Consider the $n=2$ states of the hydrogen atom, which we label by $|n\,l\,m_l\rangle$. I want to calculate whether or not there should be a sign difference in these specific matrix elements of $x$:

$$\langle211|x|200\rangle \;=\; \pm \langle21-1|x|200\rangle\;?$$

Explicit Calculation

Of course, one can show that the integral itself is the same for both signs:

$$\begin{align}\langle21\pm1|x|200\rangle &\;=\;\int\psi_{21\pm1}^\ast(\textbf{r})\;x\;\psi_{200}(\textbf{r})\;d^3\textbf{r} \\ &\;=\;\iint\left(\text{functions of }r\text{, }\theta\right)\;drd\theta\int_0^{2\pi}e^{\mp i\phi}\cos\phi \;d\phi\;.\end{align}$$

where $(r,\theta,\phi)$ are the usual spherical coordinates, and the $r$ and $\theta$ integrals are the same for both signs. One can easily verify that the $\phi$ integral gives the same result for both signs, hence $\langle211|x|200\rangle = \langle21-1|x|200\rangle$.

By Wigner-Eckart Theorem

Using $\langle j_1\, m_1 ;j_2\, m_2|J\,M\rangle$ to denote the Clebsch-Gordan coefficients, the Wigner-Eckart Theorem states that:

$$\langle j\, m|\, T_q^{(k)}|j'\,m'\rangle\;=\;\langle j'\,m';k\,q|j\, m\rangle\langle j||T^{(k)}||j'\rangle\,.$$

This is where I'm not certain, and I would like to check if I've made a mistake. The $x$-component of the position vector can be written in terms of its spherical components by:

$$x \;=\; \frac{1}{\sqrt{2}}r_{-1}-\frac{1}{\sqrt{2}}r_{+1}$$

which allows us to evaluate:

$$\begin{align}\langle211|x|200\rangle &\;=\;\frac{1}{\sqrt{2}}\langle 11|r_{-1}|00\rangle \;-\;\frac{1}{\sqrt{2}}\langle 11|r_{+1}|00\rangle\\ &\;=\;\frac{1}{\sqrt{2}}\langle 00;1-1|11\rangle\langle 1||\textbf{r}||0\rangle - \frac{1}{\sqrt{2}}\langle 00;11|11\rangle\langle 1||\textbf{r}||0\rangle\\ &\;=\;-\frac{1}{\sqrt{2}}\langle 1||\textbf{r}||0\rangle\; ;\\ \\ \langle21-1|x|200\rangle &\;=\; \frac{1}{\sqrt{2}}\langle 1-1|r_{-1}|00\rangle \;-\;\frac{1}{\sqrt{2}}\langle 1-1|r_{+1}|00\rangle\\ &\;=\;\frac{1}{\sqrt{2}}\langle 00;1-1|1-1\rangle\langle 1||\textbf{r}||0\rangle - \frac{1}{\sqrt{2}}\langle 00;11|1-1\rangle\langle 1||\textbf{r}||0\rangle\\ &\;=\;\frac{1}{\sqrt{2}}\langle 1||\textbf{r}||0\rangle\; ;\end{align}$$

i.e. wrongly concluding that $\langle211|x|200\rangle = -\langle21-1|x|200\rangle$. Where did I go wrong? I'm sure I've made a careless mistake somewhere (or have fundamentally misunderstood something) but I've been staring at this for so long that I'm going crazy...


1 Answer 1


Your mistake is the definition of the $Y_{1,\pm 1}$ spherical harmonic.

It should be $Y_{1,\pm 1} = \mp \sqrt\frac{3}{8\pi} \sin \theta e^{\pm i \phi}$ with the $\mp$ out the front changing the sign in your "Explicit calculation".

p.s. Good luck in AQP.

  • 1
    $\begingroup$ Hahahaha that obvioius, huh? Thank you very much, knew I did something silly! $\endgroup$
    – DWayne
    Commented May 28, 2023 at 16:46

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