Spinors for Beginners 4: Quantum Spin States (Stern-Gerlach Experiment)

[@kikivoorburg]

Just these four videos have already given me a much better understanding of spinors, quantum spin states, polarization, and even O(n), SO(n), U(n), and SU(n) matrices/groups. You are doing genuinely amazing work!!

[@eigenchris]

Glad to hear it. The next one is the last one for the "first step" in the staircase I showed at the beginning, so I hope you enjoy that one too.

[@kikivoorburg]

@@eigenchris I seem to have missed this reply a while ago, but I did indeed enjoy it and the following videos! This series has been wonderful, thanks for all the great work!

[@charbinger3803]

The part where you added the "also called: dual vector, covector ect.." was incredibly helpful. Using different terms in different contexts for what are basically the same general object fucks me up.

[@eigenchris]

Yeah, it's a bit unfortunate the math/physics community haven't come up with a consistent name for such a common object.

[@alejrandom6592]

Never had I seen such a simple breakdown on the SG experiment

[@KipIngram]

I do have to say that this is the best treatment of this material I've seen. I mean, it's the same material, of course, but you're just giving a concise clarity that more rare out in the world than one might hope. You clearly put a lot of work into all this - thanks very much for your efforts!

[@pathikrith]

Wow.. awesome.. cleared so many doubts and questions in few minutes. Specially those animations are so brilliant. Thank you sir.

[@taibilimunduan]

Incredibly clear explanation, as usual!

[@devalapar7878]

Your videos are one of the best. I think this would be too much and too fast for a layman. But it's perfect as a refresher for mathematicians and physicists.

[@alegian7934]

great video :) I always skip these sections where the results are super similar for the X,Y,Z axes, but I think its really important to include them in the video, as you do. It's reassuring and clears up misunderstandings 😄

[@wmstuckey]

Very pretty account of spin-1/2 (spin) in this video! Quantum information theorists have a similar way to understand quantum mechanics (QM) whence a “principle” understanding of what’s going on with spin that provides a reason for the spin outcomes we observe. We explain this in our book, "Einstein's Entanglement: Bell Inequalities, Relativity, and the Qubit" Oxford UP (2024) and I’ll summarize it here. According to Einstein, special relativity (SR) is a "principle theory," i.e., a theory whose formalism follows from an empirically discovered fact. For SR that empirically discovered fact is the light postulate - everyone measures the same value for the speed of light c, regardless of their relative motions. Since c is a constant of Nature according to Maxwell's electromagnetism, the relativity principle (the laws of physics are the same in all inertial reference frames) says it must be the same in all inertial reference frames. And, since inertial reference frames are related by uniform relative motions (boosts), the relativity principle tells us the light postulate must obtain, whence the Lorentz transformations of SR. Likewise, quantum information theorists have rendered QM a principle theory and its empirically discovered fact is called Information Invariance & Continuity (Brukner & Zeilinger, 2009). In more physical terms, Information Invariance & Continuity entails that everyone measures the same value for Planck's constant h, regardless of their relative spatial orientations or locations (let me call that the "Planck postulate"). Since h is a constant of Nature according to Planck's radiation law, the relativity principle says it must be the same in all inertial reference frames. And, since inertial reference frames are related by relative orientations or locations in space (rotations or translations), the relativity principle tells us the Planck postulate must obtain, whence the finite-dimensional Hilbert space of QM. Quantum superposition is one consequence of the Planck postulate and here is how it makes perfect sense using spin as shown in this video. Suppose you send a vertical spin up electron to Stern-Gerlach (SG) magnets oriented at 60 deg relative to the vertical. Since spin is a form of angular momentum, classical mechanics says the amount of the vertical +1 angular momentum that you should measure at 60 deg is +1*cos(60) = 1/2 (in units of hbar/2). But the SG measurement of electron spin constitutes a measurement of h (Weinberg, 2017), so everyone has to get the same +/- 1 for a spin measurement in any SG spatial orientation, which means you can't get what you expect from common sense classical mechanics. Instead, QM says the measurement of a vertical spin up electron at 60 deg will produce +1 with a probability of 0.75 and it will produce -1 with a probability of 0.25, so the average is (+1 + 1 + 1 - 1)/4 = 1/2. In other words, QM says you get the common sense classical result on 'average only' because of the observer-independence of h. Give up your dynamical bias for QM (just as is done for SR) and the physics of spin makes perfect sense. And if you explore the Bell states for spin-entangled pairs using this approach, you find it solves the mystery of entanglement without violating locality (as in Bohm’s pilot wave), statistical independence (as in superdeterminism or retrocausality), intersubjective agreement (as in QBism), or the uniqueness of experimental outcomes (as in Many Worlds).

[@locusf2]

Quantum Sense channel is also making a series on Quantum Mechanics in general, this is an excellent supplement to it and its really nice to see independent work from two really great creators!

[@BangkokBubonaglia]

I'm also watching both of these series. I have to say, between these 2 channels I am understanding this better than I did 30 years ago when I studied at Caltech. If I had great explanations like this back then, I might not have dropped my desire to be a physicist. Have to give a big thank you to both channels for producing these.

[@ytpah9823]

🎯 Key Takeaways for quick navigation: 00:13 🧲 Quantum spin states are described using the same mathematics as the polarizations of classical light waves. 00:40 🌀 The Stern-Gerlach (SG) experiment involving neutral atoms in an inhomogeneous magnetic field reveals quantized spin angular momentum. 03:18 🪙 The magnetic dipole in silver atoms is due to the electron's internal angular momentum, called spin angular momentum. 05:00 icon Internal angular momentum (Intrinsic magnetic moment) 07:00 icon Bra-ket notation 09:00 icon State "collapse" aka measurement, or projection onto real values via fourier transform 11:00 icon Z-oriented S.G. example 13:00 icon X-oriented S.G. example 14:53 🔀 Quantum superposition is represented as a linear combination of states with probability amplitudes. 17:00 icon Y-oriented S.G. example 18:51 🌐 Quantum spin states can be arranged on a block sphere, similar to polarizations of light on a Poincaré sphere. 21:00 icon 23:52 🔄 SU(2) matrices are used to rotate quantum spin states on the block sphere, taking into account the determinant's phase factor.

[@CarlosRodriguez-mx2xy]

Thank you for giving us this chance !! "You're the top!"

[@bobnob4393]

thank you for these 5 videos. They've really helped clarify some of the math my textbook has thrown at me and will help very much with my quantum computing assignment. All i have to do now is figure out the bell inequality. Can't wait for the rest of this series!

[@manabranjanghosh]

i think before studying quantum mechanics one should really know what even complex no can do inside matrices, if that picture is somewhat clear then i think it really oils up the path towards spookiness, i am an undergrad student taking q.m. at univ, and boi its really sad that they really dont know how to teach or they just love to hang in the level they they have been taught. whatever, ur vids(tensor, relativity and this) and many more generous ppl and the effort u guys put in make me feel math and physics are not worthless things to do. much much love to u guys all.

[@KipIngram]

8:11 - Ok, this is something I feel pretty strongly about. We should stop saying things like "a particle can be in several states." A superposition is not a particle being in multiple states at once. The particle is in ONE state - the one we write |v>. We can *express* that state as a linear combination of basis vectors using any basis we wish - but the particle is not IN any of those states. Usually this basis is defined by a measurement we're just *thinking about* making at some point in the future, so those basis vectors cannot possibly have any physical significance to the particle *before* we make the measurement. Their physical importance is that they define the set of *possible* states that the particle *may* go into post-measurement. Before the measurement the particle's state is |v> - a single quantum state. After the measurement the particle's state is *one of* those measurement eigenvectors. This may seem nit-picky, but I think statements like "the electron is in multiple states at the same time" and "the electron is in multiple positions at the same time" are just very confusing to a lot of people and paint an incorrect picture of what's going on. Ok, I'm getting down off of my soapbox now.

[@archilkitiashvili1227]

What about the actual observable interference between the two states?

[@KipIngram]

@@archilkitiashvili1227 There is only one STATE (at any given time). It can contain multiple terms, and since all of these terms are complex and they get added, yes, they can interfere with one another. But it's still one state. And that state can contain the POSSIBILITY of multiple future measurement outcomes. We can get the probability of each one, and we can confirm those probabilities by running the same experiment a large number of times. But a quantum system has one quantum state at any instant, and that state evolves via the Schrodinger equation so long as we don't measure anything. Then when we do measure something, one of those possible outcomes is realized. Those outcomes ARE NOT REAL until you measure and get one of them, and then ONE OF THEM is real. Prior to measurement they are mere potentialities.

[@Roxyhehe532]

I'm not a regular clz student.. I faced many problems to learn maths and physics.. but now I'm feeling good about its all 🥺🇮🇳

[@MissPiggyM976]

The best videos on this topic !

[@theronsosachavez2757]

You should give a course, it doesn't matter if you ask some money for it. I certainly pay it. Furthermore you can give a constance. Your videos and way to explain are amazing 👏

[@raulsimon2218]

Very clear, and at the right pace. Thanks a lot.

[@joeimbesi99]

Great Clarity esp Bra Ket

[@cmilkau]

In the literature I know, the notation is a lot different (I guess this is physics vs mathematics?): - the transpose is a superscript top (⊤) rather than T - the complex conjugate is denoted by a line over the symbol - the asterisk (*) denotes the adjoint vector/operator/space aka Hermitian conjugate - the dagger denotes a pseudoinverse, i.e. f f† f = f

[@eigenchris]

Yeah, this is a physics/math convention difference. It causes a lot of headaches. The notation I'm using is what you'd see in a quantum class in most physics departments.

[@biblebot3947]

The conjugate can be denoted by either a line or star. The hermitian conjugate restricted to the complex numbers(when thought of as multiplication objects) is the same as the complex conjugate. As for the pseudo inverse, I’ve only seen a plus sign.

[@greenguo1424]

"just remember quantum superposition is a fancy way of saying linear combination" 🤣

[@shaguftanaseem4067]

Thank you for providing such an awesome understanding.

[@karkunow]

It is satisfying to see the connection between polarization of light and spin states so clearly. Interesting, what are non-pure states are for in the light polarization world? We will have some mixed state described by a density matrix there... UPD. While writing the comment it seems I've found the answers here, we can describe Unpolarized Light using those: - en.wikipedia.org/wiki/Density_matrix#Example:_light_polarization - en.wikipedia.org/wiki/Stokes_parameters

[@nthumara6288]

this vidio is really good it is also gives idea of quntum formalisem

[@pannegoleyn9734]

Thank you! I've often wondered where the extra degree of freedom of using complex coefficients in QM comes from, and the idea of preserving orthogonality makes perfect sense. Which, for me, begs the question: what kind of mathematical system results if instead of using complex coefficients, you use quaternionic ones (or even octonionic ones)? What physical system would it represent? You'd add extra degrees of freedom, but lose commutativity in relationships, or even associativity, with octonions.

[@angeldude101]

I'm not sure quaternionic coefficients would be necessary, because the SU(2) matrices, from what I can tell, already _are quaternions._ I guess you could say they're 1D vectors with quaternion coefficients.

[@pannegoleyn9734]

@@angeldude101 Oh, in the current context, absolutely not, but my maths background just has me wondering what kind of structures you get by extending the system. Presumably you could have additional dimensions that way, but the changes in the structure of multiplication would do exciting things to some of the invariants

[@jethomas5]

Remember that for light polarization, you throw away some of the terms. The Jones vector itself is only a three dimensional vector, because of what we threw away.

[@parreiraleonardo4189]

9:34 Just a quick correction: in linear algebra, the euclidian inner product is not exactly the shadow. The euclidian inner product (or dot product) v⃗ • w⃗ is the length of v⃗ times the shadow of w⃗ in v⃗ , giving the same result as the product of the length of w⃗ times the shadow of v⃗ in w⃗ . The shadow itself can be of different sizes depending on v⃗ and w⃗, it is what we call the “component” of a vector along another. The shadow will only be the same size if v⃗ and w⃗ have the same length. We denote the “component of w⃗ along v⃗” as comp w⃗ (v⃗) = (v⃗ • w⃗)/ ||w⃗||.

[@dyachenkotimofey6682]

I love ur videos

[@nicolasPi_]

How do we know the states |x> and |y> have to be distinguished? So far the experiments on axis X and Y are symmetrical. And also, why not going further by introducing more states |d0>, |d1>, etc... for any random axis D0, D1, ...? As far as I understand the Stern-Gerlach experiment, the axis of measurement is orthogonal to the trajectory of the particle. How do we know the spin is not related to the particle's trajectory in the first place? Can we measure the spin on an axis parallel to the trajectory or on a non moving particle?

[@stefanosvasileiadis2732]

We see that in the case of Jones vectors a full turn in physical space corresponds to two full turns in spinor space (polarization space). But when it comes to quantum spin states we see the opposite behaviour. A full turn in spinor space (spin state space) corresponds to two full turns in physical space. Therefore, the angle-doubling relationship between physical space and spinor space depends on the situtation? By this I mean that either we can have a full turn (physical space) -> two full turns (spinor space) or the opposite.

[@eigenchris]

Yes, it depends on the situation. Spinors don't come up too often, so I don't have any other samples to give right now. But what's important is that there's an "angle doubling" between two spaces.

[@Wielorybkek]

Slowly all the pieces are coming together, it's like watching some crime fiction. Who is the Spinor? You will find out soon!

[@karkunow]

Also, while looking at diagrams like at 11:02, I had a spooky feeling of seeing the Quantum Circuits there :) It would be great if you can mention Quantum Computations somewhere in the future. Because we're also talking about the heart of those secretly - qubits.

[@eigenchris]

I think quantum circuits normally involve matrices as the gates. The boxes in this diagram indicate state collapse, not matrix multiplication. But yes, 2-state quantum systems are indeed qubits.

[@karkunow]

@@eigenchris yeah, true

[@hydraslair4723]

Eagerly waiting for a mathematical investigation of CP¹ in the next video!

[@angeldude101]

Thinking in terms of Clifford algebras, projective space is one dimension up from what it's embedding, and the Complex numbers are Cl(0,1). So the Complex projective line is one of Cl(1,1), Cl(0,2), or Cl(0,1,1) depending on whether it's hyperbolic, elliptic, or Euclidean projective geometry respectively. If these, it's worth mentioning that Cl(0,2) is the quaternions, which are also known for representing SU(2) matrices.

[@cmilkau]

26:10 Would've preferred "we CAN set the coefficients", as this choice is convention. You can swap the coefficients of |+x⟩ and |-x⟩ without violating the constraints, and you can multiply all the coefficients with any complex number on the unit circle as well.

[@eigenchris]

You're right. We can always change coordinates and get different coefficients.

[@mmer1687]

Great series! One question, at 15:48 should't coefficients of -x be switched? Looking at the picture above, it is - (1/sqrt(2))*|+z> + (1/sqrt(2))*|-z>

[@alessandroc.4543]

Bro this video is so good that if you were a cat, I would smell the back of your ears in sign of my affection!

[@ws8288]

U r too real for this 😭 M so grateful for this video God bless the person

[@physics2022]

Tribute to great experimental physicist Otto Stern, who was born OTD in 1888 and who along with Gerlach discovered spin quantization 100 years ago in this month i.e. in Feb. 1922.

[@reinerwilhelms-tricarico344]

The Stern Gerlach experiment in an older B&W video from 1967 can be found on KZbin (see url below). It's still fascinating to see, especially with the discussion about what the physics world expected in 1922, when this was first tried. It's really the nice kind of hands-on thing, and of course it looks a lot more complicated than one might expect, but it's well explained there. The film has this mechanics shop feel, and I loved it. They did however not try the extension with a 2nd SG apparatus with different orientation. Now that would be extra cool to see also experimentally confirmed. URL: kzbin.info/www/bejne/d5S3opavq5KJZ6ssi=JZllw6dhQmRgh3CF

[@orktv4673]

It may be noted that the fsct that any state is equivalent to itself times a complex phase is not some supernatural correspondence, but is actually by design. The only thing that's *real* are the probabilities, and we've introduced states to calculate with them. But anytime you calculate a probability, you multiply a ket by a bra, and whatever phase was stanfing in front of it gets lost. The phase is nothing but a redundancy coming from our manner of calculating that we need to keep track of.

[@sshh163]

i am woundering what u study or how are u doing to have all that huge experience at math at every bransh of it ?? who are u ??? how u explan all things that easy ?? iam watching the series u do 5 years ago called tensor for beginner and it is awesome thanks for your work u are the best as always

[@eigenchris]

Thanks. I don't have experience in every branch of math. I have an undergrad degree in engineering physics. I work as a programmer but I study physics in my free time. I make videos on topics I feel are difficult and/or poorly explained in books I've read.

[@dyachenkotimofey6682]

FINALLY

[@tonertoner7586]

Thank you vers much for your videos

[@sf6860]

Thank you for your wonderful videos! QQ: in the stern gerlach experiment, it’s not clear to me what the stern gerlach experiment would look like pointed along the y-axis (around 16:10 in the video). Z is vertical and x is horizontal. So y would point along the beam of silver atoms. This doesn’t seem like a reasonable physical possibility for a stern gerlach device. I understand that this dimension exists and is analogous to the circularly polarized light analogy. In that sense, y represents a phase shift in the other dimensions, not an independent device dimension. Or am I missing something?

[@eigenchris]

You could just spin the SG device 90-degrees in the horizontal plane so that you do a ZY experiment instead of a ZX experiment. I was just trying to say the experiment results are the same no matter how you orient the device.

[@AsrafAli-bq5ww]

Awesome chris ❤, I'm a very big fan of you due to your knowledge, I have seen all of your videos in spinners and general relativity & others, Now I have a doubt, hope you clear, In quantum machenics the expectation value of any observable say "x" is given by ( < psi | x | psi > / ), the probability of "x" finding in a state psi. But ( < psi | x | phi > / < psi | phi > ), what does it means, does it means that expectation value of "x" from psi state to phi state OR probability of "x" to collapse from psi state to phi state. You please clear my doubt...

[@eigenchris]

With pauli spinors we usually do / where S_x= (hbar/2)*σ_x is the "spin-x" operator. This gives you the expectation value of the x-component of angular momentum.

[@krishanusengupta7710]

I would request you to please make some videos on QFT specially on Feynman diagrams.

[@eigenchris]

I unfortunately don't know much QFT and it's unlikely I'm make videos on it.

[@MGoebel-c8e]

For what you called in the video “famously difficult to understand”, ie the individual magnetic momentum of the 47th electron: I picture it from two perspectives: From the perspective of the atom, the electron is on every point of the outer layer simultaneously and with equal probability, so there is no resulting momentum of the atom. From the perspective of the electron however, there is rotation in one specific direction (i.e. spin) as opposed to all directions with equal probability. Hence a total momentum yields. This doesn’t seem too difficult- is there something i am missing?

[@eigenchris]

There are a few confusing concepts with spin. The first is that it is quantized (you can only measure 1 of 2 values along a given axis). It is also probabilistic, so that an spin-left particle has a 50% chance of being measure to have spin-up or spin-down. Finally, there is the concept of what it means for an electron to rotate, as it is a poont particle. It cannot be a tiny ball because the ball would be spinning faster than light. The angular momentum that an electron has comes from the rotation of the spinor used to describe it. All 3 of these ideas are new to quantum and not found it classical physics.

[@stephanecouvreur1377]

At 14:40 the statement “In order to get the expected probabilities, we need to set the coefficients to so and so” should rather be “…, it is *sufficient* to set the coefficient to…” Can the coefficients to change basis from the z observable to the x observable actually be obtained from experimental observations?

[@eigenchris]

From experiments, after you pick an xyz coordinate system, you could deduce the existence of 3 pairs of pairwise-orthogonal basis vectors, one pair for each of the x,y,z axes. Within each pair, the probability of one collapsing onto the other is zero (e.g. = 0). Between each pair the probability of one collapsing onto the other is 50% (e.g. = 1/sqrt(2)). From here, there are an infinite number of possible solutions for choosing the components of these vectors. You could always pick a solution, then rotate it by 10 degrees and get another solution. But coefficients shown in this video are the easiest solution, and also the one used in almost every textbook.

[@karkunow]

Love this explanation at 5:14 about Internal Angular Momentum as kinda "Polarization". Never saw this analogy before clearly stated. There is a small mistake at 25:15 - labels on the top row should be switched, right?

[@eigenchris]

I'm looking at the part at 25:15, but I think it's correct? In physical space, a vertically polarized wave can have the + version or the - version. But in polarization space (poincare sphere), the difference disappears. Do you agree?

[@karkunow]

@@eigenchris yeah, seems you're right V and -V should go to one point on the sphere. Could you please explain what -V means for a physical space? I just thought before that V and -V are equal in physical space, because they just move along the same vertical axis. Is it this difference - when you rotate a wave 180 degrees?

[@eigenchris]

@@karkunow Yes, that's right. Just imagine taking a sine wave and spinning it a half-turn around its axis. You end up turning all the peaks to valleys and all the valleys to peaks. Both are vertically polarized, but a 1/2-cycle apart.

[@karkunow]

@@eigenchris It is kinda unexpected for me that CP1 plays different roles in those two contexts. I've thought before that it will/should play the same role. But it seems it has switched places for some reason :)

[@ws8288]

You are fab thank you so much

[@sebastiandierks7919]

At 25:08, you try to explain the parallels between electromagnetic wave polarization and spin-1/2 states. But isn't it exactly the other way round for em waves and spin-1/2 particles? As the "physical space" graphs are not both on the left/ the "state space" graphs are not both on the right? In physical space, polarizations are 90° apart, spin states of the same direction, e.g. z, are 180° apart. In state space, polarizations are 180° apart, spin states are 90° apart/orthogonal quantum states.

[@eigenchris]

Yes, the "physical spaces" are on opposite sides in either case. But the important thing is that we find an angle-doubling relationship between two spaces. It doesn't necessarily matter which side ends up being physical space.

[@copernicus633]

What is the orientation of the SG apparatus for the Y direction ? The Y direction discussion begins at about 16:00. How does the apparatus rotate to the Y direction without blocking the beam? Maybe a dumb question, but I don’t see it.

[@eigenchris]

It's a fair question. You would have to rotate the entire SG experiment so that beam deflections in Z and then Y are possible. I didn't bother showing this in my 2D drawings. I was more focussed on the math.

[@copernicus633]

⁠@@eigenchrisI was thinking that, but I was also thinking “why does the experiment care about the absolute orientation? The relative relationship of the axes are the same.” In the 2 dimensions discussed, the beam is flowing || to the Y axis, Z is “vertical” and X is left - right. If the apparatus is re-oriented so that the beam is flowing in the X direction, the possible deflections are in the Z or Y directions. But wouldn’t the outcomes be just the same, except for the labels of the axes ? Actually not! think the answer is that it actually reverses the sign of the outcome because the new orientation changes the direction of the axis Y from the previous X axis direction.

[@RFQuantumLab]

Could you please consider making a video on the mathematical model of the vacuum as a superfluid?

[@eigenchris]

I've never heard of that and know nothing about it, so unfortunately I can't.

[@briabrou]

Is there any possible correlation to gravitation?

[@BleachWizz]

19:13 - lol, the +x and -x are in the wrong points xD anyway I was waiting for this video for a while. btw it has become very clear what I was wondering about in my last comment. I'm liking a lot these videos.

[@karkunow]

why wrong? seems ok to me

[@BleachWizz]

@@karkunow they're on the diagonals. the spin -x is at spin -x+z. look, from z to -x there's not a quarter turn on the sphere. (btw sorry for the delay only now youtube sent me a notification for this, lol)

[@sebastiandierks7919]

15:47 |-x> state should point to the upper left, not the lower right

[@eigenchris]

You're right. My bad.

[@martinnjoroge6006]

I am having trouble with the formula for the Born rule. If state a can collapse into b, do we write or ? Both are used in the video and a bit confusing. Also, with |+z> being at the top of the Bloch sphere , why do we represent it with column [1 0] and not column [0 1]

[@gcewing]

1. The probability is the same either way: = *, so ||^2 = ||^2. 2. It's just a matter of convention which vectors you use to represent |+z> and |-z>. Remember that [1 0] and [0 1] are not coordinates in 3D physical space, they're orthogonal directions in the abstract 2D spin space.

[@shaguftanaseem4067]

Very well understanding,can you share the videos in a sequence I mean I have not found the lec after the introduction, thank you

[@narfwhals7843]

Here is the spinors for beginners playlist kzbin.info/www/bejne/oGbWoKSbrdeqp7s

[@shaguftanaseem4067]

Thank you.

[@u1f6f9]

hello again, I'm a bit confused by the visual of |-x> because the minus in front of |-z> and the plus in front of |+z> it should point in the second quadrant instead of the fourth?

[@eigenchris]

I don't quite get what you mean. Is there a certain timestamp in the video you're confused about?

[@br3nto]

Is there an experiment where the measurement apparatus aligns with the spin axis of the particle rather than the particle align with the measurement apparatus?

[@ArtemisiaSayakaRandazzo]

Amazing

[@otterlyso]

Would it have been worth saying a little more about the odd fact that on the Bloch sphere up and down are antipodal (physically 'correct') while on the Poincar̉é sphere vertical and horizontal are (unphysically) antipodal (when physically they are orthogonal)?

[@eigenchris]

I don't have a deep explanation for this, other than saying the in QM, the abstract state space must be "angle doubled" to get physical space, but with polarization, the abstract polarization space must be "angle halved" to get to physical space. Spinors pop up in both cases, but for "opposite" reasons, in a sense. But both situations involve an angle-doubling between two spaces.

[@noahr9539]

23:48 small mistake like in an earlier video of the series: exp(i * pi/4) ≠ i = exp(i * pi/2)

[@loose4bet]

Is there a chance you could do a series on quantum information theory?

[@eigenchris]

Probably not. I don't know amything about quantum information theory.

[@krishnaraoragavendran7592]

5:40 state

[@peterpetrofff]

Thank you! 16:20 How to put magnets for S.G. Y experiment?

[@eigenchris]

I realize I've drawn the diagrams in the same was as with Z and X, but for Y you'd just to orient the entire diagram so that the magnetic field points in the Y direction.

[@tomgraupner171]

Is it really possible and that easy to put two magnetic fields in a row? As far as I understood QM, the "decision for spin up or spin down" is done with a measurement only - not by the magnet. Thus, the magnetic field changes the wave function of the AG atoms, but you will not get two dedicated trajectories for spin up and for spin down between the magnet and the screen. IMHO: there needs to be some measurement device in place to split up the beam before you can act with the second field. Please help me to gain a better understanding. Thanks a lot, Chris!

[@eigenchris]

You've hit on an interesting issue, which is: does the state collapse happen at the magnetic field or at the detector. I admit I didn't think too hard about that when making this video. Looking at this Stack Exchange question: "The key point is that splitting out into the two Sx components and merging again without ever measuring will not collapse the state. Superposition can persist over distance as long as no measurement is made,". physics.stackexchange.com/questions/495185/sequential-stern-gerlach-experiment

[@tomgraupner171]

@@eigenchris maybe it goes like this: the first magnet creates the superposition/linear combination of "up" and "down" and the "particle/wave" is on both "trajectories" at the same time with 50% each. Next the "down" trajectory is led to a screen. Thus, now we get a measurement and this measurement (of just one path) would lead to the collapse and the "up" trajectory has indeed the "spin up" only. Would nevertheless be interesting to see how this goes w/o such measurement in the second path. What would a second magnet in both "trajectories" produce?

[@IronLotus15]

@@eigenchris I'm not sure if this experiment is possible, but what if instead of merging the two Sx+ and Sx- beams, we pass them through 2 separate SGz fields, and have 4 detectors? Would we have an even spread across all 4, or only across the 2 corresponding to Sz+? Edit: I think this is also Tom's last question

[@tomgraupner171]

@@IronLotus15 "The Feynman Lectures on Physics" describes that "stacking of SG devices" and that a measurement is needed to break the superposition. @eigenchris "thanks for clarification in the video #5"

[@LL-mq7gj]

17:11 Are the coefficients for the plus and minus X states (or Y) only different by a minus sign in the coefficient for their -Z state due to the fact that it makes the plus and minus X state’s orthogonality condition hold?

[@eigenchris]

Yes, that's right.

[@LL-mq7gj]

@@eigenchris thank you

[@wadelamble5462]

There's something I'm not getting about the use of the word "physical" at 25:21 in the table of spaces for polarization and spin. For polarization, we rotate twice in "polarization" space to rotate once in "physical" space; whereas for spin we rotate twice in "physical" space to rotate once in "state" space. Thus it seems that the relationship of "physical" to "non-physical" is reversed for polarization and spin. Maybe I'm reading too much into the word "physical" in the slide. If I were to choose the words "observable," where "observable" means "observable polarization" or "observable spin" and "internal," where "internal" means "includes phase," could I then say that for both polarization and spin the "observable" state rotates twice to return the original "internal" state? In KZbinr sudgylacmoe's excellent video on geometric algebra (kzbin.info/www/bejne/bGHdkJumeqanepo), the object that rotates half as fast is the "rotor" (per my muddled understanding, a bivector written as a complex exponential with a phase angle), applied once on the left and once on the right to a vector, and the object that rotates twice as fast is the vector being multiplied. Relating this to your explanation (so far in the series), two questions arise for me. First, in your explanation, there are two separate spaces; whereas in sudgylacmoe's geometric algebra explanation there is only one space, and the rotors "act on" the vectors. Is the quantum state space, then, in a sense, a space of rotors ??? The second curiosity is that sudgylacmoe's explanation doesn't involve any physics. Is there some intuition that would motivate mapping vectors to "observables," while mapping rotors to "internal" states?

[@eigenchris]

By "physical", I just mean it matches the standard XYZ coordinates we experience in everyday life. For wave polarizations, if you had a plastic model of a sine wave in your hand, and rotated it along its axis by a half-turn, you'd end up with the negative version of the sine wave. This is why "physical" is on the left for wave polarizations. The "polarization space" is a more abstracted mathematical space. In the case of quantum states, it's reversed. The "quantum state space" is the abstract mathematical space (where +z and -z are at right angles), and the ordinary physical space we're used to (where +z and -z point in opposite directions) is on the right side of the chart. If you prefer, you could instead refer to them as spaces which "include phase" and those that don't. Yes, SU(2) matrices are equivalent to rotors that only rotate and don't change size in geometric algebra (also equivalent to quaternions). I'll be talking about this quite a lot in later videos, starting with video #6. Vectors are objects that transform with a two-sided rotor transformation (a pair of SU(2) matrices), whereas spinors transform with a single rotor (single SU(2) matrix). I'll be talking about all of this in due time. I wanted to spend the first 5 or so videos connecting this to physics because many advanced physics students end up having to deal with spinors, but are often not given great explanations for them. As my videos go on, I will touch on a lot of the stuff sudgy talks about (notice the 3rd step in my staircase is about Clifford Algebras). For your questions: 1. The reason I have 2 spaces will become clear in the next video (#5). One space is just a projection of the other. The poincare sphere and bloch sphere are spaces called "the complex projective line". 2. I'm afraid I don't understand this question about "observables" and "internal states". Can you explain it more? What do you mean by "observable"?

[@wadelamble5462]

​@@eigenchris Thanks. By "observables" I mean, in the examples so far, the polarization of the light or the deflection of the electron. I think the question is too vague at this point. I'll see if it makes sense to ask again after watching future videos.

[@franks.6547]

But what about the relative phase of different spinors in quantum superposition? Is it, that I have first to pick one specific complex representation for spinors including a definite phase and then sort of stick to that when composing superpositions? For a crucial negative sign can't be distinguished from the phase thas is abstracted away in spinor space? Or isn't that a problem for some reason?

[@gcewing]

I don't think it's something you have to worry about. Certainly relative phases do matter, so you can't go arbitrarily ignoring phases in intermediate calculations. But you can pick whatever phase you want for your initial state, and as long as you're consistent about it from there on, everything will work out.

[@nthumara6288]

this is where i first learnd boha modle is wrong

[@sebastiandierks7919]

In quantum field theory, spin-1/2 particles like the electron (or silver atoms) are described by spinors, while spin-1 photons are described by vectors, they are explicitely NOT described by spinors. I find it interesting, and don't know the solution yet, why on the level of "normal" non-relativistic quantum mechanics and classical electromagnetism, respectively, silver atoms and electromagnetic waves are both described by spinors. Could you explain this transition to QFT?

[@eigenchris]

I'm still studying QFT and I don't know the answer yet. The Proca Lagrangian for massive vector fields tells us that massive vector waves can have 3 independent polarizations (2 transverse and 1 logintudinal). If the vector field is massless (like light), then we only get the 2 transverse polarizations; the longitudinal one becomes "no allowed". I'm not sure if this restriction to 2 polarizations for massless vector fields is what gives rise to a spinor-like description. Unfortunately I can't say.

[@karkunow]

There is a spin-decomposition theorem about which Penrose talks in his book Road To Reality. Any high order spin n/2 "particle" could be describe using n spin1/2 "particles". So vector is secretly just a 2-spinor in sense that it is composed of 2 spin 1/2 components.

[@sebastiandierks7919]

@@karkunow unfortunately that's not the answer to my question. Indeed, if you put several particles into one composite particle, like electrons in the atomic shell, nucleons into the nucleus, or quarks into hadrons, you have to follow the rules of quantum mechanical coupling of angular momenta. (If you know what you're doing, that practically amounts to looking up Clebsch-Gordon coefficients.) Mathematically, this is about the representation theory of the group SU(2). So the theorem of Penrose you were talking about exists. However, in regards to my question, a photon is an elementary particle and not made up of 2 spin 1/2 particles (, which indeed you could couple to a spin 1 particle). And even if it was composite, it then is a spin 1 particle described by a vector, not a spinor. Furthermore, "vector" and "spinor" has to do with the representation theory of the Lorentz group SO(1,3), in which rotations SO(3) ~= SU(2) is only a subgroup. Vectors live in the fundamental representation of the Lorentz group, while spinors live in the spinor representation of the corresponding spin.

[@karkunow]

​@@sebastiandierks7919 Ah, yeah saw those tables with Clebsch-Gordon coefficients. Got your point about photon. Thanks! Talking about SO(1,3)+ - it will consist of two copies of SU(2) actually, right? (or SL(2,C)). One for boosts and other for rotations. What do you mean by fundamental vs spinor rep?? If we think of SU(2)xSU(2) and its reps. They will be described in terms of j-numbers. So (1/2, 1/2) rep will be a vector representation or bispinor and (1/2, 0) and (0, 1/2) will be spinor ones.

[@neopalm2050]

Is the SU(2) symmetry of spinors just the global part of the SU(2) guage symmetry of the standard model? (i.e. is a local SU(2) symmetry just a local "twisting" of space, combined with an offset to the appropriate guage fields?)

[@eigenchris]

They are the same group, but they come up for different reasons. Spinors in 3D (Pauli spinors) will transform with SU(2), and in spacetime spinors (Weyl spinors) transform with SL(2,C). These come up because they are the "double cover" of rotation groups, which are symmetries of spacetime (I'll talk about this in later videos). The SU(2) that comes up in the standard model is related to the weak force. Specifically, U(1)xSU(2) is the symmetry group of the electroweak force. It is 4-dimensional, which relates to the 4 electroweak force-carrying particles (photon, W+ boson, W- boson, Z0 boson). I'm not sure if there's any sort of deep reason why SU(2) appears in both cases. I haven't studied the standard model much so I can't say anything more.

[@francislinkin1535]

Hello, video author, I have a lingering question that what the difference between spinor and 4-vector, it has similar form and structure. Is there some advantages that spinor has and 4-vector doesn't?

[@eigenchris]

They are different mathematical objects that object different transformation rules. As seen in this video, a Pauli spinor will transform with a single SU(2) matrix. A 3D vector transforms with a pair of SU(2) matrices. In special relativity, we use Weyl Spinors, which transform with a single SL(2,C) matrix. A 4D spacetime vector will transform with a pair of SL(2,C) matrices. Vectors will always transform twice as much as spinors do, which is why it takes spinors 2 full rotations to get back to their starting point. In Quantum Field Theory, spinors represent the spin-1/2 "fermion" type particles that obey the Pauli Exclusion Principle, whereas vectors represent spin-1 "boson" type particles that don't obey the Pauli Exclusion Principle.

[@mroygl]

Right but for the election spin angular momentum is its spin direction equal to the little magnet direction? That is... Does the spin up direction really represent its say north pole direction as on your images of SGE?

[@eigenchris]

I've seen different textbooks describe it different ways. "Up" and "Down" are ultimately just conventions.

[@mroygl]

@@eigenchris no no no ;) We've ultimately got an angular momentum. Let it be the spin angular momentum whatever but it is still the angular momentum. Right? The angular momentum is a vector and the vector here is somehow up or down but still it's got a direction. Right? Unexpectedly I just realized thru your video that SGE provided an obvious direction S-N and the spin should point somewhere namely up or down or may be left out right.. Hence the questions... Where does the spin point in SGE? How does the spin interacts with the magnetic field of the magnets in SGE?

[@eigenchris]

@@mroygl A spinning proton will have its angular momentum vector and magnetic moment vector point in the same direction, according to the right-hand rule. But since we're talking about electron spin in the SGE, they will point in opposite directions--the magnetic moment will get an extra negative sign because of the negative charge, so it follows the "left-hand rule", basically. Does that answer your question?

[@mroygl]

@@eigenchris That's much better. Thanks! Still I got much confusion... SGE was about the spin only, i.e. there should be no magnetic moment I guess. Then how will spin interact with the magnetic field in SGE?

[@eigenchris]

@@mroygl The spinning electrons do have a magnetic moment, and they act like tiny bar magnets. If the magnets have "north" pointing upward, the magnetic field from the SGE will deflect them upward. Vice versa if the spinning electron dipole has north pointing down.

[@PeterH1900]

I understand how you can do the z and x experiment by rotating 90 degrees. But I do not understand the y. It seems to me that you cannot do that with SG. Correct?

[@eigenchris]

You can take the x version of the experiment and rotate the whole thing 90 degrees so that the x magnet becomes a y magnet.

[@PeterH1900]

@@eigenchris I understand but now for the 3rd axis? If you rotate another 90 degrees you end up 180 degrees which is -x. So how do you get the z then?

[@angeldude101]

It's it just me, or did the vectors shown look just like quaternions? I'm pretty sure quaternions are themselves specifically a representation of SU(2) matrices, except they actually look like they belong in 3D, rather than hiding the 3rd dimension in a complex scalar.

[@eigenchris]

SU(2) matrices are equivalent ("isomorphic" in technical terms) to the quaternions of length 1. This will become more clear as the series goes on.

[@angeldude101]

@@eigenchris I thought it was pretty clear here, but that's probably just because quaternions are usually taught in a terrible way that obscures what's really going on. How you showed the spheres here is basically how I've come to picture quaternions.

[@sirnukesalot24]

The fact that phase can not be distinguished either physically or mathematically must drive plenty of folks nuts. Still, something makes me think that this is somehow related to quantum chromodynamics. Would I be out of my mind to suspect that spin phase states within the scope of any particle interaction is the main event in color interactions, and that's where QCD's triplicate sets of the same particles come from?

[@eigenchris]

It's an important property related to gauge theory. There is a similar property in electromagnetism where you can make a change to the electromagnetic potential without changing the electromagnetic fields. One of the important ideas in QED is that you can make local changes to the phases and compensate for it by making local changes to the EM potential; the changes end up cancelling out and are not detectable. Something similar happens in QCD, but I'm not as familiar with that.

[@sirnukesalot24]

Paying attention to these concepts off and on from the outside, it seemed like the "total" set of particles were triplicated as a consequence of breaking down realspace into three orthogonal directions. While that's little more than an amateur attempt at an interpretation on my own part, it's now starting to look like something way cooler is happening. Do you know if anyone ever tried to examine the outer product of spin vector components to see if any additional insights could be gained?

[@gcewing]

As far as I know, there's no known theoretical reason for triplication of the particles. It's not a nice clean symmetry, because they all have different, seemingly random masses. Maybe there could be some kind of broken symmetry behind it, but I haven't heard of such a theory.

[@sirnukesalot24]

@@gcewing Thanks for the heads up. It does make me wonder if these odd differences could be telling us something that we won't understand for several more decades.

[@victor_anik]

Looks like this double angle feature comes from complex numbers..

[@ceoofracism5713]

19:05 but isnt I -z > and I z > are 90 degree apart instead of 180 degree as both are orthogonal?

[@ceoofracism5713]

only in physical space they are 180 degree apart

[@doscoroll2366]

At 17:00, why can't we reuse the same coefficients with y-state as x-state?

[@eigenchris]

Because they are different states. The +y state can collapse to the +x or -x states.

[@MasterHigure]

Is there a real point to the magnetic field in the SG experiment being inhomogeneous (apart from the fact that you can't actually make a 100% homogeneous field)? Wouldn't you, at least in principle, get the same result with a more homogeneous magnetic field? Also, I'm disappointed that you didn't point out the marvelous pun that lives at the core of the bra and ket notational convention.

[@eigenchris]

Magnets (which are dipoles) wouldn't get deflected in a homogeneous field, as there is no net force. You need an inhomogeneous magnetic field to get a net force. (This is different compared to monopoles, which would feel a force in a homogeneous field.)

[@MasterHigure]

@@eigenchris Right, you need the "tidal" forces associated with inhomogeneity, a stronger pull on one pole than on the other, so to speak. I was halfway thinking charged particles traveling in a magnetic field for some reason.

[@allanc3945]

@@eigenchris I don't understand why the particles don't just rotate to become aligned with the external magnetic field, then move as you described due to the inhomogeneous field. This would result in all the particles hitting one spot on the detector. What causes the separation? Intuitively I would think at the end all the 'norths' would be pointing in the same direction, but half would be spinning left handed and half right. (thinking classically, I realize) I must have some serious misconceptions regarding the SG experiment because I can't wrap my head around how this works, and it seems like most people have no problem with it. Fantastic video, by the way. Like the person said in a previous post, I wish I had your lectures 30 years ago!

[@eigenchris]

@@allanc3945 With a homogeneous magnetic field, there would be no net linear force (because the force on the upper end of the dipole and lower end of the dipole are equal and opposite), but there would be a torque applied to the magnet, which would give results similar to what you describe. In the case of an inhomogeneous magnetic field, there ends up being a net force (because the force on the upper end of the dipole is stronger than the force on the lower end of the dipole), and this is what causes the deflection. I assume there would be some torque too but for this experiment I don't think that's relevant, because we only care about the final position.

[@allanc3945]

Maybe I get it... If the inhomogeneous field is very strong and for a BRIEF duration, the particle will feel the force from the stronger side, either attraction or repulsion, and be pushed (bumped) to a different trajectory. (in the classical model) It wouldn't have time rotate into alignment. If I'm wrong, don't tell me. I've been baffled by this for far too long

[@davidwilkie9551]

Why?

[@nthumara6288]

why in the experiment we use homogenuos magntic filed

[@eigenchris]

A homogeneous magnetic field would not pull a dipole in one direction or the other. We only get the "splitting" into two directions if the magnetic field is inhomogeneous.

[@KipIngram]

5:24 - You've drawn your atom beams as curving even well outside the magnetic field. 🙂 I think a better drawing would have them coming in straight lines out of the apparatus, having already been sent into their final direction state. Not a big deal - I guess I'm just being pedantic. 😐

[@3zdayz]

I did some digging, found a actual experiment... kzbin.info/www/bejne/d5S3opavq5KJZ6s they even include the 'what if they were bar magnets, in Helmholtz coil... (earlier in the video too). Your bar magnets aren't really limited to 2D; they could be oriented in any direction in 3D space (though twist around its n/s axis doesn't matter). The odds of it being in front or back don't really change if you simplify it to a 2D representation either, just a nit-pick. The assumption about the overall behavior demonstrated with physical rotating bar magnets doesn't actually apply though; a electron is a charged particle, and moving in a magnetic field is deflected by that field depending on its direction; unless it's already in the plane of the field, then it goes in the circle it's going; but there's no reason it would have momentum effects as demonstrated with a physical bar magnet; the deflection is work-free (At least I think I remember someone saying that) There's no force on the magnets for the deflection of the electron. The resulting spin will be either mostly aligned or mostly unaligned - also the SG magnets are very long; there's lots of time to interact, it's not like the atoms are just going through a magnetic slit. Your magnets; 1:22 arguably the points on the north pole would be hot-spot norths, and there would be very little north field in the V itself... the SG experiment above used curved magnets, which would distribute the field more smoothly; the video towards the beginning has a 'asymmetric' field demonstration with iron filings which is a good visualization of the sort of field it is... and then also related a bar magnet in a Helmholtz with such a field; I'm surprised there isn't torque applied once the magnet is inverted - it is merely repelled in the same orientation if opposing poles are aligned, and attracted otherwise, but not turned. stacks of SG work as expected - if you consider that the magnetic field of the test particle will align first, and then be what it is... up-up down-down up ->left/right because at up it's not 100% exactly up, and even if it is it's just slightly more tilted left or right when it enters a sideways SG separator. Just light polarized light filters - the UD or LR orientation is of course within that when it enters, but when it leaves, the only requirement is that the emitted photon is somewhere within the allowed input range;; and this is just a dot product of its spin axis and the detector alignment. The spin of correlated particles is formed at the last point it interacted with non-free space (birefringence crystal for example), and remains the same until detected. But these silver atoms are going at less than the speed of light; although their internal spins are still going very nearly the speed of light ( S^2+V^2=C^2; if they were going with a high V near C, then they wouldn't have much S ). So probably silver moving in a magnetic field ends up with a magnetic field; but I would expect that any time between SG detectors whatever pole it had will wander somewhat from what it was sorted into.

[@gcewing]

The SG experiment didn't use isolated electrons, it used neutral atoms, which are not deflected by a magnetic field unless they have a magnetic moment. As for torque, in the bar magnet analogy I think it's assumed that the magnets fly through quickly enough that they don't have time to rotate appreciably due to the torque. Also a better analogy would be spinning tops that are magnetised along their axis. The torque will cause them to precess gyroscopically, which will take a lot longer to flip them over that if they weren't spinning. My guess is that in the actual SG experiment the precession rate is low enough compared to the speed of the atoms that it doesn't affect the result much. It's worth noting that the real-life results aren't two perfectly sharp spots, they're spread-out blobs, and part of the reason for that might be precession effects.

[@-datolith2775]

😀

[@tomersvirsky5807]

anyone else groan when he revealed that ket and bra are put together to form a bra-ket? physics is just one big dad joke i swear to god

[@weimullerjohann9118]

Man sieht nur eine Stromflußdrehung im Plasma mit verteilten Silberionen und bestenfalls eine vorpolarisierte Drehung. kzbin.info/www/bejne/qJiXd6GXnJmJY6c

[@loose4bet]

In case of light polarization the overall phase doesn't influence what happens in polarization domain/space but it physically is a different wave, the difference can be measured at least in principle. In quantum description you've mentioned that the overall phase has no physical significance upon measurements. Is there anything that can physically happen before the collapse/measurement, like interaction between different waves where the phase can influence the measurement after that interaction happened?

[@eigenchris]

I'm not aware of any case where a "global" phase in front of the entire state can be measured. There are various entanglement experiments where you change the phase of one of the sub-components of the system, and that can be detected, but that's not a global phase.

[@loose4bet]

@@eigenchris Thanks for the reply. So if I understand this correctly, if we have 2 wave functions interacting, the individual phases can play a measurable role. But if we look at the wave function of the whole system whatever phase can be factored out is irrelevant.

[@loose4bet]

@@eigenchris BTW. Quality content. This is the best explanation I've seen. You even include small exercises. You're a very good and efficient educator. Keep it up.

[@mikel4879]

Quantum state of my as

[@shoutitallloud]

bra-vector and knicker-vector combine to lingerie product