In Sakurai Quantum Mechanics, problem 3.25b we imagine $J_z^2$ as the component of a tensor with components $T_{ij} = J_iJ_j$.

$J_z^2 = \frac{1}{3}\pmb{J}^2 + (J_z^2 - \frac{1}{3}\pmb{J}^2) $

The first term on the right side corresponds to the scalar part of the tensor, the second term corresponds to the symmetric part of tensor. In the solutions, they immedietly invoke that the second term transforms as a second rank tensor with magnetic quantum number $0$. (Notation as defined in Sakurai and Wikipedia)

$J_z^2 = \pmb{T}^{(0)}_0 + \pmb{T}^{(2)}_0$

I understand why this is rank two (because the symmetric part of the tensor is a rank two tensor), but why is the magnetic quantum number assumed to be $0$? Secondly, in general, how do I represent $T_{ij}$ in terms of the five spherical tensors, ${T}^{(0)}_0$, ${T}^{(1)}_{\pm1,0}$, and ${T}^{(2)}_{\pm2,\pm1,0}$?

  • $\begingroup$ You mean nine spherical tensors: 3 x 3 = 1 + 3 + 5. $\endgroup$ – JEB Apr 28 at 3:41

There are several approaches you can take for converting between cartesian and spherical tensors. You can start with the definition of spherical vectors:

$$ \hat e^{\pm} = \frac 1 2 \mp[\hat x \pm i\hat y]$$ $$ \hat e^0 = \hat z $$

and then make a correspondence with vector spins, $|j,m\rangle$:

$$ \hat e^{\pm} \rightarrow |1,\pm 1\rangle$$ $$ \hat e^0 \rightarrow |1, 0\rangle$$

and use Clebsch-Gordan coefficients to compute the addition of two vector spins, and then map those to spherical tensors:

$$ |J,M\rangle \rightarrow T_M^J $$

For example:

$$ |2, 2\rangle = |1,1\rangle|1,1\rangle$$

which means we can express a unit $T_2^2$ as a dyad:

$$ \hat T_2^2 = e^+e^+=\frac 1 2 (-\hat x+i\hat y)(-\hat x+i\hat y) = \frac 1 2[\hat x\hat x - \hat y\hat y - i(\hat x\hat y+\hat y\hat x)] $$

So you can do that for every combination and invert the equations.

Or: you can look at Spherical harmonics in cartesian coordinates and then convert that to dyad (but only for natural form tensors):

$$ T_1^2 \rightarrow Y_2^1 \propto -(x-iy)z \rightarrow -\hat xz + i\hat y z$$

and go from there, but normalizations and signs might no be as clear.

The result is that cartesian to spherical is as follows: The isotropic part is the trace of the cartesian tensor:

$$ T_0^0 = \frac 1 3 T_{ii} $$

The J=1 are all from the antisymmetric part:

$$ T_0^1 = \sqrt{\frac 1 2}[T_{xy}-T_{yx}] $$

$$ T_{\pm 1}^1 = \mp \sqrt{\frac 1 2}[(T_{yz}-T_{zy})\pm i(T_{zx}-T_{xz}) ] $$

For the pure rank-2, aka natural form, you start with the symmetric trace free part:

$$ S_{ij} = \frac 1 2 [T_{ij}-T_{ji}] - \frac 1 3 T_{ii}\delta_{ij}$$

as follows:

$$ T_{\pm 2}^2 = \frac 1 2 [S_{xx}-S_{yy}\pm 2iS_{xy}]$$ $$ T_{\pm 1}^2 = \frac 1 2 [S_{xz}+i S_{yz}]$$ $$ T_0^2 = \sqrt{\frac 3 2}S_{zz}$$

The inversion of the pure rank 2 should be:

$${\bf S} = \left [ \begin{array}{ccc}T_2^2+T_{-2}^2-\sqrt{\frac 2 3}T_0^2 & -i(T_2^2-T_{-2}^2) & -T_1^2+T_{-1}^2 \\ -i(T_2^2-T_{-2}^2) & -T_2^2-T_{-2}^2-\sqrt{\frac 2 3}T_0^2 & i(T_1^2+T_{-1}^2)\\ -T_1^2+T_{-1}^2 & i(T_1^2+T_{-1}^2) & \sqrt{\frac 8 3}T_0^2 \end{array} \right ]$$

At higher rank, $N$, it gets difficult: you have to use standard young tableaux with $N$ boxes to get the irreducible representations of the permutation group on $N$ letters. Each of those, when applied to the indices, corresponds to a unique $T_M^J$ subspace (Schur-Weyl Duality).

At rank 3, you find:

$$ \bf 3 \otimes 3 \otimes 3 = 7_S \oplus 3_S \oplus 5_M \oplus 3_M \oplus 5_M \oplus 3_M \oplus 1_A$$

where the subscripts refer to symmetric, mixed, and antisymmetric in indices, and the numbers are the dimension of the spherical tensor: $2J+1$.

So, for instance, $\bf 1_A$, is isotropic an proportional to $\epsilon_{ijk}$.

Meanwhile, the total symmetric part has 10 dimensions, 7 of which are pure rank, and 3 of which transform like a vector. To work that out, you start with the totally symmetric combinations of indices:

$$ S_{ijk} = \frac 1 6 [T_{ijk} +T_{ski} +T_{kij} +T_{kji} +T_{ikj} +T_{jik}]$$

and subtract the trace:

$$ N_{ijk} = S_{ijk} - \frac 3 5 \delta_{ij}V_k$$


$$ V_k = S_{iij} = S_{iji} = S_{jii} $$

becomes one of the $T_M^1$ components.

The pure rank 3 solution is:

$$T_{\pm 3}^3 = \frac 1 {\sqrt 8}[(-N_{xxx}+3N_{xyy})\mp iN_{yyy}] $$ $$T_{\pm 2}^3 = \frac 1 2[N_{xxz}-N_{yyz})\mp 2iN_{xyz}] $$ $$T_{\pm 1}^3 = \frac {\sqrt 15}3\big [ \frac 1 {\sqrt 2}[\mp N_{xzz}-iN_{yzz}]+ \frac 1 {\sqrt 8} [\mp(N_{xxx}-N_{xxy})+ i(N_{yyy}\pm N_{xxy})]\big ] $$ $$T_0^3 = \frac{\sqrt{10}} 3[\frac 1{\sqrt 2}(N_{xzz}-iN_{yzz})+ N_{zzz}]$$

The (anti-)symmetric tensor correspond to the index permutation derived from the standard Young Tableaux from the integer partitions of $(1+1+1=3)$ and $3=3$, which is similar to the rank two case.

The mixed symmetry rotationally invariant subspaces correspond to $3=2+1$, for which there are two standard Tableaux, leading to:

$$T^{(0,1;2)}=\frac 1 3 [T_{ijk}+T_{jik}-T_{kji}-T_{kij}] $$ $$T^{(0,2;1)}=\frac 1 3 [T_{ijk}+T_{kji}-T_{jik}-T_{jki}] $$

Each of these has 8 = 3 + 5 degrees of freedom, corresponding to a $J=2$ and $J=1$ part. The vector part ($J=1$) is found by taking the non-zero trace:

$$ v^1_i = T^{(0,1;2)}_{ijj} = - T^{(0,1;2)}_{jji} $$ $$ v^2_i = T^{(0,2;1)}_{ijj} = - T^{(0,2;1)}_{jij} $$

Those can be subtracted (with suitable a outer product with $\delta_{ij}$) to leave a 5 DoF object that transforms like a rank-2 tensor.

  • $\begingroup$ Thank you so much! This is super helpful $\endgroup$ – Shep Bryan Apr 28 at 20:46

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