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Let $\mathcal{A}$ be a unital $\mathrm{C}^*$-algebra with $\varphi\in\mathcal{S}(\mathcal{A})$ a state. Where $$\sigma_\varphi(a):=\sqrt{\varphi((a-\varphi(a)1_{\mathcal{A}})^2)}\qquad (a\in \mathcal{A})$$ and $[a,b]=ab-ba$ the commutator, we have Robertson's inequality (uncertainty principle):

$$\sigma_\varphi(a)\sigma_\varphi(b)\geq \frac12|\varphi([a,b])|\qquad(a,\,b\in\mathcal{A}).$$

Now consider projections $p,\,q\in\mathcal{A}$. If $\varphi$ is a state such that $\varphi(p)=0$ or $\varphi(p)=1$, then $\sigma_\varphi(p)=0$. It follows then that whether or not $p$ and $q$ commute, that $\varphi(pq)=\varphi(qp)$. This follows from the inequality above but also, in this case, if $\varphi(p)=1$ $$\varphi(a)=\varphi(pa)=\varphi(ap)=\varphi(pap)\qquad (a\in\mathcal{A}),$$ with a similar statement for with $1_{\mathcal{A}}-p$ if $\varphi(p)=0$.

However in this case the inequality cannot tell us anything about $\sigma_\varphi(q)$.

Is there an alternative to Robertson's inequality that can be used to estimate $\sigma_\varphi(q)=\varphi(q)(1-\varphi(q))$ in terms of $[p,q]$ in the degenerate case of $\sigma_\varphi(p)=0$?

Note $\sigma_\varphi(q)\in[0,1/4]$. For example, consider $\mathcal{A}=M_2(\mathbb{C})$ with $$p=\left(\begin{array}{cc} \frac12 & \frac12 e^{-i\pi/4} \\ \frac12 e^{+i\pi/4} & \frac12 \end{array}\right),$$ and $q=p^T$. Let $\varphi_0\in \mathcal{S}(M_2(\mathbb{C}))$ be any state such that $\varphi_0(p)>0$. Then it can be conditioned to a state with $\varphi(p)=1$, $$\varphi(a)=\dfrac{\varphi_0(pap)}{\varphi_0(p)}\qquad (a\in M_2(\mathbb{C})).$$ It follows that $$\varphi(q)=\frac12\implies \sigma_\varphi(q)=\sigma_\max=\frac14.$$

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  • $\begingroup$ If $A$ is commutative and $\varphi$ is diffuse, $[p,q]=0$ does not depend on $q$, but $\sigma_\varphi(q)$ can take any value between $0$ and $1/4$. So there is nothing to say about $\sigma_\varphi(q)$ by just looking at $[p,q]$. $\endgroup$
    – Mikael de la Salle
    Commented Dec 22, 2023 at 15:14
  • $\begingroup$ @MikaeldelaSalle thank you, in that case. But perhaps something can be said when the commutator is non-zero. $\endgroup$
    – JP McCarthy
    Commented Dec 22, 2023 at 16:10
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    $\begingroup$ I doubt anything can be said when $[p, q] \neq 0$ either. For example, choose $p', q' \in \mathcal{A}_1$ with $[p', q']$ whatever you wish. For any $0 \leq r \leq 1/4$, choose $q'' \in \mathcal{A}_2$ and a state $\phi$ on $\mathcal{A}_2$ with $\sigma_\phi^2(q'') = r$. Then let $\mathcal{A} = \mathcal{A}_1 \oplus \mathcal{A}_2$, $\varphi = 0 \oplus \phi$, $p = p' \oplus 1$, $q = q' \oplus q''$. Then $[p, q] = [p', q'] \oplus 0$ but $\sigma_\varphi^2(p) = 0$ and $\sigma_\varphi^2(q) = r$. (If you want $\varphi(p) = 0$ instead of $1$, just change $p = p' \oplus 1$ to $p = p' \oplus 0$.) $\endgroup$
    – David Gao
    Commented Dec 23, 2023 at 3:42
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    $\begingroup$ It may be more meaningful to ask about $pqp$ (when $\varphi(p) = 1$) or $(1 - p)q(1 - p)$ (when $\varphi(p) = 0$) instead of the commutator. $\endgroup$
    – David Gao
    Commented Dec 23, 2023 at 3:44
  • $\begingroup$ @DavidGao so, arbitrary commutator, zero variance in $p$, arbitrary variance in $q$. I think this is an answer and could be converted so and I will accept. All I can say about $pqp$ is that it does detect if the commutator vanishes as $pqp=qpq$ iff zero commutator. And $\varphi(pqp)=\varphi(q)$. $\endgroup$
    – JP McCarthy
    Commented Dec 23, 2023 at 7:44

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As requested, this is my comment converted into an answer:

Not much can be said even assuming $[p, q] \neq 0$. For example, choose $p', q' \in \mathcal{A}_1$ with $[p', q']$ whatever you wish. For any $0 \leq r \leq 1/4$, choose $q'' \in \mathcal{A}_2$ and a state $\phi$ on $\mathcal{A}_2$ with $\sigma^2_\phi(q'') = r$. Then let $\mathcal{A} = \mathcal{A}_1 \oplus \mathcal{A}_2$, $\varphi = 0 \oplus \phi$, $p = p' \oplus 1$, $q = q' \oplus q''$. Then $[p, q] = [p', q'] \oplus 0$ but $\sigma^2_\varphi(p) = 0$ and $\sigma^2_\varphi(q) = r$. (If you want $\varphi(p) = 0$ instead of $1$, just change $p = p' \oplus 1$ to $p = p' \oplus 0$.)

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