# Computing a matrix element with the Wigner-Eckart-theorem

I learned about the Wigner-Eckart theorem and want to apply it to the following matrix element $$$$\langle j \, m | r_kr_l | j' \, m'\rangle.$$$$ I know this can be done by writing the involved tensor operator as a sum of irreducible tensors $$$$r_kr_l = \left(\frac{1}{3}\delta_{kl}\vec{r}^2\right) + \left(r_kr_l - \frac{1}{3}\delta_{kl}\vec{r}^2\right)$$$$ which leads to \begin{align} \langle j \, m | r_kr_l | j' \, m'\rangle &= \langle j \, m | \frac{1}{3}\delta_{kl}\vec{r}^2 | j' \, m'\rangle + \langle j \, m | r_kr_l - \frac{1}{3}\delta_{kl}\vec{r}^2 | j' \, m'\rangle\\ &= \frac{1}{3}\lambda(j)\delta_{jj'}\delta_{mm'}\delta_{kl} + \langle j \, m | r_kr_l - \frac{1}{3}\delta_{kl}\vec{r}^2 | j' \, m'\rangle, \end{align} as $$\vec{r}^2$$ is a scalar. But I don't know how to proceed. How can I compute the second matrix element? Do I have to write the tensor in spherical coordinates so that I can apply Wigner-Eckart theorem to $$Y_l^m(\vartheta,\phi)$$ here?

• do $k,l$ run over $x,y,z$? – InertialObserver Jan 3 '19 at 20:33
• Why don't you just compute the matrix element directly? – InertialObserver Jan 3 '19 at 21:05
• @InertialObserver: Yes, the index denotes cartesian components. Regarding the direct computation: I wanted to deepen my understanding for the theorem, but if there is a way to compute this element directly, I would also be interested in that. – physicist23 Jan 3 '19 at 21:40

I am assuming here that with $$r_i$$ you mean the Cartesian components of the position operator.

In general the decomposition of $$u_i \, v_j$$, with $$\vec{u}$$ and $$\vec{v}$$ vectors, includes an additional antisymetric component $$\frac{1}{2} \left(u_i \, v_j - u_j \, v_i\right)$$ with 3 additional DOFs. In this case we don't have to worry about that.

Writing everything explicitly in terms of spherical harmonics reveals that the job is not fully done by just splitting this object in the trace and the traceless symmetric component. A bit of algebra leads us to $$$$r_i \, r_j \, = \, \frac{\sqrt{4 \pi}}{3} \, Y^{m=0}_{l=0} \, r^2 \, \delta_{ij} \, + \,r^2 \, S_{ij} \, ,$$$$ with $$$$S = \begin{bmatrix} -\frac{2}{3}\sqrt{\frac{\pi}{5}} Y^{m=0}_{l=2} + 4 \sqrt{\frac{\pi}{30}} \mathrm{Re}\left[Y^{m=2}_{l=2}\right] & 4 \sqrt{\frac{\pi}{30}} \mathrm{Im}\left[Y^{m=2}_{l=2}\right] & - 2 \sqrt{\frac{2\pi}{15}} \mathrm{Re}\left[Y^{m=1}_{l=2}\right] \\ 4 \sqrt{\frac{\pi}{30}} \mathrm{Im}\left[Y^{m=2}_{l=2}\right] & -\frac{2}{3}\sqrt{\frac{\pi}{5}} Y^{m=0}_{l=2} - 4 \sqrt{\frac{\pi}{30}} \mathrm{Re}\left[Y^{m=2}_{l=2}\right] & - 2 \sqrt{\frac{2\pi}{15}} \mathrm{Im}\left[Y^{m=1}_{l=2}\right] \\ - 2 \sqrt{\frac{2\pi}{15}} \mathrm{Re}\left[Y^{m=1}_{l=2}\right] & - 2 \sqrt{\frac{2\pi}{15}} \mathrm{Im}\left[Y^{m=1}_{l=2}\right] & \frac{4}{3}\sqrt{\frac{\pi}{5}} Y^{m=0}_{l=2} \end{bmatrix} .$$$$ This shows that the traceless symmetric part contains only components of a rank 2 spherical tensor. But the relation between Cartesian and Spherical components certainly mixes distinct values of $$m$$ for a specific Cartesian component. In particular, the real and imaginary parts of spherical harmonics will always mix $$Y^{l=l_i}_{m=|m_i|}$$ and $$Y^{l=l_i}_{m=-|m_i|}$$.

We see here that when it comes to applying the Wigner-Eckart theorem to a Cartesian tensor like this, we may be able to establish the rank $$l$$ of the distinct pieces directly and deduce the contribution to the matrix element that depends solely on $$l$$. On the other hand, like $$S$$, each piece clearly contains different irreducible components with different $$m$$'s in (generically) non-trivial combinations and the Wigner-Eckart formula depends on $$m$$ via a Clebsch-Gordan coefficient. Hence, in order to write the full explicit result, including the $$m$$ dependent CG coefficient, we would have to go through each irreducible term in each Cartesian component.

It should be mentioned that here we had the advantage of having the position vector directly involved and we were able to write things in terms of the Spherical Harmonics. In general, the relation between irreducible Spherical components and (products of) general Cartesian vectors can be computed as an extrapolation of the one you find for spherical harmonics (see section 3.10 Tensor Operators in Sakurai's Modern Quantum Mechanics, revised edition).

Finally, after you are done with the decomposition, you are free to use the Wigner-Eckart theorem in its general form to compute the matrix elements. Or, alternatively, look directly at the integrals which end up having integrands with 3 spherical harmonics. For these, there is a formula (eq 3.7.73 in Sakurai's book), which is nothing but an explicit application of the Wigner-Eckart theorem.

• First of all, it became clear to me how to write the tensor in spherical harmonics. But I do not understand what you mean by "$m$ still appears in the formular". Is it not the general case that the Clebsch-Gordan coefficients in the Wigner-Eckart theorem depend on $m$? For the following computation: Is it true that I have to compute the six independent components of the tensor separately (i.e. $xx$, $xy$, $xz$,...) and can use Wigner-Eckart for the matrix element $\langle j\, m | Y_l^m | j'\, m'\rangle$ afterwards? – physicist23 Jan 3 '19 at 22:54
• Oh maybe the context in which I wrote this was not clear. I don't think that we disagree. I simply meant to say that we know $S$ only has $l=2$ components (even before computing the matrix $S$ explicitly) and so the part of the WE formula that depends only on $l$ can be deduced directly. On the other hand, $S$ clearly contains different irreducible components with different $m$'s. Hence, in order to write the full explicit result, including the $m$ dependent CG coefficient, we would have to go term by term. – secavara Jan 3 '19 at 23:04
• So you mean that because we know $S$ is traceless and symmetric, we can deduce that it transforms like $Y_2^m$? Is the "part of the WE formula that depends only on $l$" the reduced matrix element? – physicist23 Jan 3 '19 at 23:24
• Yes, I mean the reduced matrix element. We know the $S$ is $l=2$ because the decomposition of the tensor product of 2 Cartesian vectors is fairly well known: with 1 DOF in a scalar, 3 DOFs in a "vector" (anti-simmetric piece) and 5 DOFs in the traceless symmetric part. These numbers match the values of $2l+1$ for $l=0,1,2$ and add up to $1+3+5=3^2$. This analysis can be done to tensor products of higher rank. – secavara Jan 4 '19 at 9:11