If youtube was a 1000 years old your video here on this topic would be A1. What a gift you have in explaining something I’ll never use in my boring job but am so interested in your topics. Bless you Chris with great health, intelligence and happiness always👏📕🏆
@@eigenchris However, your pronunciation of Pauli is still a bit off. The diphthong "au" is pronounced roughly the same as the diphthong in the English word "pound".
Chris is the kind of guy who comes across a vector and square roots it into spinors, and says it's more complicated than that and proceeds to make a brilliant video on that
Funny how the most interesting thing that i learned from this video is the one you didn't mention. Got me curious and then fell into a "AHA!" moment when i realised you sneakly can actually materialize "i^2" as "1 = (-1)*(-1)" and then use one of these in your own calculations, only then to obtain "b = -c", which becomes trivial. Sorry i graduated as a chem student and didn't do much maths, so this was new to me that you didn't need to "divide by i" out of a sudden and wondered for a minute where did it go. The moment in question is 9:08
Thank you for this wonderful series! It was immensely helpful for studying for my Quantum Mechanics II test :) to me it seems that you have very profound knowledge and understanding of both physics and geometry, do you mind answering what path (degrees, courses, books etc) you took in order to reach such level? I'm a 3rd year undergraduate physics student with very little mathematical background beyond what is required for a physics degree by my university.. but after getting into classical field theory and introduction to general relativity i was really mesmerized by all the geometry involved! now I am very curious to learn more about the mathematical machinery behind all these beautiful ideas in general relativity, quantum field theory and string theory, but when i tried looking it up i was very overwhelmed by the shear amount of advanced mathematical topics that stand behind these fields.. (topology, differential geometry and deferential forms, Riemannian geometry, semplectic geometry, spectral geometry, algebraic geometry, algebraic topology, differentiable manifolds and exterior calculus, calculus of variation, tensor calculus, functional analysis, harmonic analysis, spectral analysis, measure theory, stochastic processes, ergodic theory, gauge theory and hodge theory, abstract algebra, lie groups and lie algebra, category theory(?) and the list goes on and on..) it would take a full course just to study either one of these subjects (some of them will probably take 2 perhaps more) and at the end of the day i still want to focus on physics and not forsake everything for the sake of mathematics.. it would make me very happy if you could share how you managed to balance all these physics and mathematics that stands behind it.
I have a bachelor's degree in "engineering physics" from 10 years ago, so that gave me a good foundation in calculus, linear algebra, and physics. But all the specific stuff you see on my youtube channel (tensors, relativity, spinors, etc.) I've learned myself in my free time over the last few years. Yeah, I find it's very hard to get started with a lot of advanced physics topics, because many advanced textbooks are not great for beginners, especially beginners who are no longer in an academic environment, like me. The reason I made this youtube channel was to try and ease people with some math background into the more complicated topics. Most of my learning has taken place online, from youtube videos, PDF course notes, and anything else that turns up on google. I'm also in a couple discord servers where I can ask questions. It's not a very efficient process. I'd been trying to learn spinors for 2-3 years before I started making these videos. And I only learn things I'm interested in. Topology, measure theory, stochastic processes, and ergodic theory are examples of things I'm not particularly interested in. I doubt that I'll ever get to string theory just because it's so much work. It's hard for me to offer advice on how you should study. I found in undergrad I have very little time to actually understand things. I was too busy doing assignments and labs and exams and I sometimes had to skip getting a proper understanding in order to keep up with academic responsibilities. I tend to prefer physics as a hobby, where I just learn what I want, when I feel like it, and stop if I lose interest. Realistically, you can't do everything, so just start with the stuff you find most appealing. If you're learning as a hobby, you don't always need to go as deep as a lot of academic texts go to get the main ideas. Leonard Susskind has a series of lectures and books called "Theoretical Minimum" that do a good job at giving you the core ideas and equations without beating you over the head with too many details.
@@eigenchris I find the clarity of these videos exceptional, especially so, given that you've been essentially self-educated. Are you leveraging your eng degree to work as a patent clerk by any chance?
The fun thing is that a lot of advanced mathematics only comes when you're curious. When I did basic QM I nwver learned spin states are spinors living on a Riemann sphere (the term Bloch sphere did pass by in 1 exercise). You'll still be able to solve basic problems. Only when you are resolved to really understanding these objects, you will learn there is a rich field of mathematical concepts behind it, and while this may take a sidestep it will end up making physics look a lot more elegant than without it.
Great video! I’ve been trying to figure out how to define spinors in an arbitrary Clifford algebra, in such a way that I make a connection with what you did here. Is it possible to take an arbitrary abstract Pauli vector (where the Pauli matrices are symbols not matrices) which is not necessarily isotopic and express it with a spinor decomposition? How would you do this? Can’t wait for the rest of the series!
So, something to note is that you can replace the column spinor with a 2x2 matrix where the right column is all zeros, and the row spinor can be replaced by a 2x2 matrix where the bottom row is all zeros, and you end up with the same 2x2 matrix result when you multiply them. Using this trick we can "upgrade" spinors to 2x2 matrices. Matrices with only the left column being non-zero belong to what is called a "left minimal ideal". This is a fancy way of saying that multiplying it by any matrix on the left will give you another matrix with only the left column being non-zero. Spinors in clifford algebras are then defined as members of left-minimal ideals. I believe the spinor up and spinor down states are 1/2 * (1 + σz) and 1/2 * (σx + σxσz). I tried to explain all this in this 2-hour video presentation I did a couple years ago: ua-cam.com/video/_5o0Bc66fik/v-deo.html Just be warned it's messy and probably has some mistakes in it. I didn't understand spinors as well as I do now. Videos 11-15 of the series I'm making will be a more complete and correct version of the above video.
Spinors algebra is a closed algebra under multiplication an isomorphic to quaternions. In Clifford algebra, they are the algebra formed by a basis of 2-vector ("bivector") which have all the properties describe in the previous lectures. And to transform a vector into a bivector, you multiply the vector by the unit pseudo-scalar, which in the Clifford of R3, is the 3-vector, or trivector. If I'm correct.
@@eigenchris Your videos on spinors are very illustrative. My interest in them is, however, somewhat skewed: it is the relationship between spinors and ideals that I seek to deeply understand, but only because ideals are complementary to filters, and I have found a deep connection between filters and cognition: whenever a pattern of its input nervous signals is recognised by a neuron, at least one filter in its working's mathematical representation (as per my own model) is hit, meaning the neuron will fire a nervous signal; whereas lack of recognition means an ideal is hit and no nervous signal is output by the neuron. My lack of mathematical knowledge makes me pursue ways to better build and express my model of nervous systems. Maybe understanding spinors I'll learn enough. Thank you a lot!
I love your thorough and methodical style, paying attention to all the details. I think the misleading part about spinors being the square-root of a vector is that it only applies to null vectors. The normal description makes it seem like you can somehow take the square-root of (3,4,5). Thanks for explaining the caveats and how the actual decomposition works. I'm wondering if it's necessary to use as many as four pairs of spinors to decompose a non-null vector. It seems like the Pauli vector ((a,b),(c,d)) could be decomposed as ((a,b),(0,0))+((0,0),(c,d)), but that's not very symmetrical. Perhaps there's a more symmetric (but more complicated) way to decompose an arbitrary Pauli vector using only two pairs of spinors.
A number of people have been asking this. I'll talk about it in the next video when I discuss the spinor spaces where spinors live. The linear combinations of column spinors and row spinors are really tensor products of spinors and dual spinors, and this tensor product space is 4-dimensional, which is why we need 4 basis vectors for it.
I'd say rank 1 matrices is a more useful description than zero-determinant 2x2 matrices, particularly if you already know that nonzero determinant means full-rank.
Rank 1 is more general, but in this case, saying det A = 0 allows us to use the formula for the determinant of a Pauli vector derived in past videos to make claims about it's factorization into spinors
@@IronLotus15 It's equivalent to saying that the (x,y,z) vector lives on the sphere (the surface) aka the Bloch sphere. This is because pure states are rank 1 density matrices (a 2-level system can be anywhere on the Bloch ball, but a pure state is on the spherical surface).
Excited to see what happens to null spacetime vectors! I wonder how the metric signature allows for different things to happen, since the components are all real (or would 1-3 of the components (depending on east or west coast metric respectively) of a null spacetime vector be complex, given the fact that their squares are what have the minus sign; do we have to use “ict” to talk about spacetime spinors?)
Spacetime vectors and their corresponding Weyl spinors will come up in video #9. While the "ict" approach is mathematically valid, I generally prefer using 4 real components for spacetime vectors, along with the metric with minus signs where needed.
Back when studying linear algebra, in my first semester in physics course. I didn't understand why would anyone even consider matrices as bases and call it an isomorphism to the vector spaces of row/column vectors, but after your course on tensors and now spinors, that seems more reasonable. On another note, I even thought you wouldn't metion that matrix decomposition trick, but seeing to how much calculation that leads, I got why you left it to the end. I was only thinking if it would be best to break it as a sum of two matrices: one with the first row/col and the second with only zeros, the other with the second row/col and the first with only zeros. Since the determinant of both will still be zero.
You'll see in the next video that Pauli vectors live in a subspace of a 4-dimensional space made up of the tensor product of spinors and dual spinors. The decomposition into 4 matrices matches up well with this 4-dimensional space.
hey man great videos I am a graduate student in physics can you suggest books with a simple explanation to study more about these topics as you know self-study is important
2x2 Hermitian matrices can always be represented as a sum of two rank-1 matrices. Four seems a bit excessive. In general, diagonalizable nxn matrices (of which Hermitians are an example) can be represented as sum of n rank-1 matrices.
I'll show in the next video that Pauli vectors live in a 3-dimensional subspace of a larger 4D space, constructed out of pairs of spinors and dual spinors. Breaking up the 2x2 matrices into 4 parts matches with this interpretation nicely.
Another excellent video; my only complaint is that I'm left wanting more! (I shall have to start on the GR course ;) What was Pauli's direct motivation for this analysis?
I think the original motivation in physics was the Stern-Gerlach experiment. I don't know if Pauli took the whole "square root of a vector" approach. Spinors were discovered earlier in mathematics by Klein and Cartan before they were used in physics.
@8.50 you are distributing square roots over complex factors ... which is the sort of thing that leads to -1 = sqrt(-1)*sqrt(-1) = sqrt(-1 * -1) = sqrt(1) = 1
So here is what I don't get: why the complicated first half of the video if it only works if x,y,z are complex, which they are not? You then split the Pauli vector it into four summands where the factoring is trivial. Why not like that from the start?
What confused me when watching this for a second time was how the squared length of a pair of spinors/the dot product of a pair of spinors can be different from zero while the underlying 3D vector cannot. It follows from the definition of course but it is a bit surprising. Perhaps worth putting a little verbal emphasis on that - squaring the spinor looks like “half” of the formula for the determinant of the Pauli matrix).
If I remember correctly, you followed the exact same procedure in your playlist about tensors to decompose (1,1)-tensors into a sum of products of the basis vectors and the basis covectors (L^i_j e_i ε^j). Is this the same thing with spinors or is there a difference?
Basically the same, except for the fact that the "trick" for turning a 3D vector into a 2x2 matrix is highly non-obvious. Video #8 goes into this in more detail.
Sorry, I can't find the solutions for the proposed problem in this particular video. Is it supposed to be in the github? By the way, the course is amazing and helpful, you're an insta subscribe
Hello Thanks for all videos. I have a question!🤔 ... a vector in R^3 have a 3 component that they are independent but in this video z is a function of x and y. as a result we are not a 3 dimensions. how to understand that x, y, and z are independent.
@@kavehmohammadian Correct. The trick only works for zero-magnitude vectors. However, you can take any Pauli vector and write it as a sum of 4 zero-magnitude vectors. I talk about this more in video #8.
Taking the 2x2 matrix with real entries as example, does it mean that a zero-determinant 2x2 matrix with real entries can be decomposed into two elements from the real projective line? Is this a more general pattern, with the non-invertible matrices from some n-dimensional vector space being decomposable into pairs of objects from the associated projective space?
I think you can loosely think of it like that, although the column and row are not independent of each other, so it's really only 1 "ratio" result you get. Also with n x n matrices, it's not enough for the matrix to have determinant zero (non-invertible). It's required that all columns be scalar multiples of each other. This is a stronger condition than the determinant being zero... it requires that the matrix have "rank 1". It's just that "rank 1" and "det = 0" are the same for a 2x2 matrix.
I'd argue it isn't, but the properties it gives are more appropriate. The matrices are just a way to encourage useful algebraic properties, but you can alternatively assert those properties on the vectors and basically get the same results, but without writing your vectors as matrices.
For the physical cases discussed before (light polarization and quantum spin) is there any physical interpretation of the Pauli vector that corresponds to the spinor?
Interesting question, but I'm not aware of any interpretation of the corresponding Pauli vectors. The vector components would be complex, so I'm not sure what that would mean.
13:35 Why aren't they the same spinor? The idea behind spinors is that we put an equivalence relation on two vectors that are rescaled versions of each other, no? Also, noting that the second spinor is (-xi2, xi1) compared to the first (xi1, xi2), is this related to the fact that the inner product of spinors in QFT sometimes has a gamma matrix between them? 15:54 So Pauli vectors that don't have determinant 0 cannot be written as a single spinor product, but it can be written as a sum of them. What does this imply? That some vectors can only be written as a linear combination? Am I a thousand miles off when I remark that this sounds like entanglement? I see your matrix got broken up into four terms. We know opposites of the Riemann sphere are orthogonal, so I would have thought you would need at most 2 (in fact, 2 of your terms seem double), or maybe 6 if you include every corner.
There are slight differences in the ways different people treat spinors. In the Jones vector and quantum state examples I showed previously, this "equivalence relation" you mentioned comes for free. But in pure mathematics, we don't necessarily have to enforce this equivalence relation. I'll talk about this in the next video #8. Spinors in QFT are called "Dirac Spinors", which are a pair of "Weyl Spinors" (one left-chiral and the other right-chiral). I'll get to this in video #9. Entanglement is done with the tensor product, which is basically what I'm doing here with column-row products (again, this will be explained in video #8). However, just because it uses the same mathematical operation as entanglement, it doesn't mean that any quantum entanglement is involved here. Also in video #8, I'll address why the space is 4-dimensional (It's because it involves the tensor product of spinors and dual spinors). Anyway, please just be a little patient, I'll address most of this over the next month or so.
The "square root" of a 4-vector is a Weyl spinor, which transforms with SL(2,C). The "square root" of a 3-vector is a Pauli spinor, which transforms with SU(2).
Sorry about my earlier response. I replied to the wrong comment. And yes, I see your point. I tried to draw parallels with the complex projective line I described in video 5, which has a "point at infinity".
When factoring a 4X4 matrix into a column and row matrices, you say there is no solution. Do you mean that there is no unique solution? Aren't there infinite number of solutions? I'll work on this some more.
If the determinant is nonzero then he's correct that there is no solution. If a 4x4 matrix is the outer product of two 4-element vectors, a column vector Y and a row vector X, then the matrix's rows must each be the vector X scaled by the corresponding element of Y. So if the matrix's rows are not scalar multiples of one another, it cannot be factored.
Check out the Eigenvalue decomposition of matrices. If it exists, it essentially tells you how to represent the matrix as a sum of vector outer products. You can also use the singular value decomposition if you're ok with the products having unrelated factors.
Good, but complex numbers are neither positive nor negative, unless real. Also, it's not easy factoring a square root of a product into a product of square roots in the complex domain.
I don't like the pulling a complex number out of a square root. This is not allowed in general, and also not in this particular case. It only works in this case because the sign doesn't matter. For instance, note 1 = √1 = √(-1 • -1) ≠ √(-1) • √(-1) = i•i = -1.
I have some idea about why. When teaching, I have to keep my voice monotone. For me, since I'm autistic, speaking and human languages are foreign to me. I think in math and pictures. Especially when I have to concentrate and focus on something difficult, I become effectively blind. In fact, it seems that I have to use my whole brain to even speak clearly. If you are neurotypical, you may not understand how difficult it is for me to speak at all. Everyone's brain works a little differently from everyone else.
In fact, but that is the rythm of the lessons you see on every (under)grad course from renowned institutions/professors' playlists on youtube. I think his more clear and paused speech, helps brain take more time to get and process the information presented. At least for me, not native to english. I prefer this way.
I'm pretty naturally monotone. But I also read from a script to record these videos, which makes me come off as extra-monotone. So I'm not trying to sound professional... it's just the natural way my voice comes out when I read from a script.
think of this as a power point presentation, if your slides are cluttered fucked with so much information on the slide, no one will listen to you. you have too much information on many slides, you need to condense down what you show and present
If youtube was a 1000 years old your video here on this topic would be A1. What a gift you have in explaining something I’ll never use in my boring job but am so interested in your topics. Bless you Chris with great health, intelligence and happiness always👏📕🏆
Love your videos. Also as a German I appreciate the fact that you pronounce Stern-Gerlach correctly :)
Good to know I got it (mostly) right. I had to look up a pronunciation.
@@eigenchris However, your pronunciation of Pauli is still a bit off. The diphthong "au" is pronounced roughly the same as the diphthong in the English word "pound".
Thanks. I have corrected this for the next video.
Chris is the kind of guy who comes across a vector and square roots it into spinors, and says it's more complicated than that and proceeds to make a brilliant video on that
I wish you could get more attention. Your videos are indeed instructive. We appretiate your efforts.
Very good explanation, thanks. In all my years, I have never seen the square root of a spinor.
The square roots are on the spinors, where comes the saying that spinors are like the square roots of matrices.
Funny how the most interesting thing that i learned from this video is the one you didn't mention. Got me curious and then fell into a "AHA!" moment when i realised you sneakly can actually materialize "i^2" as "1 = (-1)*(-1)" and then use one of these in your own calculations, only then to obtain "b = -c", which becomes trivial. Sorry i graduated as a chem student and didn't do much maths, so this was new to me that you didn't need to "divide by i" out of a sudden and wondered for a minute where did it go.
The moment in question is 9:08
Thank you for this wonderful series! It was immensely helpful for studying for my Quantum Mechanics II test :)
to me it seems that you have very profound knowledge and understanding of both physics and geometry, do you mind answering what path (degrees, courses, books etc) you took in order to reach such level?
I'm a 3rd year undergraduate physics student with very little mathematical background beyond what is required for a physics degree by my university.. but after getting into classical field theory and introduction to general relativity i was really mesmerized by all the geometry involved!
now I am very curious to learn more about the mathematical machinery behind all these beautiful ideas in general relativity, quantum field theory and string theory, but when i tried looking it up i was very overwhelmed by the shear amount of advanced mathematical topics that stand behind these fields..
(topology, differential geometry and deferential forms, Riemannian geometry, semplectic geometry, spectral geometry, algebraic geometry, algebraic topology, differentiable manifolds and exterior calculus, calculus of variation, tensor calculus, functional analysis, harmonic analysis, spectral analysis, measure theory, stochastic processes, ergodic theory, gauge theory and hodge theory, abstract algebra, lie groups and lie algebra, category theory(?) and the list goes on and on..)
it would take a full course just to study either one of these subjects (some of them will probably take 2 perhaps more) and at the end of the day i still want to focus on physics and not forsake everything for the sake of mathematics..
it would make me very happy if you could share how you managed to balance all these physics and mathematics that stands behind it.
I have a bachelor's degree in "engineering physics" from 10 years ago, so that gave me a good foundation in calculus, linear algebra, and physics. But all the specific stuff you see on my youtube channel (tensors, relativity, spinors, etc.) I've learned myself in my free time over the last few years.
Yeah, I find it's very hard to get started with a lot of advanced physics topics, because many advanced textbooks are not great for beginners, especially beginners who are no longer in an academic environment, like me. The reason I made this youtube channel was to try and ease people with some math background into the more complicated topics.
Most of my learning has taken place online, from youtube videos, PDF course notes, and anything else that turns up on google. I'm also in a couple discord servers where I can ask questions. It's not a very efficient process. I'd been trying to learn spinors for 2-3 years before I started making these videos. And I only learn things I'm interested in. Topology, measure theory, stochastic processes, and ergodic theory are examples of things I'm not particularly interested in. I doubt that I'll ever get to string theory just because it's so much work.
It's hard for me to offer advice on how you should study. I found in undergrad I have very little time to actually understand things. I was too busy doing assignments and labs and exams and I sometimes had to skip getting a proper understanding in order to keep up with academic responsibilities. I tend to prefer physics as a hobby, where I just learn what I want, when I feel like it, and stop if I lose interest. Realistically, you can't do everything, so just start with the stuff you find most appealing. If you're learning as a hobby, you don't always need to go as deep as a lot of academic texts go to get the main ideas. Leonard Susskind has a series of lectures and books called "Theoretical Minimum" that do a good job at giving you the core ideas and equations without beating you over the head with too many details.
@@eigenchris I find the clarity of these videos exceptional, especially so, given that you've been essentially self-educated. Are you leveraging your eng degree to work as a patent clerk by any chance?
The fun thing is that a lot of advanced mathematics only comes when you're curious. When I did basic QM I nwver learned spin states are spinors living on a Riemann sphere (the term Bloch sphere did pass by in 1 exercise). You'll still be able to solve basic problems. Only when you are resolved to really understanding these objects, you will learn there is a rich field of mathematical concepts behind it, and while this may take a sidestep it will end up making physics look a lot more elegant than without it.
@@eigenchris It is the same for me. So thank you very much for your channel!
Great video! I’ve been trying to figure out how to define spinors in an arbitrary Clifford algebra, in such a way that I make a connection with what you did here. Is it possible to take an arbitrary abstract Pauli vector (where the Pauli matrices are symbols not matrices) which is not necessarily isotopic and express it with a spinor decomposition? How would you do this? Can’t wait for the rest of the series!
So, something to note is that you can replace the column spinor with a 2x2 matrix where the right column is all zeros, and the row spinor can be replaced by a 2x2 matrix where the bottom row is all zeros, and you end up with the same 2x2 matrix result when you multiply them. Using this trick we can "upgrade" spinors to 2x2 matrices.
Matrices with only the left column being non-zero belong to what is called a "left minimal ideal". This is a fancy way of saying that multiplying it by any matrix on the left will give you another matrix with only the left column being non-zero. Spinors in clifford algebras are then defined as members of left-minimal ideals. I believe the spinor up and spinor down states are 1/2 * (1 + σz) and 1/2 * (σx + σxσz).
I tried to explain all this in this 2-hour video presentation I did a couple years ago: ua-cam.com/video/_5o0Bc66fik/v-deo.html
Just be warned it's messy and probably has some mistakes in it. I didn't understand spinors as well as I do now. Videos 11-15 of the series I'm making will be a more complete and correct version of the above video.
Spinors algebra is a closed algebra under multiplication an isomorphic to quaternions. In Clifford algebra, they are the algebra formed by a basis of 2-vector ("bivector") which have all the properties describe in the previous lectures.
And to transform a vector into a bivector, you multiply the vector by the unit pseudo-scalar, which in the Clifford of R3, is the 3-vector, or trivector.
If I'm correct.
@@eigenchris Your videos on spinors are very illustrative. My interest in them is, however, somewhat skewed: it is the relationship between spinors and ideals that I seek to deeply understand, but only because ideals are complementary to filters, and I have found a deep connection between filters and cognition: whenever a pattern of its input nervous signals is recognised by a neuron, at least one filter in its working's mathematical representation (as per my own model) is hit, meaning the neuron will fire a nervous signal; whereas lack of recognition means an ideal is hit and no nervous signal is output by the neuron. My lack of mathematical knowledge makes me pursue ways to better build and express my model of nervous systems. Maybe understanding spinors I'll learn enough. Thank you a lot!
I love your thorough and methodical style, paying attention to all the details. I think the misleading part about spinors being the square-root of a vector is that it only applies to null vectors. The normal description makes it seem like you can somehow take the square-root of (3,4,5). Thanks for explaining the caveats and how the actual decomposition works. I'm wondering if it's necessary to use as many as four pairs of spinors to decompose a non-null vector. It seems like the Pauli vector ((a,b),(c,d)) could be decomposed as ((a,b),(0,0))+((0,0),(c,d)), but that's not very symmetrical. Perhaps there's a more symmetric (but more complicated) way to decompose an arbitrary Pauli vector using only two pairs of spinors.
A number of people have been asking this. I'll talk about it in the next video when I discuss the spinor spaces where spinors live. The linear combinations of column spinors and row spinors are really tensor products of spinors and dual spinors, and this tensor product space is 4-dimensional, which is why we need 4 basis vectors for it.
I'd say rank 1 matrices is a more useful description than zero-determinant 2x2 matrices, particularly if you already know that nonzero determinant means full-rank.
Rank 1 is more general, but in this case, saying det A = 0 allows us to use the formula for the determinant of a Pauli vector derived in past videos to make claims about it's factorization into spinors
You're correct, yeah.
@@IronLotus15 It's equivalent to saying that the (x,y,z) vector lives on the sphere (the surface) aka the Bloch sphere. This is because pure states are rank 1 density matrices (a 2-level system can be anywhere on the Bloch ball, but a pure state is on the spherical surface).
eigenchris the spin doctor!
Excited to see what happens to null spacetime vectors! I wonder how the metric signature allows for different things to happen, since the components are all real (or would 1-3 of the components (depending on east or west coast metric respectively) of a null spacetime vector be complex, given the fact that their squares are what have the minus sign; do we have to use “ict” to talk about spacetime spinors?)
Spacetime vectors and their corresponding Weyl spinors will come up in video #9. While the "ict" approach is mathematically valid, I generally prefer using 4 real components for spacetime vectors, along with the metric with minus signs where needed.
@@eigenchris how exciting! I’m looking forward to it
Back when studying linear algebra, in my first semester in physics course. I didn't understand why would anyone even consider matrices as bases and call it an isomorphism to the vector spaces of row/column vectors, but after your course on tensors and now spinors, that seems more reasonable.
On another note, I even thought you wouldn't metion that matrix decomposition trick, but seeing to how much calculation that leads, I got why you left it to the end.
I was only thinking if it would be best to break it as a sum of two matrices: one with the first row/col and the second with only zeros, the other with the second row/col and the first with only zeros. Since the determinant of both will still be zero.
You'll see in the next video that Pauli vectors live in a subspace of a 4-dimensional space made up of the tensor product of spinors and dual spinors. The decomposition into 4 matrices matches up well with this 4-dimensional space.
@@eigenchris I get it. Them, I'll wait for the next video.
hey man great videos I am a graduate student in physics can you suggest books with a simple explanation to study more about these topics as you know self-study is important
2x2 Hermitian matrices can always be represented as a sum of two rank-1 matrices. Four seems a bit excessive. In general, diagonalizable nxn matrices (of which Hermitians are an example) can be represented as sum of n rank-1 matrices.
I'll show in the next video that Pauli vectors live in a 3-dimensional subspace of a larger 4D space, constructed out of pairs of spinors and dual spinors. Breaking up the 2x2 matrices into 4 parts matches with this interpretation nicely.
Spinnor for Beginnor! Nice!
Another excellent video; my only complaint is that I'm left wanting more! (I shall have to start on the GR course ;)
What was Pauli's direct motivation for this analysis?
I think the original motivation in physics was the Stern-Gerlach experiment. I don't know if Pauli took the whole "square root of a vector" approach. Spinors were discovered earlier in mathematics by Klein and Cartan before they were used in physics.
@8.50 you are distributing square roots over complex factors ... which is the sort of thing that leads to -1 = sqrt(-1)*sqrt(-1) = sqrt(-1 * -1) = sqrt(1) = 1
Thanks for pointing that out. I'll add a correction in the next video.
I also so astonished that mathetical power is very powerful.chris explained so nice in Hindu scriptures vishwRup darshanam❤❤❤
Very clear, thanks!
So here is what I don't get: why the complicated first half of the video if it only works if x,y,z are complex, which they are not? You then split the Pauli vector it into four summands where the factoring is trivial. Why not like that from the start?
I wanted to introduce the more complicated ideas later, and go step-by-step.
What confused me when watching this for a second time was how the squared length of a pair of spinors/the dot product of a pair of spinors can be different from zero while the underlying 3D vector cannot. It follows from the definition of course but it is a bit surprising. Perhaps worth putting a little verbal emphasis on that - squaring the spinor looks like “half” of the formula for the determinant of the Pauli matrix).
If I remember correctly, you followed the exact same procedure in your playlist about tensors to decompose (1,1)-tensors into a sum of products of the basis vectors and the basis covectors (L^i_j e_i ε^j). Is this the same thing with spinors or is there a difference?
Basically the same, except for the fact that the "trick" for turning a 3D vector into a 2x2 matrix is highly non-obvious. Video #8 goes into this in more detail.
Sorry, I can't find the solutions for the proposed problem in this particular video. Is it supposed to be in the github? By the way, the course is amazing and helpful, you're an insta subscribe
I guess I neglected to upload them. They should be there now.
@@eigenchris thanks a lot!
Hello
Thanks for all videos.
I have a question!🤔 ... a vector in R^3 have a 3 component that they are independent but in this video z is a function of x and y. as a result we are not a 3 dimensions. how to understand that x, y, and z are independent.
If the vector has magnitude zero, only x and y are independent. z is determined by x and y.
@@eigenchris so we can't for all (x,y,z) find (€1 , €2)
as a rseult this is not "one to one maping"
@@kavehmohammadian Correct. The trick only works for zero-magnitude vectors. However, you can take any Pauli vector and write it as a sum of 4 zero-magnitude vectors. I talk about this more in video #8.
Taking the 2x2 matrix with real entries as example, does it mean that a zero-determinant 2x2 matrix with real entries can be decomposed into two elements from the real projective line? Is this a more general pattern, with the non-invertible matrices from some n-dimensional vector space being decomposable into pairs of objects from the associated projective space?
I think you can loosely think of it like that, although the column and row are not independent of each other, so it's really only 1 "ratio" result you get. Also with n x n matrices, it's not enough for the matrix to have determinant zero (non-invertible). It's required that all columns be scalar multiples of each other. This is a stronger condition than the determinant being zero... it requires that the matrix have "rank 1". It's just that "rank 1" and "det = 0" are the same for a 2x2 matrix.
I wonder whether this is a more appropriate representation of the three real spatial degrees of freedom than real-valued 3-vectors.
I'd argue it isn't, but the properties it gives are more appropriate. The matrices are just a way to encourage useful algebraic properties, but you can alternatively assert those properties on the vectors and basically get the same results, but without writing your vectors as matrices.
For the physical cases discussed before (light polarization and quantum spin) is there any physical interpretation of the Pauli vector that corresponds to the spinor?
Interesting question, but I'm not aware of any interpretation of the corresponding Pauli vectors. The vector components would be complex, so I'm not sure what that would mean.
So an invertible matrix cannot be factored into a a pair of column and row vectors.
How many videos are there in this series?
This is the latest one. There will probably be around 25.
Thanks again!
13:35 Why aren't they the same spinor? The idea behind spinors is that we put an equivalence relation on two vectors that are rescaled versions of each other, no? Also, noting that the second spinor is (-xi2, xi1) compared to the first (xi1, xi2), is this related to the fact that the inner product of spinors in QFT sometimes has a gamma matrix between them?
15:54 So Pauli vectors that don't have determinant 0 cannot be written as a single spinor product, but it can be written as a sum of them. What does this imply? That some vectors can only be written as a linear combination? Am I a thousand miles off when I remark that this sounds like entanglement? I see your matrix got broken up into four terms. We know opposites of the Riemann sphere are orthogonal, so I would have thought you would need at most 2 (in fact, 2 of your terms seem double), or maybe 6 if you include every corner.
There are slight differences in the ways different people treat spinors. In the Jones vector and quantum state examples I showed previously, this "equivalence relation" you mentioned comes for free. But in pure mathematics, we don't necessarily have to enforce this equivalence relation. I'll talk about this in the next video #8.
Spinors in QFT are called "Dirac Spinors", which are a pair of "Weyl Spinors" (one left-chiral and the other right-chiral). I'll get to this in video #9.
Entanglement is done with the tensor product, which is basically what I'm doing here with column-row products (again, this will be explained in video #8). However, just because it uses the same mathematical operation as entanglement, it doesn't mean that any quantum entanglement is involved here.
Also in video #8, I'll address why the space is 4-dimensional (It's because it involves the tensor product of spinors and dual spinors).
Anyway, please just be a little patient, I'll address most of this over the next month or so.
@@eigenchris I'll just await future videos then. Great job as always!
I think it makes more sense to say that the square root of a 4-vector is a spinor.
The "square root" of a 4-vector is a Weyl spinor, which transforms with SL(2,C). The "square root" of a 3-vector is a Pauli spinor, which transforms with SU(2).
I would feel more at ease with normalizing to |a|² + |b|² = 1 rather than a = 1, because the latter is not possible when a = 0.
Sorry about my earlier response. I replied to the wrong comment. And yes, I see your point. I tried to draw parallels with the complex projective line I described in video 5, which has a "point at infinity".
When factoring a 4X4 matrix into a column and row matrices, you say there is no solution. Do you mean that there is no unique solution? Aren't there infinite number of solutions? I'll work on this some more.
Not a unique solution, because a pair of spinors can have reciprocal factor that cancel out when multiplied
If the determinant is nonzero then he's correct that there is no solution. If a 4x4 matrix is the outer product of two 4-element vectors, a column vector Y and a row vector X, then the matrix's rows must each be the vector X scaled by the corresponding element of Y.
So if the matrix's rows are not scalar multiples of one another, it cannot be factored.
Check out the Eigenvalue decomposition of matrices. If it exists, it essentially tells you how to represent the matrix as a sum of vector outer products.
You can also use the singular value decomposition if you're ok with the products having unrelated factors.
@@fixed-point Thank you for that correction.
spinors for beginors
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Good, but complex numbers are neither positive nor negative, unless real. Also, it's not easy factoring a square root of a product into a product of square roots in the complex domain.
I don't like the pulling a complex number out of a square root. This is not allowed in general, and also not in this particular case. It only works in this case because the sign doesn't matter. For instance, note 1 = √1 = √(-1 • -1) ≠ √(-1) • √(-1) = i•i = -1.
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I feel like you're intentionally keeping your voice as monotone as possible because you think its professional ? I don't get it
I have some idea about why. When teaching, I have to keep my voice monotone. For me, since I'm autistic, speaking and human languages are foreign to me. I think in math and pictures. Especially when I have to concentrate and focus on something difficult, I become effectively blind. In fact, it seems that I have to use my whole brain to even speak clearly. If you are neurotypical, you may not understand how difficult it is for me to speak at all. Everyone's brain works a little differently from everyone else.
In fact, but that is the rythm of the lessons you see on every (under)grad course from renowned institutions/professors' playlists on youtube. I think his more clear and paused speech, helps brain take more time to get and process the information presented. At least for me, not native to english. I prefer this way.
Huh? That's your problem. I like his voice
I'm pretty naturally monotone. But I also read from a script to record these videos, which makes me come off as extra-monotone. So I'm not trying to sound professional... it's just the natural way my voice comes out when I read from a script.
@eigenchris this suits me 100% to focus on learning and considering all possibilities.
think of this as a power point presentation, if your slides are cluttered fucked with so much information on the slide, no one will listen to you. you have too much information on many slides, you need to condense down what you show and present