# I Is an electron's spin always in some definite direction?

#### Hans de Vries

Science Advisor
Gold Member
A spin 1/2 particle has total angular momentum √3/2. At most 1/2 points in any given direction, so its spin is most assuredly not in some definite direction.

Indeed, the socalled "spin-direction" is actually the axis of precession of the total spin, where the total spin is given by $s=s^x+s^y+s^z$ as shown in the figure below for a single spinor $\xi$

The three spin vectors $s^x$, $s^y$ and $s^z$ can be calculated with:

$$\begin{matrix} s^z~~=&\{&\xi^\dagger\sigma^x\xi~&\xi^\dagger\sigma^y\xi~&\xi^\dagger\sigma^z\xi&\} \\ s^x~~=&\{&\xi^\dagger \sigma^2\sigma^x\xi^*&\xi^\dagger \sigma^2\sigma^y\xi^*&\xi^\dagger \sigma^2\sigma^z\xi^*&\}\\ s^y~~=&\{&\xi^\dagger i\sigma^2\sigma^x\xi^*&\xi^\dagger i\sigma^2\sigma^y\xi^*&\xi^\dagger i\sigma^2\sigma^z\xi^*&\} \end{matrix}$$

The three spin vectors $s^x$, $s^y$ and $s^z$ can also be considered as defining the local reference frame of the spinor which can be oriented in any arbitrary direction. The expressions above are quite horrible and nontransparent. They can be written in a very elegant way using the Euler Parameter representation:

$$\begin{matrix} s^x~~=&\{&\xi^\intercal\sigma^x_x\,\xi~&\xi^\intercal\sigma^y_x\,\xi~&\xi^\intercal\sigma^z_x\,\xi&\} \\ s^y~~=&\{&\xi^\intercal\sigma^x_y\,\xi~&\xi^\intercal\sigma^y_y\,\xi~&\xi^\intercal\sigma^z_y\,\xi&\} \\ s^z~~=&\{&\xi^\intercal\sigma^x_z\,\xi~&\xi^\intercal\sigma^y_z\,\xi~&\xi^\intercal\sigma^z_z\,\xi&\} \end{matrix}$$

Spinors are much better and much easier understood in the older, real valued Euler Parameter representation. Many essential spinor properties are hopelessly lost and obscured in the complex Cayley-Klein representation that Pauli adopted and which is generally the only representation that physicist are familiar with. As a result (quantum) physicist have an incomplete understanding of spinors.

The (real) Euler parameter representation of a SPINOR is obtained as below in such a way that 3 of the 4 elements can be directly associated with $x$, $y$ and $z$. This already makes interpretation much simpler.

$$\left(\begin{array}{r} a +ib \\ c+id \end{array}\right) ~\longrightarrow~ \left(\begin{array}{r} a \\ b \\ c \\ d \end{array}\right) ~\longrightarrow~ \left(\begin{array}{r} u \\ x \\ y \\ z \end{array}\right)$$

The Euler parameter representation of a PAULI MATRIX and any other 2x2 complex matrix is obtained by replacing each complex element by a real valued 2x2 matrix.

$$a+ib ~\longrightarrow~ \left(\begin{array}{rr} a & -b \\ b & a \end{array}\right) ~~~~~~~~~~ * ~\longrightarrow~ \left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right)$$

Note that the complex conjugate operator becomes simply another real valued matrix further simplifying interpretations. Group theoretically: SU(2) combined with $*$ gives us SO(4):

The three $\mathsf{j}$ and the three $\mathsf{i}$ form the matrix representation of the left-acting quaternions and the right-acting quaternions respectively.

Note that the colored rectangles correspond with the three SO(3) generators for x, y and z. Simplifying things again because the $\mathsf{j_x,j_y,j_z}$ are in fact just the three spinor rotation generators!

The standard Pauli matrices are $\sigma^1$, $\sigma^2$ and $\sigma^3$. The indices x, y and z in the $(\mathsf{j_x,j_y,j_z})$, the $(\mathsf{i_x,i_y,i_z})$ and $(\sigma^x,\sigma^y,\sigma^z)$ are chosen so that they correspond with the SO(3) rotation matrices.

With this we now get a very elegant definition of the spinor vectors combining the above matrices into 9 different Pauli matrices using $\sigma^a_b=\mathsf{i_b\,j^a}$ where $\xi^\intercal$ is a row-vector, the transpose of $\xi$.

$$\begin{matrix} s^x~~=&\{&\xi^\intercal\sigma^x_x\,\xi~&\xi^\intercal\sigma^y_x\,\xi~&\xi^\intercal\sigma^z_x\,\xi&\} \\ s^y~~=&\{&\xi^\intercal\sigma^x_y\,\xi~&\xi^\intercal\sigma^y_y\,\xi~&\xi^\intercal\sigma^z_y\,\xi&\} \\ s^z~~=&\{&\xi^\intercal\sigma^x_z\,\xi~&\xi^\intercal\sigma^y_z\,\xi~&\xi^\intercal\sigma^z_z\,\xi&\} \end{matrix}$$
But it gets even better. All nine expressions above can be calculated with a single(!) matrix multiplication. The following very important formula to calculate the spinor's local coordinate system is completely absent from the (quantum) physics textbooks.

For an arbitrary spinor (u,x,y,z) we can calculate its local coordinate system with the following matrix multiplication:

$$\check{\xi}\,\hat{\xi} ~=~ (\xi\cdot\mathsf{i})(\mathsf{j}\cdot\xi)~=~\big|\xi \big|^2\!\!\left(\begin{smallmatrix}1~&0~&0~&0~\\0~&\mathsf{X}^{^x}&\mathsf{X}^{^y}&\mathsf{X}^{^z} \\0~&\mathsf{Y}^{^x}&\mathsf{Y}^{^y}&\mathsf{Y}^{^z} \\0~&\mathsf{Z}^{^x}&\mathsf{Z}^{^y}&\mathsf{Z}^{^z}\end{smallmatrix}\right)$$

Here $\mathsf{X,Y,Z}$ are the normalized versions of the three spin-vectors $s^x,s^y,s^z$. Together they also form a rotation matrix that defines the orientation of the spinor. The spinor operator matrices $\hat{\xi}$ and $\check{\xi}$ are written out as:

$$\begin{array}{c}\hat{\xi} ~~=~~ (\xi\cdot\mathsf{j})~~=~~(u\mathsf{j}_o+x\mathsf{j}_x+y\mathsf{j}_y+z\mathsf{j}_z) ~~=~~\left(\begin{smallmatrix} ~u & ~x & ~y & ~z \\\!\!-x & ~u &\!\!-z & ~y \\\!\!-y & ~z & ~u &\!\!-x \\\!\!-z &\!\!-y & ~x & ~u\end{smallmatrix}\right)\\ \\\check{\xi} ~~=~~(\xi\cdot\mathsf{i})~~=~~(u\mathsf{i}_o+x\mathsf{i}_x+y\mathsf{i}_y+z\mathsf{i}_z) ~~=~~\left(\begin{smallmatrix} ~u &\!\!-x &\!\!-y &\!\!-z \\ ~x & ~u &\!\!-z & ~y \\ ~y & ~z & ~u &\!\!-x \\ ~z &\!\!-y & ~x & ~u\end{smallmatrix}\right)\end{array}$$

Although the matrix multiplication is mine, the result is exactly identical to the Euler-Rodrigues formula

Now $(\xi^\intercal\sigma^y_x\,\xi)$ is the y-component of $s^x$. There is a very simple geometric explanation for this: The Pauli matrix $\sigma^y_x$ acting on $\xi$ performs a 180$^o$ rotation of the spinor around the y-axis and a -180$^o$ rotation of the spinor around its own $\mathsf{X}$-axis. We will explain this further on.

- If the $\mathsf{X}$-axis of the spinor is parallel to the y-axis then $~~~~~~~\xi^\intercal\sigma^y_x\,\xi~=~~~\xi^\intercal\xi~=~~~|\xi|^2$
- If the $\mathsf{X}$-axis of the spinor is anti-parallel to the y-axis then $\xi^\intercal\sigma^y_x\,\xi~=-\xi^\intercal\xi~=-|\xi|^2$
- If the $\mathsf{X}$-axis of the spinor is orthogonal to the y-axis then $~\xi^\intercal\sigma^y_x\,\xi~=~~0$

Now we come to the remarkable fact that the $(\mathsf{j_x,j_y,j_z})$ rotate the spinor around the coordinate axis while the $(\mathsf{i_x,i_y,i_z})$ rotate the spinor in its own local reference frame. There is a relation with the fact that one is left-acting and the other is right-acting. Consider the matrix shown below which corresponds with the (transposed) rotation matrix.

$$\left(\begin{matrix}1~&0~&0~&0~\\0~&\mathsf{X}^{^x}&\mathsf{X}^{^y}&\mathsf{X}^{^z} \\0~&\mathsf{Y}^{^x}&\mathsf{Y}^{^y}&\mathsf{Y}^{^z} \\0~&\mathsf{Z}^{^x}&\mathsf{Z}^{^y}&\mathsf{Z}^{^z}\end{matrix}\right)$$

Acting from one way rotates the columns of the matrix which corresponds with a rotation in the (world-) coordinate system $x,y,z$ while operating from the other direction rotates the rows which corresponds with a rotation in the spinors own reference frame based on the unit-vectors $\mathsf{X,Y,Z}$.

See the sections 1.2, 1.3 and 1.4 of chapter 1 and the whole of chapter 2 here:

To see how QED (1 boson field, 1 fermion generation) can be perfectly extended to the full Electroweak Theory (extra boson triplet and 3 fermion generations) using the Euler parameter representation see here: https://thephysicsquest.blogspot.com/

Last edited:

#### vanhees71

Science Advisor
Gold Member
Well, SO(4) is not SO(3) nor SU(2). The fundamental representation of SU(2) or its extensions via complexification to $\mathrm{SL}(2,\mathbb{C})$ are still the fundamental building blocks of the representations of the proper orthochronous Lorentz group, i.e., one of the fundamental corner stones of the Standard Model of HEP physics.

#### Hans de Vries

Science Advisor
Gold Member
Well, SO(4) is not SO(3) nor SU(2). The fundamental representation of SU(2) or its extensions via complexification to $\mathrm{SL}(2,\mathbb{C})$ are still the fundamental building blocks of the representations of the proper orthochronous Lorentz group, i.e., one of the fundamental corner stones of the Standard Model of HEP physics.
The representations of the proper orthochronous Lorentz group of course stay the same in the sense that they are replaced by equivalent real matrices.

Note that one can represent all generators of $\mathrm{SU}(N)$ or $\mathrm{SL}(2,\mathbb{C})$ by real matrix generators using the standard replacement of each complex matrix element by a real 2x2 matrix:

$$a+ib ~\longrightarrow~ \left(\begin{array}{rr} a & -b \\ b & a \end{array}\right)$$

So $\mathrm{SL}(2,\mathbb{C})$ is represented by the real matrices $\mathsf{j_x,j_y,j_z}$ and $\mathsf{i_x}$ as defined in my post. These matrices are all square roots of $-I$. The Pauli matrices are represented by $\sigma^a\rightarrow\mathsf{i_x\,j_a}$ while $i\rightarrow\mathsf{i_x}$ wherever you would typically write $i$ as a simplified notation of $i\sigma^o$.

But now it comes: The remaining matrices $\mathsf{i_y}$ and $\mathsf{i_z}$ are also used in standard QFT but not recognized as such. They arise when you start using the complex conjugate operator $*$. For instance in the single spinor Majorana equation which becomes:

$$\Big(~\mathsf{i_x}\partial_t + \mathsf{i_y}m~\Big) \xi~~=~~\Big(~\mathsf{j_x}\partial_x + \mathsf{j_y}\partial_y+\mathsf{j_z}\partial_z~\Big)\xi$$

Without calculation you can now directly see that taking the squares of both sides gives you the Klein Gordon equation:

$$\Big(~\partial^2_t - \partial^2_x -\partial^2_y - \partial^2_z+ m^2~\Big) \xi~~=~~ 0$$

because the $\mathsf{i_x,i_y,i_z}$ also form an anti commuting triplet. Much simpler isn't it?

#### vanhees71

Science Advisor
Gold Member
This is of course well known in the standard representation of $\mathrm{sl}(2,\mathbb{C})$ or $\mathrm{SL}(2,\mathbb{C})$: The conjugate complex of one of the two-dimensional representations is not equivalent to it. Thus you have two inequivalent irreducible two-dimensional representations, usually labelled as (1/2,0) and (0,1/2). They are known as Weyl spinors. Majorana fermions are represented by quantized Weyl-spinor fields.

The Dirac representation is $(1/2,0) \oplus (0,1/2)$, i.e., reducible as a representation of the proper orthochronous Lorentz group but irrducible for the orthochronous Lorentz group, including space reflections. In QED and QCD you need Dirac fermions since both the electromagnetic and the strong interaction are P conserving and thus you need to represent space reflections.

### Want to reply to this thread?

"Is an electron's spin always in some definite direction?"

### Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving