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1.7: 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

 

$\displaystyle \vert B \rangle \langle A\vert~?$ (39)

 

 

This clearly depends linearly on the bra $ \langle A\vert$ and the ket $ \vert B\rangle$ . Suppose that we right-multiply the above product by the general ket $ \vert C\rangle$ . We obtain

 

$\displaystyle \vert B \rangle\,\langle A \vert C\rangle = \langle A \vert C\rangle \,\vert B \rangle,$ (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

 

$\displaystyle (\vert B \rangle \langle A\vert)^{\dag } = \vert A \rangle \langle B\vert.$ (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.

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