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\SetTitle{Physics 312 -- Assignment 4} \SetAuthor{} \Setdate{}
\TitlePage{}

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\textbf{1.a.} Given the complex number $z=1-i/2$, find $z^{2},$ $z^{3},$ $%
z^{4},$ $z^{5}$ and draw them as ``vectors'' in the complex plane. Find $%
1/z, $ $1/z^{2},$ $1/z^{3},$ $1/z^{4},$ $1/z^{5}$ and draw them as
``vectors'' in the complex plane.

\textbf{b.} Given the complex number $z=\exp (i/5)$, find its
first twelve powers and draw them as ``vectors'' in the complex
plane. Do the same for $z=\exp (-i/5)$.

\textbf{c.} Find the real and imaginary parts of the complex ratio $%
(7+3i)/(5-4i)$. Find its magnitude (absolute value) and phase
angle. Draw numerator, denominator, and ratio as ``vectors'' in
the complex plane.

\vspace{0.5cm}

\textbf{a.} If $z=1-i/2$, then
\begin{eqnarray}
z^{2} &=&\left( 1-\frac{i}{2}\right) ^{2}=1+2\left( -\frac{i}{2}\right)
+\left( -\frac{i}{2}\right) ^{2}  \nonumber \\
&=&1-i-\frac{1}{4}=\frac{3}{4}-i \\
&&  \nonumber \\
z^{3} &=&\left( 1-\frac{i}{2}\right) ^{3}=1+3\left( -\frac{i}{2}\right)
+3\left( -\frac{i}{2}\right) ^{2}+\left( -\frac{i}{2}\right) ^{3}  \nonumber
\\
&=&1-\frac{3i}{2}-\frac{3}{4}+\frac{i}{8}=\frac{1}{4}-\frac{11i}{8} \\
&&  \nonumber \\
z^{4} &=&(z^{2})^{2}=\left( \frac{3}{4}-i\right) ^{2}=\left( \frac{3}{4}%
\right) ^{2}+2\left( \frac{3}{4}\right) (-i)+(-i)^{2}  \nonumber \\
&=&\frac{9}{16}-\frac{3i}{2}-1=-\frac{7}{16}-\frac{3i}{2} \\
&&  \nonumber \\
z^{5} &=&(z^{2})(z^{3})=\left( \frac{3}{4}-i\right) \left( \frac{1}{4}-\frac{%
11i}{8}\right)  \nonumber \\
&=&\left( \frac{3}{4}\right) \left( \frac{1}{4}\right) +\left( \frac{3}{4}%
\right) \left( \frac{-11i}{8}\right) +\left( -i\right) \left( \frac{1}{4}%
\right) +\left( -i\right) \left( \frac{-11i}{8}\right)  \nonumber \\
&=&\frac{3}{16}-\frac{33i}{32}-\frac{i}{4}-\frac{11}{8}=-\frac{19}{16}-\frac{%
41i}{32}
\end{eqnarray}

The numbers $z$, $z^{2}$, $z^{3}$, $z^{4}$, and $z^{5}$ are plotted below.

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"T";tempfilename 'C:/Temp/EOH6P300.wmf';tempfile-properties "XP";}}

To find an expression for $1/z$ multiply by $z^*/z^*$, where $z^*$
is the \textit{complex conjugate} of $z$, as follows
\begin{eqnarray}
\frac{1}{z}&=&\frac{1}{1-i/2} = \left(\frac{1}{1-i/2}\right)\left(\frac{1+i/2%
}{1+i/2} \right)  \nonumber \\
&=& \frac{1+i/2}{1+1/4} = \frac{4}{5}+ \frac{2i}{5}
\end{eqnarray}

The rest, $z^{-2}$, $z^{-3}$, etc., are obtained from $z^{-1}$:
\begin{eqnarray}
\frac{1}{z^{2}} &=&\left( \frac{4}{5}+\frac{2i}{5}\right) ^{2}=\left( \frac{4%
}{5}\right) ^{2}+2\left( \frac{4}{5}\right) \left( \frac{2i}{5}\right)
+\left( \frac{2i}{5}\right) ^{2}  \nonumber \\
&=&\frac{16}{25}+\frac{16i}{25}-\frac{4}{25}=\frac{12}{25}+\frac{16i}{25} \\
&&  \nonumber \\
\frac{1}{z^{3}} &=&\left( \frac{4}{5}+\frac{2i}{5}\right) \left( \frac{12}{25%
}+\frac{16i}{25}\right)  \nonumber \\
&=&\left( \frac{4}{5}\right) \left( \frac{12}{25}\right) +\left( \frac{4}{5}%
\right) \left( \frac{16i}{25}\right) +\left( \frac{2i}{5}\right) \left(
\frac{12}{25}\right) +\left( \frac{2i}{5}\right) \left( \frac{16i}{25}\right)
\nonumber \\
&=&\frac{48}{125}+\frac{64i}{125}+\frac{24i}{125}-\frac{32}{125}=\frac{16}{%
125}+\frac{88i}{125} \\
&&  \nonumber \\
\frac{1}{z^{4}} &=&\left( \frac{12}{25}+\frac{16i}{25}\right) ^{2}=\left(
\frac{12}{25}\right) ^{2}+2\left( \frac{12}{25}\right) \left( \frac{16i}{25}%
\right) +\left( \frac{16i}{25}\right) ^{2}  \nonumber \\
&=&\frac{144}{625}+\frac{384i}{625}-\frac{256}{625}=-\frac{112}{625}+\frac{%
384i}{625} \\
&&  \nonumber \\
\frac{1}{z^{5}} &=&\left( \frac{4}{5}+\frac{2i}{5}\right) \left( -\frac{112}{%
625}+\frac{384i}{625}\right)  \nonumber \\
&=&-\left( \frac{4}{5}\right) \left( \frac{112}{625}\right) +\left( \frac{4}{%
5}\right) \left( \frac{384i}{625}\right) +-\left( \frac{2i}{5}\right) \left(
\frac{112}{625}\right) +\left( \frac{2i}{5}\right) \left( \frac{384i}{625}%
\right)  \nonumber \\
&=&-\frac{448}{3125}+\frac{1536i}{3125}-\frac{224i}{3125}-\frac{768}{3125}=-%
\frac{1216}{3125}+\frac{1312i}{3125}
\end{eqnarray}

The numbers $1/z$, $1/z^{2}$, $1/z^{3}$, $1/z^{4}$, and $1/z^{5}$ are
plotted below.

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"T";tempfilename 'C:/Temp/EOH6PA01.wmf';tempfile-properties "XP";}}

\textbf{b.} Given the form of $z$ in this part, it is convenient to use
\textit{Euler's identity}: $\exp (i\theta )=\cos (\theta )+i\sin (\theta )$
(where $\theta $ is expressed in radians).
\begin{eqnarray}
z &=&\exp (i/5)=\cos (1/5)+i\sin (1/5)\approx 0.980+0.199~i  \nonumber \\
z^{2} &=&\exp (2i/5)=\cos ^{2}(1/5)-\sin ^{2}(1/5)+2i\cos (1/5)\sin (1/5)
\nonumber \\
&=&\cos (2/5)+i\sin (2/5)  \nonumber \\
z^{3} &=&\exp (3i/5)=\cos (3/5)+i\sin (3/5)  \nonumber \\
z^{4} &=&\exp (4i/5)=\cos (4/5)+i\sin (4/5)  \nonumber \\
z^{5} &=&\exp (i)=\cos (1)+i\sin (1)  \nonumber \\
z^{6} &=&\exp (6i/5)=\cos (6/5)+i\sin (6/5)  \nonumber \\
z^{7} &=&\exp (7i/5)=\cos (7/5)+i\sin (7/5)  \nonumber \\
z^{8} &=&\exp (8i/5)=\cos (8/5)+i\sin (8/5)  \nonumber \\
z^{9} &=&\exp (9i/5)=\cos (9/5)+i\sin (9/5)  \nonumber \\
z^{10} &=&\exp (2i)=\cos (2)+i\sin (2)  \nonumber \\
z^{11} &=&\exp (11i/5)=\cos (11/5)+i\sin (11/5)  \nonumber \\
z^{12} &=&\exp (12i/5)=\cos (12/5)+i\sin (12/5)
\end{eqnarray}

\newpage The first seven powers of $z$ are plotted below

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\TEXUX{$0.921t,0.389t$};linecolor "blue";linestyle 1;linethickness
1;pointstyle "point";function \TEXUX{$0.825t,0.565t$};linecolor
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10;axesstyle "normal";xis \TEXUX{t};var1name \TEXUX{$x$};valid_file
"T";tempfilename 'C:/Temp/EOH6PB02.wmf';tempfile-properties "XP";}}

And for $z=\exp (-i/5)$ we find
\begin{eqnarray}
z &=&\exp (-i/5)=\cos (1/5)-i\sin (1/5)  \nonumber \\
z^{2} &=&\exp (-2i/5)=\cos ^{2}(1/5)-\sin ^{2}(1/5)-2i\cos (1/5)\sin (1/5)
\nonumber \\
&=&\cos (2/5)-i\sin (2/5)  \nonumber \\
z^{3} &=&\exp (-3i/5)=\cos (3/5)-i\sin (3/5)  \nonumber \\
z^{4} &=&\exp (-4i/5)=\cos (4/5)-i\sin (4/5)  \nonumber \\
z^{5} &=&\exp (-i)=\cos (1)-i\sin (1)  \nonumber \\
z^{6} &=&\exp (-6i/5)=\cos (6/5)-i\sin (6/5)  \nonumber \\
z^{7} &=&\exp (-7i/5)=\cos (7/5)-i\sin (7/5)  \nonumber \\
z^{8} &=&\exp (-8i/5)=\cos (8/5)-i\sin (8/5)  \nonumber \\
z^{9} &=&\exp (-9i/5)=\cos (9/5)-i\sin (9/5)  \nonumber \\
z^{10} &=&\exp (-2i)=\cos (2)-i\sin (2)  \nonumber \\
z^{11} &=&\exp (-11i/5)=\cos (11/5)-i\sin (11/5)  \nonumber \\
z^{12} &=&\exp (-12i/5)=\cos (12/5)-i\sin (12/5)
\end{eqnarray}

\newpage The first seven of these numbers are plotted below:

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\TEXUX{$0.921t,-0.389t$};linecolor "blue";linestyle 1;linethickness
1;pointstyle "point";function \TEXUX{$0.825t,-0.565t$};linecolor
"brown";linestyle 1;linethickness 1;pointstyle "point";function
\TEXUX{$0.697t,-0.717t$};linecolor "cyan";linestyle 1;linethickness
1;pointstyle "point";function \TEXUX{$0.540t,-0.841t$};linecolor
"gray";linestyle 1;linethickness 1;pointstyle "point";function
\TEXUX{$0.362t,-0.932t$};linecolor "green";linestyle 1;linethickness
1;pointstyle "point";function \TEXUX{$0.170t,-0.985t$};linecolor
"magenta";linestyle 1;linethickness 1;pointstyle "point";xmin "0";xmax
"1";xviewmin "-0.0196";xviewmax "1";yviewmin "-1.005";yviewmax
"0.02009";rangeset"X";phi 45;theta 45;plottype 19;constrained TRUE;numpoints
10;axesstyle "normal";xis \TEXUX{t};var1name \TEXUX{$x$};valid_file
"T";tempfilename 'C:/Temp/EOH6PF03.wmf';tempfile-properties "XP";}}

\textbf{c.} Multiply by the \textit{complex conjugate} of the denominator
over itself to obtain
\begin{eqnarray}
\frac{7+3i}{5-4i} &=& \left(\frac{7+3i}{5-4i}\right) \left(\frac{5+4i}{5+4i}%
\right)  \nonumber \\
&=& \frac{35+15i+28i-12}{25+16}  \nonumber \\
&=& \frac{23+43i}{41}
\end{eqnarray}
So the real part is $23/41$ and the imaginary part is $43/41$.

Re-express the complex number in the following way
\begin{eqnarray}
r\mathrm{e}^{i\theta } &=&\frac{23}{41}+\frac{43i}{41}  \nonumber \\
r\mathrm{e}^{i\theta } &=&r\cos (\theta )+ir\sin (\theta )
\end{eqnarray}
Equating real and imaginary parts gives $r\cos (\theta )=\frac{23}{41}$ and $%
r\sin (\theta )=\frac{43}{41}$. Squaring these and adding yields
\begin{eqnarray}
r^{2}\cos ^{2}(\theta )+r^{2}\sin ^{2}(\theta ) &=&\left( \frac{23}{41}%
\right) ^{2}+\left( \frac{43}{41}\right) ^{2}  \nonumber \\
r^{2} &=&\frac{2378}{1681}=\frac{58}{41}  \nonumber \\
r &=&\sqrt{\frac{58}{41}}\approx 1.189.
\end{eqnarray}
If we take the ratio instead, we find
\begin{eqnarray}
\frac{r\sin (\theta )}{r\cos (\theta )} &=&\frac{43/41}{23/41}  \nonumber \\
\tan (\theta ) &=&\frac{43}{23}  \nonumber \\
\theta &=&\mathrm{arctan}\left( \frac{43}{23}\right) \approx 1.080\ \mathrm{%
radians}=61.86\ \mathrm{degrees}.
\end{eqnarray}

Plotted below are $7+3i$, $5-4i$, and $(7+3i)/(5-4i)$.

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Word";type "MAPLEPLOT";width 3in;height 2.0003in;depth 0pt;display
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1;linethickness 1;pointstyle "point";function \TEXUX{$5t,-4t$};linecolor
"green";linestyle 1;linethickness 1;pointstyle "point";function
\TEXUX{$23t/41,43t/41$};linecolor "red";linestyle 1;linethickness
1;pointstyle "point";xmin "0";xmax "1";xviewmin "-0.14";xviewmax
"7.143";yviewmin "-4.14";yviewmax "3.143";rangeset"X";phi 45;theta
45;plottype 19;constrained TRUE;numpoints 49;axesstyle "normal";xis
\TEXUX{t};var1name \TEXUX{$x$};valid_file "T";tempfilename
'C:/Temp/EOH6PH04.wmf';tempfile-properties "XP";}}

\vspace{0.5cm} \hrule
\vspace{0.5cm}

\textbf{2.} Two coupled oscillators (pendulums in the class demo) obey the
equations
\[
m \frac{d^2x_1}{dt^2} = - m \omega_0^2 x_1 + K (x_2-x_1)
\]
\[
m \frac{d^2x_2}{dt^2} = - m \omega_0^2 x_2 + K (x_1-x_2) .
\]
Verify that these equations have a solution (normal mode)
\[
x_1 = \cos (\omega_s t) \ \ \ \ \ \ x_2 = \cos(\omega_s t)
\]
for the appropriate value of $\omega_s$, and find $\omega_s$.
Describe in words the motion corresponding to this solution.

Similarly, verify that the equations also have a solution
\[
x_1 = \cos (\omega_a t) \ \ \ \ \ \ x_2 = -\cos(\omega_a t)
\]
for the appropriate value of $\omega_a$, and find $\omega_a$.
Describe in words the motion corresponding to this solution.

The most general motion is described by a linear combination of the normal
modes we have found:
\[
x_1 = a \cos( \omega_s t) + b \cos (\omega_a t)
\]
\[
x_2 = a \cos( \omega_s t) - b \cos (\omega_a t) .
\]

In particular, consider the solution with $a=b$. Show that this describes
the situation where, initially, pendulum $1$ is held at distance $2a$ from
the vertical, and released with zero speed; what are the initial position
and speed of pendulum $2$? Plot and describe in words the resulting motion
of each pendulum when $K/m\omega_0^2=0.1$ (weak coupling) and $%
K/m\omega_0^2=1$ (strong coupling).

For full credit, you must give accurate plots, by MAPLE or otherwise, and
they must extend over several oscillations of each pendulum.

\vspace{0.5cm}

First let us show that there is a solution to the coupled
oscillator equations of the form $x_1=\cos (\omega_s t)$ and
$x_2=\cos(\omega_s t)$. Substituting these forms into the first
equations yields
\begin{equation}
- m \omega_s ^2 \cos(\omega_s t) = - m \omega_0^2 \cos (\omega_s t) + K
\left[ \cos(\omega_s t) - \cos(\omega_s t) \right] .
\end{equation}
Note that all of the terms are proportional to $\cos(\omega_s t)$
and have no other time dependence. Hence, we can divide through by
$m \cos(\omega_s t) $ finding
\begin{equation}
\omega_s^2 = \omega_0^2 .
\end{equation}
Substituting the forms into the second equation yields exactly the same
result. Therefore, $x_1=x_2 =\cos (\omega_s t)$ is a solution, provided $%
\omega_s=\omega_0$.

In this first solution the positions of the pendulums move
together, i.e. they have the same velocity and the same
acceleration, and the distance between them never changes.

Now let us show that $x_1=-x_2=\cos (\omega_a t)$ is also a
solution. Substituting these forms into the first equations yields
\begin{equation}
- m \omega_a ^2 \cos(\omega_a t) = - m \omega_0^2 \cos (\omega_a t) + K
\left[ -\cos(\omega_a t) - \cos(\omega_a t) \right].
\end{equation}
The $\cos(\omega_a t)$ is common and can be factored out, leaving
\begin{equation}
\omega_a^2 = \omega_0^2+2~K/m .
\end{equation}
Once again, substituting these forms into the second equation yields the
same result. Consequently, $x_1=-x_2 =\cos (\omega_a t)$ is a solution,
provided $\omega_a=\sqrt{\omega_0^2+2~K/m}$.

In this second solution the pendulums move in opposite directions; their
velocity and acceleration are in the opposite directions.

A \textit{normal mode} is a motion of a system in which each piece
moves sinusoidally with the same frequency. Thus, the solutions
above are normal modes. In this particular case, the first normal
mode corresponds to the
\textit{center-of-mass} motion while the second corresponds to the \textit{%
relative} motion.

Next we asked to consider a solution of the form
\[
x_{1}(t)=a\left[ \cos \left( \omega _{0}t\right) +\cos \left( \sqrt{\omega
_{0}^{2}+2~K/m}~t\right) \right]
\]
\[
x_{2}(t)=a\left[ \cos \left( \omega _{0}t\right) -\cos \left( \sqrt{\omega
_{0}^{2}+2~K/m}~t\right) \right] ,
\]
where we have substituted in the expressions we found for $\omega _{s}$ and $%
\omega _{a}$. The initial position of pendulum $2$ is $x_{2}(t=0)=0$. The
velocity of pendulum $2$ is $v_{2}(t)=dx_{2}(t)/dt$
\begin{equation}
v_{2}(t)=a\left[ -\omega _{0}\sin \left( \omega _{0}t\right) +\sqrt{\omega
_{0}^{2}+2k/m}\sin \left( \sqrt{\omega _{0}^{2}+2K/m}~t\right) \right] ,
\end{equation}
so the initial velocity of pendulum $2$ is $v_{2}(t=0)=0$.

Using the trigonometric identities
\begin{eqnarray}
\cos A+\cos B &=&2\cos \left( \frac{A+B}{2}\right) \cos \left( \frac{A-B}{2}%
\right)  \nonumber \\
\cos A-\cos B &=&2\sin \left( \frac{A+B}{2}\right) \sin \left( \frac{B-A}{2}%
\right) ,
\end{eqnarray}
we find
\begin{eqnarray}
x_{1}(t) &=&2a\cos \left[ \frac{\left( \omega _{0}+\sqrt{\omega _{0}^{2}+2K/m%
}\right) t}{2}\right] \cos \left[ \frac{\left( \omega _{0}-\sqrt{\omega
_{0}^{2}+2K/m}\right) t}{2}\right]  \nonumber \\
x_{2}(t) &=&2a\sin \left[ \frac{\left( \omega _{0}+\sqrt{\omega _{0}^{2}+2K/m%
}\right) t}{2}\right] \sin \left[ \frac{\left( \sqrt{\omega _{0}^{2}+2K/m}%
-\omega _{0}\right) t}{2}\right] .
\end{eqnarray}
Let us express time in units of $T_{0}$ where $T_{0}$ is the natural period
of the uncoupled oscillator, i.e. $T_{0}=2\pi /\omega _{0}$. Let $s=t/T_{0}$%
, then
\begin{eqnarray}
x_{1}(s) &=&2a\cos \left[ \frac{2\pi \left( 1+\sqrt{1+2q}\right) s}{2}%
\right] \cos \left[ \frac{2\pi \left( 1-\sqrt{1+2q}\right) 2}{2}\right]
\nonumber \\
x_{2}(s) &=&2a\sin \left[ \frac{2\pi \left( 1+\sqrt{1+2q}\right) s}{2}%
\right] \sin \left[ \frac{2\pi \left( \sqrt{1+2q}-1\right) s}{2}\right] ,
\end{eqnarray}
where $q=K/m\omega _{0}^{2}$.

Below is a plot of $x_{1}(t)=2\,\cos (\pi (1+\sqrt{1.2})s)\,\cos (\pi (1-%
\sqrt{1.2})s)$\FRAME{dtbpF}{6.2552in}{2.0003in}{0pt}{}{}{}{\special{language
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45;theta 45;plottype 4;numpoints 150;axesstyle "normal";xis
\TEXUX{s};var1name \TEXUX{$x$};valid_file "T";tempfilename
'C:/Temp/EOH6Q207.wmf';tempfile-properties "XP";}}and a plot of $%
x_{2}(t)=2\sin (\pi (1+\sqrt{1.2})s)\sin (\pi (\sqrt{1.2}-1)s)$\FRAME{dtbpF}{%
6.2638in}{2.0064in}{0pt}{}{}{}{\special{language "Scientific Word";type
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(1+\sqrt{1.2})s)\sin (\pi (\sqrt{1.2}-1)s)$};linecolor "black";linestyle
1;linethickness 1;pointstyle "point";xmin "0";xmax "20";xviewmin
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\TEXUX{$x$};valid_file "T";tempfilename
'C:/Temp/EOH6UC0F.wmf';tempfile-properties "XP";}}for the weak coupling case
$q=0.1$.

For the strong coupling case ($q=1.0$), here are plots of $x_{1}(t)=2\,\cos
(\pi (1+\sqrt{3})s)\cos (\pi (1-\sqrt{3})s)$\FRAME{dtbpF}{6.2552in}{2.0003in%
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150;axesstyle "normal";xis \TEXUX{s};var1name \TEXUX{$x$};valid_file
"T";tempfilename 'C:/Temp/EOH6QB0B.wmf';tempfile-properties "XP";}}and $%
x_{2}(t)=2\sin (\pi (1+\sqrt{3})s)\sin (\pi (\sqrt{3}-1)s)$\FRAME{dtbpF}{%
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150;axesstyle "normal";xis \TEXUX{s};var1name \TEXUX{$x$};valid_file
"T";tempfilename 'C:/Temp/EOH6TK0E.wmf';tempfile-properties "XP";}}

This should make you recall various concepts connected to waves like \textit{%
interference}, \textit{superposition}, \textit{beats}, \textit{phase velocity%
}, \textit{group velocity}, etc.

\vspace{0.5cm} \hrule
\vspace{0.5cm}

\textbf{3.} The coupled pendulums of problem $2$ behave in a way similar to
two coupled quantum states. The wavefunction of two coupled quantum states
is
\[
\psi = c_1(t) \phi_1 + c_2(t) \phi_2
\]
and the time evolution is given by
\[
i \hbar \frac{d c_1}{dt} = E_1 c_1 + V c_2
\]
\[
i \hbar \frac{d c_2}{dt} = E_2 c_2 + V c_1 .
\]
Consider only the ``degenerate" case when $E_1=E_2$ (identical atoms, for
instance), analogously to the identical pendulums of problem $2$. Use $E_0$
to denote the common value of $E_1$ and $E_2$.

Verify that these equations have the solution (eigenstate)
\[
c_1= \exp(-i \omega_s t) \ \ \ \ \ \ c_2 = \exp(-i \omega_s t)
\]
for the appropriate value of $\omega_s$ and find $\omega_s$. As a
function
of time, what is the probability that the quantum system is in the state $%
\phi_1$ (if it is in this eigenstate)? In the state $\phi_2$?

Similarly, verify that the equations have the solution
\[
c_1= \exp(-i \omega_a t) \ \ \ \ \ \ c_2 = -\exp(-i \omega_a t)
\]
for the appropriate value of $\omega_a$, and find $\omega_a$. As a
function of time, what is the probability that the quantum system
is in the state $\phi_1$? In state $\phi_2$?

The most general solution is described by a linear combination of the
eigenstates we have found:
\[
c_{1}=a\exp (-i\omega _{s}t)+b\exp (-i\omega _{a}t)
\]
\[
c_{2}=a\exp (-i\omega _{s}t)-b\exp (-i\omega _{a}t).
\]
In particular, consider the solution with $a=b$. Show that this describes
the situation where, initially, the system is in the state $\phi _{1}$. Plot
(accurately) and describe in words the probabilities $|c_{1}|^{2}$ and $%
|c_{2}|^{2}$, as they vary with time, when $V/E_{0}=0.1$ (weak coupling) and
$V/E_{0}=1.0$ (strong coupling).

\vspace{0.5cm}

Let us substitute the proposed solution $c_1=c_2=\exp(-i \omega_s t)$ into
the time evolution equations
\begin{equation}
\hbar \omega_s \exp(-i \omega_s t) = E_0 \exp(-i \omega_s t) + V \exp(-i
\omega_s t).
\end{equation}
Note that $\exp(-i \omega_s t)$ is common to all of the terms and
constitutes the full time dependence of each. We can divide through by $%
\exp(-i \omega_s t)$ which gives us
\begin{equation}
\hbar \omega_s = E_0 + V.
\end{equation}
Therefore, $c_1=c_2=\exp(-i\omega_s t)$ is a solution provided $\omega_s
=(E_0+V)/\hbar$. (From the units of $\hbar$ and $\omega$ we can determine
that $E_0$ and $V$ are energies.)

When we write the expression for a wavefunction in terms of states
\[
\psi =c_{1}(t)\phi _{1}+c_{2}(t)\phi _{2},
\]
it is somewhat analogous to writing a position vector in terms of its
components $\mathbf{r(t)}=x(t)\mathbf{\hat{\imath}}+y(t)\mathbf{\hat{\jmath}}
$. In particular, we are assuming here that $\phi _{1}$ is \textit{normalized%
} (somewhat like $\mathbf{\hat{\imath}}\cdot \mathbf{\hat{\imath}}=1$) and
furthermore that $\phi _{1}$ and $\phi _{2}$ are \textit{orthogonal}
(somewhat like $\mathbf{\hat{\imath}}\cdot \mathbf{\hat{\jmath}}=0$). In
this analogy the $c$'s are amplitudes in the ``directions'' $\psi _{1}$ and $%
\psi _{2}$; and they are often called \textit{probability amplitudes}. If we
are in the ``state'' $\psi $ with $c_{1}=c_{2}=\exp [-i(E_{0}+V)t/\hbar ]$,
then the probability that the quantum system is in the state $\phi _{1}$ at
time $t$ is proportional to $|c_{1}(t)|^{2}$. The sum of all the distinct
probabilities must add up to one, so in this case, the probability to be in
state $\phi _{1}$ is
\begin{equation}
P_{\phi _{1}}=\frac{|c_{1}(t)|^{2}}{|c_{1}(t)|^{2}+|c_{2}(t)|^{2}}=\frac{1}{2%
}.
\end{equation}
And similarly the probability to be in state $\phi _{2}$ is
\begin{equation}
P_{\phi _{2}}=\frac{|c_{2}(t)|^{2}}{|c_{1}(t)|^{2}+|c_{2}(t)|^{2}}=\frac{1}{2%
}.
\end{equation}
Note that we have divided by $|c_{1}(t)|^{2}+|c_{2}(t)|^{2}$ so that the
probabilities add up to one; this procedure is called \textit{normalization}.

Now we consider another solution. Putting $c_1=-c_2= \exp(-i \omega_a t)$
into the evolution equation yields
\begin{equation}
\hbar \omega_a \exp(-i \omega_a t) = E_0 \exp(-i \omega_a t) - V \exp(-i
\omega_a t),
\end{equation}
leading to $\omega_a =(E_0-V)/\hbar$. The probability that the quantum
system is in the state $\phi_1$ is then given by
\begin{equation}
P_{\phi_1}=\frac{|c_1(t)|^2}{|c_1(t)|^2 + |c_2(t)|^2}=\frac{1}{2}.
\end{equation}
And the probability to be in state $\phi_2$ is
\begin{equation}
P_{\phi_2}=\frac{|c_2(t)|^2}{|c_1(t)|^2 + |c_2(t)|^2}=\frac{1}{2}.
\end{equation}

\textit{The subscripts $s$ and $a$ were not chosen arbitrarily; $s$ stands
for ``symmetric" and $a$ for ``antisymmetric." Note that if we were to
``exchange" the labels $1 \leftrightarrow 2$ that $\psi \leftrightarrow \psi$
in the symmetric case and $\psi \leftrightarrow - \psi$ in the antisymmetric
case. }

Now we consider the linear combination of the eigenstates with the
following coefficients
\begin{eqnarray}
c_{1} &=&a\left[ \exp \left( -i\frac{(E_{0}+V)t}{\hbar }\right) +\exp \left(
-i\frac{(E_{0}-V)t}{\hbar }\right) \right]  \nonumber \\
c_{2} &=&a\left[ \exp \left( -i\frac{(E_{0}+V)t}{\hbar }\right) -\exp \left(
-i\frac{(E_{0}-V)t}{\hbar }\right) \right] .
\end{eqnarray}
The probability that the quantum system is in the state $\phi _{1}$ is given
by
\begin{equation}
P_{\phi _{1}}=\frac{|c_{1}(t)|^{2}}{|c_{1}(t)|^{2}+|c_{2}(t)|^{2}},
\end{equation}
where
\begin{eqnarray}
|c_{1}(t)|^{2} &=&|a|^{2}\left[ 2+2\cos \left( \frac{2Vt}{\hbar }\right)
\right]  \nonumber \\
|c_{2}(t)|^{2} &=&|a|^{2}\left[ 2-2\cos \left( \frac{2Vt}{\hbar }\right)
\right]
\end{eqnarray}
Note that $|c_{1}(t)|^{2}+|c_{2}(t)|^{2}=|a|^{2}$ at all times. The
probability to be in state $\phi _{1}$ is
\begin{equation}
P_{\phi _{1}}(t)=\frac{1+\cos (2Vt/\hbar )}{2};
\end{equation}
and that to be in state $\phi _{2}$ is
\begin{equation}
P_{\phi _{2}}(t)=\frac{1-\cos (2Vt/\hbar )}{2}.
\end{equation}

It is the $V$ term which ``couples'' the two states $\phi _{1}$ and $\phi
_{2}$. The period in the absence of any coupling is $T_{0}=2\pi \hbar /E_{0}$%
, so let us express $t$ in terms of $T_{0}$, i.e. $s=t/T_{0}$. The
probabilities are then
\begin{eqnarray}
P_{\phi _{1}}(s) &=&\frac{1+\cos [4\pi s(V/E_{0})]}{2}  \nonumber \\
P_{\phi _{2}}(s) &=&\frac{1-\cos [4\pi s(V/E_{0})]}{2}.
\end{eqnarray}

For weak coupling ($V/E_{0}=0.1$) we have $P_{1}(s)=(1+\cos (0.4\pi s))/2$%
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And for the strong coupling case ($V/E_{0}=1.0$) $P_{1}(s)=(1+\cos (4\pi
s))/2$\FRAME{dtbpF}{6.4204in}{2.0003in}{0pt}{}{}{}{\special{language
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"T";tempfilename 'C:/Temp/EOH6UU0I.wmf';tempfile-properties "XP";}}and $%
P_{2}(s)=(1-\cos (4\pi s))/2$

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\end{document}
