# magnetic moment of proton

I just tried to calculate the magnetic moment of a proton. I took the

proton g-Factor of $g=5.585694$
nuclear magneton of $\mu_k = 5.050783 * 10^{−27}$ J/T
proton spin of $I=1/2$

At first I calculated the norm of the proton spin $|\vec{I}|=\hbar \sqrt{I*(I+1)}=\hbar \frac{1}{2}\sqrt{3}$

And then I put everything together in $\mu=g\mu_k\frac{|\vec{I}|}{\hbar}$ and obtain 2,44134228 × 10^-26 instead of 1.410606 × 10^-26 ...

Interesting enough, I obtain the correct value if I devide by $\sqrt{3}$. But I see no reason to do this...

It would be great if you could help me.

ftiaronsem

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If you could measure "total spin" then your answer is correct, but the point is that for measuring spin one necessarily need to choose some particular direction (the axis along which you want to measure spin). Similarly for magnetic moment. – user10001 Jul 18 '12 at 20:33
As dushya said, this is a question of definition: the nuclear magneton times g-factor is the ratio of magnetic moment to the magnitude of spin in a direction, not to the length of the angular momentum vector. – Ron Maimon Jul 19 '12 at 3:07
Thanks very much for your comments. Unfortunately I do not see how the underlying rule should look like. Is it similar to the total angular momentum of an electron? Like $J_z\Phi=m_j\hbar\Phi$, just with $I$ instead of $J$? In this case it would give the correct numeric result. Or do I have to consider some integral to derive the solution in this (in the general) case? – ftiaronsem Jul 19 '12 at 9:13
$I_z\Phi=m_I\hbar\Phi$ would really make sense, considering the angle that $I$ has to the z-Axis. Am I correct to assume that wikipedia therefore lists the projection of $I$ on the z-Axis (or better said the axis of my magnetic field)? – ftiaronsem Jul 19 '12 at 9:52

It is only a question of definition. There is the operator of interaction of particle with an externally-produced magnetic field:

$\hat{H}_{int}=-\hat{\boldsymbol{\mu}}\cdot\mathbf{H}$,

where $\mathbf{H}$ is a magnetic field and $\hat{\boldsymbol{\mu}}$ is an operator:

$\hat{\boldsymbol{\mu}}=\displaystyle \frac{g e}{2 m} \hat{\mathbf{s}}$

By the «value» of the magnetic moment of particle, people usually imply the maximum of the following diagonal matrix element:

$\mu=\langle\psi\vert \hat{\boldsymbol{\mu}}_{z} \vert\psi\rangle,$

which is of course $g \mu_N/2$

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Thank for this answer. Unfortunatelly I cant follow you on the matrix element. To what matrix are you referring to, or better what are examples of the wave functions you are using to calculate your matrix elements? Thanks in advance. – ftiaronsem Jul 19 '12 at 14:34
Sorry I forget to add the index to the operator. The wave functions are simply normalized spinors. – Grisha Kirilin Jul 19 '12 at 14:42

No spin measurement of proton can give a value more or less than $\hbar/2$. But what do we mean when we say that spin of proton is $\hbar/2$ ? Spin is a 'vector' quantity (at least this is what it is classically). So one should also specify its direction. The thing is that in this case direction doesn't matter much. If you think of proton as some sphere and choose any arbitrary axis passing through its center and carry our an experiment to measure spin along that axis you will see that it is always $\hbar/2$ (or negative of it). You will never find spin of any proton to be $\hbar\sqrt{3}/2$ no matter along which axis you measure it. For the same reason you will never find a proton with magnetic moment $2.44134228 × 10^{-26}$ $JT^{-1}$ along any axis.

Edit : vector addition in QM

Suppose you choose three perpendicular directions $i,j,k$; and suppose in three successive experiments to measure spin of proton you find its value to be :

$\hbar/2$ along $i$ axis in experiment 1.

$\hbar/2$ along $j$ axis in experiment 2.

$\hbar/2$ along $k$ axis in experiment 3.

Now you may tend to conclude (using vector addition) that "total spin" of proton should be $\hbar/2(i+j+k)$ or equivalently $\hbar\sqrt{3}/2$ along unit direction $(i+j+k)/\sqrt{3}$. But if you carry out a fourth experiment to measure spin along this direction somewhat magically you will again find that spin is $\hbar/2$ (or - $\hbar/2$). Hence ordinary rules of vector addition do not apply in this case.

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thank you very very much. This explains a lot. Do you happen to know the general formula for an arbitray nukleus, lets say a nukleus with a total spin of $1\hbar$? Would I then have two possible magnetic moments? One for $m_I=0$ and one for $m_I=1$? – ftiaronsem Jul 19 '12 at 14:29
You are right. When spin is 1, a spin measurement can give either 1 (or -1), or 0. So in that case you can have two possible magnetic moments. Formally Grisha Kirilin's answer gives the necessary general formula. – user10001 Jul 19 '12 at 14:38