Why do $\{Q,P\}$ form an irreducible set for a particle with no internal degrees of freedom?

In deriving expressions for the generators of Galilean symmetries $$\mathbf{J}, \mathbf{P}, \mathbf{G}, H$$, Ballentine (Quantum Mechanics: A Modern Development) uses that $$\{\mathbf{Q},\mathbf{P}\}$$ form an irreducible set for a isolated (free) particle with no internal degrees of freedom ($$\mathbf{Q}$$ is the position operator). Why is this? If they are irreducible then (by definition of irreducibility) there is no subspace of the abstract vector space which is left invariant under the action of both (sets of, since they are vector operators) operators. Should this be clear to me or is $$\{\mathbf{Q},\mathbf{P}\}$$ irreducible a definition of such an abstract space?

Edit: I here attach Ballentine's proof of their irreducibility (by showing that the equivalent condition from Schur's Lemma is respected by the set). It seems to only establish that the action of $$M$$ on the function $$\psi (x) =1$$ is to multiply by a constant. But how does it follow that its action on every function is to multiply by a constant?

• Have you checked the table of contents (this is actually the second time I've given you this advice)? Hint: Appendix A and B. Jan 26, 2023 at 17:57
• @TobiasFünke I didn't follow his derivation in Appendix B. In particular, it seems to only establish that the action of $M$ on the function $\psi(x) = 1$ is to multiply by a constant. But how does it follow that its action on every function is to multiply by a constant?
– EE18
Jan 26, 2023 at 18:02
• Could you please add this information, i.e. what exactly you don't understand in this derivation etc., to the question. Jan 26, 2023 at 18:03
• @TobiasFünke My apologies for the delay; I was en route to class. I've updated now.
– EE18
Jan 26, 2023 at 18:43
• Please stop posting screenshots and use MathJax instead. These are only a very few equations, so this really should be no problem. Jan 26, 2023 at 18:44

I am not sure to understand your issue. However irreducibility is here understood in the usual sense you pointed out. If there were an internal space, described by further degrees of freedom, there would be observables commuting with $$P$$ and $$Q$$ which are non trivial. The Schur's lemma would prove, from irreducibility, that these further observables are scalar multiples of the identity, namely trivial.

• Hi Prof. Moretti -- I've updated my question to reflect the conclusion. I would also be interested in a proof that does not use Schur's Lemma to establish irreducibility; namely, a proof that directly shows no nontrivial invariant subspace exists.
– EE18
Jan 26, 2023 at 18:44
• @EE18 IIRC, this is done in F. Strocchi chapter 3.3. Note that if you want to do it in a mathematical proper way, there is a lot of more work to do (you have to care about domains and so on). Jan 26, 2023 at 19:01
• I've searched Strocchi on Google but can't figure out what you refer to here -- can you mention the name of the book if you don't mind? @TobiasFünke Edit: Do you mean An Introduction to the Mathematical Structure of Quantum Mechanics: A Short Course for Mathematicians?
– EE18
Jan 26, 2023 at 19:09
• Oh, sorry, my bad. I referred to: Introduction To The Mathematical Structure Of Quantum Mechanics, An: A Short Course For Mathematicians. Jan 26, 2023 at 19:10
• Roughly speaking, the proof is made of two steps. If $A$ commutes with the spectral measure of $X$ then it is of the form $f(X)$. If a function of $X$ commutes with the spectral measure of $P$, then it is a multiple of the identity. I think a proof appears also in a book of mine, but I cannot check now. Jan 26, 2023 at 19:21

It seems to only establish that the action of $$M$$ on the function $$\psi(x)=1$$ is to multiply by a constant.

No, that's not true. The wording is a bit ambiguous, but the author is saying that the function $$m(x)$$ is what you get if you operate on the function $$\mathbb 1(x) = 1$$ with $$M$$, i.e. $$\mathbb 1(x) = 1 \implies \big(M\mathbb 1\big)(x) = m(x)$$

The claim the author is making is that if $$[M,X]=0$$, then $$M$$ must take the form of a so-called multiplication operator $$\big(M\psi\big)(x) = m(x)\psi(x)$$ for some function $$m:\mathbb R\rightarrow \mathbb C$$. If that is the case, then commutation with $$P:= -i\frac{d}{dx}$$ requires that $$m$$ be a constant function: $$\big([M,P]\psi\big)(x) = m(x)\psi'(x) - \big(m'(x)\psi(x) + m(x)\psi'(x)\big) = -m'(x)\psi(x) = 0$$ $$\implies m'(x) = 0$$ $$\implies m(x) = C$$ for some constant $$C$$.

• I think I follow you but want to check. Are you saying that after we establish $\big(M\mathbb 1\big)(x) = m(x) = c$, $c$ a constant, then we can go back to equation (B.2) and observe that it holds for all $f(x)$ (even though this $f$ was introduced as a coordinate representation for the action of some operator function of the operator $\hat{x}$)?
– EE18
Jan 26, 2023 at 19:51
• @EE18 Not quite, I've updated my answer. Jan 26, 2023 at 20:06
• Lovely, thank you. I think in fact I was understanding in my original answer (your $\psi$ is Ballentine's $f$ in (B.2)) but I appreciate you fleshing out the argument in detail nevertheless.
– EE18
Jan 26, 2023 at 20:10
• @EE18 That's not quite right - the $f$ in Ballentine's passage is part of the argument that $M$ must take the form of a multiplication operator, which I did not prove. He says that if $[M,X]=0$, then $Mf(x) \psi(x) = f(x)M\psi(x)$ for any function $f$. In particular, $M f(x) \mathbb 1(x) = f(x) M\mathbb 1(x) \equiv f(x) m(x)$, from which it follows that for arbitrary $\psi$, $M\psi(x) = m(x)\psi(x)$. Jan 26, 2023 at 20:39
• And in this latest comment you're saying that Ballentine is using "let $f = \psi$" in the last step, no (since we just established $M f(x) \mathbb 1(x) \equiv Mf(x) = f(x) m(x)$, where I've used $\equiv$ to emphasize that we can now think of M ia acting on the c-function $f$ even though it was introduced as an operator)? And using that (as they're now c-numbers) $f(x) m(x) = m(x) f(x)$?
– EE18
Jan 26, 2023 at 20:49