I'm trying to follow Nielsen and Chuang Book on Quantum Computation and Quantum Information. There is Box 2.4 on the Heisenberg Uncertainty Principle. I got stuck pretty fast. In that box they define: $$ \left<\psi | AB|\psi \right> = x+iy $$ where $x$ an $y$ are real. They note that the commutator $\left<\psi \left | [A,B]\right |\psi \right> = 2iy$ and the anticommutator is $\left<\psi |\{ A,B \}|\psi \right> = 2x$. Thus, the book says that this implies the following. $$ \left |\left<\psi \left | [A,B]\right |\psi \right>\right|^2 + \left| \left<\psi | \left\{ A,B\right\}|\psi \right>\right|^2=4\left| \left<\psi | AB|\psi \right>\right|^2 $$ I'm trying to proof that statement, but I cannot figure it out. I tried two ways.

1. I expand the lhs of the equation to obtain the rhs.

$$ \begin{matrix} \left |\left<\psi \left | [A,B]\right |\psi \right>\right|^2 + \left| \left<\psi | \left\{ A,B\right\}|\psi \right>\right|^2=\\ \left |\left<\psi \left | AB-BA\right |\psi \right>\right|^2 + \left| \left<\psi | AB+BA|\psi \right>\right|^2= \\ \left |\left<\psi \left | AB\right |\psi \right>-\left<\psi \left | BA\right |\psi \right>\right|^2 + \left |\left<\psi \left | AB\right |\psi \right>+\left<\psi \left | BA\right |\psi \right>\right|^2 =\\ \left<\psi \left | AB\right |\psi \right>^2-2\left<\psi \left | AB\right |\psi \right>\left<\psi \left | BA\right |\psi \right> + \left<\psi \left | BA\right |\psi \right>^2+\left<\psi \left | AB\right |\psi \right>^2+2\left<\psi \left | AB\right |\psi \right>\left<\psi \left | BA\right |\psi \right> + \left<\psi \left | BA\right |\psi \right>^2 =\\ 2\left<\psi \left | AB\right |\psi \right>^2+2\left<\psi \left | BA\right |\psi \right>^2 \end{matrix} $$

Which doesn't seem equal to $4\left| \left<\psi | AB|\psi \right>\right|^2$ (unless it commutes but I guess it is not the case, is it?).

2. I expand from $x$ and $y$'s definition.

Expanding lhs: $$ \begin{matrix} \left |\left<\psi \left | [A,B]\right |\psi \right>\right|^2 + \left| \left<\psi | \left\{ A,B\right\}|\psi \right>\right|^2=\\ |2iy|^2+|2x|^2\\ -4y^2+4x^2 \end{matrix} $$ Expanding rhs: $$ \begin{matrix} 4\left| \left<\psi | AB|\psi \right>\right|^2=\\ |x+iy|^2=\\ x^2+2ixy-y^2 \end{matrix} $$

Maybe I'm missing something, but it the lhs, the expansion seems real and the rhs the expansion seems complex.

I feel like missing something obvious, but I'm failing to find an answer.


1 Answer 1


What you seem to be missing is that $|\cdot |$ refers to the modulus of a complex number. For a complex number $z=x+iy$ we have $|z|=\sqrt{x^2+y^2}=\sqrt{z\cdot\overline{z}}$ where $\overline{z}=x-iy$ is the conjugate of $z$. So (in your second attempt for example): $$|x+iy|^2=(x+iy)(x-iy)=x^2-ixy+iyx+y^2=x^2+y^2 \color{red}{\neq} x^2+2ixy-y^2$$ The same thing went wrong in your first calculation since you treated the modulus the same way you would treat the absolute value for real numbers.

EDIT: To see why $\langle\psi|BA\psi\rangle=x-iy$, assuming $A$ and $B$ are Hermitian operators, write: $$\langle\psi|BA|\psi\rangle=\langle A^\dagger B^\dagger \psi|\psi\rangle = \langle AB\psi|\psi\rangle = \overline{\langle\psi|AB|\psi\rangle}=x-iy$$

  • $\begingroup$ Ok, my complex sense didn't kick in and I treated the things as real. Thanks a lot @Giorgos. So, for the first expansion $2\left<\psi \left | AB\right |\psi \right>\left<\psi \left | BA\right |\psi \right>$ should be $\overline{\left<\psi \left | AB\right |\psi \right>}\left<\psi \left | BA\right |\psi \right>+\left<\psi \left | AB\right |\psi \right>\overline{\left<\psi \left | BA\right |\psi \right>}$ which is going to cancel anyway (Maybe I'm doing another silly mistake). $\endgroup$
    – silgon
    Commented Sep 18, 2022 at 6:35
  • $\begingroup$ Thus, the only thing missing for me is why $\left<\psi | BA|\psi \right> = x-iy$ without expanding from $\left<\psi \left | [A,B]\right |\psi \right>$, any thoughts on that? $\endgroup$
    – silgon
    Commented Sep 18, 2022 at 6:35
  • $\begingroup$ $(\langle \psi| BA|\psi\rangle)^{\dagger} = (|\psi\rangle)^{\dagger} A^{\dagger} B^{\dagger} (\langle \psi|)^{\dagger} = \langle \psi | AB |\psi\rangle$ which assumes that $A$ and $B$ are both operators. $\endgroup$ Commented Sep 18, 2022 at 9:38
  • $\begingroup$ @silgon I have edited my answer. These are basically properties of the inner product and Hermitian operators. $\endgroup$ Commented Sep 18, 2022 at 11:03
  • $\begingroup$ Thanks @NAMcMahon for the comment. Also, Giorgos, I accepted the answer. I was having a hard time with that for both silly mistakes and simple reasons. Thank you both! $\endgroup$
    – silgon
    Commented Sep 18, 2022 at 11:22

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