Magnetic Moments

We are interested in asking what these static moments are for systems of nucleons. In order to answer this question we need to break down the contribution into parts arising from the orbital motion of charges (giving a contribution from the velocity of the proton charge) and a part coming from the intrinsic spin of the nucleons. Let us begin by forming the contribution to the magnetic moment arising from the orbital motion of the proton charge. The current density of a charge distribution with velocity ${\bf v}$ is given by ${\bf j}(x)~=~\rho (x) {\bf v}(x)$. Therefore, our definition of ${\bf M}$ leads to

 

$${\bf M_{\rm orb} } = {1\over 2}\int\ d^3x\ {\bf x}\times {\bf j}$$

$$= {1\over 2}\int\ d^3x\ \rho (x) {\bf x}\times {\bf v}$$

$$= {e\over 2 M_N}\int\ d^3x\ \rho^\prime (x) {\bf x}\times {\bf p}$$

$$= {e\over 2 M_N}\int\ d^3x\ \sum_{j=1}^Z\ \psi^*\ l^{(j)} \psi \ \ \ ,$$ (18)

where $\rho^\prime$ is the probability density associated with the charge density $\rho$, $\psi$ is the wavefunction of the state and $l^{(j)}$ is the orbital angular momentum operator. Further, being good quantum mechanics we realize that forming an operator will be the way to go and we can define

 

$${\bf {\hat M}_{\rm orb} } = \mu_0\ \sum_{j=1}^Z\ l^{(j)} \ \ \ ,$$ (19)

where $\mu_0~=~e/(2 M_N)$ is the Dirac moment of a nucleon. It is convenient to decompose this object into it irreducible representations of isospin as we will be using isospin to classify nuclear states, and recalling that a proton was projected from a nucleon spinor by $p~=~{1\over 2}~(1+\tau_3)N$ and similarly and neutron $n~=~{1\over 2}~(1-\tau_3)N$. We can then write

 

$${\bf {\hat M}_{\rm orb} } = \mu_0\ \sum_{j=1}^A\ {1\over 2} (1+\tau_3^{(j)}) l^{(j)} \ \ \ ,$$ (20)

where the sum has been extended from only a sum over protons to a sum over all nucleons in the system. The electric charge of each is projected out with the isospin operator.

It is easy to include the contribution to the magnetic moment from the intrinsic magnetic moments of the neutron and proton. You will recall the the magnetic moment of the proton is $\mu_p~=~+2.792847386\pm 0.000000063~\mu_0~=~\kappa_p\mu_0$ and magnetic moment of the neutron is $\mu_n~=~-1.91304275\pm 00000045 \mu_0~=~\kappa_n\mu_0$. Recalling that there is a "g-factor" of 2 for spin ${1\over 2}$ particles we have that

 

$${\bf {\hat M}_{\rm spin} } = 2 \mu_0\ \left( \sum_{j=1}^Z\ \kappa_p S^{(j)} + \sum_{j=Z+1}^A\ \kappa_n S^{(j)} \right)$$

$$= 2 \mu_0\ \sum_{j=1}^A\ \left( {1\over 2}(1+\tau_3^{(j)}) \left[... ...ppa_n\right) + {1\over 2}\left(\kappa_p-\kappa_n\right) \right] S^{(j)} \right.$$

$$\left. + {1\over 2}(1-\tau_3^{(j)}) \left[ {1\over 2}\left(\kappa... ...ppa_p\right) + {1\over 2}\left(\kappa_n-\kappa_p\right) \right] S^{(j)} \right)$$

$$= \mu_0 \left( (\kappa_p+\kappa_n) S + (\kappa_p-\kappa_n) \sum_{j=1}^A \tau_3^{(j)} S^{(j)} \right) \ \ \ ,$$ (21)

where $S$ is the total spin operator. We sum the orbital with with spin contribution to find the total magnetic moment operator ${\bf\hat M} = {\bf {\hat M}_{\rm spin} } + {\bf {\hat M}_{\rm orb} }$.

Let me make some comments here. We have made some big assumptions getting to this point. Firstly, we have assumed that the only charged currents in the nucleus are those from the motion of the nucleons. We have neglected the contributions from mesons, which we know are present in a field theory description of nucleons, and as we saw last time are responsible for the long-range part of the interaction. Secondly, we have assumed that the nucleons retain their free-space ``identity'', by which I mean that their magnetic moments in a nuclear environment are the same as those in free-space. Corrections to this are a bit harder as we can only ever measure S-matrix elements and not the individual contributions. Clearly, these assumptions will be exact in the limit that the nuclear interactions ``turn-off'', as they then would not ``know'' about each other. So one might guess that corrections to these results behave like $E_{\rm binding}/M_N$. I will leave this subject at this point, to return to later. This is just a warning, NOT to stop thinking about this!

We will be wanting to form matrix elements of the above operators between states of good $J$ and $S$, and not good $L$ (tensor type interactions). In order to proceed we need to know how to take such matrix elements and we will need a simple application of the Wigner-Eckhart theorem, that we will briefly derive.

The Wigner-Eckhart theorem tells us that the matrix element of a tensor operator $T_k^{(\mu )}$ of rank $2k+1$ has matrix elements between states of good total angular momentum $J$of the form

 

$$\langle J^\prime m^\prime \vert T^{(\mu)}_K \vert J m\rangle = \l... ...rime m^\prime\rangle \langle J^\prime\vert\vert T_K \vert\vert J\rangle \ \ \ .$$ (22)

This demonstrates that the matrix element can be decomposed into a reduced matrix element (denoted with the double vertical lines) , that is independent of the alignment of the states in the subspace, and a Clebsch-Gordan coefficient that factors in the orientation in the subspace. If we have an arbitrary vector operator, i.e. $k~=~1$, then this becomes

 

$$\langle J^\prime m^\prime \vert T^{(\mu)}_1 \vert J m\rangle = \l... ...rime m^\prime\rangle \langle J^\prime\vert\vert T_1 \vert\vert J\rangle \ \ \ ,$$ (23)

and hence in the ratio of matrix elements for two different vector operators with same alignment between states of the same alignment, the Clebsch-Gordan factor cancels

 

$${ \langle J^\prime m^\prime \vert A^{(\mu)} \vert J m\rangle \ove... ...vert J\rangle \over \langle J^\prime\vert\vert B_1 \vert\vert J\rangle} \ \ \ ,$$ (24)

where I have dropped the subscript denoting vector operator. Imagine now that one of the operators is the angular momentum generator itself

 

$$\langle J , m=J \vert J^z \vert J , m=J\rangle = \langle J J 1 0 \vert J J\rangle \langle J\vert\vert J\vert\vert J\rangle$$

$$J = \sqrt{J\over J+1} \langle J\vert\vert J\vert\vert J\rangle$$

$$\langle J^\prime\vert\vert J\vert\vert J\rangle = \sqrt{J(J+1)}\ \delta_{JJ^\prime} \ \ \ .$$ (25)

To continue, we note that the matrix element of a scalar operator formed by contracting the angular momentum generators with a vector operator $A$, must be proportional to the reduced matrix element of $A$. We can then determine the constant of proportionality by setting $A~=~J$ as follows

 

$$\langle J m^\prime=J\vert J\cdot A \vert J m=J\rangle = C\ \langle J\vert\vert A \vert\vert J \rangle$$

$$J(J+1) = C\ \langle J\vert\vert J \vert\vert J \rangle \ =\ C \sqrt{J(J+1)}$$

$$\langle J m^\prime=J\vert J\cdot A \vert J m=J\rangle = \sqrt{J(J+1)} \langle J\vert\vert A \vert\vert J \rangle \ \ .$$ (26)

Hence, it follows from the above that

 

$$\langle J m^\prime=J\vert A_z \vert J m=J\rangle = {\langle J m^\... ...rt J m=J\rangle \over J(J+1) } \langle J m^\prime=J\vert J_z \vert J m=J\rangle$$

$$= {J\over J(J+1)} \langle J m^\prime=J\vert J\cdot A \vert J m=J\rangle \ \ \ .$$ (27)

As an application of this result to the magnetic moment, consider an operator of the form

 

$${\cal O} = a\sum_{j=1}^A l^{(j)} + b \sum_{j=1}^A s^{(j)} + c\sum_{j=1}^A \tau_3^{(j)} l^{(j)} + d\sum_{j=1}^A \tau_3^{(j)} s^{(j)} \ \ \ ,$$ (28)

where $l$ and $s$ are the orbital and spin angular momentum generators, and the constants a,b,c,d are known. Matrix elements of operators involving $\tau_3$ will depend upon the $I_z$ quantum number of the state acted on. Therefore if we add the matrix elements between states of equal and opposite $I_z$ (denoted by $\langle M_\pm\rangle$), theses contributions vanish, leaving only contributions from $a$ and $b$,

 

$$\langle M_+\rangle + \langle M_-\rangle = \langle J, m=J , I, I_Z... ..._Z\rangle + \langle J, m=J , I, -I_Z\vert{\cal O} \vert J, m=J , I, -I_Z\rangle$$

$$= 2 \langle J, m=J , I \vert a\sum_{j=1}^A l^{(j)} + b \sum_{j=1}^A s^{(j)} \vert J, m=J , I \rangle$$

$$= 2 \langle J, m=J , I \vert {1\over 2}(a+b) \sum_{j=1}^A (l^{(j)... ...\over 2}(a-b) \sum_{j=1}^A (l^{(j)}- s^{(j)} ) \vert J, m=J , I \rangle \ \ \ .$$ (29)

The first operator is the total angular momentum of the state $\sum_{j=1}^A (l^{(j)}+ s^{(j)} )=J$, while the second requires use of the relation we proved above. Looking at the $z-$component

 

$$\langle J, m=J , I \vert \sum_{j=1}^A (l^{(j)}- s^{(j)} )_z \vert J, m=J , I \rangle = \langle J, m=J , I \vert( L-S )_z \vert J, m=J , I \rangle$$

$$= {1\over J+1}\langle J, m=J , I \vert J\cdot (L-S) \vert J, m=J , I \rangle \ \ \ ,$$ (30)

and so we have that

 

$$\langle M_+\rangle + \langle M_-\rangle = (a+b) J + {1\over J+1} \langle J, m=J , I \vert J\cdot (L-S) \vert J, m=J , I \rangle \ \ \ ,$$ (31)

and by using $S^2=(J-L)^2$ and $L^2=(J-S)^2$ we find that

 

$$\langle M_+\rangle + \langle M_-\rangle = (a+b) J + {a-b\over J+1} \langle J, m=J , I \vert L^2-S^2 \vert J, m=J , I\rangle \ \ \ .$$ (32)

Applying this relation to the magnetic moments of Mirror nuclei, e.g. $^{25}Mg$ and $^{25}Al$, where the nuclei only differ by the nucleon on top of the $^{24}Mg$ closed shell being either a proton or a neutron (the nuclear wavefunctions are identical except for the $I_z$ quantum number) ,

 

$$\mu_+ + \mu_- = \mu_0\left[ \left({1\over 2}+\kappa_p+\kappa_n\ri... ...r 2}-\kappa_p-\kappa_n\right) {\langle L^2-S^2\rangle\over J+1} \right] \ \ \ .$$ (33)

In forming this relation we have used

 

$$a={1\over 2}\mu_0 \ \ \ ,\ \ \ b=\left(\kappa_p+\kappa_n\right)\mu_0 \ \ \ .$$ (34)

Lets get back to our nucleus of interest for the moment, the deuteron. We want to find out what we expect its magnetic moment is for the possible spin-space wavefunctions that it can possible have. It is clear from eq. (33) that is magnetic moment is ($J=1$, $I=0$)

 

$$\mu_d= {\mu_0\over 2} \left[ \left({1\over 2}+\kappa_p+\kappa_n\r... ...er 2}-\kappa_p-\kappa_n\right) {\langle L^2-S^2\rangle\over 2 } \right] \ \ \ .$$ (35)

If the deuteron had space-spin configurations given by $^{2s+1} L_J$, then we find that

 

$$^1 P_1 \ \ :\ \ \mu_d = {1\over 2}\ \mu_0$$

$$^3 S_1 \ \ :\ \ \mu_d = \kappa_p+\kappa_n = 0.879\ \mu_0$$

$$^3 P_1 \ \ :\ \ \mu_d = {1\over 2}\left({1\over 2} + \kappa_p+\kappa_n\right) \ \mu_0 = 0.689\ \mu_0$$

$$^3 D_1 \ \ :\ \ \mu_d = {1\over 2}\left({3\over 2} - \kappa_p-\kappa_n\right) \ \mu_0 = 0.310\ \mu_0 \ \ \ .$$ (36)

As we know that the deuteron wavefunction is a linear combination of $S$ and $D$ states, we realize that the contributions to the magnetic moments will add incoherently, as they are in a different partial wave and this, for a mixing angle $\omega$ as defined last time we have that

 

$$\mu_d = \left(0.879 \cos^2\omega \ +\ 0.310 \sin^2\omega\right)\ \mu_0 \ =\ \left(0.879 - 0.569 \sin^2\omega\right)\ \mu_0 \ \ \ ,$$ (37)

and in order to reproduce the experimentally observed $\mu_d = 0.857$ we require that $\sin^2\omega\sim 4\%$. If the deuteron was a combination of the P-states, then there would be an interference term (dependent upon the exact form of the radial wavefunction) as they are the same partial wave.

Notice that the result is independent of the form of the radial wavefunction. This is the most naive estimate of the D-state admixture, in the most naive single particle model of the nucleus. There are corrections arising from meson exchange, for example, we might imagine attaching a photon to any lines in the graph we drew last time responsible for the potential between nucleons. The graphs we are keeping here are those where the photon attaches to a nucleon, but there are other graphs we have not retained where the photon attaches to the exchanged meson.