Magnetic Dipole Moment & Gyromagnetic Ratio

January 16, 2016

GoLuckyRyan Basic, theory , , , , , Leave a comment

I always confuses on the definition, and wiki did not have any summary. so,

The Original definition is the Hamiltonian of a magnetic dipole under external magnetic field \vec{B},

H = -\vec{\mu}\cdot \vec{B},

where \vec{\mu} is magnetic dipole moment (MDM). It is

\vec{\mu} = g \frac{q}{2 m} \vec{J} = g \frac{\mu}{\hbar} \vec{J} = \gamma \vec{J}.

Here, the g is the g-factor, \mu is magneton, and \vec{J} is the total spin, which has a intrinsic factor m\hbar / 2 inside. \gamma is gyromegnetic ratio.

We can see, the g-factor depends on the motion or geometry of the MDM. For a point particle, the g-factor is exactly equal to 2. For a charged particle orbiting, the g-factor is 1.

Put everything into the Hamiltonian,

H = -\gamma \vec{J}\cdot \vec{B} = -\gamma J_z B = -\gamma \hbar \frac{m}{2} B [J],

Because energy is also equal E = \hbar f , thus, we can see the \gamma has unit of frequency over Tesla.

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Take electron as an example, the MDM is Bohr magneton \mu_{e} = e\hbar/(2m_e). The MDM is,

\vec{\mu_e} = g_e \frac{e}{2 m_e} \vec{S} = g_e \frac{\mu_e}{\hbar}\vec{S} = \gamma_e \vec{S}.

The magnitude of MDM is,

|\vec{\mu_e}|= g_e \frac{e}{2 m_e} \frac{\hbar}{2} = \gamma_e \frac{\hbar}{2} [JT^{-1}],

The gyromagnetic ratio is,

\gamma_e = g_e \frac{\mu_e}{\hbar} [rad s^{-1} T^{-1}].

Since using rad s^{-1} is not convenient for experiment. The gyromagnetic ratio usually divided by 2\pi,

\gamma_e = g_e \frac{\mu_e}{2\pi\hbar} [Hz T^{-1}].

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To evaluate the magnitude of  MDM of  single particle state, which has orbital angular momentum and spin, the total spin \vec{J} = \vec{L} + \vec{S}. However, the g-factor for \vec{L} is difference from that for \vec{S}. Thus, the MDM is not parallel to total spin. We have to use Landé Formula,

\left< JM|\vec{V}|JM'\right> = \frac{1}{J(J+1)} \left< JM|(\vec{J}\cdot\vec{V})|JM\right> \left<JM|\vec{J}|JM'\right>

or see wiki, sorry for my laziness.

The result is

g=g_L\frac{J(J+1)+L(L+1)-S(S+1)}{2J(J+1)}+g_S\frac{J(J+1)-L(L+1)+S(S+1)}{2J(J+1)}

For J = L \pm 1/2,

g = J(g_L \pm \frac{g_S-g_L}{2L+1})

Electromagnetic multi-pole moment

March 7, 2011

GoLuckyRyan Basic, Porperties, theory , , , , , , , , , 5 Comments

Electromagnetic multipole comes from the charge and current distribution of the nucleons.

Magnetic multipole in the nucleus has 2 origins, one is the spin of the nucleons, and another is the relative orbital motion of the nucleons.  the magnetic charge or monopoles either do not exist or are very small. the next one is the magnetic dipole, which causes by the current loop of protons.

Electric multipole is solely by the proton charge.

From electromagnetism, we knew that the multipole has  different radial properties, from the potential of the fields:

\displaystyle \Psi(r) = \frac{1}{4\pi\epsilon_0} \int\frac{\rho(r')}{|r-r'|}d^3r'

\displaystyle A(r) = \frac{\mu_0}{4\pi}\int\frac{J(r')}{|r-r'|}d^3r'

and expand them into spherical harmonic by using:

\displaystyle \frac{1}{|r-r'|} = 4\pi\sum_{l=0}^{\infty}\sum_{m=-l}^{m=l} \frac{1}{2l+1}\frac{r_{<}^l}{r_>^{l+1}} Y_{lm}^*(\theta',\phi')Y_{lm}(\theta,\phi)

we have

\displaystyle \Psi(r) = \frac{1}{\epsilon_0} \sum_{l,m}\frac{1}{2l+1}\int Y_{lm}^*(\theta',\phi') r'^l\rho(r')d^3r' \frac{Y_{lm}(\theta,\phi)}{r^{l+1}}

\displaystyle A(r)=\mu_0 \sum_{l,m}\frac{1}{2l+1}\int Y_{lm}^*(\theta',\phi') r'^l J(r') d^3r' \frac{Y_{lm}(\theta,\phi)}{r^{l+1}}

we can see the integral gives us the required multipole moment. the magnetic and electric are just different by the charge density and the current density. we summarize in this way :

\displaystyle q_{lm} = \int Y^*_{lm}(\theta',\phi') r'^l \O(r') d^3 r'

where O can be either charge or current density. The l determines the order of multipole. and the potential will be simplified :

\displaystyle M(r)=\sum_{l,m}\frac{1}{2l+1} q_{lm} \frac{Y_{lm}(\theta,\phi)}{r^{l+1}}

where M can be either electric or magnetic potential, and I dropped the constant. since the field is given by 1st derivative, thus we have:

  1. monopole has 1/r^2 dependence
  2. dipole has 1/r^3
  3. quadrapole has 1/r^4
  4. and so on

The above radial dependences are the same for electric or magnetic. for an easy name of the multipole, we use L-pole, in which L can be 0 for the monopole, 1 for the dipole, 2 for the quadrupole, etc.. and we use E0 for the electric monopole, M0 for the magnetic monopole.

Since the nucleus must preserve parity, the parity for the electric and magnetic moment is diffident. the difference comes from the charge density and current density have different parity. The parity for charge density is even, but the current density is odd. 1/r^2 has even parity, 1/r^3 has odd parity. therefore

  • electric L-pole — (-1)^{L}
  • magnetic L-pole — (-1)^{L+1}

for easy comparison:

  • E0, E2, E4… and M1,M3, M5 … are even
  • E1,E3,E5…. and M0, M2, M4…. are odd

The expectation value for L-pole, we have to calculate :

\displaystyle \int \psi^* Q_{lm} \psi dx

where Q_{lm} is a multipole operator ( which is NOT q_{lm}), and its parity is follow the same rule. the parity of the wave function will be canceled out due to the square of itself. thus, only even parity are non-Zero. those are:

  • E0, E2, E4…
  • M1,M3, M5 …

that makes sense, think about a proton orbits in a circular loop, which is the case for E1, in time-average, the dipole momentum should be zero.