Wigner-Eckart theorem: what is $q $? How does $T^{(k)}_{q}$ relate to an operator?

Question

We saw in class that the Wigner-Eckart theorem is,

$$ \langle \alpha', j',m'|T^{(k)}_{q}|\alpha, j, m \rangle = \langle j,k;m,q|j',m'\rangle \frac{\langle \alpha', j'||T^{(k)}||\alpha, j \rangle}{\sqrt{2j + 1}} $$

In the context of the perturbation theroy, if we consider a perturbation of the Hamiltonian,

$$ \Delta V = eEz $$

For the hydrogen atom inside an electric field along $z$, neglecting the spin of the electron, we saw that we could write the spherical tensor as $$ T^{(k)}_{q} = T^{(1)}_0 = eEz $$ I was wondering, how can we know that $q$ here is equals to $0$ ? It's maybe obvious but I'm a little bit lost about this spherical tensor and Wigner-Eckart Theorem.

Answer 1

In your notation $k$ is the degree and $q$ the component of the tensor. You can easily verify that $z$ is proportional to the $0$th component since the components satisfy $[L_z,T^{(k)}_q]=q T^{(k)}_q$. Moreover $z\sim Y_{10}(\theta, \phi)$, confirming the value $q=0$ for the component (and for that matter that $k=1$).