0
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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.

--- ADDED

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.