13.1 The Electromagnetic Hamiltonian

This section describes very basically how electro­magnetism fits into quantum mechanics. However, electro­magnetism is fundamentally relativistic; its carrier, the photon, readily emerges or disappears. To describe electro­magnetic effects fully requires quantum electro­dynamics, and that is far beyond the scope of this text. (However, see addenda {A.15} and {A.23} for some of the ideas.)

In classical electro­magnetics, the force on a particle with charge $q$ in a field with electric strength $\skew3\vec{\cal E}$ and magnetic strength $\skew2\vec{\cal B}$ is given by the Lorentz force law

$$\fbox{$\displaystyle m \frac{{\rm d}\vec v}{{\rm d}t} = ... ...3\vec{\cal E}+ \vec v\times \skew2\vec{\cal B}\right) $} %$$ (13.1)

where $\vec{v}$ is the velocity of the particle and for an electron, the charge is $q$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\vphantom0\raisebox{1.5pt}{$-$}$$e$.

Unfortunately, quantum mechanics uses neither forces nor velocities. In fact, the earlier analysis of atoms and molecules in this book used the fact that the electric field is described by the corre­sponding potential energy $V$, see for example the Hamiltonian of the hydrogen atom. The magnetic field must appear differen­tly in the Hamiltonian; as the Lorentz force law shows, it couples with velocity. You would expect that still the Hamiltonian would be relatively simple, and the simplest idea is then that any potential corre­sponding to the magnetic field moves in together with momentum. Since the momentum is a vector quantity, then so must be the magnetic potential. So, your simplest guess would be that the Hamiltonian takes the form

$$\fbox{$\displaystyle H = \frac{1}{2m}\left({\skew 4\wide... ...kew{-.5}\vec p}}- q \skew3\vec A\right)^2 + q \varphi $} %$$ (13.2)

where $\varphi$ $\vphantom0\raisebox{1.5pt}{$=$}$ $V$$\raisebox{.5pt}{$/$}$$q$ is the “electric potential” and $\skew3\vec A$ is the “magnetic vector potential.” And this simplest guess is in fact right.

The relationship between the vector potential $\skew3\vec A$ and the magnetic field strength $\skew2\vec{\cal B}$ will now be found from requiring that the classical Lorentz force law is obtained in the classical limit that the quantum uncertainties in position and momentum are small. In that case, expec­tation values can be used to describe position and velocity, and the field strengths $\skew3\vec{\cal E}$ and $\skew2\vec{\cal B}$ will be constant on the small quantum scales. That means that the derivatives of $\varphi$ will be constant, (since $\skew3\vec{\cal E}$ is the negative gradient of $\varphi$), and presumably the same for the derivatives of $\skew3\vec A$.

Now according to chapter 7.2, the evolution of the expec­tation value of position is found as

$$\frac{{\rm d}\langle {\skew 2\widehat{\skew{-1}\vec r}}\ra... ...langle \frac{{\rm i}}{\hbar} [H,{\skew0\vec r}] \right\rangle$$

Working out the commutator with the Hamiltonian above, {D.72}, you get,

$$\frac{{\rm d}\langle {\skew 2\widehat{\skew{-1}\vec r}}\ra... ...skew 4\widehat{\skew{-.5}\vec p}}- q\skew3\vec A\right\rangle$$

This is unexpected; it shows that ${\skew 4\widehat{\skew{-.5}\vec p}}$, i.e. $\hbar\nabla$$\raisebox{.5pt}{$/$}$${\rm i}$, is no longer the operator of the normal momentum $m\vec{v}$ when there is a magnetic field; ${\skew 4\widehat{\skew{-.5}\vec p}}-q\skew3\vec A$ gives the normal momentum. The momentum represented by ${\skew 4\widehat{\skew{-.5}\vec p}}$ by itself is called “canonical” momentum to distin­guish it from normal momentum:

The canonical momentum $\hbar\nabla$$\raisebox{.5pt}{$/$}$${\rm i}$ only corre­sponds to normal momentum if there is no magnetic field involved.

(Actually, it was not that unexpected to physicists, since the same happens in the classical description of electro­magnetics using the so-called Lagrangian approach, chapter 1.3.2.)

Next, Newton’s second law says that the time derivative of the linear momentum $m\vec{v}$ is the force. Since according to the above, the linear momentum operator is ${\skew 4\widehat{\skew{-.5}\vec p}}-q\skew3\vec A$, then

$$m \frac{{\rm d}\langle\vec v\rangle}{{\rm d}t} = \frac{... ...langle \frac{\partial \skew3\vec A}{\partial t} \right\rangle$$

The objective is now to ensure that the right hand side is the correct Lorentz force (13.1) for the assumed Hamiltonian, by a suitable definition of $\skew2\vec{\cal B}$ in terms of $\skew3\vec A$.

After a lot of grinding down commutators, {D.72}, it turns out that indeed the Lorentz force is obtained,

$$m \frac{{\rm d}\langle \vec v \rangle}{{\rm d}t} = q \le... ...l E}+ \langle \vec v \rangle \times \skew2\vec{\cal B}\right)$$

provided that:

$$\fbox{$\displaystyle \skew3\vec{\cal E}= -\nabla \varphi... ...quad \skew2\vec{\cal B}= \nabla \times \skew3\vec A $} %$$ (13.3)

So the magnetic field is found as the curl of the vector potential $\skew3\vec A$. And the electric field is no longer just the negative gradient of the scalar potential $\varphi$ if the vector potential varies with time.

These results are not new. The electric scalar potential $\varphi$ and the magnetic vector potential $\skew3\vec A$ are the same in classical physics, though they are a lot less easy to guess than done here. Moreover, in classical physics they are just convenient mathematical quantities to simplify analysis. In quantum mechanics they appear as central to the formul­ation.

And it can make a difference. Suppose you do an experi­ment where you pass electron wave functions around both sides of a very thin magnet: you will get a wave inter­ference pattern behind the magnet. The classical expec­tation is that this inter­ference pattern will be independent of the magnet strength: the magnetic field $\skew2\vec{\cal B}$ outside a very thin and long ideal magnet is zero, so there is no force on the electron. But the magnetic vector potential $\skew3\vec A$ is not zero outside the magnet, and Aharonov and Bohm argued that the inter­ference pattern would therefore change with magnet strength. So it turned out to be in experi­ments done subsequently. The conclusion is clear; nature really goes by the vector potential $\skew3\vec A$ and not the magnetic field $\skew2\vec{\cal B}$ in its actual workings.