# LOG#098. Group theory(XVIII).

This and my next blog post are going to be the final posts in this group theory series. I will be covering some applications of group theory in Quantum Mechanics. More advanced applications of group theory, extra group theory stuff will be added in another series in the near future.

Angular momentum in Quantum Mechanics

Take a triplet of linear operators, $J=(J_x,J_y,J_z)$. We say that these operators are angular momentum operators if they are “observable” or observable operators (i.e.,they are hermitian operators) and if they satisfy

$\boxed{\displaystyle{\left[J_i,J_j\right]=i\hbar\sum_k \varepsilon_{ijk}J_k}}$

that is

$\left[J_x,J_y\right]=i\hbar J_z$

$\left[J_y,J_z\right]=i\hbar J_x$

$\left[J_z,J_x\right]=i\hbar J_y$

The presence of an imaginary factor $i$ makes compatible hermiticity and commutators for angular momentum. Note that if we choose antihermitian generators, the imaginary unit is absorbed in the above commutators.

We can determine the full angular momentum spectrum and many useful relations with only the above commutators, and that is why those relations are very important. Some interesting properties of angular momentum can be listed here:

1) If $J_1,J_2$ are two angular momentum operators, and they sastisfy the above commutators, and if in addition to it, we also have that $\left[J_1,J_2\right]=0$, then $J_3=J_1+J_2$ also satisfies the angular momentum commutators. That is, two independen angular momentum operators, if they commute to each other, imply that their sum also satisfy the angular momentum commutators.

2) In general, for any arbitrary and unitary vector $\vec{n}=(n_x,n_y,n_z)$, we define the angular momentum in the direction of such a vector as

$J_{\vec{n}}=n\cdot J=n_xJ_x+n_yJ_y+n_zJ_z$

and for any 3 unitary and arbitrary vectos $\vec{u},\vec{v},\vec{w}$ such as $\vec{w}=\vec{u}\times\vec{v}$, we have

$\left[J_{\vec{u}},J_{\vec{u}}\right]=i\hbar J_{\vec{w}}$

3) To every two vectors $\vec{a},\vec{b}$ we also have

$\left[\vec{a}\cdot\vec{J},\vec{b}\cdot\vec{J}\right]=i\hbar (\vec{a}\times \vec{b})\cdot \vec{J}$

4) We define the so-called “ladder operators” $J_+,J_-$ as follows. Take the angular momentum operator $J$ and write

$J_+=J_x+iJ_y$

$J_-=J_x-iJ_y$

These operators are NOT hermitian, i.e, ladder operators are non-hermitian operators and they satisfy

$J_+^+=J_-$

$J_-^+=J_+$

5) Ladder operators verify some interesting commutators:

$\left[J_x,J_+\right]=J_+$

$\left[J_x,J_-\right]=-J_-$

$\left[J_+,J_-\right]=2J_z$

6) Commutators for the square of the angular momentum operator $J^2=J_x^2+J_y^2+J_z^2$

$\left[J^2,J_k\right]=0,\forall k=x,y,z$

$\left[J^2,J_+\right]=\left[J^2,J_-\right]=0$

$J^2=\dfrac{1}{2}\left(J_+J_-+J_-J_+\right)+J_z^2$

$J_-J_+=J^2-J_z(J_z+I)$

$J_+J_-=J^2-J_z(J_z-I)$

8) Positivity: the operators $J_i^2,J_\pm,J_{+}J_{.},J_-J_+,J^2$ are indefinite positive operators. It means that all their respective eigenvalues are positive numbers or zero. The proof is very simple

$\langle \Psi \vert J_i^2\vert \Psi\rangle =\langle \Psi\vert J_iJ_i\vert\Psi\rangle =\langle \Psi\vert J^+_iJ_i\vert\Psi\rangle =\parallel J_i\vert\Psi\rangle\parallel\geq 0$

In fact this also implies the positivity of $J^2$. For the remaining operators, it is trivial to derive that

$\langle \Psi\vert J_-J_+\vert\Psi\rangle\geq 0$

$\langle \Psi\vert J_+J_-\vert \Psi\rangle\geq 0$

since

$\langle\Psi\vert J_-J_+\vert\Psi\rangle =\langle\Psi\vert J_+^+J_+\vert\Psi\rangle=\parallel J_+\vert\Psi\rangle\parallel\geq 0$

$\langle\Psi\vert J_+J_-\vert\Psi\rangle =\langle\Psi\vert J_-^+J_-\vert\Psi\rangle =\parallel J_-\vert\Psi\rangle\parallel\geq 0$

The general spectrum of the operators $J^2, J_z$ can be calculated in a completely general way. We have to search for general eigenvalues

$J^2\vert\lambda,\mu\rangle=\lambda\vert\lambda,\mu\rangle$

$J_z\vert\lambda,\mu\rangle=\mu\vert\lambda,\mu\rangle$

The general procedure is carried out in several well-defined steps:

1st. Taking into account the positivity of the above operators $J^2,J_i^2,J_+J_-,J_-J_+$, it means that there is some interesting options

A) $J^2$ is definite positive, i.e., $\lambda \geq 0$. Then, we can write for all practical purposes

$\lambda=j(j+1)\hbar^2$ with $j\geq 0$

Specifically, we define how the operators $J^2$  and $J_z$ act onto the states, labeled by two parameters $j,m$ and $\vert j,m\rangle$ in the following way

$J^2\vert j,m\rangle =j(j+1)\hbar^2\vert j,m\rangle$

$J_z\vert j,m\rangle =m\hbar \vert j,m\rangle$

B) $J_+,J_-,J_+J_-$ are positive, and we also have

$J_-J_+\vert j,m\rangle =\left(J^2-J_z(J_z+I)\right)\vert j,m\rangle =(j-m)(j+m+1)\hbar^2\vert j,m\rangle$

$J_+,J_-\vert j,m\rangle =\left(J^2-J_z(J_z-I)\right)\vert j,m\rangle =(j+m)(j-m+1)\hbar^2\vert j,m\rangle$

That means that the following quantities are positive

$(j-m)(j+m+1)\geq 0 \leftrightarrow \begin{cases}j\geq m;\;\; j\geq -m-1\\ j\leq m;\;\; j\leq -m-1\end{cases}$

$(j+m)(j-m+1)\geq 0 \leftrightarrow \begin{cases}j\geq -m;\;\; j\geq m-1\\ j\leq -m;\;\; j\leq m-1\end{cases}$

Therefore, we have deduced that

(1) $\boxed{-j\leq m\leq j \leftrightarrow \vert m\vert \leq j}$ $\forall j,m$

(2) $\boxed{m=\pm j\leftrightarrow \parallel J_\pm \vert j,m\rangle \parallel^2=0}$

2nd. We realize that

(1) $J_+\vert j,m\rangle$ is an eigenstate of $J^2$ and eigenvalue $j(j+1)$. Check:

$J^2\left(J_+\vert j,m\rangle \right)=J_+\left(J^2\vert j,m\rangle\right)=j(j+1)\hbar^2\left(J_+\vert j,m\rangle\right)$

(2) $J_+\vert j,m\rangle$ is an eigentstate of $J_z$ and eigenvalue $(m+1)$. Check (using $\left[J_z,J_+\right]=J_+$:

$J_z\left(J_+\vert j,m\rangle \right)=J_+(J_z+I)\vert j,m\rangle =(m+1)\hbar \left(J_+\vert j,m\rangle\right)$

(3) $J_-\vert j,m\rangle$ is an eigenstate of $J^2$ with eigenvalue $j(j+1)$. Check:

$J^2\left(J_-\vert j,m\rangle \right)=J_-\left(J^2\vert j,m\rangle\right)=j(j+1)\hbar^2\left(J_-\vert j,m\rangle\right)$

(4) $J_-\vert j,m\rangle$ is an eigenvector of $J_z$ and $(m-1)$ is its eigenvalue. Check:

$J_z\left(J_-\vert j,m\rangle \right)=J_-(J_z-I)\vert j,m\rangle =(m-1)\hbar \left(J_-\vert j,m\rangle\right)$

Therefore, we have deduced the following conditions:

1) if $m\neq j$, equivalently if $m\neq -j$, then the eigenstates $J_+\vert j,m\rangle$, equivalently $J_-\vert j,m\rangle$, are the eigenstates of $J^2,J_z$. The same situation happens if we have vectors $J_+^p\vert j,m\rangle$ or $J_-^q\vert j,m\rangle$ for any $p,q$ (positive integer numbers). Thus, the sucessive action of any of these two operators increases (decreases) the eigenvalue $m$ in one unit.

2) If $m=j$ or respectively if $m\neq -j$, the vectors $J_+\vert j,m\rangle$, respectively $J_-\vert j,m\rangle$ are null vectors:

$\exists p\in \mathbb{Z}/\left\{J_+^p\vert j,m\rangle\neq 0,J_+^{p+1}\vert j,m\rangle=0\right\}$, $m+p=j$.

$\exists q\in \mathbb{Z}/\left\{J_-^q\vert j,m\rangle\neq 0,J_-^{q+1}\vert j,m\rangle=0\right\}$, $m-q=-j$.

If we begin by certain number $m$, we can build a series of eigenstates/eigenvectors and their respective eigenvalues

$m-1,m-2,\ldots,m-q=-j$

$m+1,m+2,\ldots,m+q=j$

So, then

$m+p= j$

$m-q=-j$

$2m=q-p$

$2j=p+q$

And thus, since $p,q\in\mathbb{Z}$, then $j=k/2,k\in \mathbb{Z}$. The number $j$ can be integer or half-integer. The eigenvalues $m$ have the same character but they can be only separated by one unit.

In summary:

(1) The only possible eigenvalues for $J^2$ are $j(j+1)$ with $j$ integer or half-integer.

(2) The only possible eigenvalues for $J_z$ are integer numbers or half-integer numbers, i.e.,

$\boxed{m=0,\pm \dfrac{1}{2},\pm 1,\pm\dfrac{3}{2},\pm 2,\ldots,\pm \infty}$

(3) If $\vert j,m\rangle$ is an eigenvector for $J^2$ and $J_z$, then

$J^2\vert j,m\rangle=j(j+1)\hbar^2\vert j,m\rangle$ $j=0,1,2,\ldots,$

$J_z\vert j,m\rangle=m\hbar\vert j,m\rangle$ $-j\leq m\leq j$

We have seen that, given an state $\vert j,m\rangle$, we can build a “complete set of eigenvectors” by sucessive application of ladder operators $J_\pm$! That is why ladder operators are so useful:

$J_+\vert j,m\rangle, J_+^2\vert j,m\rangle, \ldots, J_-\vert j,m\rangle, J_-^2\vert j,m\rangle,\ldots$

This list is a set of $(2j+1)$ eigenvectors, all of them with the same quantum number $j$ and different $m$. The relative phase of $J^p_\pm\vert j,m\rangle$ is not determined. Writing

$J_\pm\vert j,m\rangle =c_m\vert j,m+1\rangle$

from the previous calculations we easily get that

$\parallel J_+\vert j,m\rangle\parallel^2=(j-m)(j+m+1) \hbar^2\langle j,m\vert j,m\rangle$

$\vert c_m\vert^2=(j-m)(j+m+1)\hbar^2=j(j+1)\hbar^2-m(m+1)\hbar^2$

$\parallel J_-\vert j,m\rangle \parallel^2=(j+m)(j-m+1)\hbar^2\langle j,m\vert j,m\rangle$

$\vert c_m\vert^2=(j+m)(j-m+1)\hbar^2=j(j+1)\hbar^2-m(m-1)\hbar^2$

The modulus of $c_m$ is determined but its phase IS not. Remember that a complex phase is arbitrary and we can choose it arbitrarily. The usual convention is to define $c_m$ real and positive, so

$J_+\vert j,m\rangle =\hbar \sqrt{j(j+1)-m(m+1)}\vert j,m+1\rangle$

$J_-\vert j,m\rangle =\hbar \sqrt{j(j+1)-m(m-1)}\vert j,m-1\rangle$

Invariant subspaces of angular momentum

If we addopt a concrete convention, the complete set of proper states/eigentates is:

$B=\left\{ \vert j,-j\rangle ,\vert j,-j+1\rangle,\ldots,\vert j,0\rangle,\ldots,\vert j,j-1\rangle,\vert j,j\rangle\right\}$

This set of eigenstates of angular momentum will be denoted by $E(j)$, the proper invariant subspace of angular momentum operators $J^2,J_z$, with corresponding eigenvalues $j(j+1)$.

The previously studied (above) features tell  us that this invariant subspace $E(j)$ is:

a) Invariant with respect to the application of $J^2,J_z$, the operators $J_x,J_y$, and every function of them.

b) $E(j)$ is an irreducible subspace in the sense we have studied in this thread: it has no invariant subspace itself!

The so-called matrix elements for angular momentum in these invariant subspaces can be obtained using the ladder opertors. We have

(1) $\langle j,m\vert J^2\vert j',m'\rangle = j(j+1)\hbar^2 \delta_{jj'}\delta_{mm'}$

(2) $\langle j,m \vert J_z\vert j',m'\rangle =m\hbar \delta_{jj'}\delta_{mm'}$

(3) $\langle j,m\vert J_+\vert j',m'\rangle =\hbar \sqrt{j(j+1)-m'(m'+1)}\delta_{jj'}\delta_{m,m'+1}$

(4) $\langle j,m\vert J_-\vert j',m'\rangle =\hbar \sqrt{j(j+1)-m'(m'-1)}\delta_{jj'}\delta_{m,m'-1}$

Example(I): Spin 0. (Scalar bosons)

If $E(0)=\mbox{Span}\left\{ \vert 0\rangle\right\}$

This case is trivial. There are no matrices for angular momentum. Well, there are…But they are $1\times 1$ and they are all equal to cero. We have

$J^2\vert 0\rangle =0\hbar^2=0(0+1)\hbar^2\cdot 1=0$

$J_x=J_y=J_z=J_+=J_-=0$

Example(II): Spin 1/2. (Spinor fields)

Now, we have $E(1/2)=\mbox{Span}\left\{\vert 1/2,-1/2\rangle,\vert 1/2,1/2\rangle\right\}$

The angular momentum operators are given by multiples of the so-called Pauli matrices. In fact,

$J^2=\dfrac{3}{4}\hbar^2\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=\dfrac{3\hbar^2}{4}I=\dfrac{3\hbar^2}{4}\sigma_0$

$J_x=\dfrac{\hbar}{2}\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}=\dfrac{\hbar}{2}\sigma_x$

$J_y=\dfrac{\hbar}{2}\begin{pmatrix} 0 & -i\\ i & 0\end{pmatrix}=\dfrac{\hbar}{2}\sigma_y$

$J_z=\dfrac{\hbar}{2}\begin{pmatrix} 1 & 0\\ 0 & -1\end{pmatrix}=\dfrac{\hbar}{2}\sigma_z$

and then $J_k=\dfrac{\hbar}{2}S_k=\dfrac{\hbar}{2}\sigma_k$ and $J^2=\dfrac{3}{4}\hbar^2I=\dfrac{3}{4}\hbar^2 \sigma_0$.

The Pauli matrices have some beautiful properties, like

i) $\sigma_x^2=\sigma_y^2=\sigma_z^2=1$ The eigenvalues of these matrices are $\pm 1$.

ii) $\sigma_x\sigma_y=i\sigma_x$, $\sigma_y\sigma_z=i\sigma_x$, $\sigma_z\sigma_x=i\sigma_y$. This property is related to the fact that the Pauli matrices anticommute.

iii) $\sigma_j\sigma_k=i\varepsilon_{jkl}\sigma_l+\delta_{jk}I$

iv) With the “unit” vector $\vec{n}=\left(\sin\theta\cos\psi,\sin\theta\sin\psi,\cos\theta\right)$, we get

$\vec{n}\cdot \vec{S}=\begin{pmatrix} \cos\theta & e^{-i\psi}\sin\theta\\ e^{i\psi}\sin\theta & -\cos\theta\end{pmatrix}$

This matrix has only two eigenvalues $\pm 1$ for every value of the parameters $\theta,\psi$. In fact the matrix $\sigma_z+i\sigma_x$ has only an eigenvalue equal to zero, twice, and its eigenvector is:

$e_1=\dfrac{1}{\sqrt{2}}\begin{pmatrix} -i\\ 1\end{pmatrix}$

And $\sigma_z-i\sigma_x$ has only an eigenvalue equal to zero twice and eigenvector

$e_2=\dfrac{1}{\sqrt{2}}\begin{pmatrix} i\\ 1\end{pmatrix}$

Example(III): Spin 1. (Bosonic vector fields)

In this case, we get $E(1)=\mbox{Span}\left\{\vert 1,-1\rangle,\vert 1,0\rangle,\vert 1,1\rangle\right\}$

The restriction to this subspace of the angular momentum operator gives us the following matrices:

$J^2=2\hbar^2\begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\end{pmatrix}=2\hbar^2I_{3\times 3}$

$J_x=\dfrac{\hbar}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0\end{pmatrix}$

$J_y=\dfrac{\hbar}{\sqrt{2}}\begin{pmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0\end{pmatrix}$

$J_z=\hbar\begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1\end{pmatrix}$

and where

A) $J^2_x,J^2_y,J^2_z$ are commutative matrices.

B) $J_x^2+J_y^2+J_z^2=J^2=2\hbar^2I=1(1+1)\hbar^2$

C) $J_x^2+J_y^2$ is a diagonarl matrix.

D) $J_3+iJ_1=\hbar\begin{pmatrix} 1 & i/\sqrt{2} & 0\\ i/\sqrt{2} & 0 & i/\sqrt{2}\\ 0 & i/\sqrt{2} & -1\end{pmatrix}$ is a nilpotent matrix since $(J_3+iJ_1)^2=0_{3\times 3}$ with 3 equal null eigenvalues and one single eigenvector

$e_1=\dfrac{1}{2}\begin{pmatrix}-1\\ -i/\sqrt{2}\\ 1\end{pmatrix}$

Example(IV): Spin 3/2. (Vector spinor fields)

In this case, we have $E(3/2)=\mbox{Span}\left\{\vert 3/2,-3/2\rangle,\vert 3/2,-1/2\rangle,\vert 3/2,1/2\rangle,\vert 3/2,3/2\rangle\right\}$

The spin-3/2 matrices can be obtained easily too. They are

$J_x=\dfrac{\hbar}{2}\begin{pmatrix}0 & \sqrt{3} & 0 & 0\\ \sqrt{3} & 0 & 2 & 0\\ 0 & 2 & 0 & \sqrt{3}\\ 0 & 0 & \sqrt{3} & 0\end{pmatrix}$

$J_y=\hbar\begin{pmatrix}0 & -i\sqrt{3} & 0 & 0\\ i\sqrt{3} & 0 & -2i & 0\\ 0 & 2i & 0 & -i\sqrt{3}\\ 0 & 0 & i\sqrt{3} & 0\end{pmatrix}$

$J_z=\hbar\begin{pmatrix}3/2 & 0 & 0 & 0\\ 0 & 1/2 & 0 & 0\\ 0 & 0 & -1/2 & 0\\ 0 & 0 & 0 & -3/2\end{pmatrix}$

$J^2=J_x^2+J_y^2+J_z^2=\dfrac{15}{4}\hbar^2I_{4\times 4}=\dfrac{3(3+2)\hbar^2}{4}\begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\end{pmatrix}$

The matrix

$Z=J_z+iJ_x=\hbar\begin{pmatrix}3/2 & i\sqrt{3}/2 & 0 & 0\\ i\sqrt{3}/2 & 1/2 & i & 0\\ 0 & i & -1/2 & i\sqrt{3}/2\\ 0 & 0 & i\sqrt{3}/2 & -3/2\end{pmatrix}$

is nonnormal since $\left[Z,Z^+\right]\neq 0$ and it is nilpotent in the sense that $Z^4=(J_z+iJ_x)^4=0_{4\times 4}$ and its eigenvalues is zero four times. The only eigenvector is the vector

$e_1=\dfrac{1}{\sqrt{8}}\begin{pmatrix}i\\ -\sqrt{3}\\ -i\sqrt{3}\\ 1\end{pmatrix}$

This vector is “interesting” in the sense that it is “entangled” and it can not be rewritten as a tensor product of two $\mathbb{C}^2$. There is nice measure of entanglement, called tangle, that it shows to be nonzero for this state.

Example(V): Spin 2. (Bosonic tensor field with two indices)

In this case, the invariant subspace is formed by the vectors $E(2)=\mbox{Span}\left\{\vert 2,-2\rangle,\vert 2,-1\rangle, \vert 2,0\rangle,\vert 2,1\rangle,\vert 2,2\rangle\right\}$

For the spin-2 particle, the spin matrices are given by the following $5\times 5$ matrices

$J_x=\hbar\begin{pmatrix}0 & 1 & 0 & 0 & 0\\ 1 & 0 & \sqrt{6}/2 & 0 & 0\\ 0 & \sqrt{6}/2 & 0 & \sqrt{6}/2 & 0\\ 0 & 0 & \sqrt{6}/2 & 0 & 1\\ 0 & 0 & 0 & 1 & 0\end{pmatrix}$

$J_y=\hbar\begin{pmatrix}0 & -i & 0 & 0 & 0\\ i & 0 & -i\sqrt{6}/2 & 0 & 0\\ 0 & i\sqrt{6}/2 & 0 & -i\sqrt{6}/2 & 0\\ 0 & 0 & i\sqrt{6}/2 & 0 & -i\\ 0 & 0 & 0 & i & 0\end{pmatrix}$

$J_z=\hbar\begin{pmatrix}2 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & -1 & 0\\ 0 & 0 & 0 & 0 & -2\end{pmatrix}$

$J^2=J_x^2+J_y^2+J_z^2=6\hbar^2I_{5\times 5}=6\hbar^2\begin{pmatrix}1 & 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{pmatrix}$

Moreover, the following matrix

$Z=J_z+iJ_x=\hbar\begin{pmatrix}2 & i & 0 & 0 & 0\\ i & 1 & i\sqrt{6}{2} & 0 & 0\\ 0 & i\sqrt{6}/2 & 0 & i\sqrt{6}/2 & 0\\ 0 & 0 & i\sqrt{6}/2 & -1 & i\\ 0 & 0 & 0 & i & -2\end{pmatrix}$

is nonnormal and nilpotent with $Z^5=(J_z+iJ_x)^5=0_{5\times 5}$. Moreover, it has 5 null eigenvalues and a single eigenvector

$e_1=\begin{pmatrix}1\\ 2i\\ -\sqrt{6}\\ -2i\\ 1\end{pmatrix}$

We see that the spin matrices in 3D satisfy for general s:

i) $J_x^2+J_y^2+J_z^2=s(s+1)I_{2s+1}$ $\forall s$.

ii) The ladder operators for spin s have the following matrix representation:

$J_+=\begin{pmatrix} 0 & \sqrt{2s} & 0 & 0 & \ldots & 0\\ 0 & 0 & \sqrt{2(2s-1)} & 0 & \ldots & 0\\ 0 & 0 & 0 & \sqrt{3(2s-2)} & \ldots & 0\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots\\ 0 & 0 & 0 & 0 & \ldots & \sqrt{2s}\\ 0 & 0 & 0 & 0 & \ldots & 0\end{pmatrix}$

Moreover, $J_-=J_+^+$ in the matrix sense and the above matrix could even be extended to the case of  a non-bounded spin particle. In that case the above matrix would become an infinite matrix! In the same way, for spin s, we also get that $Z=J_z+iJ_1$ would be (2s+1)-nilpotent and it would own only a single eigenvector with Z having $(2s+1)$ null eigenvalues. The single eigenvector can be calculated quickly.

Example(VI): Rotations and spinors.

We are going to pay attention to the case of spin 1/2 and work out its relation with ordinary rotations and the concept of spinors.

Given the above rotation matrices for spin 1/2 in terms of Pauli matrices, we can use the following matrix property: if M is a matrix that satisfies $A^2=I$, then we can write that

$e^{iAt}=\cos t I+i\sin t A$

Then, we write

$e^{i\sigma_x t}=\cos t I+i\sin t\sigma_x=\begin{pmatrix} \cos t & i\sin t\\ i\sin t & \cos t\end{pmatrix}$

$e^{i\sigma_y t}=\cos t I+i\sin t\sigma_y=\begin{pmatrix} \cos t & \sin t\\ -\sin t & \cos t\end{pmatrix}$

$e^{i\sigma_z t}=\cos t I+i\sin t\sigma_z=\begin{pmatrix} \cos t+i\sin t & 0\\ 0 & \cos t-i\sin t\end{pmatrix}=\begin{pmatrix}e^{it} & 0 \\ 0 & e^{-it}\end{pmatrix}$

From these equations and definitions, we can get the rotations around the 3 coordinate planes (it corresponds to the so-called Cayley-Hamilton parametrization).

a) Around the plance (XY), with the Z axis as the rotatin axis, we have

$R_z(\theta)=\exp\left(-i\theta \dfrac{J_z}{\hbar}\right)=\exp\left(-i\dfrac{\theta\sigma_z}{2}\right)=\begin{pmatrix}e^{-i\frac{\theta}{2}} & 0\\ 0 & e^{-i\frac{\theta}{2}}\end{pmatrix}$

b) Two sucessive rotations yield

$R(\theta,\phi)=\exp\left(-i\dfrac{\phi\sigma_z}{2}\right)\exp\left(-i\dfrac{\theta\sigma_y}{2}\right)=\begin{pmatrix}e^{-i\frac{\phi}{2}}\cos\frac{\theta}{2} & e^{i\frac{\phi}{2}}\sin\frac{\theta}{2}\\e^{-i\frac{\phi}{2}}\sin\frac{\theta}{2} & e^{i\frac{\phi}{2}}\cos\frac{\theta}{2}\end{pmatrix}$

Remark: $R_z(2\pi)=-I$!!!!!!!

Remark(II):   $R(\phi=0,\theta)=\begin{pmatrix}\cos\frac{\theta}{2} & -\sin\frac{\theta}{2}\\ \sin\frac{\theta}{2} & \cos\frac{\theta}{2}\end{pmatrix}$

This matrix has a strange $4\pi$ periodicity! That is, rotations with angle $\beta=2\pi$ don’t recover the identity but minus the identity matrix!

Imagine a system or particle with spin 1/2, such that the wavefunction is $\Psi$:

$\Psi=\begin{pmatrix}\Psi_1\\ \Psi_2\end{pmatrix}$

If we apply a $2\pi$ rotation to this object, something that we call “spinor”, we naively would expect that the system would be invariant but instead of it, we have

$R(2\pi)\Psi=-\Psi$

The norm or length is conserved, though, since

$\vert \Psi\vert^2=\vert\Psi_1\vert^2+\vert\Psi_2\vert^2$

These objects (spinors) own this feature as distinctive character. And it can be generalized to any value of j. In particular:

A) If $j$ is an integer number, then $R(2\pi)=I$. This is the case of “bosons”/”force carriers”.

B) If $j$ is half-integer, then $R(2\pi)=-I$!!!!!!!. This is the case of “fermions”/”matter fields”.

Rotation matrices and the subspaces E(j).

We learned that angular momentum operators $J$ are the infinitesimal generators of “generalized” rotations (including those associated to the “internal spin variables”). A theorem, due to Euler, says that every rotation matrix can be written as a function of three angles. However, in Quantum Mechanics, we can choose an alternative representation given by:

$U(\alpha,\beta,\gamma)=\exp\left(-\alpha\dfrac{iJ_x}{\hbar}\right)\exp\left(-\beta\dfrac{iJ_y}{\hbar}\right)\exp\left(-\gamma\dfrac{iJ_z}{\hbar}\right)$

Given a representation of J in the subspace $E(j)$, we obtain matrices $U(\alpha,\beta,\gamma)$ as we have seen above, and these matrices have the same dimension that those of the irreducible representation in the subspace $E(j)$. There is a general procedure and parametrization of these rotation matrices for any value of $j$. Using a basis of eigenvectors in $E(j)$:

$\boxed{\langle j',m'\vert U\vert j,m\rangle =D^{(j)}_{m'm}\delta_{jj'}}$

and where we have defined the so-called Wigner coefficients

$D^{(j)}_{m'm}(\alpha,\beta,\gamma)=\langle j'm'\vert e^{-\alpha\frac{iJ_z}{\hbar}}e^{-\beta\frac{iJ_y}{\hbar}}e^{-\gamma\frac{iJ_x}{\hbar}}\vert jm\rangle\equiv e^{-i(\alpha m'+\beta m)}d^{(j)}_{m'm}$

The reduced matrix only depends on one single angle (it was firstly calculated by Wigner in some specific cases):

$\boxed{d^{(j)}_{m' m}(\beta)=\langle j'm'\vert \exp\left(-\beta \dfrac{i}{\hbar}J_y\right)\vert jm\rangle}$

Generally, we will find the rotation matrices when we “act” with some rotation operator onto the eigenstates of angular momentum, mathematically speaking:

$\boxed{\displaystyle{U(\alpha,\beta,\gamma)\vert j,m\rangle=\sum_{j',m'}\vert j',m'\rangle \langle j',m'\vert U\vert j,m\rangle=\sum_{m'}D^{(j)}_{m'm}\vert j,m'\rangle}}$

See you in my final blog post about  basic group theory!