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Let $\mathcal{A}$ be a unital $\mathrm{C}^*$-algebra with $\varphi\in\mathcal{S}(\mathcal{A})$ a state. Where $$\sigma_\varphi(a):=\sqrt{\varphi((a-\varphi(a)1_{\mathcal{A}})^2)}\qquad (a\in \mathcal{A})$$ is a variance, 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 $$\varphi(q)=\frac12\implies \sigma_\varphi(q)=\sigma_\max=\frac14.$$

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 $$\varphi(q)=\frac12\implies \sigma_\varphi(q)=\sigma_\max=\frac14.$$

Let $\mathcal{A}$ be a unital $\mathrm{C}^*$-algebra with $\varphi\in\mathcal{S}(\mathcal{A})$ a state. Where $$\sigma_\varphi(a):=\sqrt{\varphi((a-\varphi(a)1_{\mathcal{A}})^2)}\qquad (a\in \mathcal{A})$$ is a variance, 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 $$\varphi(q)=\frac12\implies \sigma_\varphi(q)=\sigma_\max=\frac14.$$

I may give further context later.

Let $\mathcal{A}$ be a unital $\mathrm{C}^*$-algebra with $\varphi\in\mathcal{S}(\mathcal{A})$ a state. Where $$\sigma_\varphi(a):=\sqrt{\varphi((a-\varphi(a)1_{\mathcal{A}})^2)}\qquad (a\in \mathcal{A})$$ is a variance, 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 $$\varphi(q)=\frac12\implies \sigma_\varphi(q)=\sigma_\max=\frac14.$$

Let $\mathcal{A}$ be a unital $\mathrm{C}^*$-algebra with $\varphi\in\mathcal{S}(\mathcal{A})$ a state. Where $$\sigma_\varphi(a):=\sqrt{\varphi((a-\varphi(a)1_{\mathcal{A}}))^2}\qquad (a\in \mathcal{A})$$ is a variance, 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 $$\varphi(q)=\frac12\implies \sigma_\varphi(q)=\sigma_\max=\frac14.$$

I may give further context later.

Let $\mathcal{A}$ be a unital $\mathrm{C}^*$-algebra with $\varphi\in\mathcal{S}(\mathcal{A})$ a state. Where $$\sigma_\varphi(a):=\sqrt{\varphi((a-\varphi(a)1_{\mathcal{A}})^2)}\qquad (a\in \mathcal{A})$$ is a variance, 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 $$\varphi(q)=\frac12\implies \sigma_\varphi(q)=\sigma_\max=\frac14.$$

I may give further context later.