also for your videos, they are being amazing!) Yes, it answers perfectly. It is very easy to see what you say if you think in row vectors for "bras" and in column vectors for "kets".
@@quantumsensechannel I am having trouble getting intuition for this yet, I will have to rewatch a few times and think about it on my own. However, if I understand correctly, is that state itself. Through this understanding, I can intuitively understand what you said about how |A_i> would be like summing |A_i> for ever A_i, which is gonna give you the sum of all projection operators, which represent ever quantum state scaled by how much |phi> aligns with it. With that understanding, it makes absolute sense that this is just decomposing and resumming all the components of |phi>, which makes it obvious that this is just the identity operator. I wish you had shown this intuition in the video, rather than just showing how we can arrive at this conclusion through the math, because when I was first watching it I didn't have that concrete of an understanding of what are (and I still don't exactly, even though I had already watched nearly everything and I am now rewatching it), and I just accepted the conclusion because it's what the math said and I could understand every step
Not the video I thought I'd get when searching what are bras. and now i know why so few people understand them... You would have to know quantum mechanics.
I'm a bit confused. At one point is seemed as if the output of a linear functional is a scalar; but then its output was expressed as a row vector - which is not a scalar. It also seemed to me that a linear functional operated on a column vector and produced its transpose, but this isn't true - it selects one component of a column vector and produces a row with the same number of elements, but all the non-selected elements set to zero? Those replacements-by-zero are what prevents the output row vector as being a transpose of the original column, am I right?
7:24 The notation comes from the closure is valid if there is only one possible x state. Otherwise, there should be an integration over x. Though if you are not making a rigorous representation then its fine.
For me this was always one of the things that made QM for workable from a purely practical sense… Coming from GR, which is still a nightmare for me to this day,
Trying to keep track of indices of your various square matrices of rank 3 tensors just makes it so much more legwork… so many component relationships to keep track of for metric tensor, et al…
Man you clear all my doubts regarding QM.Thakns a lot from pakistan
Really into this series, really good job.
Your vedios are really good.
Please make vedios on general relativity also
this music is really beautiful ❤
THANK YOUUU LEGEND makes so much sense
The present paper consists of two parts. In the first part, we prove a
noncommutative analogue of the Riesz(-Markov-Kakutani) theorem on representation of
functionals on an algebra of continuous functions by regular measures on the underlying
space.
In order obtaining an identity matrix I there should be a double sum on two indexes.
thank you so much...I really appreciate it... keep going..
can you recommend a good text book for quantum mechanics?
Hello, thank you for watching!
There are a ton of great textbooks for learning quantum mechanics. If you are learning it for the very first time, I don't think you can beat "Introduction to Quantum Mechanics" by Griffiths. He doesn't go too deep into the mathematical formalism (and some people despise the book for that), but I think that's good for your first pass at quantum mechanics -- I think you need to build intuition into quantum mechanics before diving deep into the math; walk before you run.
After your first quantum mechanics pass, I'd suggest either Shankar or Sakurai. Both are classic graduate level quantum texts, with Sakurai maybe being a bit more advanced and modern than Shankar.
That being said, my all time favorite quantum mechanics textbook is "Quantum Mechanics: A Modern Development" by Ballentine. This is the quantum textbook you read after you've read all the other ones. He really digs into every single nook and cranny of quantum mechanics, and doesn't leave any stone unturned. In the book, he has a habit of saying "Here's what other quantum mechanics textbooks say, and here's why they are wrong." It is a sufficiently advanced textbook, so I would only recommend it if you feel quite comfortable with quantum mechanics. This is also the textbook used when I took graduate quantum mechanics courses, so it has a good reputation.
Hopefully this gave you a starting point of where to search!
-QuantumSense
@@quantumsensechannel
oh than you very much for your responding and giving this tips, I am gonna do as you have just told me... you are the best...and keep going ...God bless you...
Why do we always assume were working in a orthonormal basis? Is this something we can always assume or are we assuming this because it makes the math easier to understand? If the basis isn't orthonormal how would you figure other the quantum state?
At time 3:35 You mentioned that a vector space can be transformed into its dual vector space consisting of linear functionals. Are you considering each set of individual vector space in R2 example of vector space or combining the sets of the vector space into one dual vector space of linear functionals as showed in example?
In the initial example a single linear functional Lx was introduced which acts on 2-D vectors, v. Then the dual space was defined as the set of "all" linear functionals. Now what are the other functionals in this set?
Great stuff! 🎉
Fantastic video!
Keep in mind If you multiply a matrix by any zero matrix, you get another zero matrix. Always a bunch of zeros, though potentially a different number of zeros.
Is a linear functional the same thing as a covector?
Oh wow that just made a bunch of stuff click with the probability function as linear functional. Awesome.
Hey
I m new to Quantum mechanics.
Is this series suitable for me if I already have a limited background in QM?
Hello, thank you for watching.
The prerequisites necessary are listed in the first episode. Essentially, just a basic knowledge of calculus and linear algebra is needed.
-QuantumSense
This, in practice.
Ket is our measurement, Bra is our DETECTOR.
What cancels a ket, it's that bra, so our ket is the antibra.
And ket bra, density matrix >< like X cross product? 🙁
Idk if someone already asked but how long is the playlist gonna be?
Hello! Thank you for watching.
In the first episode, I lay out all the chapters I have planned to release, with additional chapters added in the future.
-QuantumSense
8:27 "let's move the inner product to the front". How is that possible? Don't you have to take the conjugate per the rule?
He's not exchanging the inputs of the inner product. He's just moving it to the other side of the vector. Since the inner product gives a scalar this makes no difference. Then he takes advantage of the Dirac notation to turn the KetBra into an operator.
thank you need review each 59 now yet Einstein 1905 did not haVE ORDER OF OPERATIONS TIME DIALATION YET
from india tq
Excellent! I’ve been struggling a bit with the notion of dual spaces, but this was clear and precise and helped a lot. Thanks! Great series!
(I’ve been working through Dirac’s book _The Principles of Quantum Mechanics_ where he introduces the bra-ket notation. _Not_ an easy read!)
check out Basil J. Hiley - he points out how Dirac was wrong. So does John G. Williamson. They have lectures on youtube.
It's funny how in very serious quantum mathematics the names are just a pun
bro I searched for quatum maths for years, i found you
Fun fact!: These are the only bras physics majors will ever see
thumbs up :)
Very different from India
Oh the bras, indeed.
I love your videos and animations! Drop me a line if you ever wanted to do a collab! : )
I love bras
i thing Einstein's notation is better
the only problem is u will quickly run out of letters for indices
I don't see much benefit to bra-key notation to some alternatives I've seen, but at least it's better than Einstein notation in my opinion. Get those ugly indices out of here! What are they even indexing anyways? I thought relativity was all about not having a privileged reference frame, not about describing components in terms of some privileged reference frame.
@@angeldude101 Well not having a privileged reference frame does not mean that the tensors (which are those "reference frame free" thing that we are indexing) do not have a certain number of components(4 or more)
@@BlackHole-qw9qg But those components literally change depending on the angle you look at them from, so they're ultimately meaningless except in a context, and ideally we'd be doing math that doesn't care about or reference that kind of context. They're necessary for computation, but not for formulating the desired relationships and symmetries.
And plus, math that doesn't even acknowledge the dimensionality of your space is easier to project down to something easier to visualize. As far as I'm concerned, the equations governing 1+2D spacetime are the same as the equations governing 1+3D spacetime.
@@angeldude101 Okay so first of all there is something called abstract index notation (maybe you didn't know about it) which make a use of indices only to specify the rank and type of tensor you're using. This notation doesn't care about reference frame and all operations done with them are right in every reference frame. And actually one of the motivation behind "replacing" Einstein notation by Abstract index notation was to recover a more reference frame-free notation.
Plus, don't forget that sometimes it is necessary to see what happens in a specific reference frame (for example when we want to know what an observer falling in a black hole do see versus what is seen by an observer watching him falling). In those cases we use Einstein greek letters (by convention latin are for abstract index notation) and it is quite useful too. Another useful application to switching to a specific reference frame is that if a tensor equation (so a reference-free thing) simplify to T=0 or T=G we know that the tensor T=0 or T=G in every other reference frame cuz it does not depend to the reference frame.
And for your 2nd paragraph, keep in mind that while 1+2D space is quite similar to 1+3D, at the end of the day we'll still have to do physics not maths, so we'll have to recover 1+3D. And as you said yourself and with all those notations it is actually not that much easier in 1+2 than in 1+3D. We're only just removing one index for naught.
Hopefully it's more clear for you that we're not using all of those indices just for fun and because we failed in algebra class haha
@@BlackHole-qw9qg From quickly looking over the Wikipedia page, abstract index notation looks like it's just Einstein notation but with variables.
I used 1+2D spacetime specifically because it's the lowest dimension that has any spacelike rotations. 1+1D spacetime only has timelike rotations, but again, the math shouldn't really care about what dimension we're in. A rotation or reflection is a rotation or reflection, whether spacelike, timelike, in 2D, 3D, 4D, or even 5D and beyond.
There are differences in how 1+2D and 1+3D work due to formulae depending on the surface area of the unit N-sphere, but that can still be generalized, even if it makes Newtonian gravity freak out in 1+2D.
2:35
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bruh
Do they actually call these "bras?" 😂
Misleading title: the video does not actually define bra and ket notation like the title implies.
Sir Why you write it that Lx of column matrix give a row matrix . In matrices the first column represent all x element then why you pick up only the first element .2ndly i think,
the resulted matrix will have some valve on first column element like [ 3 0] or [2 0] and the second element always be zero because Lx only only pick x element and in examples you also write some element in 2nd column like [ 2 7] . 3rdly why we use duel space (R²) why not R³ or R⁴ because in quantum mechanics vector space can be form from infinite bases .
Sir please answer or give me your email address