3
$\begingroup$

I'm reading Mathematical Methods in Quantum Mechanics by Gerald Teschl and I came across the following exercise whose statement is causing me some troubles. It goes like this:

Let $A$ and $B$ two self-adjoints operators. Then $A$ and $B$ commute if and only if the corresponding spectral projections $P_{A}(\Omega)$ and $P_{B}(\Omega)$ commute fore every Borel set $\Omega$.

He defines that two self adjoint operators $A$ and $B$ commute if and only if their resolvents $R_A(z_1):=(A-z_1)^{-1}$ and $R_B(z_2):=(B-z_2)^{-1}$ commute for at least one $z_1$ in the resolvent of $A$ and one $z_2$ in the resolvent of $B$. After this, he proves that this is equivalent to $[f(A),g(B)]=0$ for arbitrary bounded Borel functions (which trivially shows one implication of the exercise).

My main problem is understanding the hypothesis of the problem: does it say that $P_B(\Omega)P_A(\Omega')=P_A(\Omega')P_B(\Omega)$ for arbitrary $\Omega, \Omega'$ or that this is only valid for $\Omega=\Omega'$? In the former case, I can see how to prove both directions (approximating bounded measurable functions by simple ones), but I can't see how to proceed in the latter case.

I've tried to look for definitions in other books, but they don't really solve this ambiguity (they state this in a similar way).

Any help is appreciated. Thanks in advance

$\endgroup$
  • $\begingroup$ They commute on the same borel set $\Omega $ $\endgroup$ – Mhenni Benghorbal Feb 13 '16 at 6:23
  • 1
    $\begingroup$ @MhenniBenghorbal Thanks! But now...could you give me a hint to solve this? Because this is just the case in which I'm stuck. I tried to prove that $[f(A),f(B)]=0$ for every bounded measurable function, but couldn't prove it for simple functions... $\endgroup$ – Brandon Feb 13 '16 at 7:41
  • $\begingroup$ When one says "the spectral projections commute" in this context, what is meant is that $P_A(M)P_B(M')=P_B(M')P_A(M)$ (what Teschl writes must be based on a typo or a slip of mind). But you've asked an interesting question, namely, whether this is already implied by the same condition for just $M'=M$. $\endgroup$ – user138530 Feb 13 '16 at 23:01
1
$\begingroup$

From your question, it seems that you already know that if $P_A (M) P_B (M') = P_B (M') P_A (M)$ for all Borel sets $M,M'$, you are done.

I will show that this holds if we assume $P_A (M) P_B (M) = P_B (M) P_A (M)$ for all Borel sets $M$. For brevity, I write $\mu := P_A$ and $\nu := P_B$.


Let us first assume that $M \cap N = \emptyset$. In this case, because of $\mu(M) \mu(N) = \mu(M\cap N) = 0$ (and likewise for $\nu$), we have \begin{align*} &\mu(M) \nu(M) + \mu(M)\nu(N) + \mu(N) \nu (M) + \mu(N) \nu(N) \\ &= [\mu(M) + \mu(N)] \cdot [\nu(M) + \nu(N)] \\ &= \mu(M \cup N) \nu(M \cup N) \\ &= \nu(M \cup N) \mu(M \cup N) \\ &= [\nu(M) + \nu(N)] \cdot [\mu(M) + \mu(N)] \\ &= \nu(M) \mu(M) + \nu(M)\mu(N) + \nu(N)\mu(M) + \nu(N)\mu(N). \end{align*} Using $\mu(N) \nu(N) = \nu(N) \mu(N)$ (and the analogous formula for $M$), we deduce $$ \mu(M) \nu(N) + \mu(N)\nu(M) = \nu(M) \mu(N) + \nu(N) \mu(M). \qquad (\ast) $$

Now, multiply this equation from the left by $\mu(N)$. Because of $\mu(N) \mu(M) = \mu(N \cap M) = 0$ and $\mu(N) \mu(N) = \mu(N)$, we get \begin{align*} \mu(N) \nu(M) &= \mu(N) \nu(M) \mu(N) + \mu(N) \nu(N) \mu(M) \\ &= \mu(N) \nu(M) \mu(N) + \nu(N) \underbrace{\mu(N)\mu(M)}_{=0} \\ &= \mu(N) \nu(M) \mu(N). \tag{$\dagger$} \end{align*}

Note that since $\mu(N)$ and $\nu(M)$ are self-adjoint, so is the right-hand side of $(\dagger)$. Hence, so is the left-hand side, i.e. $$ \mu(N) \nu(M) =[\mu(N) \nu(M)]^\ast = \nu(M) \mu(N). $$ This establishes what we want, but only under the assumption $N \cap M = \emptyset$.


Now, let $M,N$ be arbitrary Borel sets. Then $M = (M \setminus N) \cup (M\cap N)$, where the union is disjoint. Analogously, $N = (N \setminus M) \cup (M \cap N)$, so that we get \begin{align*} &\mu(M) \nu(N) \\ &= [\mu(M \setminus N) + \mu(M \cap N)] \cdot [\nu(N \setminus M) + \nu(M \cap N)] \\ &= \mu(M \setminus N) \nu (N \setminus M) + \mu(M \setminus N) \nu(M \cap N)\\ &\quad + \mu(M \cap N)\nu(N \setminus M) + \mu(M \cap N) \nu(M \cap N) \\ &\overset{(\ast)}{=} \nu(N \setminus M) \mu (M \setminus N) + \nu(M \cap N)\mu(M \setminus N) \\ &\quad + \nu(N \setminus M)\mu(M \cap N) + \nu(M \cap N) \mu(M \cap N) \\ &= \nu(N) \mu(M) \end{align*} Here, the step marked with $(\ast)$ used the assumption on the last term and the step from above (for disjoint Borel sets) on all the other terms. Finally, the last step used a similar computation as the first two steps.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.