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Suppose we get the Bell state $$ |\Phi ^{+}\rangle ={\frac {1}{{\sqrt {2}}}}(|0\rangle _{A}\otimes |0\rangle _{B}+|1\rangle _{A}\otimes |1\rangle _{B}). $$ If we now apply a unitary operator $U$ only to the first qubit then this operator is given as follows: $$ U \otimes I $$ but if we consider a POVM with the elements ($E_1, E_2$), and since we get the same result when we measure the second qubit, then we should have $$ E_1 \otimes I = I \otimes E_1 $$ But this is obviously not the case. Where is my reasoning error here?

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    $\begingroup$ It might be easier to understand where your misunderstanding comes from if you try to work through an example. Note that $\langle \psi| (X\otimes I) |\psi \rangle = \langle \psi| (I\otimes X) |\psi \rangle$ does not imply $(X\otimes I) =(I\otimes X) $. $\endgroup$
    – Rammus
    Dec 5, 2021 at 0:00
  • $\begingroup$ @Rammus I was able to do it with an example as long as $E_1 = |v_1\rangle \langle v_1|$ and $v_i$ is an orthonormal basis. But this is not general enough. $\endgroup$
    – Johny Dow
    Dec 5, 2021 at 18:45
  • $\begingroup$ @JohnyDow The question makes an impression that you expect that the operator equality $A=B$ follows from the equality of a single matrix element $\langle v|A|v\rangle=\langle v|B|v\rangle$ for some vector $v$. However, this is obviously false and your question includes a counterexample (set $v=|\Phi^+\rangle$, $A=E_1\otimes I$ and $B=I\otimes E_1$), so it is not clear what it is that you're asking :-) $\endgroup$ Dec 5, 2021 at 20:38
  • $\begingroup$ @AdamZalcman For a maximally entangled state measuring the first or the second qubit should give us the same result, correct? So is it correct to say that $\langle \psi|(I \otimes X)|\psi\rangle = \langle \psi|(X \otimes I)|\psi\rangle$? $\endgroup$
    – Johny Dow
    Dec 5, 2021 at 23:53
  • $\begingroup$ Re question 1: No. This depends on which maximally entangled state you use. For example, measuring the first or the second qubit of $|\Psi^+\rangle=(|01\rangle+|10\rangle)/\sqrt2$ in the computational basis will give opposite results. Re question 2: No. For example if $|\psi\rangle=|0+\rangle$ then $\langle\psi|I\otimes X|\psi\rangle = 1$ but $\langle\psi|X\otimes I|\psi\rangle = 0$. (I'm voting to close because the post remains unclear. I encourage you to edit to clarify what the question is.) $\endgroup$ Dec 6, 2021 at 0:29

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Your assumption that performing the same POVM on first and second qubit always gives the same probability outcome (even sticking to maximally entangled states).

You might be assuming that this is the case because you are thinking of a maximally entangled state written as $|00\rangle+|11\rangle$. Such a state can equivalently be written, for any pair of orthonormal vectors $|u\rangle,|v\rangle$, as $$|0,0\rangle+|1,1\rangle = |u,\bar u\rangle+|v,\bar v\rangle.$$ For such a state, your assumption does indeed clearly hold. However, this is not the most general form of a maximally entangled state, which in general has the form $$|u_1,v_1\rangle + |u_2,v_2\rangle$$ for some pair of orthonormal bases $\{|u_1\rangle,|u_2\rangle\}$ and $\{|v_1\rangle,|v_2\rangle\}$. This is the type of state you also get after applying a local operation to your initial maximally entangled state. For these, the same POVM might lead to different outcome probabilities on the two systems.

It is, however, true that given some measurement $E_1$ on the first system, you can find an "equivalent" measurement $E_1'$ which applied on the second system gives same outcome probabilities. This will simply be the operator which acts in the basis $\{|v_1\rangle,|v_2\rangle\}$ in the same way $E_1$ acts in the basis $\{|u_1\rangle,|u_2\rangle\}$.

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    $\begingroup$ |0,0> + |1,1> isn't equal to |i,i> + |-i,-i>. That equals |0,0> - |1,1>. The state that's invariant to basis is the singlet state |0,1> - |1,0>. $\endgroup$ Dec 6, 2021 at 2:08
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    $\begingroup$ @CraigGidney you are right, thanks. I should have written $|u,\bar u\rangle+|v,\bar v\rangle$. The invariance I was thinking of is $(U\otimes\bar U)\sum_i |i,i\rangle=\sum_i |i,i\rangle$ for any unitary $U$ (which is equivalent to the statement $UIU^\dagger=I$ with $I$ identity). Using your example, this amounts to |00>+|11>=|i,-i>+|-i,i>. In this notation, the invariance of the singlet you refer to corresponds to the identity $UJU^T=J$ with $J$ standard symplectic matrix, which is true because every unit-determinant 2x2 matrix is symplectic $\endgroup$
    – glS
    Dec 6, 2021 at 9:00
  • $\begingroup$ @glS notably, the related measurement can always be expressed as $I\otimes E_1^T$ when the state is maximally entangled $\endgroup$ Dec 6, 2021 at 15:05
  • $\begingroup$ @gIS Technically, $UJU^T=(\det U)J$, but we can of course ignore the global phase (or assume that $U\in SU(2)$). $\endgroup$ Dec 7, 2021 at 21:38

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