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

[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?

 

[joneall]

Is there a way to preview a post, or do I have to post it and then do an edit?

 

[PeterDonis]

Is there a way to preview a post

Click the Preview icon at the upper right corner of the post window.

 

[PeterDonis]

a rotation through an angle of 2π along the x, y or z axes returns the negative of the spinor (10).

Yes, this is correct.

And this is the difference between a 2-d spinor and a 3-d vector, their different properties under rotation.

As a general statement, this is correct, yes: the properties of spinors and vectors under rotation are different. But of course that statement by itself is not very informative.

what is the connection between a spinor and a vector under rotation?

And the answer to that question is, there isn't one. That is, you can't start from the properties of a vector under rotation and apply some simple rule to get the properties of a spinor under rotation, or vice versa. You have to go through a separate analysis of each case, as is done, for example, in Chapter 7 of Ballentine that I referred to before.

It's what I said in the lst paragraph.

No, it isn't. It may have been what you meant to say, but it's not what you said.

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?

Not just from the spin properties alone, no. You have to work out the effects of the total angular momentum operator.