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What is the difference between these two?

$\langle x|x\rangle$ and $|x\rangle\langle x|$

Are they the same? If they're the same, why are they used in these two different forms?

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  • $\begingroup$ One's a number, the other an operator. If this refers to a state-vector, the first is "related" to normalization and the second to the density operator of such state. $\endgroup$ – Vendetta Jun 18 at 17:32
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They are not the same. One is the outer product and one is the inner product.

In the finite-dimensional real-number case for example, if $$|x\rangle = \begin{bmatrix}1\\2\\3\end{bmatrix},$$ then $$|x\rangle\langle x| = \begin{bmatrix}1\\2\\3\end{bmatrix} \begin{bmatrix}1&2&3\end{bmatrix} = \begin{bmatrix}1&2&3\\2&4&6\\3&6&9\end{bmatrix},$$ while $$\langle x | x\rangle = \begin{bmatrix}1&2&3\end{bmatrix} \begin{bmatrix}1\\2\\3\end{bmatrix} = 1 + 4 + 9 = 14.$$

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In Bra-Ket notation $| x \rangle$ is a vector and $\langle x |$ is a covector. Formally, a covector is a map that goes from $V \rightarrow \mathbb{R}$, i.e. it “eats a vector” and gives you a number. When we write $$ \langle x | x \rangle $$ we are saying that the covector $\langle x |$ is acting on the vector $|x\rangle$, so $\langle x | x\rangle$ is a real number.

On the other hand, $$ | x \rangle \langle x| $$ isn’t acting on anything. If we toss some vector at it $|a\rangle$, we would have $$ |x\rangle \langle x | a \rangle $$ which is just a real number times $|x \rangle$. We have given the above a vector and we got a vector back out so it is a map between vectors. To summarize then, $|x\rangle \langle x|$ and $\langle x | x \rangle $ represent different types of objects: the first is a map from $V \rightarrow V$, wheras the second is an element $b \in \mathbb{R}$.

To be concrete, let $$ |x\rangle = \begin{pmatrix} 1\\ 0 \end{pmatrix} $$ . Then we would write $$ \langle x | = \begin{pmatrix} 1&0\end{pmatrix} $$ which is the Hermitian conjugate of $|x\rangle$, in this case just the transpose. So $$ \langle x | x \rangle = \begin{pmatrix} 1&0 \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix} = 1 $$ And $$ | x \rangle \langle x | = \begin{pmatrix} 1\\ 0 \end{pmatrix} \begin{pmatrix}1&0\end{pmatrix} = \begin{pmatrix} 1&0\\ 0&0 \end{pmatrix} $$ A number and a map, as expected.

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  • $\begingroup$ Internalizing this is what made me realize how clever of a "hack" the bra-ket notation really is. Also, for the record, the names "bra" and "ket" are suitably horrible puns. $\endgroup$ – Arthur Jun 14 at 9:14
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In general $\langle A\vert B\rangle$ is a scalar product whereas $\vert A\rangle\langle B\vert$ is an operator. To this this last suppose $\vert \psi\rangle$ is an arbitrary vector in your space. Then $$ \vert A\rangle\langle B\vert \psi\rangle $$ is a vector proportional to $\vert A\rangle$ since $\langle B\vert \psi\rangle \in \mathbb{C}$ is a (complex) number, so that $\vert A\rangle\langle B\vert$ maps a vector to another vector.

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