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The outer product

So far we have formed the following products: $\langle B\vert A \rangle$, $X\vert A\rangle$, $\langle A\vert X$, $X Y$, $\langle B\vert X\vert A \rangle$. Are there any other products we are allowed to form? How about
\begin{displaymath}
\vert B \rangle \langle A\vert ?
\end{displaymath} (39)

This clearly depends linearly on the ket $\vert A\rangle$ and the bra $\vert B\rangle$. Suppose that we right-multiply the above product by the general ket $\vert C\rangle$. We obtain
\begin{displaymath}
\vert B \rangle \langle A \vert C\rangle = \langle A \vert C\rangle \vert B \rangle,
\end{displaymath} (40)

since $\langle A \vert C\rangle$ is just a number. Thus, $\vert B\rangle \langle A\vert$ acting on a general ket $\vert C\rangle$ yields another ket. Clearly, the product $\vert B\rangle \langle A\vert$ is a linear operator. This operator also acts on bras, as is easily demonstrated by left-multiplying the expression (39) by a general bra $\langle C\vert$. It is also easily demonstrated that
\begin{displaymath}
(\vert B \rangle \langle A\vert)^{\dag } = \vert A \rangle \langle B\vert.
\end{displaymath} (41)

Mathematicians term the operator $\vert B\rangle \langle A\vert$ the outer product of $\vert B\rangle$ and $\langle A\vert$. The outer product should not be confused with the inner product, $\langle A\vert B \rangle$, which is just a number.


next up previous
Next: Eigenvalues and eigenvectors Up: Fundamental concepts Previous: Operators
Richard Fitzpatrick 2006-02-16