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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.
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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.

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  • $\begingroup$ You are certainly right. I miss formulated what I meant, I apologize. I was asking about the conditions in which the Schrödinger's Relation turns to be the same as the Robertson's Relation (by necessity, not just because it is a lower bound). For example, for $[\hat{Q}, \hat{P}] = i\hbar$, Gaussian states reach the minimum uncertainty $\frac{\hbar}{2}$, so I know that the second term with the anti-commutator must be $0$, but I don't know how. Also, thanks for the reference, I'll check it out. $\endgroup$
    – Gilgamesh
    Commented Sep 27, 2021 at 5:31
  • $\begingroup$ @SomeUser I edited my answer. $\endgroup$ Commented Sep 27, 2021 at 6:26
  • $\begingroup$ I see this relations as giving conditions on operators (commutator, anticommutator, etc.) and getting a lower bound for all states. So, while I agree on what you added, I can't relate it to the question. I have corrected it, I hope is clearer. $\endgroup$
    – Gilgamesh
    Commented Sep 27, 2021 at 6:54
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    $\begingroup$ @SomeUser that can be done by substituting $Q=Q$ and $P=i\hbar\frac{\partial}{\partial x}$ and gaussian wavefunction. The average value of Q becomes zero. And calculating others that term will be zero. The other commutator term is a constant and is independent of wavefuntion. $\endgroup$ Commented Sep 27, 2021 at 7:22
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    $\begingroup$ Ok, the commutator relation implies those representations on the q-space. Then since we previously know a particular state (Gaussian) which makes the second term $0$, we are certain that we have found the minimal bound. This proves it and answer's my question. Thank you. $\endgroup$
    – Gilgamesh
    Commented Sep 27, 2021 at 9:18

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