0
$\begingroup$

I'm reading Quantum Computation and Quantum Information by Nielsen and Chuang (http://mmrc.amss.cas.cn/tlb/201702/W020170224608149940643.pdf). Equation 2.92 on page 85 is a bit confusing to me

if the state of the quantum system is $|\psi\rangle$ immediately before the measurement, then the probability that result $m$ occurs is given by $$p(m) = \langle\psi| M^†_m M_m |\psi\rangle$$

On page 62, they define $\langle\phi | A | \psi\rangle$ as the inner product between $|\phi\rangle$ an $A|\psi\rangle$

So, in this case, A = $M^†_m M_m$, which means the operator $M_m$ is pre-multiplied by it's Hermitian adjoint (the conjugate of the transpose) Then it is matrix-multiplied by the vector $|\psi\rangle$. Then the result of that is inner-producted with $|\psi\rangle$

is that correct?

$\endgroup$
2
  • 1
    $\begingroup$ Note that $\langle \psi | M_m^\dagger M_m | \psi \rangle = \langle M_m \psi|M_m \psi\rangle$ i.e. you're taking the inner product of $M_m |\psi\rangle$ with itself. $\endgroup$ – Beta Decay Feb 21 '20 at 19:08
  • 2
    $\begingroup$ For future reference: use \langle \rangle instead of < > $\endgroup$ – Superfast Jellyfish Feb 21 '20 at 19:11
1
$\begingroup$

In this case they are taking the inner product of $M_m | \psi \rangle$ with itself. The adjoint of that is $\langle \psi | M_m^\dagger$, so the inner product is $\langle \psi | M_m^\dagger M_m | \psi \rangle$.

$\endgroup$
1
  • $\begingroup$ ah, and it seems equation 2.18 on page 67 defines the inner product of |a> and |b> as the conjugate-transpose-of-~| a> times |b>. I think the problem is that I did so much linear algebra in R instead of C. That makes it more clear, thanks! $\endgroup$ – Mohammad Athar Feb 21 '20 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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