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We know that during measurement quantum system evolves as: $$ |\psi_f\rangle = \frac{M_r |\psi_i\rangle}{\sqrt{\langle\psi_i|M_r^\dagger M_r|\psi_i\rangle}} $$ where $M_r$ is the measurement operator corresponding to the outcome $r$ and it satisfies the constraint such that $\sum_r M_r^\dagger M_r=1$. Now consider a simple case where I have only two measurement operators: $M_0$ and $M_1$. Then my questions are as follows:

  1. Are $M_0$ and $M_1$ commutative always or they can be non-commutative? I guess they can be non commutative.

  2. Suppose I divide my total evolution time into $n$ steps and I measure randomly either $M_0$ or $M_1$ at each time step. How physics is different when $M_0$ and $M_1$ are commutative and when they are non commutative?

Qmechanic
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Parveen
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  • possible duplicate https://physics.stackexchange.com/q/184524/ –  Oct 25 '18 at 13:19
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    "How physics is different when M0 and M1 are commutative and when they are non commutative?" -- There are probably a million answers for that. Do you have something concrete in mind? – Norbert Schuch Oct 25 '18 at 16:14
  • @NorbertSchuch: Well, I just wanted to know how the evolution will be different. I agree with you that there are probably million answer to it but I am interested only in the measurement evolution of the system. – Parveen Oct 26 '18 at 05:57
  • As I said, this is too broad. – Norbert Schuch Oct 26 '18 at 08:17

2 Answers2

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It is possible that $M_0$ and $M_1$ do not commute.

As an example let us consider a qubit with states $|0\rangle$ and $|1\rangle$ in the computational basis and let us define the states $|\pm\rangle:=\frac{1}{\sqrt{2}}(|0\rangle\pm|1\rangle)$.

Let us define the measurement operators $M_0:=|1\rangle\langle+|$ and $M_1:=|1\rangle\langle-|$. $$M^\dagger_0M_0+M^\dagger_1M_1=|+\rangle\langle+|+|-\rangle\langle-|=1$$ $$M_0M_1=\frac{1}{\sqrt{2}}|1\rangle\langle-|\qquad M_1M_0=-\frac{1}{\sqrt{2}}|1\rangle\langle+| $$ These operators $M_0$ and $M_1$ are indeed valid measurement operators and do not commute.

user1677907
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I think you have a slight misunderstanding.

$M$ is the measurement, $M_0$ and $M_1$ correspond the the outcomes of the measurement. For a projector in some computational basis, $M_0=\vert 0\rangle\langle 0\vert$ and $M_1=\vert 1\rangle\langle 1\vert$. It has nothing to do with commuting or not commuting - it is only one measurement that you make and the state after measurement is either $\vert 0\rangle$ or $\vert 1\rangle$ and the term you wrote in the question takes care of the normalization.

As posted in the comments, if $M_0$ and $M_1$ are orthogonal, then you can say something about making $n$ successive measurements namely, that the state stays in whatever outcome you got after the first measurement. Beyond that, I can't think of anything specific that applies to the state after applying a sequence of $M$ with non orthogonal $M_i$.

user1936752
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  • What you have written is known as Projective measurement. There is another type of measurement known as "Generalized measurement" and this is what my question deals with. – Parveen Oct 25 '18 at 12:51
  • The answer is valid for general measurements too, the projector operators just help you see it more clearly. The different outcomes of a single POVM are not the same different measurements. The question of commuting or not would only apply if you had a second measurement - not on the different outcomes of one measurement. For example, one cannot choose if the outcome is $M_0$ or $M_1$ - only one of these outcomes happen and they happen with a probability given by $Tr(\rho M_i^{\dagger}M_i)$ – user1936752 Oct 25 '18 at 13:30
  • Well ok. I have mentioned in point 2 that I make measurement at each time step, i.e. I am having sequence of measurements. It means either $M_0$ or $M_1$ acts on the system at each time step and in the end I will be having a final state resulting from $n$ measurements. – Parveen Oct 25 '18 at 15:30
  • Yes, and here I cannot think of result for the general case. If $M_0$ is orthogonal to $M_1$, the state stays in whatever outcome you got after the first measurement but I'm not sure if there is something more general that can be said for arbitrary $M_0$ and $M_1$. – user1936752 Oct 25 '18 at 16:47
  • It is not true that the state stays the same for commuting M0 and M1. Take, e.g., M0=diag(1/3 2/3) and M1=diag(2/3 1/3). – Norbert Schuch Oct 26 '18 at 08:18
  • I made the claim for orthogonal $M_i$, not commuting $M_i$. Commuting has no physical meaning here - you are talking about different outcomes of the same measurement – user1936752 Oct 26 '18 at 10:45
  • Guess I implicitly assumed you meant commuting since this is what the question asked. It is BTW also wrong for orthogonal (at least Hilbert-Schmidt-orthogonal), take e.g. M0=|0><1| and M1=|1><0|. – Norbert Schuch Oct 30 '18 at 11:24
  • The elements of a generalized measurement are positive operators i.e. $\langle M_i x, x\rangle \geq 0$ for all $x$. I didn't explicitly state this because the context of the question makes it clear. Your counterexample consists of non-positive operators. Try $x = \frac{1}{\sqrt{2}}(\vert 0\rangle - \vert 1\rangle)$ and it fails the positivity condition. The statement remains true for all such orthogonal elements. – user1936752 Oct 30 '18 at 15:49
  • True, but the $M_i$ in the question (which makes the context clear) is not the positive operator -- the positive operators are the $M_i^\dagger M_i$. Also, would you mind using @[username] to notify me of your comments? – Norbert Schuch Oct 31 '18 at 02:07