Measuring the product is the same as first measuring with one matrix then with the other

Question

I want to show that if one measures $M_1$ on $|\psi\rangle$ then measures $M_2$ on the resulting state and then the associated probability space will be the same as the one for measuring the product $M_2 M_1$ on that pure state $|\psi\rangle$ (also assuming that the $M_i$ are Hermitian and commute). So in our first measurement procedure, we have two measurement operators, the first one yields,

$$p(1)=\langle \psi|M_1^\dagger M_1|\psi\rangle\tag{1}$$

and the resulting state is,

$$\frac{M_1|\psi\rangle}{\sqrt{\langle \psi|M_1^\dagger M_1|\psi\rangle}}.\tag{2}$$

If we then measure it using $M_2$ then we get,

$$p(2)=\left\langle\frac{M_1|\psi\rangle}{\sqrt{\langle \psi|M_1^\dagger M_1|\psi\rangle}} \bigg| M_2^\dagger M_2 \bigg| \frac{M_1|\psi\rangle}{\sqrt{\langle \psi|M_1^\dagger M_1|\psi\rangle}}\right\rangle.\tag{3}$$

With $M_2M_1$ we get,

$$p(1)=\langle\psi|(M_2M_1)^\dagger M_2M_1|\psi\rangle.$$

However, we only have one measurement operator here being $M_2M_1$, so how can the probability spaces be the same?

Answer 1

For the case of "multiplying the outcomes" the probability spaces are not equivalent. However, looking simply at cascading measurements versus single measurements, the probability spaces are the same because the resulting states are the same:

The state of the system $| \psi_1 \rangle$ after the measurement $M_1$ is,

$$ | \psi_1 \rangle = \frac{M_1 | \psi \rangle}{\sqrt{\langle \psi | M_1^\dagger M_1 | \psi \rangle}} .$$

Then, the state of the system $| \psi_2 \rangle$ after the measurement $M_2$ is,

$$ | \psi_2 \rangle = \frac{M_2 | \psi_1 \rangle}{\sqrt{\langle \psi_1 | M_2^\dagger M_2 | \psi_1 \rangle}} $$

$$ \hspace{55mm}= \frac{M_2 M_1 | \psi \rangle}{\sqrt{\langle \psi | M_1^\dagger M_1 | \psi \rangle}} \cdot \frac{\sqrt{\langle \psi | M_1^\dagger M_1 | \psi \rangle}}{\langle \psi | M_1^\dagger M_2^\dagger M_2 M_1 | \psi \rangle} $$

$$ \hspace{22mm} = \frac{M_2 M_1 | \psi \rangle}{\sqrt{\langle \psi | M_1^\dagger M_2^\dagger M_2 M_1 | \psi \rangle}} .$$

On the other hand, the state of the system $| \psi_3 \rangle $ after the measurement $ M_2 M_1 $ is,

$$ | \psi_3 \rangle = \frac{M_2 M_1 | \psi \rangle}{\sqrt{\langle \psi | M_1^\dagger M_2^\dagger M_2 M_1 | \psi \rangle}} = | \psi_2 \rangle .$$

Answer 2

This isn't true. Here's a counterexample.

Note the density matrix after the product measurement is not the same as the density matrix after the two individual measurements.