Schrödinger’s equation in a magnetic field
This lecture is only for entertainment. I would like to give the
lecture in a somewhat different style—just to see how it works
out. It’s not a part of the course—in the sense that it is not
supposed to be a last minute effort to teach you something new. But,
rather, I imagine that I’m giving a seminar or research report on the
subject to a more advanced audience, to people who have already been
educated in quantum mechanics. The main difference between a seminar
and a regular lecture is that the seminar speaker does not carry out
all the steps, or all the algebra. He says: “If you do such and such,
this is what comes out,” instead of showing all of the details. So in
this lecture I’ll describe the ideas all the way along but just give
you the results of the computations. You should realize that
you’re not supposed to understand everything immediately, but believe
(more or less) that things would come out if you went through the
steps. All that aside, this is a subject I want to talk about. It is
recent and modern and would be a perfectly legitimate talk to give at
a research seminar. My subject is the Schrödinger equation in a
classical setting—the case of superconductivity. Ordinarily, the wave function which appears in the Schrödinger
equation applies to only one or two particles. And the wave function
itself is not something that has a classical meaning—unlike the
electric field, or the vector potential, or things of that kind. The
wave function for a single particle is a “field”—in the
sense that it is a function of position—but it does not generally
have a classical significance. Nevertheless, there are some situations
in which a quantum mechanical wave function does have classical
significance, and they are the ones I would like to take up. The
peculiar quantum mechanical behavior of matter on a small scale
doesn’t usually make itself felt on a large scale except in the
standard way that it produces Newton’s laws—the laws of the
so-called classical mechanics. But there are certain situations in
which the peculiarities of quantum mechanics can come out in a special
way on a large scale. At low temperatures, when the energy of a system has been reduced
very, very low, instead of a large number of states being involved,
only a very, very small number of states near the ground state are
involved. Under those circumstances the quantum mechanical character
of that ground state can appear on a macroscopic scale. It is the
purpose of this lecture to show a connection between quantum mechanics
and large-scale effects—not the usual discussion of the way that
quantum mechanics reproduces Newtonian mechanics on the average, but a
special situation in which quantum mechanics will produce its own
characteristic effects on a large or “macroscopic” scale. I will begin by reminding you of some of the properties of the
Schrödinger equation.1 I want to describe the behavior of a particle
in a magnetic field using the Schrödinger equation, because the
superconductive phenomena are involved with magnetic fields. An
external magnetic field is described by a vector potential, and the
problem is: what are the laws of quantum mechanics in a vector
potential? The principle that describes the behavior of quantum
mechanics in a vector potential is very simple. The amplitude that a
particle goes from one place to another along a certain route when
there’s a field present is the same as the amplitude that it would go
along the same route when there’s no field, multiplied by the
exponential of the line integral of the vector potential, times the
electric charge divided by
Planck’s constant2 (see Fig. 21–1):
\begin{equation}
\label{Eq:III:21:1}
\braket{b}{a}_{\text{in $\FLPA$}}=\braket{b}{a}_{A=0}\cdot
\exp\biggl[\frac{iq}{\hbar}\int_a^b\FLPA\cdot d\FLPs\biggr].
\end{equation}

It is a basic statement of quantum mechanics. Now without the vector potential the Schrödinger equation of a
charged particle (nonrelativistic, no spin) is

\begin{equation}
\label{Eq:III:21:2}
-\frac{\hbar}{i}\,\ddp{\psi}{t}=\Hcalop\psi=
\frac{1}{2m}\biggl(\frac{\hbar}{i}\,\FLPnabla\biggr)\cdot
\biggl(\frac{\hbar}{i}\,\FLPnabla\biggr)\psi+q\phi\psi,
\end{equation}


\begin{gather}
\label{Eq:III:21:2}
-\frac{\hbar}{i}\,\ddp{\psi}{t}=\Hcalop\psi=\\[1ex]
\frac{1}{2m}\biggl(\frac{\hbar}{i}\,\FLPnabla\biggr)\cdot
\biggl(\frac{\hbar}{i}\,\FLPnabla\biggr)\psi+q\phi\psi,\notag
\end{gather}

where $\phi$ is the electric potential so that $q\phi$ is the
potential energy.3 Equation (21.1) is equivalent
to the statement that in a magnetic field the gradients in the
Hamiltonian are replaced in each case by the gradient minus $q\FLPA$,
so that Eq. (21.2) becomes

\begin{equation}
\label{Eq:III:21:3}
-\frac{\hbar}{i}\,\ddp{\psi}{t}=\Hcalop\psi=
\frac{1}{2m}\biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\cdot
\biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi+q\phi\psi.
\end{equation}


\begin{gather}
\label{Eq:III:21:3}
-\frac{\hbar}{i}\,\ddp{\psi}{t}=\Hcalop\psi=\\[1ex]
\frac{1}{2m}\biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\cdot
\biggl(\frac{\hbar}{i}\,\FLPnabla-q\FLPA\biggr)\psi+q\phi\psi.\notag
\end{gather}

This is the Schrödinger equation for a particle with charge $q$
moving in an electromagnetic field $\FLPA,\phi$ (nonrelativistic, no
spin). To show that this is true I’d like to illustrate by a simple example
in which instead of having a continuous situation we have a line of
atoms along the $x$-axis with the spacing $b$ and we have an
amplitude $iK\!/\hbar$ per unit time for an electron to jump from one atom to another when there is no
field.4  Now according to Eq. (21.1) if
there’s a vector potential in the $x$-direction $A_x(x,t)$, the
amplitude to jump will be altered from what it was before by a
factor $\exp[(iq/\hbar)\,A_xb]$, the exponent being $iq/\hbar$ times the vector
potential integrated from one atom to the next. For simplicity we will
write $(q/\hbar)A_x\equiv f(x)$, since $A_x$, will, in general, depend
on $x$. If the amplitude to find the electron at the atom “$n$”
located at $x$ is called $C(x)\equiv C_n$, then the rate of change of
that amplitude is given by the following equation:
\begin{align}
-\frac{\hbar}{i}\ddp{}{t}C(x)=E_0C(x)&\!-\!Ke^{-ibf(x+b/2)}C(x\!+\!b)\notag\\
\label{Eq:III:21:4}
&-\!Ke^{+ibf(x-b/2)}C(x\!-\!b).
\end{align}
 There are three pieces. First, there’s some energy $E_0$ if the
electron is located at $x$. As usual, that gives the
term $E_0C(x)$. Next, there is the term $-KC(x+b)$, which is the amplitude
for the electron to have jumped backwards one step from
atom “$n+1$,” located at $x+b$. However, in doing so in a vector
potential, the phase of the amplitude must be shifted according to the
rule in Eq. (21.1). If $A_x$ is not changing
appreciably in one atomic spacing, the integral can be written as just
the value of $A_x$ at the midpoint, times the spacing $b$. So
$(iq/\hbar)$ times the integral is just $ibf(x+b/2)$. Since the
electron is jumping backwards, I showed this phase shift with a minus
sign. That gives the second piece. In the same manner there’s a
certain amplitude to have jumped from the other side, but this time we
need the vector potential at a distance $(b/2)$ on the other side
of $x$, times the distance $b$. That gives the third piece. The sum gives
the equation for the amplitude to be at $x$ in a vector potential. Now we know that if the function $C(x)$ is smooth enough (we take the
long wavelength limit), and if we let the atoms get closer together,
Eq. (21.4) will approach the behavior of an electron in
free space. So the next step is to expand the right-hand side
of (21.4) in powers of $b$, assuming $b$ is very small. For
example, if $b$ is zero the right-hand side is just $(E_0-2K)C(x)$, so
in the zeroth approximation the energy is $E_0-2K$. Next comes the terms
in $b$. But because the two exponentials have opposite signs, only even
powers of $b$ remain. So if you make a Taylor expansion of $C(x)$,
of $f(x)$, and of the exponentials, and then collect the terms in $b^2$,
you get

\begin{align}
-\frac{\hbar}{i}\,\ddp{C(x)}{t}&=E_0C(x)-2KC(x)\notag\\
\label{Eq:III:21:5}
&\quad-Kb^2\{C''(x)-2if(x)C'(x)-if'(x)C(x)-f^2(x)C(x)\}.
\end{align}


\begin{align}
\label{Eq:III:21:5}
-\frac{\hbar}{i}\,\ddp{C(x)}{t}&=E_0C(x)-2KC(x)\\[1ex]
&-Kb^2\bigl\{C''(x)-2if(x)C'(x)\,-\notag\\[1ex]
&\phantom{-Kb^2\{}\;if'(x)C(x)-f^2(x)C(x)\bigr\}.\notag
\end{align}

(The “primes” mean differentiation with respect to $x$.) Now this horrible combination of things looks quite complicated. But
mathematically it’s exactly the same as

\begin{equation}
\label{Eq:III:21:6}
-\frac{\hbar}{i}\,\ddp{C(x)}{t}=(E_0-2K)C(x)-Kb^2
\biggl[\ddp{}{x}-if(x)\biggr]
\biggl[\ddp{}{x}-if(x)\biggr]C(x).
\end{equation}


\begin{align}
\label{Eq:III:21:6}
-\frac{\hbar}{i}\ddp{C(x)}{t}&=(E_0-2K)C(x)\\
&-Kb^2 \biggl[\ddp{}{x}-if(x)\biggr]
\biggl[\ddp{}{x}-if(x)\biggr]C(x).\notag
\end{align}

The second bracket operating on $C(x)$ gives $C'(x)$
minus $if(x)C(x)$. The first bracket operating on these two terms gives the
$C''$ term and terms in the first derivative of $f(x)$ and the first
derivative of $C(x)$. Now remember that the solutions for zero
magnetic field5 represent a
particle with an effective mass $m_{\text{eff}}$ given by
\begin{equation*}
Kb^2=\frac{\hbar^2}{2m_{\text{eff}}}.
\end{equation*}

If you then set $E_0=2K$, and put back $f(x)=(q/\hbar)A_x$, you can
easily check that Eq. (21.6) is the same as the first
part of Eq. (21.3). (The origin of the potential energy
term is well known, so I haven’t bothered to include it in this
discussion.) The proposition of Eq. (21.1) that the
vector potential changes all the amplitudes by the exponential factor
is the same as the rule that the momentum operator,
$(\hbar/i)\FLPnabla$ gets replaced by
\begin{equation*}
\frac{\hbar}{i}\,\FLPnabla-q\FLPA,
\end{equation*}

as you see in the Schrödinger equation of (21.3).