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\begin{document}

\section{Multiple Compatible Operators}

Since operators are represented as matrices, the product of two
operators isn't necessarily commutative, i. e., in general $AB \ne
BA$.  The {\em commutator} of two operators is defined as
\begin{equation}
	[A,B] \equiv AB - BA .
\end{equation}

If two operators commute, i. e., if $[A,B] = 0$, then they share
eigenvectors.  This is easy to show: If $A \psi = a \psi$ where $A$ is
an operator, then $\psi$ is an eigenvector of $A$ with eigenvalue $a$.
Note that $AB \psi = BA \psi = B a \psi = a B \psi$.  If we write
$\phi = B \psi$, then $A \phi = a \phi$.  In other words, both $\psi$
and $\phi$ are eigenvectors of the operator $A$ with the same
eigenvalue, $a$.  If there is no degeneracy, then $\psi$ and $\phi$
are parallel, i. e., $\phi = b \psi$, where $b$ is just a number.
However, recalling that $\phi = B \psi$, we see that $B \psi = b
\psi$, and $\psi$ is also an eigenvector of $B$.

If degeneracy exists such that multiple eigenvectors of $A$ have the
same eigenvalue $a$, then one cannot immediately infer that $\phi$ is
parallel to $\psi$.  However, all is not lost.  If there are $N$
eigenvectors of $A$, $\psi_i ,~i = 1, \ldots n$, then each of the $N$
$\phi_i = B \psi_i$ can be constructed as a linear superposition of
the $\psi_i$.  This is easily shown since $A \phi_i = AB \psi_i = BA
\psi_i = Ba \psi_i = aB \psi_i = a \phi_i$, implying that each of the
$\phi_i$ is itself an eigenvector of $A$.  However, any eigenvectors
of $A$ with eigenvalue $a$, including the $\phi_i$, can be constructed
as linear superpositions of the $\psi_i$.  All that remains is to
solve the restricted eigenvalue problem $B \chi = b \chi$, where
$\chi$ is an arbitrary linear superposition of the $\psi_i$.  There
will in general be $N$ such eigenvectors, $\chi_i ,~i = 1, \ldots N$,
with associated eigenvalues $b_i$.  The $\chi_i$ are eigenvectors of
both $A$ and $B$, and the theorem is proved for the degenerate case.

The following is also true: If all eigenvectors of $A$ are
eigenvectors of $B$, and if all eigenvectors of $B$ are eigenvectors
of $A$, then $[A,B] = 0$.  (See if you can show this!)

Two or more operators that mutually commute are said to be {\em
compatible}.  A quantum mechanical system becomes completely specified
when all operators compatible with the Hamiltonian and with each other
are found.

\subsection{Position and Momentum}

The position operator is just $x$, while the momentum operator is
\begin{equation}
	p = - i \hbar \frac{\partial}{\partial x} .
\end{equation}
Applying $xp$ and $px$ to an arbitrary function $\psi (x)$ results in
\begin{equation}
	xp \psi = - i \hbar x \frac{\partial \psi}{\partial x} ~~~~~
	px \psi = - i \hbar \frac{\partial}{\partial x} (x \psi ) =
		-i \hbar \psi - i \hbar x \frac{\partial \psi}{\partial x} ,
\end{equation}
so that $[x,p] \psi = i \hbar \psi$.  Since this is true for any
$\psi$, we infer that
\begin{equation}
	[x,p] = i \hbar ,
\end{equation}
and $x$ and $p$ are incompatible operators.

If operators are incompatible, then exact simultaneous measurements of
the corresponding physical quantities are impossible.  This is because
a measurement of a quantity exhibits no quantum mechanical uncertainty
only when the state vector of the system is an eigenvector of the
operator in question.  If two operators are incompatible, then if the
state vector is an eigenvector of the first operator, it cannot be an
eigenvector of the second operator.  The value of the physical
variable associated with the second operator is thus uncertain.  This
is the origin of the {\em uncertainty principle}, which states (among
other things) that the position and the momentum of a particle cannot
be simultaneously measured with infinite accuracy.

\subsection{Energy and Momentum of a Free Particle}

The Hamiltonian of a free particle is just
\begin{equation}
	H = \frac{p^2}{2m} =
	-\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} .
\end{equation}
Applying $pH$ to an arbitrary function results in
\begin{equation}
	pH \psi = i \frac{\hbar^3}{2m} \frac{\partial^3 \psi}{\partial x^3} .
\end{equation}
However, application of $Hp$ to $\psi$ yields the same result.  Thus,
$[H,p] = 0$, and the Hamiltonian and momentum operators are
compatible.

Eigenvectors of the free particle Hamiltonian can be obtained by
solving the equation
\begin{equation}
	H \psi =
	-\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} =
	E \psi ,
\label{freepart}
\end{equation}
where $E$ is the particle energy.  There are two solutions to this
equation for each value of $E$:
\begin{equation}
	\psi_{1,2} = \exp [ \pm i (2mE)^{1/2} x/ \hbar - iEt/ \hbar ] .
\label{alt1}
\end{equation}
The Hamiltonian is thus degenerate.  However, these solutions are also
eigenvectors of the momentum operator, with the momentum eigenvalue
being $+(2mE)^{1/2}$ in the first case and $-(2mE)^{1/2}$ in the
second.

Alternate solutions to equation (\ref{freepart}) are
\begin{eqnarray}
	\psi_a & = & \cos [ (2mE)^{1/2} x/ \hbar ] \exp (-iEt/ \hbar )
		\nonumber \\
	\psi_b & = & \sin [ (2mE)^{1/2} x/ \hbar ] \exp (-iEt/ \hbar ) .
\label{alt2}
\end{eqnarray}
These can be obtained as a linear combination of the solutions given
by equation (\ref{alt1}) since $\cos(a) = [ \exp (ia) + \exp
(-ia)]/2$, etc.  However, notice that these solutions are {\em not}
eigenvectors of the momentum operator since $p \psi_a \ne const \times
\psi_a$ and $p \psi_b \ne const \times \psi_b$.  The actual
relationships in this case are $p \psi_a = - (2mE)^{1/2} \psi_b$ and
$p \psi_b = (2mE)^{1/2} \psi_a$.

\subsection{Parity and Energy}

The parity operator makes the transformation $x \rightarrow -x$.  If,
for example, $x$ takes on the possible values $-1, 0, 1$, then the
parity operator $P$ takes the form
\begin{equation}
	P = \left( \begin{array}{ccc}
		0 & 0 & 1 \\
		0 & 1 & 0 \\
		1 & 0 & 0
	\end{array} \right) .
\end{equation}
If we have a state vector
\begin{equation}
	\psi = \left( \begin{array}{c}
		\psi (-1) \\
		\psi (0) \\
		\psi (1)
	\end{array} \right) ,
\end{equation}
then
\begin{equation}
	P \psi = \left( \begin{array}{c}
		\psi (1) \\
		\psi (0) \\
		\psi (-1)
	\end{array} \right) .
\end{equation}
Generalization to more $x$ values is clear.

Notice that application of the parity operator twice yields the
identity operation, i. e., $P^2 = I$.  The only possible eigenvalues
of the parity operator are thus $\pm 1$.  In general this means that
the parity operator is extremely degenerate, with many eigenvectors
for each of these eigenvalues.  In fact, all even functions of $x$ are
eigenvectors with eigenvalue $+1$, while all odd functions of $x$ are
eigenvectors with eigenvalue $-1$. The former are said to have {\em
even parity}, while the later have {\em odd parity}.

The parity operator commutes with the Hamiltonian for the free
particle.  This may be shown as follows:
\begin{equation}
	PH \psi (x) = P \left[ -\frac{\hbar^2}{2m}
		\frac{\partial^2 \psi (x)}{\partial x^2} \right] =
	-\frac{\hbar^2}{2m}
		\frac{\partial^2 \psi (-x)}{\partial (-x)^2} .
\end{equation}
However,
\begin{equation}
	HP \psi (x) = H \psi (-x) =
	-\frac{\hbar^2}{2m} \frac{\partial^2 \psi (-x)}{\partial x^2} =
	-\frac{\hbar^2}{2m} \frac{\partial^2 \psi (-x)}{\partial (-x)^2} ,
\end{equation}
where the chain rule operation $\partial \psi (-x) / \partial x =
- \partial \psi (-x) / \partial (-x)$ has been invoked twice.

Notice that whereas $\psi_{1,2}$ in equation (\ref{alt1}) are
eigenvectors of the momentum operator, the alternate solutions to the
free particle problem, $\psi_{a,b}$, defined in equation (\ref{alt2}),
are eigenvectors of the parity operator, with $\psi_a$ having even
parity and $\psi_b$ having odd parity.

\subsection{Parity and Momentum}

The parity and momentum operators do not commute.  In particular it
is easy to show that
\begin{equation}
	Pp \psi (x) = +i \hbar \frac{\partial \psi (-x)}{\partial (-x)} ~~~~~
	pP \psi (x) = -i \hbar \frac{\partial \psi (-x)}{\partial (-x)} ,
\end{equation}
which means that parity and momentum are incompatible operators.
Thus, parity, momentum, and the free particle Hamiltonian are jointly
incompatible even though the Hamiltonian is individually compatible
with both.  This is an example of {\em alternate incompatible
representations} of the state of a system.  We will encounter other
examples of this situation as we learn more about quantum mechanics.

\subsection{Generalized Uncertainty Relations}

We showed that operators are compatible if they commute.  In this
section we show that the observables associated with two non-commuting
operators cannot be simultaneously measured to arbitrary accurary.

Suppose $\psi$ is the state vector of some system while $A$ and $B$
are Hermitian operators associated with measurable variables on the
system.  We wish to compute the RMS uncertainty in the expectation
values of the two variables, $\sigma_A^2 = <(A - <A>)^2 >$,
$\sigma_B^2 = <(B - <B>)^2 >$.  Without loss of generality we can
set $<A> = 0$ and $<B> = 0$ for any case simply by redefining the
origin of the coordinate system for each variable.  Let us assume that
this has been done for both variables, so that $\sigma_A^2 = <A^2 >$
and $\sigma_B^2 = <B^2 >$.

Let us now define two auxiliary vectors, $\chi_A = A \psi$ and $\chi_B
= B \psi$.  The \emph{Schwartz inequality} states that
\begin{equation}
	(\chi_A , \chi_A ) (\chi_B , \chi_B ) \ge
	| (\chi_A , \chi_B )|^2 .
\end{equation}
For vectors in three-dimensional space this just states that the
square of the dot product of two vectors $(\bold{A} \cdot \bold{B})^2$
is less than the product of the magnitudes of the vectors $|\bold{A}|
|\bold{B}|$.  However, the dot product is just $|\bold{A}| |\bold{B}|
\cos \alpha$ where $\alpha$ is the angle between the vectors, so the
Schwartz inequality is simply equivalent to stating that the cosine
is less than or equal to unity in three-dimensional space.

We now make the following manipulations starting with the Schwartz
inequality applied to the Hilbert space vectors $\chi_A$ and $\chi_B$:
\begin{eqnarray}
	(\chi_A , \chi_A ) (\chi_B , \chi_B ) & \ge &
		| (\chi_A , \chi_B ) |^2 \nonumber \\
	(A \psi , A \psi ) (B \psi , B \psi ) & \ge &
		| (A \psi , B \psi ) |^2 \nonumber \\
	(\psi , A^2 \psi ) ( \psi , B^2 \psi ) & \ge &
		| ( \psi , AB \psi ) |^2 ,
\end{eqnarray}
where the last step works because $A$ and $B$ are Hermitian.  We now
note that these dot products are in the form of expectation values,
so we continue as follows:
\begin{eqnarray}
	| <AB> |^2 & = & |<AB + BA>/2 + <AB - BA>/2|^2 \nonumber \\
	& = & |<AB + BA> + <[A, B]>|^2 /4 \nonumber \\
	& \ge & |<[A,B]>|^2 /4 .
\end{eqnarray}
The last step depends on the following argument: The operator $AB +
BA$ is Hermitian, and so produces real expectation values.  The
expectation value $<AB + BA>$ is therefore real.  However, the
operator $[A,B]$ is anti-Hermitian, i. e., $[A,B]' = (AB)' - (BA)' =
B'A' - A'B' = BA - AB = -[A,B]$, so it produces imaginary expectation
values.  Thus, $|<AB + BA>/2 + <AB - BA>/2|^2$ is in the form $|\alpha
+ i \beta |^2 = \alpha^2 + \beta^2$, where $\alpha$ and $\beta$ are
real.  However, $\alpha^2 + \beta^2 \ge \beta^2$.

Taking the square root of the above result shows that
\begin{equation}
	\sigma_A \sigma_B \ge |<[A,B]>|/2 .
\end{equation}
Thus, the product of the uncertainties in the expectation values of
$A$ and $B$ is at a minimum half the expectation value of the
commutator of $A$ and $B$.  If the commutator is non-zero, the
uncertainty in the two variables cannot be simulaneously reduced to
zero.

Let us apply this to the position and momentum operators, $A = x$ and
$B = -i \hbar \partial / \partial x$.  We find that
\begin{equation}
	<[x,p]> = i \hbar .
\label{comexp}
\end{equation}
Thus, we have
\begin{equation}
	\sigma_x \sigma_p \ge \hbar / 2 .
\end{equation}
This is a precise expression of the well-known uncertainty relationship
between position and momentum.

\end{document}
