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Physics LibreTexts

11.5: Electron Spin

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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,$ ???

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}).$ ???

However,

$ ({\bf p}+e\,{\bf A})\times ({\bf p}+e\,{\bf A}) = e\,{\bf p}\time...
...\bf A} -{ \rm i}\,e\,\hbar\,{\bf A}\times \nabla = -{\rm i}\,e\,\hbar\,{\bf B},$ ???

where (Hc+ecϕ)2=(p+eA)2+m2ec2+eB.

???

Consider the non-relativistic limit. In this case, we can write

δH is small compared to δH2 , and other terms involving δHeϕ+12me(p+eA)2+e2meB. ???

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

=e2me. ???

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, i˙Lx=[Lx,H]. ???

However, it is easily demonstrated that

=0, ??? =0, ??? =ipz, ??? =ipy. ???

Hence, we obtain

˙Lx=cγ5(Σ2pzΣ3py). ???

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(Σ3pyΣ2pz), ???

which implies that

˙Lx+2˙Σ1=0. ???

Since there is nothing special about the L+(/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)


This page titled 11.5: Electron Spin is shared under a not declared license and was authored, remixed, and/or curated by Richard Fitzpatrick.

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