The vector $\mid \phi \rangle \langle \phi \mid \psi \rangle$ is the projection of a vector $\mid \psi \rangle$ along the vector $\mid \phi \rangle$?

Question

I am currently studying the textbook Mathematical methods of quantum optics by Ravinder R. Puri. When going over some basic facts associated with bra-ket notation, the author says the following:

The vector $\mid \phi \rangle \langle \phi \mid \psi \rangle$ is the projection of a vector $\mid \psi \rangle$ along the vector $\mid \phi \rangle$.

I'm unsure of this. The scalar product $\langle \phi \mid \psi \rangle$ is a measure of the overlap between the vectors $\mid \psi \rangle$ and $\mid \phi \rangle$. So how is it then the case that $\mid \phi \rangle \langle \phi \mid \psi \rangle$ is the projection of a vector $\mid \psi \rangle$ along the vector $\mid \phi \rangle$? I would greatly appreciate it if people would please take the time to explain this.

Answer 1

You are already on the right track

The scalar product $\langle \phi | \psi \rangle$ is a measure of the overlap between the vectors $| \psi \rangle$ and $| \phi \rangle$

Indeed, and overlap can also be read as projection, this is just a simple inner product of two vectors. And you know that is the way of getting a projection: via the inner product.

Now if you multiply that scalar times the vector $|\phi \rangle$ you will get the projection along $|\phi \rangle$