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Suppose we have a single qubit operator $U$ with eigenvalues $±1$, so that $U$ is both Hermitian and unitary, so it can be regarded both as an observable and a quantum gate. Suppose we wish to measure the observable $U$. That is, we desire to obtain a measurement result indicating one of the two eigenvalues, and leaving a post-measurement state which is the corresponding eigenvector. How can this be implemented by a quantum circuit? Show that the following circuit implements a measurement of $U$:

qc

Ref. to Exercise 4.34 in QC and QI by Nelsen and Chuang

$$ |\psi_0\rangle =|0\rangle\otimes|\psi_{in}\rangle\\ |\psi_1\rangle =\frac{|0\rangle+|1\rangle}{\sqrt{2}}\otimes|\psi_{in}\rangle\\ |\psi_2\rangle=\frac{1}{\sqrt{2}}[|0\rangle\otimes |\psi_{in}\rangle+|1\rangle\otimes U|\psi_{in}\rangle]\\ |\psi_3\rangle=\frac{1}{2}[|0\rangle\otimes(I+U)|\psi_{in}\rangle+|1\rangle\otimes(I-U)|\psi_{in}\rangle] $$ Note that,

$U(I+U)|\psi_{in}\rangle=(U+U^2)|\psi_{in}\rangle=(U+I)|\psi_{in}\rangle=1(I+U)|\psi_{in}\rangle$ and $U(I-U)|\psi_{in}\rangle=-1(I-U)|\psi_{in}\rangle$

$\implies (I+U)|\psi_{in}\rangle$ and $(I-U)|\psi_{in}\rangle$ are eigenvectors of the operator $U$ with corresponding eigenvalues $+1$ and $-1$, respectively.

And by projecting the first qubit the second qubit is projected to either $(I+U)|\psi_{in}\rangle$ or $(I-U)|\psi_{in}\rangle$, therefore this circuit implements a measurement of $U$.


It is proved that projective measurements together with unitary dynamics are sufficient to implement a general measurement, by introducing an ancilla system having an orthonormal basis $|m\rangle$ in one-to-one correspondence with the possible outcomes of the measurement we wish to implement.

Please check Page 94, QC and QI by Nelsen and Chuang for the proof of this statement.

Are we using this statement to implement the circuit above ?


Can't we have a circuit like : qc2

with $U'=\sum |m \rangle\langle\psi_m |=|0\rangle\langle \psi|(I+U)+|1\rangle\langle \psi|(I-U)$ where $|\psi_i\rangle$ are the eigenvectors of the operator $U$ ?

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    $\begingroup$ If I understand correctly, the problem with using the circuit $U|\psi\rangle$ is that upon measuring you destroy the state in the sense that you get the outcome $m$ but you don't obtain a state left in the eigenvector of $U$. The first circuit you mentioned does achieve this. $\endgroup$
    – Condo
    Oct 7 '21 at 17:22
  • $\begingroup$ @Condo That makes sense. How about its connection with the statement in the middle section of my post where I mentioned about introducing an ancilla system to equate a general measurement with a unitary dynamics together with a projective measurement ? $\endgroup$
    – Sooraj S
    Oct 7 '21 at 20:19
  • $\begingroup$ I don't think you need the statements about general measurements $\equiv$ unitary dynamics + projective measurements is used/needed to explain the first circuit you discuss. Where do you think there would be any sort of general measurement? $\endgroup$
    – Condo
    Oct 8 '21 at 14:18
  • $\begingroup$ @Condo Thanks I think your first comment gives the reasoning, though it could be posted as an answer. $\endgroup$
    – Sooraj S
    Oct 14 '21 at 16:30
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I don't think you need the statements about general measurements ≡ unitary dynamics + projective measurements as it's not used/needed to explain the first circuit you discuss.

If I understand correctly, the problem with using the circuit $U|ψ⟩$ is that upon measuring you destroy the state in the sense that you get the outcome $m$ but you don't obtain a state left in the eigenvector of $U$. The first circuit you mentioned does achieve this.

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