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

$$\vert B \rangle \langle A\vert~?$$ (1.42)

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

$$\vert B \rangle\,\langle A \vert C\rangle = \langle A \vert C\rangle \,\vert B \rangle,$$ (1.43)

because $\langle A \vert C\rangle$ 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 (1.42) by a general bra $\langle C\vert$ . It is also readily shown that

$$(\vert B \rangle \langle A\vert)^{\dag } = \vert A \rangle \langle B\vert.$$ (1.44)

(See Exercise 4.) 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.