Formalism: Observables and Hermitian Operators

I think I mentioned back in Chapter 1 that the whole fight between the Einstein camp and the Copenhagen camp was over this one word — “observation.”

So in classical mechanics you had a plain old physical quantity called $Q$. In quantum mechanics, that same $Q$ becomes an operator:

$$\hat{Q}$$

And what this operator means is “measuring the physical quantity.”

What you actually get out of it is the expectation value.

So “observation,” written as a formula, is:

$$\langle Q \rangle \quad = \quad \int \psi^{*} Q \psi \quad = \quad \langle \psi^{*} | Q\psi \rangle$$

Earlier I was waving my hands about this like, “yeah, it’s the physical-quantity-measuring guy, you know, like a request for the energy measurement…” — just spouting nonsense, basically. heh.

The thing above? That’s the definition. Definition definition definition. heh heh.

So what does it actually mean to measure? Mathematically — it’s multiplying by an $n \times n$ matrix.

But hold on. The expectation value you get out of this — common sense says it has to be a real number, right!?!?

OK so now we’re gonna prove that.

“Whoa~ how is it just taken for granted that everything comes out real? Surely at least once you’d get a complex number sneaking out, right??”

If we write “the result is always real” as a formula, it looks like this:

$$\langle Q \rangle \quad = \quad \langle Q \rangle^{*}$$

Now let’s just write that out line by line:

$$\langle Q \rangle \quad = \quad \langle Q \rangle^{*} \\ \langle \psi | Q\psi \rangle \quad = \quad \langle \psi | Q\psi \rangle^{*} \\ \langle \psi | Q\psi \rangle \quad = \quad \langle Q\psi | \psi \rangle$$

So the punchline is:

$$\langle \psi | Q\psi \rangle \quad = \quad \langle Q\psi | \psi \rangle$$

The operator $Q$ that was sitting over in the right ket — you can move it over into the left bra, and the calculation has to give you the same thing. That’s what this is saying.

Aha ha ha — and the math people, who go off and study operators that satisfy that red-boxed condition separately, they have a name for it:

“Hermitian Operator”! heh heh heh.

Hermitian means:

$$Q = Q^{\dagger}$$

Anything that satisfies this? Hermitian matrix. Done.

$\dagger$

This little guy is called a dagger.

(And yeah — from the shape alone, “dagger” is 100% the right call. It’s a knife. Literally a knife. hahahahahaha)

Taking the dagger of a matrix means:

transpose it AND complex-conjugate it!!

But wait — in Dirac notation, the relationship between a bra vector and a ket vector is… transpose + complex conjugate, isn’t it?!?!?! Right?!!? (Go look at what I wrote in the earlier post, it’s in there.)

So, putting it together:

“Aha!!! Measuring = multiplying by a matrix.

And the result only makes physical sense when it comes out real.

Whoa — but that means the operator doing the measuring isn’t just any matrix. It has to be a Hermitian matrix~~~~~~!!~~!~!!!!!”

“Aha — so in quantum mechanics, all~~~~~ operators~~~~~ are Hermitian Operators.”

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Originally written in Korean on my Naver blog (2015-08). Translated to English for gdpark.blog.