Deriving Uncertainty Relation for Canonical Commuting Observables from Schrödinger Uncertainty Relation

Question

In case the names are not standard: \begin{equation} \sigma_{\hat{A}}^{2}\sigma_{\hat{B}}^{2} \geq \left\vert \frac{1}{2i} \left\langle \left[\hat{A},\hat{B}\right] \right\rangle \right\vert^{2} + \left\vert \frac{1}{2} \left\langle \left\{\hat{A},\hat{B}\right\}\right\rangle - \left\langle\hat{A}\right\rangle \left\langle\hat{B}\right\rangle \right\vert^{2} \label{eq:1} \tag{Scrhödinger's Uncertainty Relation} \end{equation} \begin{equation} \sigma_{\hat{A}}\sigma_{\hat{B}} \geq \frac{\left\vert \left\langle \left[\hat{A},\hat{B}\right]\right\rangle \right\vert}{2} \label{eq:2} \tag{Robertson's Uncertainty Relation} \end{equation} Given $\hat{Q},\hat{P}$ observables such that $\left[\hat{Q},\hat{P}\right] = i\hbar$ \begin{equation} \sigma_{\hat{Q}}\sigma_{\hat{P}} \geq \frac{\hbar}{2} \label{eq:3} \tag{Uncertainty Relation for Cannonical Commuting Observables} \end{equation}

For \ref{eq:3}, we know that the equality is reached, meaning $\left[\hat{Q},\hat{P}\right] = i\hbar$ is a sufficient condition for \begin{equation} \left\vert \frac{1}{2} \left\langle \left\{\hat{A},\hat{B}\right\}\right\rangle - \left\langle\hat{A}\right\rangle \left\langle\hat{B}\right\rangle \right\vert^{2} = 0 \end{equation} on \ref{eq:1}. This is what I'm not able to prove.

An ideal answer would contain the necessary conditions for this - that the \ref{eq:1} turns out to be \ref{eq:2} and the equality is reached - to happen.

PD:

• I'm aware of how to derive \ref{eq:3} from the commutation relation by other means. I'm interested in what is mentioned above.
• In case it sounds like homework, hints or references are highly appreciated.

Answer 1

Robertson's Uncertainty Relation will hold even if $$\begin{equation} \left\vert \frac{1}{2} \left\langle \left\{\hat{A},\hat{B}\right\}\right\rangle - \left\langle\hat{A}\right\rangle \left\langle\hat{B}\right\rangle \right\vert^{2} \neq 0 \end{equation}$$ Because $$\sigma _{A}^{2}\sigma _{B}^{2}\geq \left|{\frac {1}{2}}\langle \{{\hat {A}},{\hat {B}}\}\rangle -\langle {\hat {A}}\rangle \langle {\hat {B}}\rangle \right|^{2}+\left|{\frac {1}{2i}}\langle [{\hat {A}},{\hat {B}}]\rangle \right|^{2}$$ $$\Rightarrow\sigma _{A}^{2}\sigma _{B}^{2}\geq \left|{\frac {1}{2i}}\langle [{\hat {A}},{\hat {B}}]\rangle \right|^{2}$$ since both of them are positive and if something is bigger than the addition of two positive real numbers then it is also bigger than each positive number seperately. $$\Rightarrow\sigma _{A}^{2}\sigma _{B}^{2}\geq \left|{\frac {1}{2}}\langle [{\hat {A}},{\hat {B}}]\rangle \right|^{2}$$ $$\Rightarrow\sigma _{A}\sigma _{B}\geq {\frac {1}{2}}\left|\langle [{\hat {A}},{\hat {B}}]\rangle \right|$$

Scrhödinger's Uncertainty Relation is a stronger condition and if it satisfied then Robertson's Uncertainty Relation directly follows from that.

But Scrhödinger's Uncertainty Relation is satisfied only if the vector $ {\hat {B}}|\Psi \rangle $ is in the domain of the unbounded operator $ {\hat {A}}$, which is not always the case. A counter example is given here in Wikipedia.

Also refer to chapter 12 of Quantum Theory for Mathematicians by Brian C. Hall 2013 for more information.

Edit: Based on the comment I understood that you are asking the condition for a state in which both inequalities will imply each other and both equalities are satisfied.

Let there be some states $|\Psi_i \rangle$ for which Scrhödinger's Uncertainty Relation satisfies the equality and let there be some states $|\psi_i \rangle$ for which Robertson's Uncertainty Relation satisfies the equality. If $|\Psi_i \rangle$ and $|\psi_i \rangle$ have a common state, then for that state both relations satisfy with equality.

Since such a state satisfies both equalities it also satisfies: $\begin{equation} \left\vert \frac{1}{2} \left\langle \left\{\hat{A},\hat{B}\right\}\right\rangle - \left\langle\hat{A}\right\rangle \left\langle\hat{B}\right\rangle \right\vert^{2} = 0 \end{equation}$

But even if $\begin{equation} \left\vert \frac{1}{2} \left\langle \left\{\hat{A},\hat{B}\right\}\right\rangle - \left\langle\hat{A}\right\rangle \left\langle\hat{B}\right\rangle \right\vert^{2} = 0 \end{equation}$ for a state that doesn't imply that for that state both uncertainty relations are satisfied with equality.