This is related to exercise 4.34,

The operation described can be written as $(H \otimes I)C^1(U)(H \otimes I)(|0\rangle \otimes |\psi\rangle)$

I can get to the point where the state of the system is given by:

$|0\rangle \otimes(I+U)|\psi\rangle + |1\rangle \otimes(I-U)|\psi\rangle$,

where $U$ is a Hermitian unitary with eigenvalues $-1$ and $+1$ with corresponding eigenvectors $|\lambda_-\rangle$ and $|\lambda_+\rangle$ respectively.

however I am stuck on the final part that when measuring $q_0$ the post-measurement state is given by the corresponding eigenvector of $U$. This reduces down to to showing that

$(I-U)|\psi\rangle = |\lambda_-\rangle$

$(I+U)|\psi\rangle = |\lambda_+\rangle$

I have tried using the spectral decomp of $U$ however I can't seem to get it to lead anywhere. My current trail of thought (not sure if correct) is if taking the density of the system, $(I\pm U)$ reduces down to projectors for $|\lambda_\pm\rangle$, s.t. $P_{\pm}|\psi\rangle = c_{\pm}\lambda_\pm $.

-- Update --

An answer using projectors (as suspected) is using the fact that for $U$ to be unitary and Hermitian then $U = (2P - I)$ for an orthogonal projector $P$ (https://math.stackexchange.com/questions/57148/matrices-which-are-both-unitary-and-hermitian), hence $(I+U)$ and $(I-U)$ reduce down to a projector $P$ and its orthogonal complement $2(I-P)$, thus projecting $\psi$ onto the eigenvectors.


1 Answer 1


Just need to write it out: \begin{align} (I-U)|\psi\rangle &= I\big(c_+|\lambda_{-}\rangle + c_{-}|\lambda_{-}\rangle\big) - U\big(c_+|\lambda_{-}\rangle + c_{-}|\lambda_{-}\rangle\big) \\ &= c_+|\lambda_{+}\rangle + c_{-}|\lambda_{-}\rangle - \big(c_+|\lambda_{+}\rangle - c_{-}|\lambda_{-}\rangle\big) \\ &= c_+|\lambda_{+}\rangle + c_{-}|\lambda_{-}\rangle - c_+|\lambda_{+}\rangle + c_{-}|\lambda_{-}\rangle \\ &= c_{-}|\lambda_{-}\rangle \end{align}

I've used $U(c_{-}|\lambda_{-}\rangle) = -c_{-}|\lambda_{-}\rangle$.

  • $\begingroup$ I'm a little confused as to how/where you decomposed $\psi$, unless you're saying that the eigenvectors of $U$ are $|\lambda_+\rangle = |0\rangle$ and $|\lambda_-\rangle = |1\rangle$ $\endgroup$
    – Sam Palmer
    Commented Apr 28, 2020 at 19:26
  • 1
    $\begingroup$ I've decomposed it in terms of the eigenvectors of U. $\endgroup$
    – Peter-Jan
    Commented Apr 28, 2020 at 19:29
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    $\begingroup$ The eigenvectors don't have to be |0> and |1>. By the spectral theorem, U has an orthonormal basis of eigenvectors. $\endgroup$
    – Peter-Jan
    Commented Apr 28, 2020 at 19:39

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