# LOG#099. Group theory(XIX).

Final post of this series!

The topics are the composition of different angular momenta and something called irreducible tensor operators (ITO).

Imagine some system with two “components”, e.g., two non identical particles. The corresponding angular momentum operators are:

$J_1\cdot J_1, J_2\cdot J_2, J_1^z, J_2^z$

The following operators are defined for the whole composite system:

$J=J_1+J_2$

$J_z^T=J_z^1+J_z^2$

$J^2=(J_1+J_2)^2$

These operators are well enough to treat the addition of angular momentum: the sum of two angular momentum operators is always decomposable. A good complete set of vectors can be built with the so-called tensor product:

$\vert j_1j_2,m_1m_2\rangle =\vert j_1,m_1\rangle \otimes \vert j_2,m_2\rangle$

This basis $\vert j_1j_2,m_1m_2\rangle$ could NOT be an basis of eigenvectors for the total angular momentum operators $J^2_T,J_z^T$. However, these vector ARE simultaneous eigenvectors for the operators:

$J_1\cdot J_1,J_2\cdot J_2,J_z^1, J_z^2$

The eigenvalues are, respectively,

$\hbar^2 j_1(j_1+1)$

$\hbar^2 j_2(j_2+1)$

$\hbar m_1$

$\hbar m_2$

Examples of compositions of angular momentum operators are:

i) An electron in the hydrogen atom. You have $J=l+s$ with $l=r\times p$. In this case, the invariant hamiltonian under the rotation group for this system must satisfy

$\left[H,J\right]=0$

ii) N particles without spin. The angular momentum is $J=l_1+l_2+\cdots+l_N$

iii) Two particles with spin in 3D. The total angular momentum is the sum of the orbital part plus the spin part, as we have already seen:

$J=l+s=l_1+l_2+s_2+s_2$

iv) Two particles with spin in 0D! The total angular momentum is equal to the spin angular momentum, that is,

$J=S=s_1+s_2$

In fact, the operators $J^2,J_1\cdot J_1,J_2\cdot J_2,J_z$ commute to each other (they are said to be mutually compatible) and it shows that we can find a common set of eigenstates

$\vert j_1j_2,JM\rangle$

The eigenstates of $J^2, J_z$, with eigenvalues $\hbar^2 J(J+1)$ and $\hbar M$ are denoted by

$\vert \Omega J,M\rangle$

and where $\Omega$ is an additional set of “quantum numbers”.

The space generated by $\vert \Omega,JM\rangle$, for a fixed number $J$, and $2J+1$ vectors, $-J\leq M\leq J$, is an invariant subspace and it is also irreducible from the group theory viewpoint. That is, if we find a vector as a linear combination of eigenstates of a single particle, the remaining vectors can be built in the same way.

The vectors $\vert j_1j_1,JM\rangle$ can be written as a linear combination of those $\vert j_1j_1,m_1m_2\rangle$. But the point is that, due to the fact that the first set of vectors are eigenstates of $J_1\cdot J_1,J_2\cdot J_2$, then we can restrict the search for linear combinations in the vector space with dimension $(2j_1+1)(2j_2+1)$ formed by the vectors $\vert j,m\rangle$ with fixed $j_1,j_2$ quantum numbers. The next theorem is fundamental:

Theorem (Addition of angular momentum in Quantum Mechanics).

Define two angular momentum operators $J_1,J_2$. Define the subspace, with $(2j_1+1)(2j_2+1)$ dimensions and $j_1\geq j_2$, formed by the vectors

$\vert j_1j_2,m_1m_2\rangle=\vert j_1,m_1\rangle \otimes \vert j_2,m_2\rangle$

and where the (quantum) numbers $j_1,j_2$ are fixed, while the (quantum) numbers $m_1,m_2$ are “variable”. Let us also define the operators $J=J_1+J_2$ and $J^2,J_z$ with respective eigenvalues $J,M$. Then:

(1) The only values that J can take in this subspace are

$J\in E=\left\{ \vert j_1-j_2\vert, \vert j_1-j_2+1\vert,\ldots,j_1+j_2-1,j_1+j_2\right\}$

(2) To every value of the number J corresponds one and only one set or series of $2J+1$ common eigenvectors to $J_z$, and these eigenvector are denoted by $\vert JM\rangle$.

Some examples:

i) Two spin 1/2 particles. $J=s_1+s_2$. Then $j=0,1$ (in units of $\hbar=1$). Moreover, as a subspaces/total space:

$E(1/2)\otimes E(1/2)=E(0)\oplus E(1)$

ii) Orbital plus spin angular momentum of spin 1/2 particles. In this case, $j=l+s$. As subspaces/total space decomposition we have

$E(l)\otimes E(1/2)=E(l+1/2)\oplus E(l-1/2)$ if $l\neq 0$

$E(l)\otimes E(1/2)=E(1/2)$ if $l=0$

iii) Orbital plus two spin parts. $j=l+s_1+s_2$. Then, we have

$E(l)\otimes E(1/2)\otimes E(1/2)=E(l)\otimes (E(0)+E(1))=E(l)\otimes E(0)\oplus E(l)\otimes E(1)$

This last subspace sum is equal to $E(l)\oplus E(l+1)\oplus E(l)\oplus E(l-1)$ if $l\neq 0$ and it is equal to $E(0)\oplus E(1)$ if $l=0$.

In the case we have to add several (more than two) angular momentum operators, we habe the following general rule…

$E=E(j_1)\otimes E(j_2)\otimes E(j_3)\otimes \ldots \otimes E(j_n)$

We should perform the composition or addition taking invariant subspaces two by two and using the previous theorem. However, the theory of the addition of angular momentum in the case of more than 2 terms is more complicated. In fact, the number of times that a particular subspace appears could not be ONE. A simple example is provided by 2 non identical particles (2 nucleons, a proton and a neutron), and in this case the total angular momentum with respect to the center of masses and the spin angular momentum add to form $j=l+s_1+s_2$. Then

$E(l)\otimes E(1/2)\otimes E(1/2)=E(l)\otimes (E(0)\oplus E(1))=E(l)\otimes E(0)\oplus E(l)\otimes E(1)$

This subspace sum is equal to $E(l)\oplus E(l+1)\oplus E(l)\oplus E(l-1)$ if $l\neq 0$ and $E(0)\oplus E(1)$ if $l=0$.

Clebsch-Gordan coefficients.

We have studied two different set of vectors and bases of eigentstates

(1) $\vert j_1j_2,m_1m_1\rangle$, the common set of eigenstates to $J_1^2,J_2^2,J_{z1},J_{z2}$.

(2) $\vert j_1j_1,JM\rangle$, the common set of eigenstates to $J_1^2,J_2^2,J^2,J_z$.

We can relate both sets! The procedure is conceptually (but not analytically, sometimes) simple:

$\displaystyle{\vert j_1j_2,JM\rangle=\sum_{m_1=-j_1}^{j_1}\sum_{m_2=-j_2;m_1+m_2=M}^{j_2}\vert j_1j_2,m_1m_2\rangle\langle j_1j_2,m_1m_2\vert JM\rangle}$

The coefficients:

$\boxed{\langle j_1j_2,m_1m_2\vert JM\rangle}$

are called Clebsch-Gordan coefficients. Moreover, we can also expand the above vectors as follows

$\displaystyle{\vert j_1j_2,m_1m_1\rangle=\sum_{J=\vert j_1-j_2\vert}^{J=j_1+j_2}\sum_{M=-J}^{M=J}\vert J M\rangle \langle J M\vert j_1j_2,m_1m_2\rangle}$

and here the coefficients

$\boxed{\langle J M\vert j_1j_2,m_1m_2\rangle}$

are the inverse Clebsch-Gordan coefficients.

The Clebsch-Gordan coefficients have some beautiful features:

(1) The relative phases are not determined due to the phases in $\vert j_1j_2,JM\rangle$. They do depend on some coefficients $c_m$. For any value of J, the phase is determined by recurrence! It shows that

$\langle j_1j_2,j_1 J-j_1\vert J,J\rangle \in \mathbb{R}^+$

This convention implies that the Clebsch-Gordan (CG) coefficients are real numbers and they form an orthogonal matrix!

(2) Selection rules. The CG coefficients $\langle j_1j_2,m_1m_2\vert J,M\rangle$ are necessarily null IF the following conditions are NOT satisfied:

i) $M=m_1+m_2$.

ii) $\vert j_1-j_2\vert \leq J\leq j_1+j_2$

iii) $j_1+j_2+J\in \mathbb{Z}$

The conditions i) and ii) are trivial. The condition iii) can be obtained from a $2\pi$ rotation to the previous conditions. The two factors that arise are:

$R(2\pi)\vert j,m\rangle=(-1)^{2j}\vert j,m\rangle \leftrightarrow (-1)^{2J}=(-1)^{( j_1+j_2)}$

(3) Orthogonality.

$\displaystyle{\sum_{m_1=-j_1}^{j_1}\sum_{m_2=-j_2}^{j_2}\langle j_1j_2,m_1m_2\vert J,M\rangle\langle j_1j_2,m_1m_2\vert J' M'\rangle=\delta_{JJ'}\delta_{MM'}}$

$\displaystyle{\sum_{J=\vert j_1-j_2\vert}^{j_1+j_2}\sum_{M=-J}^{J}\langle j_1j_2,m_1m_2\vert J,M\rangle\langle j_1j_2,m'_1m'_2\vert J M\rangle=\delta_{m_1m'_1}\delta_{m_2m'_2}}$

(4) Minimal/Maximal CG coefficients.

In the case $J,M$ take their minimal/maximal values, the CG are equal to ONE. Check:

$\vert j_1j_2,J=j_1+j_2, J=M\rangle=\vert j_1j_2,m_1=j_1,m_2=j_2\rangle$

(5) Recurrence relations.

5A) First recurrence:

$C_J=\sqrt{J(J+1)-M(M-1)}\langle m_1m_2\vert J,M-1\rangle=$

$=\sqrt{j_1(j_1+1)-m_1(m_1+1)}\langle m_1+1,m_2\vert J,M\rangle+$

$+\sqrt{j_2(j_2+1)-m_2(m_2+1)}\langle m_1,m_2+1\vert J,M\rangle$

5B) Second recurrence:

$C'_J=\sqrt{J(J+1)-M(M+1)}\langle m_1,m_2\vert J,M+1\rangle=$

$=\sqrt{j_1(j_1+1)-m_1(m_1-1)}\langle m_1-1,m_2\vert J,M\rangle+$

$+\sqrt{j_2(j_2+1)-m_2(m_2-1)}\langle m_1,m_2-1\vert J,M\rangle$

These relations 5A) and 5B) are obtained if we apply the ladder operators $J_\pm$ in both sides of the equation defining the CG coefficients and using that

$J_\pm \vert JM\rangle=(J_{1\pm}+J_{2\pm})\vert JM\rangle$

$J_\pm \vert JM\rangle=\hbar \sqrt{J(J+1)-M(M\pm1)}\vert J,M\pm 1\rangle$

Irreducible tensor operators. Wigner-Eckart theorem.

There are 4 important previous definitions for this topic:

1st. Irreducible Tensor Operator (ITO).

We define $(2k+1)$ operators $T^{(k)}_q$, with $q\in \left[-k,k\right]$ the standard components of an irreducible tensor operator (ITO) of order $k$, $T^{(k)}$, if these components transform according to the following rules

$\displaystyle{U(\alpha,\beta,\gamma)T^{(k)}_qU^{-1}(\alpha,\beta,\gamma)=\sum_{q=-k}^{k}D^{(k)}_{qq'}(\alpha,\beta,\gamma)T^{(k)}_{q'}}$

2nd. Irreducible Tensor Operator (II): commutators.

The $(2k+1)$ operators $T^{(k)}_q$, $q\in \left[-k,k\right]$, are the components of an irreducible tensor operator (ITO) of order k, $T^{(k)}$, if these components satisfy the commutation rules

$\left[J_{\pm},T^{(k)}_q\right]=\hbar \sqrt{k(k+1)-q(q\pm 1)}T^{(k)}_{q\pm 1}$

$\left[ J_z,T^ {(k)}_q\right]=q\hbar T^{(k)}_q$

The 1st and the 2nd definitions are completely equivalent, since the 2nd is the “infinitesimal” version of the 1st. The proof is trivial, by expansion of the operators in series and identification of the involved terms.

3rd. Scalar Operator (SO).

We say that $S=T^0_0$ is an scalar operator, if it is an ITO with order k=0. Equivalently,

$U(\alpha,\beta,\gamma)SU^{-1}(\alpha,\beta,\gamma)=S$

One simple way to express this result is the obvious and natural statement that scalar operators are rotationally invariant!

4th. Vector Operator (VO).

We say that $V$ is a vector operator if

$\displaystyle{U(\alpha,\beta,\gamma)V^{(1)}_qU^{-1}(\alpha,\beta,\gamma)=\sum_{q=-1}^1D^{(1)}_{qq'}(\alpha,\beta,\gamma) V^{(1)}_{q'}}$

Equivalently, a vector operator is an ITO of order k=1.

The relation between the “standard components” (or “spherical”) and the “cartesian” (i.e.”rectangular”) components is defined by the equations:

$V_1=-\dfrac{1}{2}(V_x+iV_y)$

$V_0=V_z$

$V_{-1}=\dfrac{1}{2}(V_x-iV_y)$

In particular, for the position operator $R=(r_1,r_0,r_{-1})$, this yields

$r_1=-\dfrac{1}{\sqrt{2}}(x+iy)$

$r_=z$

$r_{-1}=\dfrac{1}{\sqrt{2}}(x-iy)$

Similarly, we can define the components for the momentum operator

$p=(p_1,p_0,p_{-1})$ or the angular momentum

$L=(L_+,L_-,L_z)\equiv (L_1,L_0,L_{-1})$

Now, two questions arise naturally:

1) Consider a set of $(2k+1)(2k'+1)$ operators, built from ITO $T^{(k)}_qT^{(k')}_{q'}$. Are they ITO too? If not, can they be decomposed into ITO?

2) Consider a set of $(2k+1)(2J+1)$ vectors, built from certain ITO, and a given base of eigenvalues for the angular momentum. Are these vectors an invariant set? Are these vectors an irreducible invariant set? If not, can these vectors be decomposed into irreducible, invariant sets for certain angular momentum operators?

Some theorems help to answer these important questions:

Theorem 1. Consider $T^{(k_1)}_{q_1}, T^{(k_2)}_{q_2}$, two irreducible tensor operators with $q_1\in \left[-k_1,k_1\right]$ and $q_2\in \left[-k_2,k_2\right]$. Take $k$ and $q\in \left[-k,k\right]$ arbitrary. Define the quantity

$\displaystyle{S^{(k)}_q\equiv \sum_{q_1=-k_1}^{k_1}\sum_{q_2=-k_2}^{k_2}T^{(k_1)}_{q_1}T^{(k_2)}_{q_2}\langle k_1 k_2,q_1 q_2\vert k q\rangle}$

Then, the operators $S^{(k)}_q$ are the “standard” components of certain ITO with order $k$. Moreover, we have, using the CG coefficientes:

$\displaystyle{T^{(k_1)}_{q_1}T^{(k_2)}_{q_2}=\sum_{q_1=-k}^{k}\sum_{q_2=\vert k_1-k_2\vert}^{k_1+k_2}S^{(k)}_q\langle k q\vert k_1 k_2, q_1 q_2\rangle}$

Theorem 2.  Let $T^{(k)}_{q_1}$ be certain ITO and $\vert j_2 m_1\rangle$ a set of $(2j_2+1)$ eigenvectors of angular momentum. Let us define

$\displaystyle{\vert \omega_{JM }\rangle =\sum_{q_1=-k_1}^{k_1}\sum_{m_2=-j_2}^{j_2}\left(T^{(k_1)}_{q_1}\vert j_2 m_2\rangle\right)\langle k_1 j_2, q_1 m_2\vert J M\rangle}$

These vectors are eigenvectors of the TOTAL angular momentum:

$J^2\vert \omega_{JM}\rangle =J(J+1)\hbar^2\vert \omega_{JM}\rangle$

$J_z\vert \omega_{JM}\rangle=M\hbar \vert \omega_{JM}\rangle$

Note that, generally, these eigenstates are NOT normalized to the unit, but their moduli do not depend on $M$. Moreover, using the CG coefficients, we algo get

$\displaystyle{T^{(k)}_{q_1}\vert j_2 m_2\rangle =\sum_{M=-J}^J\sum_{J=\vert k_1-j_2\vert}^{k_1+j_2}\vert J M\rangle \langle J M\vert k_1 j_2, q_1 m_2\rangle}$

Theorem 3 (Wigner-Eckart theorem).

If $T^{(k)}_q$ is an ITO and some bases for angular momentum are provided with $\vert j_1 m_1\rangle$ and $\vert j_2 m_2\rangle$, then

$\boxed{\langle j_2 m_2\vert T^{(k)}_{q}\vert j_1 m_1\rangle = \langle j_1 j_2, m_1 m_2\vert k q\rangle \dfrac{1}{2k+1}\langle j_2\vert \vert \mathbb{T}^{(k)}_q\vert\vert j_1\rangle}$

and where the quantity

$\boxed{\langle j_2\vert \vert \mathbb{T}^{(k)}_q\vert\vert j_1\rangle}$

is called the reduced matrix element.  The proof of this theorem is based on (4) main steps:

1st. Use the $(2k+1)(2j+1)$ vectors (varying $q, m$),  $T^{(k)}_q\vert j m\rangle$.

2nd. Form the linear combination/superposition

$\displaystyle{\vert \omega_{JM}\rangle=\sum_{m,q}\left( T^{(k)}_q\vert j m\rangle\right)\langle k j, q m\vert J M\rangle}$

and use the theorem (2) above to obtain

$\langle J' M'\vert J M\rangle=\delta_{JJ}\delta_{MM}F(J)$

3rd. Use the CG coefficients and their properties to rewrite the vectors in the base with J and M. Then, irrespectively the form of the ITO, we obtain

$\displaystyle{T^{(k)}_q\vert j m\rangle=\sum_{J,M}\langle J M\vert k j, q m\rangle \vert \omega_{JM}\rangle}$

4th. Project onto some other different state, we get the desired result

$\displaystyle{\langle \omega_{J'M'}\vert T_q^{(k)}\vert j m\rangle=\sum_{J,M}\langle \omega_{J' M'}\vert \langle J M\vert k j, q m\rangle \vert \omega_{JM}\rangle}$

or equivalently

$\displaystyle{\langle \omega_{J'M'}\vert T_q^{(k)}\vert j m\rangle=\sum_{J,M}\langle \delta_{J' M'}\delta_{J'M'}F(J)\langle J M\vert k j,q m\rangle}$

i.e.,

$\displaystyle{\langle \omega_{J'M'}\vert T_q^{(k)}\vert j m\rangle=F(J)\langle J M\vert k j, q m\rangle}$

Q.E.D.

The Wigner-Eckart theorem allows us to determine the so-called selection rules. If you have certain ITO and two bases $\vert j_1,m_1\rangle$ and $\vert j_2, m_2\rangle$, then we can easily prove from the Wigner-Eckart theorem that

(1) If $m_1-m_1\neq q$, then $\langle j_1 m_1\vert T^{(k)}_q\vert j_2 m_2\rangle=0$.

(2) If $\vert j_1-j_2\vert < k < j_1+j_2$ does NOT hold, then $\langle j_1 m_1\vert T^{(k)}_q\vert j_2 m_2\rangle=0$.

These (selection) rules must be satisfied if some transitions are going to “occur”. There are some “superselection” rules in Quantum Mechanics, an important topic indeed, related to issues like this and symmetry, but this is not the thread where I am going to discuss it! So, stay tuned!

I wish you have enjoyed my basic lectures on group theory!!! Some day I will include more advanced topics, I promise, but you will have to wait with patience, a quality that every scientist should own! 🙂

See you in my next (special) blog post ( number 100!!!!!!!!).

# 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!