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Section 5.6 Spin Magnetic Moment in a Magnetic Field

We are now ready to apply the major ideas of this chapter to some interesting examples which have important applications to medical physics, chemistry, and astronomy.

Both protons and electrons have intrinsic spin, with spin quantum number \(s = 1/2\) (both are fermions). The proton also has a charge \(q = +e\text{.}\) Classically speaking, the proton is viewed as a tiny spinning charged sphere that has current loops associated with the circulating charge. In the previous unit on electricity and magnetism, we learned that a magnetic moment \(\vec{\mu}\) can be associated with these circulating currents. When these concepts are applied to the proton as a classical spinning charged sphere, we find that the magnetic moment of the proton is proportional to the proton's intrinsic angular momentum \(\vec{S}\text{.}\) In fact, the magnetic moment for the proton is found to be

\begin{equation} \vec{\mu} = \frac{2 \mu_\text{p} }{\hbar} \vec{S}\label{eq_magMoment}\tag{5.26} \end{equation}

where \(\mu_\text{p} = 1.41 \times 10^{-26}\) \mathrm{J/T}. Since measuring any component of \(\vec{S}\) results in a value of \(\pm \frac{\hbar}{2}\text{,}\) equation (5.26) tells us that, for a proton, \(\vec{\mu}\) has a magnitude of \(\mu_\text{p}\) and points in the direction of \(\vec{S}\text{.}\)

We learned in the unit on electricity and magnetism that if a magnetic moment \(\vec{\mu}\) is placed in a constant magnetic field \(\vec{B}\text{,}\) there is a magnetic potential energy associated with the orientation of the magnetic moment relative to the magnetic field given as

\begin{equation} U = - \vec{\mu} \cdot \vec{B}\text{.}\label{eq_magPotential}\tag{5.27} \end{equation}

This energy is greatest when \(\vec{\mu}\) and \(\vec{B}\) are pointing in opposite directions and least when pointing in the same direction. If we take \(\vec{B}\) to be pointing in the \(z\)-direction such that \(\vec{B} = B_0 \hat{z}\text{,}\) then Eqs. (5.26) and (5.27) can be written as

\begin{equation} U = - \vec{\mu} \cdot \vec{B} = - \frac{2 \mu_\text{p} }{\hbar} B_0 S_z = - 2 \mu_\text{p} B_0 m_s\label{eq_magPotential2}\tag{5.28} \end{equation}

where we have used equation (5.12) for the \(z\)-component of the proton's spin. Since \(m_s = \pm 1 / 2\) for a spin-\(1/2\) particle, each of the orientations of the proton (spin-up or spin-down) corresponds to a different energy of the proton in the magnetic field

\begin{align*} U_+ = - \mu_\text{p} B_0 \amp \mbox{\((m_s = +\frac{1}{2}\), spin-up)}\\ U_- = + \mu_\text{p} B_0 \amp \mbox{\((m_s = -\frac{1}{2}\), spin-down)} \text{.} \end{align*}

Consequently, the proton has a different energy depending on whether it is in the spin-up (\(| +z \rangle\)) state or the spin-down (\(| -z \rangle\)) state. Therefore the states \(| +z \rangle\) and \(| -z \rangle\text{,}\) in addition to being states of definite \(z\)-component of spin \(S_z\text{,}\) are also states of definite energy \(U_+\) and \(U_-\text{,}\) respectively. Thus when placed in a magnetic field, the state of the proton splits into two possible energy states, as shown in Figure 5.11 separated in energy by

\begin{equation} \Delta E_\text{ proton } \equiv |U_+ - U_-| = 2\mu_\text{p} B_0\text{.}\label{eq_deltaE}\tag{5.29} \end{equation}
Figure 5.11. Splitting of energy levels of a proton in a magnetic field

Figure 5.11 suggests that for a spin-up proton \(| +z \rangle\) placed in a magnetic field \(\vec{B} = B_0 \hat{z}\text{,}\) if the proton then absorbs a photon of energy \(E_\text{ photon } = \Delta E_\text{ proton } = 2 \mu_\text{p} B_0\text{,}\) then the proton will make a transition to the spin-down state \(| -z \rangle\text{,}\) which has higher energy. The reversal of spin component \(S_z\) is called spin-flipping.

Consider a proton that is placed in a magnetic field such that it is in the spin-up state \(| +z \rangle\text{.}\) If the magnitude of the magnetic field is \(B_0 = 24.8 \Xunits{mT}\text{,}\) determine the frequency of the photon that would be absorbed by the proton causing it to make a transition to the spin-down state?

Solution.

For the magnetic field strength given, the energy difference between the spin-up and spin-down states is

\begin{align*} \Delta E_\text{ proton } =\mathstrut \amp 2 \mu_\text{p} B_0\\ =\mathstrut \amp 2 (1.41 \times 10^{-26} \Xunits{J/T}) (24.8 \times 10^{-3} \Xunits{T})\\ \amp \approx\amp 6.99 \times 10^{-28} \Xunits{J} \text{.} \end{align*}

Therefore the frequency associated with a photon of this energy is

\begin{equation} f_\text{ photon } = \frac{E_\text{ photon } }{h} = \frac{\Delta E_\text{ proton } }{h} = \frac{6.99 \times 10^{-28} \Xunits{J}}{6.63 \times 10^{-34} \Xunits{J \cdot s}} = 1.05 \times 10^{6} \Xunits{Hz} = 1.05 \Xunits{MHz}\text{.}\tag{5.30} \end{equation}

This frequency lies in the region of the electromagnetic spectrum associated with radio waves.

Like the proton, an electron also has intrinsic spin with spin quantum number \(s = 1/2\text{.}\) In contrast to the proton, the negative charge of an electron means its magnetic moment \(\vec{\mu}\) points opposite the direction of \(\vec{S}\text{.}\) The electron has \(\mu_\text{e} = -9.28 \times 10^{-24} \Xunits{J/T}\text{,}\) where the negative sign denotes the opposing directions of \(\vec{\mu}\) and \(\vec{S}\text{.}\) Replacing \(\mu_\text{p}\) with \(\mu_\text{e}\) in equation (5.28), we find that the higher energy level for an electron occurs when the electron is spin-up and the lower energy level occurs when the electron is spin-down (for the case where \(\vec{B}\) is pointing in the positive \(z\)-direction).

In a hydrogen atom, the nucleus (a proton) and the electron both have spin magnetic moments. Thus, the energy levels of the hydrogen atom are slightly split by the interaction between the electron spin and the nuclear spin, a phenomenon known as hyperfine splitting. When the electron's spin flips, the atom transitions between two hyperfine energy levels and a photon can be emitted or absorbed in the process. Because so much of the gas composing galaxies is neutral hydrogen, this spin-flip transition allowed astronomers to map the spiral structure of the Milky Way for the first time. In your homework, you'll take a closer look at hyperfine splitting.