Undergrad Pauli spin matrices under inversion and 180 degree rotation?

[zmth]

TL;DR

not applicable

How do the Pauli spin matrices transform under an inversion ? I think I mean to say the 3 dimensional improper rotation which is just in 3 dimensional matrix notation minus the identity - so exactly how are the 2 dimensional Pauli spin matrices changed. And under a 180 rotation do the 'y' and 'z' matrices just get multiplied by sqrt(-1) ? And if so how do we determine when they will be multiplied by minus sqrt(-1) = -i or plus i ? Would a counterclockwise rotation of 180 degrees mean to multiply the y and z components by +i and the x component unchanged - actually I mean to say the three 2 dimensional matrices regarded as an operator - guess I mean to say in the same sense that I have read under a 360 degree rotation they are multiplied by -1 no matter what the axis of rotation(why the axis of rotation does not matter is a bit confusing). Would they be considered as vector or pseudo(axial) vector operators or neither one ?

 

[vanhees71]

Since spin is an angular momentum, it's a pseudovector, i.e., it doesn't change under space reflection.

A rotation in the 1/2-representation is given by
$$\hat{R}_{\vec{n}}(\varphi)=\exp(-\mathrm{i} \vec{n} \cdot \hat{\vec{\sigma}} \varphi/2)=\hat{1} \cos(\varphi/2)-\mathrm{i} \vec{n} \cdot \hat{\vec{\sigma}} \sin(\varphi/2).$$
For $\varphi=\pi$ you thus get $\hat{R}_{\vec{n}}(\pi)=-\mathrm{i} \vec{n} \cdot \hat{\vec{\sigma}}$ and for $\varphi=2 \pi$ you have $\hat{R}_{\vec{n}}(\varphi)=-\hat{1}$.

A rotation acts on the spin matrices as to any other vector operator
$$\hat{R} \hat{\sigma}_j \hat{R}^{\dagger} =R_{jk} \sigma_k,$$
where $R_{jk}$ is the usual 3D orthogonal rotation matrix,
$$(R_{jk})=\begin{pmatrix} \cos \varphi & \sin \varphi & 0 \\
-\sin \varphi & \cos \varphi & 0 \\
0 & 0 & 1 \end{pmatrix}$$
For $\varphi=\pi$ we thus have $(R_{jk})=\mathrm{diag}(-1,-1,1)$ and thus
$$\hat{R} \hat{\sigma}_1 \hat{R}^{\dagger}=-\hat{\sigma}_1, \quad \hat{R} \hat{\sigma}_2 \hat{R}^{\dagger}=-\hat{\sigma}_2, \quad \hat{R} \hat{\sigma}_3 \hat{R}^{\dagger}=\hat{\sigma}_3.$$

 

[zmth]

I see by the matrix multiplication how you get that but here is what seems to disagree. Under a rotation about the x-axis a vector along the x-axis does not change ? Using the 3 dimensional R matrices operating on the two dimensional spin, denoting them here as ' vector' $\sigma$ matrices, and your $R_{jk}$ matrix looks like a rotation about the z - axis. HOw do you explain it as being a 180 degree rotation about the x-axis ? Okay as a vector operator you could use a 3 dimensional real matrix, NOT irreducible, representation of O3, (or call it SO3 ? ), as in the context of a 3 dimensional real orthogonal group a 180 degree rotation about the x-axis , $$\begin{array}{ccc}1&0&0\\0&-1&0\\0&0&-1 \end{array}.$$ Maybe disagreement is simply I neglected or forgot to state in my original question a rotation specifically about the X-axis and NOT what may apparently be taken about the z axis and further mistake of including phrase "... will be multiplied by minus sqrt(-1) = -i or plus i ? " which is not actually directly applicable at that point ?
In two dimensional spin(operator) form I get the transformation of $\sigma_{1,2,3}$ under 180 degree counterclockwise rotation about x-axis:
\begin{eqnarray*}e^{-i\frac \pi 2\cdot \sigma_1} \sigma_1 ({e^{-i\frac \pi 2\cdot \sigma_1}})^{(-1)}
= \begin{array}{cc}0&-i\\-i&0 \end{array}\left [ \begin{array}{cc}0&1\\1&0 \end{array}\right ] \begin{array}{cc}0&i\\i&0 \end{array}=\left [ \begin{array}{cc}0&1\\1&0 \end{array}\right ]\\
e^{-i\frac \pi 2\cdot \sigma_1} \sigma_2 ({e^{-i\frac \pi 2\cdot \sigma_1}})^{(-1)}
= \begin{array}{cc}0&-i\\-i&0 \end{array}\left [ \begin{array}{cc}0&-i\\i&0 \end{array}\right ] \begin{array}{cc}0&i\\i&0 \end{array}=\left [ \begin{array}{cc}0&i\\-i&0 \end{array}\right ]\\
e^{-i\frac \pi 2\cdot \sigma_1} \sigma_3 ({e^{-i\frac \pi 2\cdot \sigma_1}})^{(-1)}
= \begin{array}{cc}0&-i\\-i&0 \end{array}\left [ \begin{array}{cc}1&0\\0&-1 \end{array}\right ] \begin{array}{cc}0&i\\i&0 \end{array}=\left [ \begin{array}{cc}-1&0\\0&1 \end{array}\right ]\end{eqnarray*}
In this case for some reason the two ended up giving equivalently the same as in your answer also- perhaps that turns out to be true in general for some reason ?. Anyway it seems to me that your expressions would be equivalent to a 180 degree rotation about the z axis ?

I guess there is no answer to what is an inversion transformation effect on the Pauli spin matrices - or it has no analog as far as they are concerned ? OR is it -i times minus the identity in a two dimensional matrix for the action of the inversion on Pauli spin matrices. Specifically to the left as $\left [\begin{array}{cc} i&0\\0&i \end{array} \right ]$ and on the right as exactly the opposite sign matrix(or do i have the signs reversed ?) because the 'adjoint', as I assume your symbol $\dagger$ means, conjugate transpose which for a unitary matrix is the inverse? And since both right post mult. and to the left pre-multiplying matrices are multiple of unit matrix in this case they cancel and all Pauli spin matrices remain unchanged ? In the case of the 3 dimensional real matrix 'vector' transform operator form could one assume that sine of pi is zero and cosine pi is -1 so the 3 dimensional real case is -1 times minus the identity = identity so also NO change for an inversion and agrees with the two dimensional answer of NO change. Is this a correct interpretation of an inversion acting on spin? And only use linear and unitary operations and not antilinear nor antiunitary operators ?

Yes I would also think that a reflection (only) about any plane would leave any pertinent axial or pseudo vector sign unchanged. Anyway an inversion by itself on the two vectors would result in no change because the 2 minus signs in the (cross or vector) product of the 2 vectors cancel? Anyway my main question is this: Aside from a real constant it is stated in Jansen and Boon "Finite Groups" , 1967, end of page 262 that the spin-orbit coupling term in the Hamiltonian acting here on the wave function $\Psi$ in age old basic(NOt using 2nd quantization etc...) quantum theory is $$ -i\sigma\cdot(\nabla V(r)\times \nabla)\Psi .$$
Next page 263 about middle it is stated regarding that spin-orbit part of H: "... but the axial vector $\nabla V(r)\times \nabla$ becomes det $D(\alpha^{-1})\alpha^{-1} (\nabla V(r)\times \nabla).$ Here $D(\alpha^{-1})$ is the orthogonal 3 dimensional matrix for point-operation $\alpha^{-1}$, and
det $D(\alpha^{-1})$ is 1 or -1 according as $\alpha^{-1}$ is proper or improper." Putting $\alpha^{-1}$ as 3 dimension real matrix as above. The precense of det $D(\alpha^{-1})$ seems a bit strange but could one state that it is used as an 'artifice' multiplier to cancel the minus sign which can be factored out of an improper, det $D(\alpha)=-1$ improper rotation matrix? And that factoring can be taken equivalently as minus the identity matrix times a real proper , det=1, rotation matrix. So for the axial vector we need another -1 factor in the improper case to cancel so that end result expressed in the chosen form, that is NOT being otherwise expressed in the form of the product of two matrices(while the direct axial vector expression IS a product) will have the correct sign ? Again we don't use any antiunitary operations ?

 

[vanhees71]

Of course, the calculation for rotations around the 1-axis is completely analogous to that around the 3-axis. For a 180-degree rotation the 1-component is of course invariant and the 2- and 3-components change sign.

There are no antiunitary group representations involved, because rotations are continuously connected to the identity operator, and thus are represented by unitary.

Also space reflections are represented by unitary transformations. Only time-reversal is represented by antiunitary representations.

I cannot make sense of your other formulae. Which book are you citing from?

 

[zmth]

Of course, the calculation for rotations around the 1-axis is completely analogous to that around the 3-axis. For a 180-degree rotation the 1-component is of course invariant and the 2- and 3-components change sign.

There are no antiunitary group representations involved, because rotations are continuously connected to the identity operator, and thus are represented by unitary.

Also space reflections are represented by unitary transformations. Only time-reversal is represented by antiunitary representations.

I cannot make sense of your other formulae. Which book are you citing from?

Jansen and Boon "Finite groups" 1967 in part esp that part in quotes but some of the other expressions are my own interpretations and questions to ask if others agree eg as to the expression for inversion mainly.

 

[zmth]

Jansen and Boon "Finite groups" 1967 in part esp that part in quotes but some of the other expressions are my own interpretations and questions to ask if others agree eg as to the expression for inversion mainly.

spin orbit interaction term is −iσ⋅(∇V(r)×∇)Ψ.