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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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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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