Formalism: Determinate States

Alright, transcribing from Griffiths today. Let’s go.

So in regular QM, if you take some physical quantity $Q$ and measure it across an ensemble of identically prepared systems — all sitting in the exact same state $\psi$ — you do not expect to get the same measurement every time. That’s just the uncertainty principle doing its thing.

Then the textbook goes:

“But is it possible that when we measure $Q$ on some state, the same value $q$ comes out every single time?”

Wait. Hold on. Two sentences ago you literally said “you do not expect the same value to come out.” And now you’re asking if it’s possible for the same value to come out? lolololol

Is this even a valid question?? lol lol ugh fine, whatever, moving on heh.

OK — continuing the transcription.

If that is possible, then this state is called a determinate state with respect to $Q$.

And actually — once I read the next line, it clicked.

We already know an example of a determinate state. A stationary state is a determinate state with respect to the Hamiltonian. If you measure the energy of a particle in stationary state $\psi_n$, you are guaranteed to get the allowed energy $E_n$. No randomness. Just $E_n$.

So a determinate state w.r.t. $Q$ is just the version of that where “something” — some specific $q$ — is guaranteed to come out as the measurement.

Now if we stop writing this in words and write it as a formula — the condition is just “the standard deviation of $Q$ is 0”:

$$ 0 \;=\; \sigma_{Q}^{2} \;=\; \left< \psi\,|\,(\hat{Q}-q)^{2}\psi \right> \;=\; \left< \psi\,|\,(\hat{Q}-q)(\hat{Q}-q)\psi \right> \\ \text{and since } \hat{Q} \text{ is Hermitian,}\\ \;=\; \left< (\hat{Q}-q)\psi \,|\, (\hat{Q}-q)\psi \right> \\ \therefore \;(\hat{Q}-q)\psi \;=\; 0 \\ \therefore \;\hat{Q}\psi \;=\; q\psi $$

And boom — out pops the eigenvalue equation. $\psi$ is the eigenfunction of $\hat{Q}$ with eigenvalue $q$.

But — and this is the point — it’s not just any $\psi$. It’s specifically a determinate-state $\psi$.

OK, let me tidy up the vocabulary real quick.

Spectrum: the set of all eigenvalues of some operator. That’s it, that’s the word.

Now — there’s no rule that different eigenfunctions have to have different eigenvalues. You could totally have:

• eigenfunction $G$ with eigenvalue $q$
• eigenfunction $D$ with eigenvalue $q$

Same $q$! Different functions! This can absolutely happen.

When it does, we say the spectrum is degenerate (or “the spectrum overlaps”).

And — following right along from everything we’ve been doing — there’s one more thing we can prove:

Eigenfunctions with different eigenvalues are orthogonal to each other.

And the cool part is this proof chains straight off the previous ones, tail biting tail. Watch:

“A measurable physical quantity has to be real” → “so the operator has to be Hermitian” → “and because of that, eigenfunctions with different eigenvalues are orthogonal.”

See how it all links up? Now let’s actually do this one.

Take operator $\hat{Q}$, and say two different eigenfunctions $f$ and $g$ have eigenvalues $q$ and $q'$ respectively:

$$\hat{Q}f \;=\; qf \qquad // \qquad \hat{Q}g \;=\; q'g$$$$ \hat{Q} \text{ is Hermitian, so}\\ \left< f\,|\,\hat{Q}g \right> \;=\; \left< \hat{Q}f\,|\,g \right> \\ \text{plug in the eigenvalue equations:}\\ \left< f\,|\,q'g \right> \;=\; \left< qf\,|\,g \right> \\ \text{and since } q \text{ and } q' \text{ are just numbers, pull them out:}\\ q'\left< f\,|\,g \right> \;=\; q\left< f\,|\,g \right> \\ \therefore \;\left< f\,|\,g \right> \;=\; 0 \\ \text{i.e., } f \text{ and } g \text{ are orthogonal.} $$

Done.

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