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Next: Spin wave functions Up: lecture_9 Previous: lecture_9

Electron spin

The quantum numbers $n,l,m$ are not sufficient to fully characterize the physical state of the electrons in an atom. In 1926, Otto Stern and Walther Gerlach carried out an experiment that could not be explained in terms of the three quantum numbers $n,l,m$ and showed that there is, in fact, another quantum-mechanical degree of freedom that needs to be included in the theory.



The experiment is illustrated in the figure below:

Figure: The Stern-Gerlach apparatus.
\includegraphics[scale=0.5]{Stern-Gerlach.eps}
A beam of atoms (e.g. hydrogen or silver atoms) is sent through a spatially inhomogeneous magnetic field with a definite field gradient toward one of the poles. It is observed that the beam splits into two beams as it passes through the field region.



It is known that a current loop in a nonuniform magnetic field experiences a net force. This is illustrated below:

Figure: A current loop in a magnetic field.
\includegraphics[scale=0.5]{electron_current_loop1.eps} \includegraphics[scale=0.5]{electron_current_loop2.eps}
This force arises from an energy $E$ given by

\begin{displaymath}
E = -{\bf M}\cdot {\bf B}
\end{displaymath}

where ${\bf M}$ is called the magnetic moment of the current loop. A current loop, as the figure suggests, is caused by circulating charges. A circulating particle has an associated angular moment ${\bf L}$. If the charge of the particle is $q$ and its mass is $m$, then the magnetic moment is given by

\begin{displaymath}
{\bf M} = {q \over 2m}{\bf L}
\end{displaymath}



The fact that the beam splits into 2 beams suggests that the electrons in the atoms have a degree of freedom capable of coupling to the magnetic field. That is, an electron has an intrinsic magnetic moment ${\bf M}$ arising from a degree of freedom that has no classical analog. The magnetic moment must take on only 2 values according to the Stern-Gerlach experiment. The intrinsic property that gives rise to the magnetic moment must have some analog to an angular momentum and hence, must be a property that, unlike charge and mass, which are simple numbers, is a vector property. This property is called the spin, ${\bf S}$, of the electron. As the expression above suggests, the intrinsic magnetic moment ${\bf M}$ of the electron must be propertional to the spin

\begin{displaymath}
{\bf M} = \gamma {\bf S}
\end{displaymath}



In quantum mechanics, spin share numerous features in common with angular momentum, which is why we represent it as a vector. In particular, spin is quantized, i.e. we have certain allowed values of spin. Like angular momentum, the value of the magnitude squared $S^2$ of spin is fixed, and one of its components $S_z$ is as well. For an electron, the allowed values of $S_z$ are

\begin{displaymath}
S_z \longrightarrow m_s\hbar\;\;\;\;\;\;\;m_s = -{1 \over 2},{1 \over 2}
\end{displaymath}

while $S^2$ has just one value $3\hbar^2/4$, corresponding to a general formula $s(s+1)\hbar^2$, where $s=1/2$. For this reason, the electron is called a spin-1/2 particle. The formula $s(s+1)\hbar^2$ is generally valid for any spin-$s$ particle. The constant $\gamma$ is called the spin gyromagnetic ratio

\begin{displaymath}
\gamma = -{ge \over 2m_e }
\end{displaymath}

where $g=2$ for electrons. Note that $\gamma <0$.



For an electron in a uniform magnetic field ${\bf B}$, the energy is determined by the spin ${\bf S}$:

\begin{displaymath}
E = -\gamma {\bf S}\cdot {\bf B}
\end{displaymath}

We choose the uniform field ${\bf B}$ to be along the $z$-direction ${\bf B} = (0,0,B)$. Since the field lines flow from the north pole to the south pole, this choice of the ${\bf B}$ field means that the north pole lies below the south pole on the $z$-axis. In this case,

\begin{displaymath}
E = -\gamma B S_z
\end{displaymath}

Since $\gamma <0$, we see that the lowest energy configuration uas ${\bf S}$ antiparallel to ${\bf B}$. Given the two values of $S_z$, we have two allowed energies or two energy levels corresponding to the two values of $m_s$:

\begin{displaymath}
E_{1/2} = -{\gamma B \hbar \over 2}
\;\;\;\;\;\;\;\;\;\;
E_{-1/2} = {\gamma B \hbar \over 2}
\end{displaymath}

Using the given expression for $\gamma$, which is negative, we obtain the two energy levels

\begin{displaymath}
E_{1/2} = {eB\hbar \over 2m_e }
\;\;\;\;\;\;\;\;\;\;
E_{-1/2} = -{eB\hbar \over 2m_e }
\end{displaymath}

Unlike position and momentum, which have clear classical analogs, spin does not. But if we think of spin in pseudoclassical terms, we can think of a spinning charged particle, which is similar to a loop of current. Thus, if the particle spins about the $z$-axis, then ${\bf S}$ points along the $z$-axis. Since the spinning charge is negative, the left-hand rule can be applied. When the fingers of the left hand curl in the direction of the spin, the thumb points in the direction of the spin. A spinning charge produces a magnetic field similar to that of a tiny bar magnet. In this case, the spin vector ${\bf S}$ points toward the south pole of the bar magnet. Now if the spinning particle is placed in a magnetic field, it tends to align the spin vector in the opposite to the magnetic field lines, as the figure above suggests.

Now, when the electron is placed in a nonuniform magnetic field, with the field increasing in strength toward the north pole of the field source, the spin-down ($m_s=-1/2$) electrons have their bar-magnet poles oriented such that the south pole points toward the north pole of the field source, and these electrons will be attracted toward the region of stronger field. The spin-up ($m_s=1/2$) electrons have their bar-magnet north poles oriented toward the north pole of the field source and will be repelled to the region of weaker field, thus causing the beam to split as observed in the Stern-Gerlach experiment.



The implication of the Stern-Gerlach experiment is that we need to include a fourth quantum number, $m_s$ in our description of the physical state of the electron. That is, in addition to give its principle, angular, and magnetic quantum numbers, we also need to say if it is a spin-up electron or a spin-down electron.



Note that we have added spin into the our quantum theory as a kind of a posteriori consideration, which seems a little contrived. In fact, the existence of the spin degree of freedom can be derived in a very natural way in the relativistic version of quantum mechanics, where it simply pops out of the relativistic analog of the Schrödinger equation, known as the Dirac equation.


next up previous
Next: Spin wave functions Up: lecture_9 Previous: lecture_9
Mark E. Tuckerman 2011-11-03