On the Robertson uncertainty relation when $\big|\dfrac{1}{2} \langle\{A,B\}\rangle - \langle A \rangle \langle B \rangle \big|^2 \neq 0$

Question

The Robertson uncertainty relation is

$\sigma^2_A \sigma^2_B \geq \big|\dfrac{1}{2} \langle\{A,B\}\rangle - \langle A \rangle \langle B \rangle \big|^2 + \big| \dfrac{1}{2i} \langle [A,B] \rangle \big|^2.$

Where $\sigma^2_X$ is the variance of the operator $X$ and $\{A,B\}$, $[A,B]$ are the anti-commutator and the commutator of the Hermitian operators $A$ and $B$, respectively.

The uncertainty relation is more commom presented in the form

$\sigma^2_A \sigma^2_B \geq \big| \dfrac{1}{2i} \langle [A,B] \rangle \big|^2.$

Where there are commom physical examples which have that satisfied, e.g. $[x,p] \geq \dfrac{\hbar}{2}$, but these examples have $\big|\dfrac{1}{2} \langle\{A,B\}\rangle - \langle A \rangle \langle B \rangle \big|^2=0$.

I am trying to find a quantum system where the term $\big|\dfrac{1}{2} \langle\{A,B\}\rangle - \langle A \rangle \langle B \rangle \big|^2 \neq 0$, so that the lowest limit of the product of the variances of $A$ and $B$ have a dependence on the latter. So, to answer my question, it is necessary to give a possible physical system where $\big|\dfrac{1}{2} \langle\{A,B\}\rangle - \langle A \rangle \langle B \rangle \big|^2 \neq 0$ for $A$ and $B$ Hermitian.

Any help or ideas are welcome.

Answer 1

You might profit from calculating a few expectation values for, e.g. the oscillator, for which $a^\dagger |n\rangle= \sqrt{n+1} | n+1\rangle$. Take $A=a$ and $B=a^\dagger$, so that $$ [a,a^\dagger ]=1, \qquad \{ a,a^\dagger \} =a a^\dagger + a^\dagger a=1+2N. $$

Look at the first excited state, $|1\rangle$, so $a|1\rangle=|0\rangle$, so your expectation values are $$ \langle 1| a|1\rangle=\langle 1|a^\dagger| 1\rangle=0,\\ \langle 1| \{ a,a^\dagger \} |1\rangle= 3,\\ \langle 1| [a,a^\dagger ]|1\rangle= 1, $$ so that, for this state $$ \sigma_a^2 \sigma^2_{a^\dagger}\geq 9/4 +1/4= 10/4. $$ The anticommutator, of course, is not a constant, unlike the commutator, nor should you expect it to be.