I What's the relation between spinor space and SO(3) vector space?

[joneall]

TL;DR Summary

I don't understand what it means to say that when a SO(3) vector space goes from 0 to 2 pi, a spinor space goes only from 0 to pi.

First, simply, how do we do an experiment to verify that when we rotate a physical system through some angle, a vector is rotated through that angle but a spinor is rotated through half that angle. How do we measure this? One book mentions using a neutron interferometer, but gives no details.

In fact, here, we are comparing the effects of Lorentz rotations on two different representations of two different but isomorphic groups. It just is not clear to me what the relation between objects in different representations may be. It might be that the word "isomorphic" tells me they are really talking about the same thing. But vectors and spinors are most definitely NOT the same thing.

Maybe my question should be about the relations (if any) between representations? Is there a consensus on such a question or its answer?

 

[PeterDonis]

when a SO(3) vector space goes from 0 to 2 pi, a spinor space goes only from 0 to pi.

Where are you getting this statement from? Please give a reference.

One book mentions using a neutron interferometer, but gives no details.

Which book? Please give a reference.

 

[Nugatory]

First, simply, how do we do an experiment to verify that when we rotate a physical system through some angle, a vector is rotated through that angle but a spinor is rotated through half that angle. How do we measure this?

Vectors and spinors are abstract mathematical objects that obey various mathematical rules so this is something that we calculate (mostly using the methods of abstract algebra) not something that we measure. Compare with the role of the abstract mathematical objects “function of position” and “gradient” in classical physics: useful results come from their mathematical properties once we’ve learned how to use them to model conservative forces.

One book mentions using a neutron interferometer, but gives no details.

Not a sure thing but if you would tell us which book, there’s a pretty good chance that someone here is familiar enough with it to explain this.

It just is not clear to me what the relation between objects in different representations may be. It might be that the word "isomorphic" tells me they are really talking about the same thing. But vectors and spinors are most definitely NOT the same thing.
Maybe my question should be about the relations (if any) between representations? Is there a consensus on such a question or its answer?

Yes, both isomorphism between groups and representations of groups have rigorous and generally accepted mathematical definitions. But for a non-rigorous example of group isomorphism, consider two groups: the four integers 0,1,2,3 under addition modulo 4; and the four 90-degree rotations of an object in a plane. Integers and 90-degree rotations are most definitely not the same thing, but the two groups have exactly the same mathematical structure.

 

[joneall]

Sorry to have forgotten to cite my sources.

In "No-nonsense QFT", by Jakob Schwichtenberg, he says in note 38 on page 91: ".. its [sic] only through the interactions with gauge fields that we can detect that a spinor indeed picks up a minus sign after a rotation by 360° (for example, in a neutron interferometer)."

The difference in angles is discussed in many books, of different levels. For instance, Sean Carroll in his latest book, about "Quanta and fields", says on page 242: "A particle of spin s is invariant under a rotation by 2 $\pi$ /s radians."

They all point out that the ratio of 2 necessary for the rotations to return to the same state is why SU(2) is the double cover of SO(3).

 

[PeterDonis]

a spinor ... picks up a minus sign after a rotation by 360°

"A particle of spin s is invariant under a rotation by 2 $\pi$ /s radians."

Neither of these statements is what you said in the OP of this thread. In the OP, you said these things:

when a SO(3) vector space goes from 0 to 2 pi, a spinor space goes only from 0 to pi.

when we rotate a physical system through some angle, a vector is rotated through that angle but a spinor is rotated through half that angle.

Did you get those statements from the references you gave? Or is it your own interpretation?

 

[joneall]

Since we are being precise, I will point out that the Lorentz transformations for a spinor or SU(2) go like $exp(i\theta/2)$ whereas those for spatial rotations or SO(3) go like $exp(i\theta)$. So when the angle for the vector changes by an angle $\phi$ that for the spinor changes only for $\phi/2$. From this, the statements I made are clearly true. These are my deductions, not interpretations.

 

[PeterDonis]

the Lorentz transformations

Lorentz transformations of what?

when the angle for the vector changes by an angle $\phi$ that for the spinor changes only for $\phi/2$.

No, that is not what the Lorentz transformation property you wrote down means. Again, what is being Lorentz transformed? Hint: it's not the "angle of rotation".

 

[PeterDonis]

Since we are picking nits

No, we're not. We are trying to get you to correct what appears to be a significant misunderstanding that you have.

 

[joneall]

I admit to not understanding this. So instead of using the Socratic method, why don't you just tell me what my error is? That is why I asked the question after all. I do not get at all what you mean by "what is being Lorentz transformed? Hint: it's not the "angle of rotation".' It seems clear to me that what is being transformed by rotation through an angle in the SU(2) or SO(3) (I am trying to pick my words carefully, but I'm not an expert...) is a vector in the SO(3) case and a spinor in the SU(2) one. My whole question is about the relation between these. Is that clearer? (I hope so.) In any case, thanks for your input.

BTW, what is an OP? It seems to be the initial post in a thread.

 

[PeterDonis]

why don't you just tell me what my error is?

I did, in the last part of my post #7. I asked you a question in that post. Can you answer it? If you have read the references you gave, you should be able to.

I do not get at all what you mean by "what is being Lorentz transformed?

What does the Lorentz transformation act on? You could say "vectors or spinors", but that just postpones the question: what is it about vectors or spinors that the Lorentz transformation changes? In fact, in this case you don't even need to use Lorentz transformations; just ordinary 3-D spatial rotations are enough. A rotation by $\theta$ changes something by $\exp(i \theta)$ in the case of a vector, i.e., a spin-1 particle, and something by $\exp(i \theta / 2)$ in the case of a spinor, i.e., a spin-1/2 particle. What is that something? Again, if you've read the references you gave, you should be able to answer that question because they should tell you.

Also, the complex exponentials should be a hint: what role do complex exponentials play in the description of spin-1 and spin-1/2 particles in QM? (Non-relativistic QM is fine, since, as I said above, you don't need to consider Lorentz transformations for this; just ordinary 3-D spatial rotations are enough.)

Hint: it's not the "angle of rotation".'

The 3-D spatial rotation I referred to above is a rotation by an angle $\theta$ in both cases, vector and spinor. It is not a rotation by $\theta$ for a vector and a rotation by $\theta / 2$ for a spinor.

My whole question is about the relation between these.

You are correct that they are two different representations of a group. If you are only considering integer spin particles, using SO(3) as the group works fine. But if you want to include half-integer spin particles, you need to use SU(2) as the group. As you have said, SU(2) is the double cover of SO(3), which means, heuristically, that for each group element in SO(3) there are two group elements in SU(2), whose actions on integer spin representations (of which "vector" is one) are the same--the action you're used to from SO(3)--but whose actions on half-integer spin representations (of which "spinor" is one) are related by a minus sign. So, for example, if we consider the group element "rotation by $2 \pi$" in SO(3), there are two group elements in SU(2) that correspond to it, which you can think of, heuristically, as "rotation by $2 \pi$" and "rotation by $4 \pi$". The first of the two is the one whose action on spinors has the minus sign (both act the same on vectors).

That still leaves the question of where, exactly, the minus sign appears, which is the question I posed above and in post #7.

 

[PeterDonis]

what is an OP? It seems to be the initial post in a thread.

Yes, that's what it is. It can also mean the person who posted the initial post in the thread.

 

[PeterDonis]

One book mentions using a neutron interferometer, but gives no details.

It is true that you can use an interferometer to detect the difference in how spatial rotations act on vectors and spinors. This is also a clue to what it is that a spatial rotation changes about a vector and spinor. What property of a wave does interference involve?

 

[Martin_Sz]

TL;DR Summary: I don't understand what it means to say that when a SO(3) vector space goes from 0 to 2 pi, a spinor space goes only from 0 to pi.

First, simply, how do we do an experiment to verify that when we rotate a physical system through some angle, a vector is rotated through that angle but a spinor is rotated through half that angle. How do we measure this? One book mentions using a neutron interferometer, but gives no details.

In fact, here, we are comparing the effects of Lorentz rotations on two different representations of two different but isomorphic groups. It just is not clear to me what the relation between objects in different representations may be. It might be that the word "isomorphic" tells me they are really talking about the same thing. But vectors and spinors are most definitely NOT the same thing.

Maybe my question should be about the relations (if any) between representations? Is there a consensus on such a question or its answer?

Spinor space and SO(3) vector space are intricately linked, but not identical. Here's a concise explanation of their relationship:
* SO(3) Vector Space: This space represents rotations in three dimensions. Elements within this space are three-dimensional vectors that define the rotation axis and angle.
* Spinor Space: This space, unlike a vector space, possesses a more complex structure. Spinors, the mathematical objects residing in this space, can be used to represent rotations as well. However, they often require a complex number representation and behave differently under rotations compared to vectors.
Connection: SO(3) transformations can be mapped to rotations using spinors. This mapping isn't a one-to-one correspondence, but it establishes a relationship between the two spaces. Spinors offer a more nuanced way to describe rotations, incorporating additional properties beyond the rotation axis and angle.

 

[joneall]

Sean Carroll's statement already mentioned, "A particle of spin s is invariant under a rotation by 2 /s radians." seems to clearly indicate he is talking about rotation of the particle, what I referred to as the vector or spinor in my posts. You ask me what it is about them that is transformed. You ask where the minus sign appears when a spinor is rotated through $\pi$, but that is exactly the question I was asking when I started this post. Apparently that was not clear.

Does it matter whether we rotate the vector/spinor or the space SU(2/3) it is living in? That also is part of my question.

I appreciate your trying to get me to answer my own question, but I don't see what else there is to answer.

 

[joneall]

You ask, 'What does the Lorentz transformation act on? You could say "vectors or spinors", but that just postpones the question: what is it about vectors or spinors that the Lorentz transformation changes?' It acts on their orientation in the vector or spinor space. Is that what you are trying to lead me to? That was implied all along... If we were talking about U(1), it would be the phase, but we are not talking about U(1).

You also say, 'The 3-D spatial rotation I referred to above is a rotation by an angle in both cases, vector and spinor. It is not a rotation by for a vector and a rotation by for a spinor.'. Then why is it divided by two in the transform of a spinor?

 

[PeterDonis]

It acts on their orientation in the vector or spinor space. Is that what you are trying to lead me to?

No.

 

[PeterDonis]

seems to clearly indicate he is talking about rotation of the particle

Mathematically, he is talking about acting on the mathematical object that describes the particle with an element of the group SU(2).

You ask me what it is about them that is transformed.

Yes. What mathematical object that describes the particle is being acted on by an element of the group SU(2)?

Don't the references you gave discuss this? Don't they write down any equations that describe the kind of actions we are talking about?

 

[PeterDonis]

Does it matter whether we rotate the vector/spinor or the space SU(2/3) it is living in?

I don't know what "rotating the vector/spinor space" would mean.

There is a distinction that is made in some sources between rotating the particle itself and rotating the coordinate system. For this discussion, both can be considered to be equivalent.

 

[joneall]

For SU(2) it's the spin, of course. It's the only thing that has any orientation in a rep of that. That's what SU(2) is about, spinors. I could quote the matrix for a rotation, but it's getting late over here and I don't have the courage to start with latex. ;) They involve exponentials of i times a parameter (our $\theta$) times a Pauli matrix.

 

[PeterDonis]

For SU(2) it's the spin, of course.

What is the spin? What are you responding to?

I could quote the matrix for a rotation, but it's getting late over here and I don't have the courage to start with latex. ;) They involve exponentials of i times a parameter (our ) times a Pauli matrix.

The complex exponentials in that case are how you get an element of a group from an element of its Lie algebra, by exponentiating the Lie algebra element. They're not the same as the complex exponentials that appear in the action of a group element on the mathematical object that represents a particle.

You still haven't answered the question: what mathematical object that represents the particle? I used the word "wave" earlier. Does that suggest anything?

 

[joneall]

I can't figure out how to quote part of a post. Is there a help on this somewhere?

PeterDonis said: 'What is the spin? What are you responding to? ... You still haven't answered the question: what mathematical object that represents the particle?'

My answer spin was the answer to your question, what mathematical object...
We do not have 36 things here, we have a spin (which IS the wave function, or at least the relevant part of it for this discussion) and we have the rotation of the space (or the electron, for instance).

PeterDonis also said: 'The complex exponentials in that case are how you get an element of a group from an element of its Lie algebra, by exponentiating the Lie algebra element. They're not the same as the complex exponentials that appear in the action of a group element on the mathematical object that represents a particle.'

Let's be careful not to confuse angles with infinitesimal generators and group parameters. The generators and parameters, 3 of them for SU(2) ($2^2 -1$) are defined by
$$T^{a} = -i \frac{\partial D}{\partial \theta_a} \Big|_{\theta_a = 0} $$
where the transformation is given by
$$D (g(\theta)) = e^{i{\theta_a T^a}}$$ so that
$$D ({\theta}_a) = I + i \delta\theta_a T^{a}$$. An SU(2) rotation has the form
$$U = e^{i \beta J_i }$$.
It looks to me as if the complex exponentials are the same.

I have no idea what you are getting after. So please stop asking me that question and please answer it.

 

[joneall]

BTW, a big deal is made out of the1/2 factor in the exponential for an SO(3) rotation or for such a rotation in Minkowski space. But the equation for a Lorentz boost has a real exponential and also includes the factor of 1/2. Seems to me this should be considered important also, altho the interpretation may be even more difficult. Under a given boost, should we say spinors come out half as fast as vectors?

 

[PeterDonis]

I can't figure out how to quote part of a post.

Highlight the part you want to quote, and click the "Reply" button that should pop up next to the highlight.

 

[PeterDonis]

which IS the wave function

Ah, now you're at least recognizing the wave function. No, the wave function is not "the spin", it is the wave function--or more precisely, as you note, the spin degrees of freedom within the wave function. The "spin" itself is an observable, or more precisely a set of observables. Observables and wave functions are different things.

Let's be careful not to confuse angles with infinitesimal generators and group parameters.

For the groups we are discussing, the group parameter is an angle--or more precisely, if we choose a specific axis of rotation, the group parameter that describes rotations about that axis is an angle. For SO(3) the angle runs from $0$ to $2 \pi$. For SU(2) it runs from $0$ to $4 \pi$. That angle is, physically, the angle through which a rotation is made about the given axis in order to induce the change in the wave function described by the rotation matrix.

I have no idea what you are getting after.

First, I was trying to get you to recognize that in QM, operators act on wave functions (or whatever you want to call the quantum state of the system). Rotations are operators.

Second, when you wrote down the complex exponentials $\exp(i \theta)$ for a vector and $\exp(i \theta / 2)$ for a spinor, you were writing down the change in a particular part of the wave function that is induced by the rotation operator for angle $\theta$ about a given axis. What part of the wave function is that? We are talking about complex exponentials with unit norm. What part of a complex number is that? Again, the fact that we use an interferometer to test for these changes is a clue: what aspect of a wave governs whether wave interference is constructive or destructive?

please stop asking me that question and please answer it.

I can't answer your question as you originally asked it because, as I've already said, it was based on misunderstandings. We're making progress in getting those misunderstandings resolved.

 

[PeterDonis]

BTW, a big deal is made out of the 1/2 factor in the exponential for an SO(3) rotation

No, for an SU(2) rotation. You can't do an SO(3) rotation on a spinor, and in an SO(3) rotation of a vector there is no 1/2 factor.

the equation for a Lorentz boost has a real exponential and also includes the factor of 1/2.

For a spinor, yes. But here the group is different; it's not SU(2).

Under a given boost, should we say spinors come out half as fast as vectors?

No. Again, the factor of 1/2 does not change the boost parameter, just as the factor of 1/2 for ordinary spatial rotations doesn't change the angle of rotation. The factor of 1/2 comes in in the change induced in the wave function by the same boost (or rotation) when applied to a spinor instead of a vector.

 

[PeterDonis]

It looks to me as if the complex exponentials are the same.

All the ones you wrote down in that post are for rotation operators, not wave functions. Again, they're not the same.

Correspondingly, the groups SO(3) and SU(2) are groups of rotation operators, not wave functions. The Hilbert spaces for the vector and spinor wave functions, or more precisely the spin degrees of freedom of those wave functions, are not SO(3) and SU(2). They are, heuristically, $\mathbb{C}^2$ for spinors and $\mathbb{C}^3$ for vectors, the Hilbert spaces of complex 2-vectors and 3-vectors with unit norm. The rotation operators act on those vectors. (Here I am using the term "vector" in a different sense from "spin-1 particle". Unfortunately I don't know of a different term to distinguish those two concepts.)

 

[joneall]

All the ones you wrote down in that post are for rotation operators, not wave functions. Again, they're not the same.

I don't believe I said they were wave functions, since I know perfectly well they represent rotations of spinors or vectors. If I expressed that badly, well, my bad.

Correspondingly, the groups SO(3) and SU(2) are groups of rotation operators, not wave functions. The Hilbert spaces for the vector and spinor wave functions, or more precisely the spin degrees of freedom of those wave functions, are not SO(3) and SU(2). They are, heuristically, $\mathbb{C}^2$ for spinors and $\mathbb{C}^3$ for vectors, the Hilbert spaces of complex 2-vectors and 3-vectors with unit norm. The rotation operators act on those vectors.

Why "heuristically"? Is it more complicated than that?

 

[PeterDonis]

Why "heuristically"?

Because I didn't want to get into the weeds of fermion vs. boson and the complications involved with a proper rigorous treatment of identical particle wave functions.

 

[joneall]

No. Again, the factor of 1/2 does not change the boost parameter, just as the factor of 1/2 for ordinary spatial rotations doesn't change the angle of rotation. The factor of 1/2 comes in in the change induced in the wave function by the same boost (or rotation) when applied to a spinor instead of a vector.

Aha, now we are getting somewhere. This was what my original question (remember that?) was about. Are you then saying that when I rotate a spinor by a given angle, the change in the spinor is what you would expect given half that angle? If so, then all this about the spinor only being rotated through half the angle is sort of ... well, not obvious. Correct?

Still, if I start with a spin-up (along, say, z) electron and rotate the whole system (i.e., the total wave function, a product of spatial, spinor and maybe other parts) thru $2\pi$ around, say, the x-axis, the spinor z-component will now be pointing down? That is the real question. I think it's even correctly formulated.

 

[joneall]

So what, finally, was my 'significant misunderstanding'? Confusing the rotation (operator) and the wave being rotated?

 

[PeterDonis]

when I rotate a spinor by a given angle, the change in the spinor is what you would expect given half that angle?

No. "What you would expect" basically assumes a vector, but you're not rotating a vector, you're rotating a spinor. Rotating a spinor by a given angle changes a particular property of the wave function by half as much as rotating a vector through the same angle.

if I start with a spin-up (along, say, z) electron and rotate the whole system (i.e., the total wave function, a product of spatial, spinor and maybe other parts) thru $2\pi$, the spinor z-component will now be pointing down?

No. You are still not thinking about what I have repeatedly asked you: what part of the wave function is affected?

In the particular case of a rotation by an angle $2 \pi$, a spinor wave function picks up a minus sign. You started with the wave function $\ket{z+}$ (z spin up). Rotating by $2 \pi$ thus gives you the wave function $- \ket{z+}$. What does that wave function represent, physically? Certainly not z spin down.

 

[joneall]

In the particular case of a rotation by an angle $2 \pi$, a spinor wave function picks up a minus sign. You started with the wave function $\ket{z+}$ (z spin up). Rotating by $2 \pi$ thus gives you the wave function $- \ket{z+}$. What does that wave function represent, physically? Certainly not z spin down.

Sorry, I just do not get it. It seems to me if wave function $\ket{z+}$ with z-spin up, with which you agree, then the rotation to get wave function $- \ket{z+}$ very definitely represents spin down. It seems to me that $- \ket{z+}$ actually IS $\ket{z-}$. I mean, we've turned something pointing up thru 180°. That must leave it pointing down. What else can it mean?

I guess this is where you ask me yet again what part of the wave function is affected? As far as I can understand, both the spinor and the spatial part are affected.

On the other hand, it's true that what is physically relevant is the modulus of the wave function squared, where the minus sign does not matter. Such a symmetry should lead to a conservation law (Noether). But spin is not conserved independently of orbital/spatial angular momentum. My confusion is increasing, not decreasing.

Please don't just ask another question or point out my incompetence again. I am tiring of that. Please just explain what is going on. Isn't that what Physics Forums is for, helping those who don't understand something to do so? I do thank you for all the attention to my questions.

 

[Mordred]

Simply giving the answers doesn't really help someone understand. It's usually better to step someone through a problem without simply giving the answers

 

[PeterDonis]

Sorry, I just do not get it.

What basic QM textbooks have you read? What I am telling you is basic QM.

It seems to me if wave function $\ket{z+}$ with z-spin up, with which you agree, then the rotation to get wave function $- \ket{z+}$ very definitely represents spin down.

It may seem that way to you, but it's wrong. If we have a state $\ket{\psi}$, then $c \ket{\psi}$, where $c$ is any complex number, represents the same state. Normally we require states to have unit norm, so we would restrict $c$ to be a complex exponential $e^{i \alpha}$, where $\alpha$ is a real number--i.e., a phase.

Again, this is basic QM, so if you are not familiar with it, I strongly suggest that you take the time to work through a QM textbook.

So we have the state $\ket{z+}$ that represents z spin up, and we have the state $\ket{z-}$ that represents z spin down. From the above it should be evident that we cannot multiply either one of these states by any complex number and get the other. Whatever operation might exist that transforms one of these states into the other, it can't be as simple as just multiplying by some complex number.

we've turned something pointing up thru 180°.

No, we haven't. We've turned it thru 360 degrees. We rotated by $2 \pi$, not $\pi$. Once again you are confusing the rotation angle with the effect on the wave function. They're not the same.

Also see further comments below.

Please just explain what is going on.

I am trying to, but you do not appear to have the necessary background knowledge of basic QM. You need to fix that if a discussion like this is going to be fruitful. For a QM textbook I would recommend Ballentine, although there are many and different people have different preferences, so others might recommend a different one.

That said, I'll expand briefly on what I said above for the case of spinors, i.e., spin-1/2 particles. (More details about what I am going to say can be found in Chapter 7 of Ballentine.) Remember that I said the Hilbert space for the spin degree of freedom of these particles is complex 2-vectors, i.e., column vectors like this:

$$
\begin{bmatrix}
a \\
b
\end{bmatrix}
$$

where $a$ and $b$ are complex numbers, and if we want to restrict to vectors of unit norm, as is common in QM, we require $|a|^2 + |b|^2 = 1$. Note that in writing down this column vector we have to choose a basis, which for spin degrees of freedom means choosing a rotation axis. We make the common choice of the $z$ axis, and this means that the two basis column vectors represent z spin up and z spin down, i.e., we have

$$
\ket{z+} = \begin{bmatrix}
1 \\
0
\end{bmatrix}
$$

$$
\ket{z-} = \begin{bmatrix}
0 \\
1
\end{bmatrix}
$$

This makes it obvious that you can't multiply one of these by any complex number and get the other.

The matrices that represent rotations of spinors are elements of SU(2) in its 2 x 2 matrix representation. You seem to be reasonably familiar with these, and their normal representation relies on the normal representation of the Pauli matrices, which is in the $z$ axis basis in which I wrote down the above column vectors. So you should be able to verify that rotating either of the above column vectors by $2 \pi$ about any axis just multiplies the vector by $-1$. You should also be able to verify that rotating either of the above column vectors by $\pi$ about either the $x$ or the $y$ axis turns it into minus the other.

So it is possible to turn $\ket{z+}$ into $\ket{z-}$ (but with a minus sign) by rotating through 180 degrees--but you have to pick a rotation axis that is orthogonal to the $z$ axis to do it. Rotating by 180 degrees about the $z$ axis itself doesn't change $\ket{z+}$ (or $\ket{z-}$) physically--although it does multiply it by $i$.

 

[Nugatory]

Sorry, I just do not get it. It seems to me if wave function $\ket{z+}$ with z-spin up, with which you agree, then the rotation to get wave function $- \ket{z+}$ very definitely represents spin down. It seems to me that $- \ket{z+}$ actually IS $\ket{z-}$. I mean, we've turned something pointing up thru 180°. That must leave it pointing down. What else can it mean?

If the wave function is $\alpha|z+\rangle+\beta|z-\rangle$, then a measurement along the $z$ axis will be up with probability $\alpha^2$ and down with probability $\beta^2$ - obviously $\alpha^2+\beta^2=1$. Google for “Born rule” for more information.

The post I’m replying to is all the $\beta=0$ case, with different values of $\alpha$. Note that both $\alpha$ and $\beta$ are in general complex.

 

[joneall]

Simply giving the answers doesn't really help someone understand. It's usually better to step someone through a problem without simply giving the answers

I agree, but we don't seem to be getting anywhere.

 

[PeterDonis]

we don't seem to be getting anywhere.

Have you read posts #34 and #35?

 

[joneall]

What basic QM textbooks have you read? What I am telling you is basic QM.

This may bore you. Short answer: I've studied and read QM a lot, including in graduate school in the 60s. You can skip the rest except the last paragraph, which is important.

I once (before 1972) was a particle physicist, post-docs at BNL and College de France. But I haven't worked in physics since and I am trying to catch up in retirement. In addition to whatever books we used in the 60s (I only remember Leighton's "Modern physics"), I have recently read Susskind's Theoretical minimum series (GR yet to be read), QM text books by Griffiths and Townsend, as well as QFT and symmetry books by Robinson, Schwichtenberg, and Lancaster and Blundell, with delvings into others. So it's not what I've read that is the problem.

I think one problem is that books about symmetry talk about Lagrangians and not much about wave functions. I can't recall ever having seen symmetry consideration applied to anything but Lagrangians or equations of motion, like Dirac's, but never to wave functions, which are always expressed as sums of plane waves, with matrix coefficients in the case of spinors. So when I am asked what part of the wave function is being changed, I don't know.

 

[joneall]

Have you read posts #34 and #35?

Reading now. Thanks.

 

[joneall]

The matrices that represent rotations of spinors are elements of SU(2) in its 2 x 2 matrix representation. You seem to be reasonably familiar with these, and their normal representation relies on the normal representation of the Pauli matrices, which is in the axis basis in which I wrote down the above column vectors. So you should be able to verify that rotating either of the above column vectors by $2\pi$ about any axis just multiplies the vector by -1. You should also be able to verify that rotating either of the above column vectors by about either the or the axis turns it into minus the other.

Yes! Using $$
R_x(\theta) =
e^{i\theta \frac{\sigma_3}{2}} =
\begin{pmatrix}
cos(\frac{\theta}{2}) & i sin(\frac{\theta}{2}) \\
i sin(\frac{\theta}{2}) & cos(\frac{\theta}{2})
\end{pmatrix}
$$
multiplies by -1 for $\theta=2\pi$. This also says that rotation thru any angle has the result equivalent to rotation by half that angle for a spinor. E.g., rotation by an angle $\pi$ just multiplies by $i$. Multiplication by a complex number is just a change in phase, not in eigenvalue.

And that answers my original question -- or what that question was intended to be. Rotation changes the eigenspinor but does not change its eigenvalue in the rotated state. (Did I get that right? I think so.) Similar result for orbital angular momentum, minus the -1 factor.

 

[joneall]

One more point. I quoted Sean Carroll, "A particle of spin s is invariant under a rotation by $2\pi/s$ radians." Would it not be more precise to say "particle wave function" instead of particle? The observable particle is the modulus square of the wave function and that is invariant under a rotation of $\pi$ or any other angle. Or am I picking nits now?

 

[Nugatory]

Would it not be more precise to say "particle wave function" instead of particle?

Yes, but natural language isn’t a precise instrument.

It would be even precise to say “phase of the state vector describing the quantum system under consideration”, but if Carroll were to be consistently that precise the resulting prose would be unreadable. So everyone using natural language will cut some corners in their phrasing and expect the context and the reader’s background understanding to close the gap.

 

[PeterDonis]

Rotation changes the eigenspinor but does not change its eigenvalue in the rotated state.

That's correct; the eigenvalues are fixed by the spin type (spin-1/2, spin-1, etc.).

 

[PeterDonis]

rotation by an angle $\pi$ just multiplies by $i$.

That depends on which axis you rotate about and what vector you are rotating. As I said before, if you take either of the spin-z eigenvectors and rotate it by $\pi$ about any axis perpendicular to the $z$ axis, you get minus the other spin-z eigenvector. Only if you rotate a spin-z eigenvector about the $z$ axis do you just multiply it by a phase.

Multiplication by a complex number is just a change in phase, not in eigenvalue.

Yes, that's correct. But not all rotations are just mutiplication by a complex number. See above.

 

[joneall]

That depends on which axis you rotate about and what vector you are rotating. As I said before, if you take either of the spin-z eigenvectors and rotate it by about any axis perpendicular to the axis, you get minus the other spin-z eigenvector. Only if you rotate a spin-z eigenvector about the axis do you just multiply it by a phase.

Rotation by $\pi$ about the x-axis multiplies by $i$. Isn't that a phase, $e^{i\pi/2}$?

 

[PeterDonis]

Rotation by $\pi$ about the x-axis

Rotation of what vector? A given rotation does not act the same way on all vectors.

 

[PeterDonis]

Using $$
R_x(\theta) =
e^{i\theta \frac{\sigma_3}{2}} =
\begin{pmatrix}
cos(\frac{\theta}{2}) & i sin(\frac{\theta}{2}) \\
i sin(\frac{\theta}{2}) & cos(\frac{\theta}{2})
\end{pmatrix}
$$

In the basis we have been using, $\sigma_3$, i.e., $\sigma_z$, is diagonal, so there should not be any off diagonal components in $\exp(i \theta \sigma_3 / 2)$. Also, $\sigma_3$ in this basis represents an infinitesimal rotation about the $z$ axis, not the $x$ axis.

 

[joneall]

Sorry, that should have been $\sigma_1$ not 3.
$$
R_1(\theta) =
e^{i\theta \frac{\sigma_1}{2}} = =
\begin{pmatrix}
cos(\frac{\theta}{2}) & i sin(\frac{\theta}{2}) \\
i sin(\frac{\theta}{2}) & cos(\frac{\theta}{2})
\end{pmatrix}
$$

 

[PeterDonis]

Sorry, that should have been $\sigma_1$ not 3.

Yes. So for rotation by $\pi$ about the $x$ axis, the matrix becomes:

$$
R_1(\pi) = \begin{bmatrix}
0 & i \\
i & 0
\end{bmatrix}
$$

How does this matrix act on the vector that represents spin-z up, i.e., on the following vector?

$$
\begin{bmatrix}
1 \\
0
\end{bmatrix}
$$

The result won't just be that vector multiplied by $i$.

 

[joneall]

See I bumbled several things there. Let's start over. A (z-up) spinor rotated about the x-axis undergoes a transformation
$$
R_1(\theta) =
e^{i\theta \frac{\sigma_1}{2}} =
\begin{pmatrix}
cos(\frac{\theta}{2}) & i sin(\frac{\theta}{2})
i sin(\frac{\theta}{2}) & cos(\frac{\theta}{2})
\end{pmatrix}
$$ which for θ=2π is just the negative of the identity. (I've gone thru all these with pencil and paper...) Ditto for
$$R_2(\theta) =e^{i\theta \frac{\sigma_2}{2}} =
\begin{pmatrix}
cos(\frac{\theta}{2}) & sin(\frac{\theta}{2}) \\
- sin(\frac{\theta}{2}) & cos(\frac{\theta}{2})
\end{pmatrix}$$
and
$$R_3(\theta) =
e^{i\theta \frac{\sigma_3}{2}} =
\begin{pmatrix}
cos(\frac{\theta}{2}) + i sin(\frac{\theta}{2}) & 0 \\
0& cos(\frac{\theta}{2}) - i sin(\frac{\theta}{2}))
\end{pmatrix}$$.
So a rotation through an angle of 2π along the x, y or z axes returns the negative of the spinor (10). And this is the difference between a 2-d spinor and a 3-d vector, their different properties under rotation.

OK now?

My original question was intended to be (but apparently was not), what is the connection between a spinor and a vector under rotation? It's what I said in the lst paragraph.

Can one deduce from this that a rotation on a system with a spinor and a vector will require conservation of the total angular momentum, spin plus orbital?