# How does an outer product really work?

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)?

• It'd be easier to use the vector/matrix representation of bra-ket/operator. From there the direct product will be more intuitive to understand – rnels12 Feb 1 '19 at 10:00

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".

• This helps clear things up. We covered this stuff in linear maths last semester so it's making some sense in my brain. Wait, so $\langle \phi |$ is a linear functional? Why did no one tell me?! And $\phi^*(\mu)$ is just an element of a field (likely the complex), right? – sangstar Feb 1 '19 at 16:54
• Yes, exactly @sangstar :) – Noiralef Feb 1 '19 at 17:49
• I have a question newly about this. To my knowledge, a dual transformation $T^*$ usually acts on the linear functional $f$ assigned to it from the departure space, and acts on it to produce $T^*(f) = f (T)$ where $T$ is the associated linear transformation of the dual, which is really only meaningful with $T^*(f) (v) = f (T(v))$. Can you translate this here for me? What I'm trying to piece together is let $\hat B = T^*$ and $\langle \phi | = \phi^*$. Then (where $| n \rangle$ is some ket) $\hat B | n \rangle = \phi^* (T(v))$ but what's $T(v)$ here? How can I think of it as $\psi(x)$? – sangstar Feb 9 '19 at 23:58
• In bra-ket notation I think that is just $( \langle \phi | \hat B ) | \psi \rangle := \langle \phi | ( \hat B | \psi \rangle )$. (But you should think about your question again, because you wrote $\hat B = T^\ast$, and $T^\ast$ acts on functionals, but $\hat B$ acts on vectors.) If that doesn't clear it up, it's probably better if you create a new question - here, no one except me will notice your comments, and I'm not always around :) – Noiralef Feb 10 '19 at 0:23

This answer is a preview of the more complete answer by Noiralef:

The equation $$\hat B=|\psi\rangle\,\langle\phi|$$ is really just an abbreviation for this statement:

$$\hat B$$ is the operator that satisfies $$\hat B|\mu\rangle=z|\psi\rangle$$ for all $$|\mu\rangle$$, with $$z=\langle\phi|\mu\rangle$$.

It's a nice abbreviation because of the way the notation fits together.