2
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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: $$\langle\psi|BA|\psi\rangle=\langle A^\dagger B^\dagger \psi|\psi\rangle = \langle AB\psi|\psi\rangle = \overline{\langle\psi|AB|\psi\rangle}=x-iy$$

$\endgroup$
5
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.