In my latest assignment I'm tasked with finding a magnetic moment $\mu$ of a hydrogen atom, whose current distribution $\mathbf{j}(\mathbf{r})$ looks like $$\mathbf{j}(\mathbf{r})=\frac{e\hbar}{3^8 \pi ma^4} \frac{r^3}{a^3}e^{-\frac{2r}{3a}}\sin\theta\cos^2\theta\mathbf{e_\varphi},$$ where $a$ is the Bohr radius and $m$ is the electron mass. It is also said that the electron orbits at a radius $r$, so I assume I need to integrate the radial component from 0 to $r$

So I got the usual formula for the magnetic moment, $$\mu={{1}\over{2}}\int d^3r'(\mathbf{r}\times\mathbf{j}(\mathbf{r}))$$

The cross product term can be expressed as $$\mathbf{r}\times\mathbf{j}(\mathbf{r})=r\cdot j(\mathbf{r})\cdot\sin\frac{\pi}{2}\mathbf{e_\theta}=rj(\mathbf{r})\mathbf{e_\theta}$$

So the moment becomes

$=\frac{e\hbar}{3^8ma^7}\int_{0}^{r}\int_{0}^{\pi}r'^4e^{-\frac{2}{3a}r'}\sin\theta\cos^2\theta dr'd\theta\mathbf{e_\theta}$ $$u:=\frac{2}{3a}r', dr'=\frac{3}{2}a\cdot du$$ $$v:=\cos\theta, d\theta=-\frac{dv}{sin\theta}$$ $=-\frac{e\hbar}{3^8ma^7}(\frac{3}{2}a)^5\int_{0}^{u(r)}u^4e^{-u}du\int_{1}^{-1}v^2dv\mathbf{e_\theta}$

(and after several layers of integration by parts)



I'm fairly certain in my integrals, but this result is extremely messy, which makes me doubt if I chose the correct approach in the first place

Am I using the correct formula? And if I am, am I integrating $dr'$ over correct boundaries?


Your basis vector $\mathbf{e}_\theta$ is angle-dependent. You have to take this into account when integrating. There are different ways of doing this. The easiest one is probably to re-express it in terms of Cartesian basis vectors. I am guessing $\mathbf{e}_\theta=\cos\theta \mathbf{\hat{z}}+\sin\theta\cos\phi \mathbf{\hat{x}}+\sin\theta\sin\phi \mathbf{\hat{y}}$. Now cartesian vectors do not depend on position, so this you can integrate.


How do you go from $\int d^3 r$ to spherical coordinates? You seem to be missing a $\sin\theta$, and maybe a factor of 2. Do it more carefully. Also what is $r$ in your final answer? How is it defined? As I understand this is an artefact from substituting an integral for $r'=0\dots r$ instead of integral over the whole space. To undo this substitution you should let $r\to\infty$ which will clean up the result.

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  • $\begingroup$ your result should be a z-polarized vector $\endgroup$ – Cryo Dec 13 '18 at 12:05
  • $\begingroup$ I believe it's $-\sin\theta\hat{z}+...$, if Wikipedia is to be trusted $\endgroup$ – Andrii Kozytskyi Dec 13 '18 at 22:17
  • $\begingroup$ So I used this and got a z-directed vector, yes, but I still am getting this exponential term in the result. Should it be an issue? $\endgroup$ – Andrii Kozytskyi Dec 13 '18 at 22:20
  • $\begingroup$ "I believe it's −sinθz^" probably right on this one. $\endgroup$ – Cryo Dec 14 '18 at 1:08
  • $\begingroup$ Ah, yes I obviously forgot to substitute the proper integration terms, thank you. When integrating from 0 to $\infty$ I got a relatively compact $-\frac{243}{16}\frac{e\hbar}{ma^2}$, which does sound plausible. I had my doubts about integrating to infinity, but the current density is defined in such a way that the integral doesn't diverge, so there's no reason for not integrating over the entire space $\endgroup$ – Andrii Kozytskyi Dec 14 '18 at 1:47

So the question may have been a bit vague ("have I done everything correctly?"), so I feel obliged to put up a proper answer now. I was wrong in a whole bunch of places

Firstly, as Cryo has pointed out in the comments, the $\hat{\theta}$ vector is position dependent and not uniquely defined, so to fix this one would transform the unit vector to cartesian coordiantes: $$\hat{\theta}=\cos\theta\cos\varphi\hat{\mathbf{x}}+\cos\theta\sin\varphi\hat{\mathbf{y}}-\sin\hat{\mathbf{z}}.$$ The second thing that was pointed out is that the triple integral in spherical coordinates obviously have an additional $r'^2\sin\theta$, which I forgot

and so the integral becomes

$=\frac{1}{2}\int_{0}^{\infty}\int_0^{2\pi}\int_0^\pi\frac{e\hbar}{3^8\pi ma^4}\frac{r'^3}{a^3}e^{-2r'/3a}\sin\theta\cos^2\theta\cdot r'\cdot\cdot(\cos\theta\cos\varphi\hat{\mathbf{x}}+\cos\theta\sin\varphi\hat{\mathbf{y}}-\sin\hat{\mathbf{z}})r'^2\sin\theta dr'd\varphi d\theta$

$=\frac{1}{2}\frac{e\hbar}{3^8}\int_0^\infty\int_0^\pi r'^6e^{-2r'/3a}\sin^2\theta\cos^2\theta([\sin\varphi]_0^{2\pi}\cos\theta\hat{\mathbf{x}}+[-\cos\varphi]_0^{2\pi}\cos\theta\hat{\mathbf{y}}-[\varphi]_0^{2\pi}\sin\theta\hat{\mathbf{z}})dr'd\theta$ $$[\sin\varphi]_0^{2\pi}=0-0=0; [-\cos\varphi]_0^{2\pi}=-1+1=0$$ $=-\frac{e\hbar}{3^8ma^7}\int_0^\infty\int_0^\pi r'^6e^{2r'/3a}\sin^3\theta\cos^2\theta dr'd\theta\hat{\mathbf{z}}$ $$u:=\frac{2}{3a}r',\space dr'=\frac{3}{2}ar'$$ $$v:=\cos\theta,\space d\theta=-\frac{dv}{\sin\theta}$$ $=\frac{e\hbar}{3^8ma^7}\int_0^\infty (\frac{3}{2}au)^6e^{-u}(\frac{3}{2}a)du\int_{1}^{-1}(1-v^2)v^2dv$,

and, after a whole lot of integration by parts,

$=\frac{e\hbar}{3\cdot 2^7m}\cdot 720\cdot[\frac{1}{3}v^3-\frac{1}{5}v^5]_1^{-1}$

$=\frac{e\hbar}{3\cdot 2^7m}\cdot 720\cdot(-\frac{4}{15})$


However, you may have noticed that I forgot to implement one of the conditions given in the question, namely that the electron orbits the proton at a distance $r$. With this in mind the calculation becomes a lot easier:

$...=-\frac{e\hbar}{3^8ma^7}\int_0^\infty\int_0^\pi r'^6e^{2r'/3a}\color{red}{\delta(r'-r)}\sin^3\theta\cos^2\theta dr'd\theta\hat{\mathbf{z}}$

$=-\frac{e\hbar}{3^8ma^7}r^6e^{-2r/3a}\int_0^\pi\sin^3\theta\cos^2\theta dr'd\theta\hat{\mathbf{z}}$

which is just




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