Born rule for a sequence of measurements: Why this particular form? If we have some observable $O=\sum_i\lambda_iP_i$ where $P_i$ are the usual projectors you get in a spectral decomposition, then the probability for a single measurement yielding an outcome $\lambda_j$ is $$p(\lambda_j) = \mathrm{Tr}\left[P_j\rho\right]$$ If at a later time the observable $O' = \sum_i \lambda'_iP'_i$ is measured, the probability of the sequence of outcomes $\lambda_j\lambda'_k$ is often given in literature as $$p(\lambda_j,\lambda'_k) = \mathrm{Tr}\left[P'_kP_j\rho P_jP'_k\right]$$ See e.g. this paper. But I have never seen this written as $$p(\lambda_j,\lambda'_k) = \mathrm{Tr}\left[P'_kP_j\rho\right]$$ Is there a reason this second expression would not hold?
 A: Using the cyclic property of the trace, and the fact that projectors are idempotent we have:
$$\mathrm{tr}[P'_kP_j\rho P_jP'_k] = \mathrm{tr}[P_jP_k'P_j\rho].$$
Thus would like to know if it is true that
$$\mathrm{tr}[P_jP_k'P_j\rho] = \mathrm{tr}[P_k'P_j\rho].$$
There are many ways to see that this is not the case, for example, let $P=|0\rangle\langle0|$ and $P'=|+\rangle\langle+|$ be the two projectors, where as usual $$|+\rangle = \frac{1}{\sqrt{2}}\big(|0\rangle+|1\rangle\big)$$
Then clearly $$P'P= \frac{1}{\sqrt2}|+\rangle\langle0|$$ and $$PP'P= \frac{1}{2}|0\rangle\langle0|$$
and if you let $\rho=|+\rangle\langle+|$ then you see that they are different.
More abstractly, the map $$(A,B)\longmapsto\mathrm{tr}\left[A^\dagger B\right]$$ is a scalar product on the real vector space of positive trace-class linear operators on a Hilbert space. As you might or might not know, this means that
$$(\forall \rho:\mathrm{tr}[P_jP_k'P_j\rho] = \mathrm{tr}[P_k'P_j\rho]) \iff P_jP_k'P_j = P_k'P_j,$$
which clearly cannot be.
