quantum operator conceptQuantumOperatorConcept2009-03-11 00:51:182010-02-14 14:23:43TopicObservables and Statescommutatorcommute% this is the default PlanetMath preamble. as your knowledge
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% define commands hereConsider the function $\frac{\partial \Psi}{\partial t}$, the derivative of $\Psi$ with respect to time; one can say that the operator $\frac{\partial}{\partial t}$ acting on the function $\Psi$ yields the function $\frac{\partial \Psi}{\partial t}$. More generally, if a certain operation allows us to bring into correspondence with each function $\Psi$ of a certain function space, one and only one well-defined function $\Psi^{\prime}$ of that same space, one says the $\Psi^{\prime}$ is obtained through the action of a given \emph{operator} $A$ on the function $\Psi$, and one writes
$$
\Psi^{\prime} = A \Psi.
$$
By definition $A$ is a \emph{linear operator} if its action on the function $\lambda_1 \Psi_1 + \lambda_2 \Psi_2$, a linear combination with constant (complex) coefficients, of two functions of this function space, is given by
$$
A\left( \lambda_1 \Psi_1 + \lambda_2 \Psi_2 \right) = \lambda_1 \left( A \Psi_1 \right ) + \lambda_2 \left ( A \Psi \right ).
$$
Among the linear operators acting on the wave functions
$$ \Psi := \Psi(\mathbf{r},t) := \Psi(x,y,z,t) $$
associated with a particle, let us mention:
\begin{enumerate}
\item the differential operators ${\partial} / {\partial} x$,${\partial} / {\partial} y$,${\partial} / {\partial} z$,${\partial} / {\partial} t$, such as the one which was considered above;
\item the operators of the form $f(\mathbf{r},t)$ whose action consists in multiplying the function $\Psi$ by the function $f(\mathbf{r},t)$
\end{enumerate}
Starting from certain linear operators, one can form new linear operators by the following algebraic operations:
\begin{enumerate}
\item multiplication of an operator $A$ by a constant $c$:
$$ (cA)\Psi := c(A\Psi) $$
\item the sum $S = A + B$ of two operators $A$ and $B$:
$$ S\Psi := A \Psi + B \Psi $$
\item the product $P=AB$ of an operator $B$ by the operator $A$:
\end{enumerate}
Note that in contrast to the sum, \emph{the product of two operators is not commutative}. Therein lies a very important difference between the algebra of linear operators and ordinary algebra.
The product $AB$ is not necessarily identical to the product $BA$; in the first case, $B$ first acts on the function $\Psi$, then $A$ acts upon the function $(B\Psi)$ to give the final result; in the second case, the roles of $A$ and $B$ are inverted. The difference $AB-BA$ of these two quantities is called the \emph{commutator} of $A$ and $B$; it is represented by the symbol $[A,B]$:
\begin{equation}
[A,B] := AB - BA
\end{equation}
If this difference vanishes, one says that the two operators commute:
$$AB = BA$$
As an example of operators which do not commute, we mention the operator $f(x)$, multiplication by function $f(x)$, and the differential operator ${\partial} / {\partial x}$. Indeed we have, for any $\Psi$,
$$ \frac{\partial}{\partial x} f(x) \Psi = \frac{\partial}{\partial x} (f \Psi) = \frac{ \partial f}{\partial x} \Psi + f \frac{\partial \Psi}{\partial x} = \left ( \frac{\partial f}{\partial x} + f \frac{\partial}{\partial x} \right ) \Psi $$
In other words
\begin{equation}
\left [ \frac{\partial}{\partial x},f(x) \right ] = \frac{\partial f}{\partial x}
\end{equation}
and, in particular
\begin{equation}
\left [ \frac{\partial}{\partial x},x \right ] = 1
\end{equation}
However, any pair of derivative operators such as ${\partial} / {\partial} x$,${\partial} / {\partial} y$,${\partial} / {\partial} z$,${\partial} / {\partial} t$, commute.
A typical example of a linear operator formed by sum and product of linear operators is the Laplacian operator
$$ \nabla^2 := \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2} $$
which one may consider as the scalar product of the vector operator gradient $\nabla := \left( \frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z}\right )$, by itself.
\subsection{References}
[1] Messiah, Albert. "Quantum Mechanics: Volume I." Amsterdam, North-Holland Pub. Co.; New York, Interscience Publishers, 1961-62.
This entry is a derivative of the Public domain work [1].