How does an outer product really work?

Question

I'm taught that, for arbitrary wavefunctions $\psi, \phi$, that:

$$\hat B = |\psi \rangle \langle \phi \mid$$

Which produces a new ket apparently when applied to a ket, as..

$$\hat B \ | \mu \rangle = |\psi \rangle \langle \phi \mid \mu \rangle = C | \psi \rangle$$

Which, following this logic makes sense following it, but how do I know that applying $| \mu \rangle$ to $\hat B$ that it naturally goes where it did other than the fact that "it just fits that way" (the way my brain would reply to this in my head) because bras and kets fit together. But that doesn't sound like a very rigorous understanding on my part. Why does it go to the bra vector and form an inner product other than "it just would" (which is what I get the urge to think)?

Answer 1

The appeal of the bra-ket notation is exactly that "it just fits that way" - everything works out like it should if you just "match" bras and kets.

But as a beginner, it is good to keep in mind what actually happens in the background. Let us go through your example:

• A "ket" $|\psi\rangle$ is just a vector in the Hilbert space. In the following, I will not use bra-ket notation and just write $\psi \in \mathcal H$.

• There is a canonical isomorphism between a Hilbert space and its dual. This is the mathsy way of saying the following: Given a vector $\phi \in \mathcal H$, the function $\phi^\ast: \mathcal H \to \mathbb C$ defined as $$ \phi^\ast(\psi) = \langle \phi, \psi \rangle \tag 1 $$ is a linear map. In physics, we write the bra $\langle \phi |$ for $\phi^\ast$.

• An operator $\hat B$ is defined by how it acts on a vector of the Hilbert space. In your case, that is $$ \hat B(\mu) = \phi^\ast(\mu)\, \psi = \psi\, \phi^\ast(\mu) . \tag 2 $$ This is a linear map $\mathcal H \to \mathcal H$ by definition.

Writing (2) as $\hat B = \psi\, \phi^\ast$ is now a quite obvious abbreviation. A physicist would write $\hat B = |\psi \rangle\!\langle \phi|$ and it "fits".