11.5: Electron Spin
( \newcommand{\kernel}{\mathrm{null}\,}\)
According to Equation ???, the relativistic Hamiltonian of an electron in an electromagnetic field is
\( \mbox{\boldmath\)⋅(p+eA)+βmec2. ???Hence,
![$ \left(\frac{H}{c}+\frac{e}{c}\,\phi\right)^2 = \left[\mbox{\boldm...
...mbox{\boldmath$\alpha$}\cdot({\bf p}+e\,{\bf A})\right]^{\,2} + m_e^{\,2}\,c^2,$](http://farside.ph.utexas.edu/teaching/qm/lectures/img2729.png)
where use has been made of Equations ??? and ???. Now, we can write
i=1,3, where Σi=(σi0[0.5ex]0σi). ???Here, 0 and 1 denote σi are conventional γ5γ5=1 , and
\( \left(\frac{H}{c}+\frac{e}{c}\,\phi\right)^2 = \left[\mbox{\boldmath\) ???Now, a straightforward generalization of Equation ??? gives
\( \mbox{\boldmath\)\( \mbox{\boldmath\)\( \mbox{\boldmath\)a and Σ . It follows that![$ \left[\mbox{\boldmath$\Sigma$}\cdot({\bf p}+e\,{\bf A})\right]^{\...
...,\mbox{\boldmath$\Sigma$}\cdot ({\bf p}+e\,{\bf A})\times ({\bf p}+e\,{\bf A}).$](http://farside.ph.utexas.edu/teaching/qm/lectures/img2741.png)
However,

where (Hc+ecϕ)2=(p+eA)2+m2ec2+eℏ⋅B.
???Consider the non-relativistic limit. In this case, we can write
δH is small compared to δH2 , and other terms involving δH≃−eϕ+12me(p+eA)2+eℏ2me⋅B. ???This Hamiltonian is the same as the classical Hamiltonian of a non-relativistic electron, except for the final term. This term may be interpreted as arising from the electron having an intrinsic magnetic moment
=−eℏ2me. ???In order to demonstrate that the electron's intrinsic magnetic moment is associated with an intrinsic angular momentum, consider the motion of an electron in a central electrostatic potential: i.e., A=0 . In this case, the Hamiltonian ??? becomes
\( \mbox{\boldmath\)x component of the electron's orbital angular momentum, iℏ˙Lx=[Lx,H]. ???However, it is easily demonstrated that
=0, ??? =0, ??? =iℏpz, ??? =−iℏpy. ???Hence, we obtain
˙Lx=cγ5(Σ2pz−Σ3py). ???It can be seen that x -component of the total angular momentum of the system must be a constant of the motion (because a central electrostatic potential exerts zero torque on the system). Hence, we deduce that the electron possesses additional angular momentum that is not connected with its motion through space. Now,
[Σ1,γ5] [Σ1,Σ1] [Σ1,Σ2] [Σ1,Σ3] [Σ1,H]=2icγ5(Σ3py−Σ2pz), ???which implies that
˙Lx+ℏ2˙Σ1=0. ???Since there is nothing special about the L+(ℏ/2)Σ is a constant of the motion. We can interpret this result by saying that the electron has a spin angular momentum Σ , which must be added to its orbital angular momentum in order to obtain a constant of the motion. According to ???, the relationship between the electron's spin angular momentum and its intrinsic (i.e., non-orbital) magnetic moment is
=−eg2meS, ???where the gyromagnetic ratio g=2.
???As explained in Section 5.5, this is twice the value one would naively predict by analogy with classical physics.
Contributors
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)