Review of Operators

First, a little review. Recall that the square integrable functions form a vector space, much like the familiar 3D vector space.

\begin{displaymath}\bgroup\color{black}\vec{r}=a\vec{v_1}+b\vec{v_2} \egroup\end{displaymath}

in 3D space becomes

\begin{displaymath}\bgroup\color{black}\vert\psi\rangle=\lambda_1\vert\psi_1\rangle+\lambda_2\vert\psi_2\rangle .\egroup\end{displaymath}

The scalar product is defined as

\begin{displaymath}\bgroup\color{black} \langle\phi\vert\psi\rangle=\int\limits_{-\infty}^\infty dx\;\phi^*(x)\psi(x) \egroup\end{displaymath}

and many of its properties can be easily deduced from the integral.

\begin{displaymath}\bgroup\color{black} \langle\phi\vert\psi\rangle^*=\langle\psi\vert\phi\rangle \egroup\end{displaymath}

As in 3D space,

\begin{displaymath}\bgroup\color{black}\vec{a}\cdot\vec{b}\leq\vert a\vert\;\vert b\vert \egroup\end{displaymath}

the magnitude of the dot product is limited by the magnitude of the vectors.

\begin{displaymath}\bgroup\color{black}\langle\psi\vert\phi\rangle\leq\sqrt{\langle\psi\vert\psi\rangle\langle\phi\vert\phi\rangle} \egroup\end{displaymath}

This is called the Schwartz inequality.

Operators are associative but not commutative.

\begin{displaymath}\bgroup\color{black} AB\vert\psi\rangle=A(B\vert\psi\rangle)=(AB)\vert\psi\rangle \egroup\end{displaymath}

An operator transforms one vector into another vector.

\begin{displaymath}\bgroup\color{black} \vert\phi'\rangle=O\vert\phi\rangle \egroup\end{displaymath}

Eigenfunctions of Hermitian operators

\begin{displaymath}\bgroup\color{black} H\vert i\rangle=E_i\vert i\rangle \egroup\end{displaymath}

form an orthonormal

\begin{displaymath}\bgroup\color{black}\langle i\vert j\rangle=\delta_{ij} \egroup\end{displaymath}

complete set

\begin{displaymath}\bgroup\color{black}\vert\psi\rangle=\sum\limits_i\langle i\v...
...\sum\limits_i \vert i\rangle \langle i\vert\psi\rangle .\egroup\end{displaymath}

Note that we can simply describe the \bgroup\color{black}$j^{th}$\egroup eigenstate at \bgroup\color{black}$\vert j\rangle$\egroup.

Expanding the vectors \bgroup\color{black}$\vert\phi\rangle$\egroup and \bgroup\color{black}$\vert\psi\rangle$\egroup,

\begin{eqnarray*}
\vert\phi\rangle &=&\sum\limits_i b_i \vert i\rangle \\
\vert\psi\rangle &=&\sum\limits_i c_i \vert i\rangle \\
\end{eqnarray*}


we can take the dot product by multiplying the components, as in 3D space.

\begin{displaymath}\bgroup\color{black}\langle\phi\vert\psi\rangle=\sum\limits_ib_i^*c_i \egroup\end{displaymath}

The expansion in energy eigenfunctions is a very nice way to do the time development of a wave function.

\begin{displaymath}\bgroup\color{black}\vert\psi(t)\rangle=\sum\limits_i\langle i\vert\psi(0)\rangle \vert i\rangle e^{-iE_it/\hbar} \egroup\end{displaymath}

The basis of definite momentum states is not in the vector space, yet we can use this basis to form any state in the vector space.

\begin{displaymath}\bgroup\color{black}\vert\psi\rangle={1\over\sqrt{2\pi\hbar}}\int\limits_{-\infty}^\infty dp \phi(p) \vert p\rangle \egroup\end{displaymath}

Any of these amplitudes can be used to define the state.

\begin{eqnarray*}
c_i=\langle i\vert\psi\rangle \\
\psi(x)=\langle x\vert\psi\rangle \\
\phi(p)=\langle p\vert\psi\rangle \\
\end{eqnarray*}


Jim Branson 2013-04-22