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Say we a basis of kets such as$$ \beta ~ := ~\left\{\left|j=1,\, m=-1\right> , ~ \left|j=1, \, m=0\right>, ~ \left|j=1, \, m=1\right>\right\} \,.$$

Then it's plain to see why in matrix representation$$J_z~=~h \begin{pmatrix} 1&0&0\\0&0&0\\0&0&-1 \end{pmatrix}$$ as this just represents $m$ being either $0,$$1$ or $-1$.

But why is$$ J^2=2h^2 \begin{pmatrix} 1&0&0\\0&1&0\\0&0&1 \end{pmatrix} \,?$$

It can't be gotten from just $J_z$ and I don't know what the matrix representation for $J_x, J_y$ nor do I feel i'm expected to know them as they weren't described in class. So how can one compute the matrix representation of $J^2$?

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The short answer is that $J^2$ is a scalar under all rotations, and the only matrix that has that property is the identity matrix, so you have to have $J^2\propto I$.

I think that the process and physics are a bit easier to understand if we start from the elemental rotation matrices in $3$-d about each axis \begin{align} R_x & = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos\theta & -\sin\theta \\ 0 & \sin\theta & \cos\theta \end{array}\right] & R_y & = \left[\begin{array}{ccc} \cos\theta & 0 & \sin\theta \\ 0 & 1 & 0 \\ -\sin\theta & 0 & \cos\theta \end{array}\right] & R_z & = \left[\begin{array}{ccc} \cos\theta & -\sin\theta & 0\\ \sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{array}\right]. \end{align} where the pattern of signs is chosen so that a positive rotation gives a rotation around the axis in the sense defined to be positive by the right hand rule. A matrix/operator is said to generate a transformation if, for an infinitesimal transformation, it satisfies $$R(\theta) = I - i \frac{J\theta}{\hbar} + \mathcal{O}(\theta^2),$$ where the transformation is $R$ and the generator is $J$. From that we get that \begin{align} J_x & = \left[\begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & -i\hbar \\ 0 & i\hbar & 0 \end{array}\right] & J_y & = \left[\begin{array}{ccc} 0 & 0 & i\hbar \\ 0 & 0 & 0 \\ -i\hbar & 0 & 0 \end{array}\right] & J_z & = \left[\begin{array}{ccc} 0 & -i\hbar & 0 \\ i\hbar & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]. \end{align}

We now look for the eigenvectors and eigenvalues of $J_z$. The defining equation for the eigen-problem is $$J_z \vec{v} = \lambda \vec{v}$$ for some eigenvalue $\lambda$ and its corresponding eigenvector $\vec{v}$. We get that the eigenvalue/eigenvector pairs of $J_z$ are \begin{align} \lambda_{-1} & = -1 \Rightarrow & \vec{v}_{-1} & = \frac{1}{\sqrt{2}}\left[\begin{array}{c} i \\ 1 \\ 0 \end{array}\right], \\ \lambda_0 & = 0 \Rightarrow & \vec{v}_0 & = \left[\begin{array}{c} 0 \\ 0 \\ i \end{array}\right],\ \mathrm{and} \\ \lambda_1 & = 1 \Rightarrow & \vec{v}_1 & = \frac{1}{\sqrt{2}}\left[\begin{array}{c} -i \\ 1 \\ 0 \end{array}\right], \end{align} where I have taken care to normalize the eigenvectors to unit length (i.e. $\vec{v}^* \cdot \vec{v} = 1$), and to set the overall phases of the unit vectors to make the results match certain other conventions.

It is a defining property of the eigenvalue problem for Hermitian matrices that if we define $V \equiv \left[\vec{v}_1,\ \vec{v}_0,\ \vec{v}_{-1}\right]$, the matrix with normalized distinct eigenvectors in the columns, then \begin{align} V & = \left[\begin{array}{ccc} -\frac{i}{\sqrt{2}} & 0 & \frac{i}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ 0 & i & 0 \end{array}\right],\ \mathrm{and} & V^\dagger J_z V &= \hbar \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array}\right]. \end{align} Note that $V^\dagger V=I$ and $\operatorname{det}(V)=1$, meaning $V$ is a special unitary matrix. That fact means that $V$ is a transformation from one orthonormal basis to another. If we say that the above matrices are just representations of a more abstract operator in a particular basis, and use a $\rightarrow$ to signify representation, then we can say that in this basis $$J_z\rightarrow \hbar \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{array}\right].$$ If we examine $J_x$ and $J_y$ in this basis, we get \begin{align} J_x &\rightarrow \frac{\hbar}{\sqrt{2}} \left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{array}\right],\ \mathrm{and} & J_y \rightarrow i\frac{\hbar}{\sqrt{2}} \left[\begin{array}{ccc} 0 & -1 & 0 \\ 1 & 0 & -1 \\ 0 & 1 & 0 \end{array}\right]. \end{align}

Now you can evaluate $J^2 = J_x^2 + J_y^2 + J_z^2$ explicitly.

Notice how our original basis was in terms of the three states that are invariant under rotations about their respective axes, the $m_x=0$, $m_y=0$, and $m_z=0$ states, respectively.

Note also that the choice of phases made won't effect $J^2$. That was done so that the $J_x$ and $J_y$ constructed will match those generated by following the development in Wikipedia's ladder operator article.

You can make a more general construction if you look at the commutation relations among $J_x$, $J_y$ and $J_z$ as defined above, and rename the $x$ through $z$ axes as $1$ through $3$ to get that $$[J_i,\, J_j] = i\hbar \sum_{k=1}^3 \epsilon_{ijk} J_k, \tag1$$ with $\epsilon_{ijk}$ the Levi-Civita symbol. We then say that any collection of three Hermitian matrices that satisfies the commutation relations in (1) are generators of the symmetry transformation we call rotations in physics, in some particular representation/basis. The next step is to prove that $$[J^2,\, J_i] = 0,$$ which shows that the $J^2$ and $J_z$ can be simultaneously digaonalized. I have forgotten the next step in the process, but this more general approach works for all possible values of orbital angular momentum and spin, making it much more powerful.

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The $J$ component matrices are essentially defined by satisfying the relations:

$ [J_i,J_j] = J_iJ_j - J_jJ_i = i\epsilon_{ijk}J_k$

Your $J_z$ is missing a negative sign, btw.

But now knowing:

$[J_x, J_z] = -iJ_y$

$[J_y, J_z] = iJ_x$

$J_x^2 + J_y^2 + \begin{pmatrix} 1&0&0\\0&0&0\\0&0&-1 \end{pmatrix}^2 = 2\begin{pmatrix} 1&0&0\\0&1&0\\0&0&1 \end{pmatrix}$

You have enough to figure out $J_x$ and $J_y$. You can check your work with answers online - If you search matrices Jx and Jy and Jz youll find answers everywhere.


Alternatively, angular momentum theory state you can define $J_+ = J_x + iJ_y$ such that $J_+|l,m> = |l, m+1>$. This is for the straightforward representation with

$|1, -1> = \begin{pmatrix} 0\\0\\1 \end{pmatrix}$

$|1, 0> = \begin{pmatrix} 0\\1\\0 \end{pmatrix}$

$|1, +1> = \begin{pmatrix} 1\\0\\0 \end{pmatrix}$

$J_+|1, -1> = |1, 0> $ and so on. You can then figure out pretty easily that

$ J_+ = \begin{pmatrix} 0&1&0 \\ 0&0&1\\ 0&0&0 \end{pmatrix}$ $ J_- = \begin{pmatrix} 0&0&0 \\ 1&0&0\\ 0&1&0 \end{pmatrix}$

And then of course $J_x = \frac{1}{2}(J_+ + J_-)$ and $J_y = \frac{1}{2}(J_+ - J_-)$

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  • $\begingroup$ If I only know one J component which as I said is $J_z$,then I can't use that relation as it requires two of them . $\endgroup$ – bhapi May 14 '18 at 19:54
  • $\begingroup$ thank you for explaining that to me, I was too quick to disregard in my original post , it's exam season right for me so I'm a little tired and grumpy XD btw I edited my post to fix the mistake you noticed :) $\endgroup$ – bhapi May 14 '18 at 20:05
  • $\begingroup$ This cannot possibly be right since the square of your matrix on the left is the unit, implying that $J_x^2=J_y^2=0$. You might be missing a factor of $2$ in front of your unit matrix on the right. $\endgroup$ – ZeroTheHero May 14 '18 at 20:10
  • $\begingroup$ @ZeroTheHero yes you're right I was being sloppy. I think all the j=1 matrices carry a $\frac{1}{\sqrt{2}}$ also $\endgroup$ – Señor O May 14 '18 at 20:13
  • $\begingroup$ Your ladder operators are incorrect by factor of $\sqrt{2}$. $\endgroup$ – ZeroTheHero May 14 '18 at 21:22
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After some thoughts and given you have not been shown much, I think the simplest approach is to realize that $J^2$ is the square of the length of the vector $\vec J$ and so does not depend on the basis chosen to evaluate it.

Thus, start with the rotation about $\hat z$ \begin{align} R_z(\theta)=\left( \begin{array}{ccc} \cos (\theta ) & \sin (\theta ) & 0 \\ -\sin (\theta ) & \cos (\theta ) & 0 \\ 0 & 0 & 1 \\ \end{array} \right) \end{align} from which one extracts $$ J_z:=-i\frac{d}{d\theta}R_z\vert_{\theta=0}=\left( \begin{array}{ccc} 0 & -i & 0 \\ i & 0 & 0 \\ 0 & 0 & 0 \\ \end{array} \right) $$ This is not your matrix $J_z$ but it does have the same eigenvalues and in fact, the eigenvectors are $z, x\pm i y$ and are proportional to the spherical harmonics $Y_{1m}(\theta,\phi)$ expressed in Cartesian coordinates. Thus, the $(x,y,z)$ basis of this calculation is just a change of basis away from the $Y_{1m}$, and the choice of basis does not matter in computing $J_x^2+J_y^2+J_z^2$. Note that the spherical harmonics are a common set of basis functions for states of $\ell=1$.

You can check immediately that $$ J_z^2=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{array} \right)\, . $$ You can work out $R_y$ and $R_x$, compute $J_y$ and $J_x$ in the same manner but it doesn't take much to conclude, by symmetry, that \begin{align} J_x^2&=\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)\, , \\ J_y^2&=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)\, , \end{align} since rotations about $\hat x$ and $\hat y$ leave the first and middle axis unchanged, respectively. Thus, it easily follows that $$ J_x^2+J_y^2+J_z^2= 2 \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)\, , $$ as required.

The nice thing about this derivation is that it shows explicitly that $J_z$ and $J_x$ and $J_y$ are in fact related to an infinitesimal transformation about their respective axes, v.g. $$ R_z(\Delta \theta)=e^{i\Delta \theta L_z}\approx 1+i \Delta \theta L_z + \ldots $$

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