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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?

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    $\begingroup$ Do you have a copy of Nielsen & Chuang? This, if I understand your question correctly, is exactly the content of exercise 2.57. Here (PDF alert) you can find the solutions (see page 22) ~ this is a rendered version of this github repo. $\endgroup$
    – JSdJ
    Mar 31 at 15:58
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    $\begingroup$ Are your $M_{1}$ and $M_{2}$ measurement observables or measurement operators? $\endgroup$
    – JSdJ
    Mar 31 at 16:02
  • $\begingroup$ @JSdJ They're measurement operators. $\endgroup$
    – user15388
    Apr 1 at 9:50
  • $\begingroup$ Your question is still a bit ambiguous to me; are $M_{1}$ and $M_{2}$ operators from the same of from different measurements? Remember - a quantum measurement in its most general form is a set of operators $\{M_{m}\}$, where the outcome $m$ is associated to operator $M_{m}$ and occurs with probability $p(m) = \langle \psi | M_{m}^{\dagger}M_{m} | \psi \rangle$. If they are operators from different measurements (a cascaded measurement), @ryanhill1 has shown that your identity holds. If they are operators from the same measurement, your identity does not hold (in general). $\endgroup$
    – JSdJ
    Apr 1 at 10:20
  • $\begingroup$ $M_1$ and $M_2$ can be observables only, because it is stated in the question "$M_i$ are Hermitian and commute". It does not make sense for measurement operators. And then OP writes formulas for measurement operators, evidently not understating what he is doing. The question as it is does not make sense to me. $\endgroup$
    – kludg
    Apr 1 at 11:04
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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,

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

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  • $\begingroup$ Why do you think that the state after measuring $M_1$ on the state $\psi$ is $| \psi_1 \rangle = \frac{M_1 | \psi \rangle}{\sqrt{\langle \psi | M_1^\dagger M_1 | \psi \rangle}} $ ? $\endgroup$
    – kludg
    Mar 31 at 15:49
  • $\begingroup$ It's the definition of the measurement operator, Nielson and Chuang Equation 2.93 $\endgroup$
    – rjh324
    Mar 31 at 15:55
  • $\begingroup$ I think there is some misunderstanding with this formalism. If $M_i$ is a "measurement operator", then you don't "measure $M_i$", but rather you measure some underlying POVM, find an outcome $i$ with probability $\langle M_i\rangle_\psi$, and have a corresponding residual post-measurement state $\propto M_i |\psi\rangle$. So you "find" an outcome corresponding to $M_i$, more than "perform the measurement $M_i$". Thus, $\sim M_2 M_1|\psi\rangle$ is the state you get in the cases in which you found the outcomes $1$ and then $2$ (given some underlying, here unspecified, POVM) $\endgroup$
    – glS
    Apr 6 at 14:37
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This isn't true. Here's a counterexample.

enter image description here

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

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  • $\begingroup$ I don't understand what question you answered. $\endgroup$
    – kludg
    Mar 31 at 15:52
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    $\begingroup$ The questions seems to be regarding measurement operators, not regarding measurement observables ~ for the operators, the cascaded measurement is actually equal to the (cartesian) product of the original operators. $\endgroup$
    – JSdJ
    Mar 31 at 16:04
  • $\begingroup$ The first words of the question clearly ask about measuring observables; and then OP uses formulas for measurement operators. I don't think that the question in its current form makes sense. $\endgroup$
    – kludg
    Mar 31 at 16:14
  • $\begingroup$ @kludg The wording of the question is kind of weird, I also don't know if it's a measurement operator or observable (but I guess it's the former) since we're asked to "Measure $AB$ on $|\psi\rangle$" $\endgroup$
    – user15388
    Mar 31 at 16:17
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    $\begingroup$ Looks like OP doesn't understand the difference between observables and measurement operators. $\endgroup$
    – kludg
    Mar 31 at 16:22

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