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My lecturer today said that when an applied magnetic field, $B_0$, is applied, the electron with spin parallel to $B_0$ is higher in energy than the electron with spin antiparallel to $B_0$. Similarly, in protons, it is the opposite; parallel spin is lower in energy and antiparallel spin is higher in energy. Why is that?

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  • $\begingroup$ Do you want to know why there is an energy difference to begin with, or why electrons and protons behave differently? $\endgroup$ – orthocresol Aug 24 '16 at 2:36
  • $\begingroup$ I understand that there exists a energy difference, but I just don't understand why when the spin is parallel to the direction of the external magnetic field it is higher in energy $\endgroup$ – Patrick Robertson Aug 24 '16 at 3:43
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In addition to the other answer, the energy of a magnetic moment in a magnetic field is $E=-\bar{\bf{\mu}}.\bar{\bf{B}}$ where the dot indicates the dot product between the two vectors $\bar{\bf{\mu}}$ and $\bar{\bf{B}}$, thus the energy is lowest when the magnetic dipole is parallel to the field $B_0$; $E=-\mu B_0$. The sign of $\mu$ is opposite for nuclear and electron spin angular momentum.
The magnetic moment of a nuclear spin is defined as $\mu=+\gamma I$, where $I$ is the nuclear angular momentum and if the field is along only the $z$ direction $\mu_z=\gamma I_z=m_z\gamma \hbar$ (Joule/Tesla) where $\gamma$ is the magnetogyric ratio, (unit radians/Tesla/sec) the components are $m_z$, the projection, magnetic or azimuthal spin quantum number on the $z$ axis, the axis parallel to the magnetic field $B_0$. $I_z$ is the $z$ component of the nuclear spin angular momentum. (The magnetogyric ratio is $\gamma = g\mu _0/\hbar$ where $\mu_0$ is the nuclear magnetron and $g$ the nuclear spin factor, different for each type of nucleus). The energy is then $E=-m_z\gamma\hbar B_0 $. For a proton $m_z=\pm 1/2$.

The spin magnetic moment of the electron is defined in a similar way to that for nuclear spin but with a negative sign; $\mu_e=-\gamma _e I_e$ where $I_e$ is the electron spin angular momentum and $\gamma _e$ the electron magnetogyric ratio. ( $\gamma _e=g_e\mu_B/\hbar$, $g_e$ is the electron $g$ factor, $g_e =2.0023..$, and $\mu _B$ the Bohr magnetron. Some authors give a negative $g_e$, in this case the negative sign is removed from $\mu _e$).
(The nuclear magnetogyric ratio can be positive or negative, e.g. $\ce{26.75.10^7 rad T^{-1} s^{-1}}$ for protons. For a few types of nuclei e.g. $\ce{^{15}N, ^{29}Si}$ $\gamma$ is negative. The magnetogyric ratio for the electron is much $\ce{larger 1.76 10^11 rad T^{-1} s^{-1}}$.)

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In short: electrons and protons have magnetic moments. Electrons are negatively charged, and the direction of their magnetic moments is reversed from that of protons.

Intuitively, the potential energy is lowest when the magnetic moment is aligned with the magnetic field, as is the case with the proton. Because electrons are negatively charged, however, their magnetic moment points in the opposite direction, and the potential energy is then lowest when the magnetic moment is anti-aligned with the magnetic field.

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