# Magnetic moment of a radially symmetric current

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)

$$=-\frac{e\hbar}{3^3\cdot2^5ma^2}[-e^{-2r/3a}\Bigl((\frac{2}{3a}r)^4+4(\frac{2}{3a}r)^3+12(\frac{2}{3a}r)^2+24(\frac{2}{3a}r)+24\Bigr)+24]\cdot[-\frac{2}{3}]\mathbf{e_\theta}$$

$$=\frac{e\hbar}{6^4ma^2}[24-e^{-2r/3a}\Bigl((\frac{2}{3a}r)^4+4(\frac{2}{3a}r)^3+12(\frac{2}{3a}r)^2+24(\frac{2}{3a}r)+24\Bigr)]\mathbf{e_\theta}$$.

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.

• your result should be a z-polarized vector – Cryo Dec 13 '18 at 12:05
• I believe it's $-\sin\theta\hat{z}+...$, if Wikipedia is to be trusted – Andrii Kozytskyi Dec 13 '18 at 22:17
• 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? – Andrii Kozytskyi Dec 13 '18 at 22:20
• "I believe it's −sinθz^" probably right on this one. – Cryo Dec 14 '18 at 1:08
• 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 – 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,

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

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

$$=-\frac{1}{2}\frac{e\hbar}{m}$$

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

$$-(-\frac{4}{15})\frac{e\hbar}{3^8ma^7}r^6e^{-2r/3a}=(\frac{4}{15})\frac{e\hbar}{3^8ma^7}r^6e^{-2r/3a}$$,

or

$$4.0644\cdot10^{-5}\frac{e\hbar}{ma^7}r^6e^{-2r/3a}$$