I'm currently studying density matrices, and have been frequently coming across the construction

$$|\psi\rangle\langle\psi| \,.$$

What is the formal meaning of this composition? I understand $\langle \psi|$ to be an element of the dual space (to that vector space for which $|\psi\rangle$ is a member) but I don't quite understand what it means to put them together.

I have been treating this object as a linear operator on the space of ket vectors, and assuming a certain associativity to its composition, such that

$$\bigg(|\psi\rangle \langle \psi|\bigg) |\phi\rangle = |\psi\rangle \bigg(\langle \psi|\phi\rangle \bigg) = \alpha |\psi\rangle \,,$$

but is there a way to make this more formal? Thank you.


It is easy. Assume that $\psi \in \cal H$ is normalized to $1$. In this case, $|\psi\rangle\langle\psi|$ it is nothing but the orthogonal projector $P_\psi$ onto the one-dimensional linear space generated by the vector $\psi$.

Putting $|\psi\rangle$ and $\langle\psi|$ together simply means to exploit the tensor product.

If $\phi' \in \cal H'$ the (topological) dual space of $\cal H$ and $\psi \in \cal H$, it is well defined $\psi \otimes \phi' \in \cal H \otimes \cal H'$ as a (bounded) linear operator from $\cal H$ to $\cal H$.

When $\psi' \in \cal H'$ is the image of $\psi \in \cal H$ under the Riesz' (anti)-isomorphism which identifies $\cal H$ with $\cal H'$, and $||\psi||=1$, then $P_\psi :=\psi \otimes \psi' \in \cal H \otimes \cal H'$ is the orthogonal projector onto the closed subspace generated by $\psi$. Another way to write down it is just $|\psi\rangle\langle\psi|$.


I suggest to think of it like this: $$\left|\psi\right.\rangle = \left(\begin{array}{c}\psi_1\\\psi_2\\\cdots\\\psi_n\end{array}\right)\;,\qquad \langle\left.\psi\right| = \left(\begin{array}{c}\psi_1^*\,\psi_2^*\,\cdots\,\psi_n^*\end{array}\right) $$ And using standard rule ("every element in row per every element in column") for matrix multiplication. This way you get: $$ \langle\left.\psi\right.\left|\,\psi\right.\rangle = \left(\begin{array}{c}\psi_1^*\,\psi_2^*\,\cdots\,\psi_n^*\end{array}\right)\left(\begin{array}{c}\psi_1\\\psi_2\\\cdots\\\psi_n\end{array}\right) =\left|\psi_1\right|^2 +\left|\psi_2\right|^2+\cdots+\left|\psi_n\right|^2 $$ And for your question: $$ \left|\,\psi\right.\rangle \langle\left.\psi\right|= \left(\begin{array}{c}\psi_1\\\psi_2\\\cdots\\\psi_n\end{array}\right) \left(\begin{array}{c}\psi_1^*\,\psi_2^*\,\cdots\,\psi_n^*\end{array}\right) =\left(\begin{array}{ccc} \psi_1\psi_1^* & \psi_1\psi_2^* & \cdots & \psi_1\psi_n^*\\ \psi_2\psi_1^* & \psi_2\psi_2^* & \cdots & \psi_2\psi_n^*\\ \cdots & \cdots & \cdots & \cdots\\ \psi_n\psi_1^* & \psi_n\psi_2^* & \cdots & \psi_n\psi_n^*\\ \end{array}\right) $$

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    $\begingroup$ It'd probably be good to remark that in your $ket=column$ expression you're abusing notation, because this equation is meaningless if understood literally: ket is a vector, while column is its representation in given basis. $\endgroup$ – Ruslan Mar 18 '14 at 13:44

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