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fixed typos along Noiralef's suggestions
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Daniel
Daniel
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Assuming the system and probe are initially uncorrelated, the initial density matrix is $$ \rho(0) = \rho_S(0) \otimes \rho_P(0). $$

After interaction for a time $t$, the system and probe are entangled $$ \rho(t) = e^{-i H_{\text{full}}t}\rho(0)e^{iH_{\text{full}}t} \equiv \rho_{SP}.$$

Then the observable $\hat Q$ is measured, which can be written $$ \hat Q = \sum_q q\hat\Pi_q,\qquad\text{with}\quad\hat\Pi_q=\sum_j|{q,j}\rangle\langle{q,j}|,$$ i.e., $\hat Q$ has a discrete spectrum (continuous involves replacing the sum with an integral), where the eigenvalue $q$ has an eigenspace spanned by $|q,j\rangle$ (it is degenerate if there is more than on $j$, which is the typical situation).

In a projective measurement of $\hat Q$, you obtain the eigenvalue $q$ with a probability $\text{Tr}[\rho(t)\hat Q]$$P(q)=\text{Tr}[\rho(t)\hat Q]$. The state of your system is then $$\rho_q=\frac{\hat\Pi_q\rho(t)\hat\Pi_q}{\text{Tr}[\rho(t)\hat\Pi_q]}. $$$$\rho_q=\frac{\hat\Pi_q\rho(t)\hat\Pi_q}{P(q)}. $$

If you're not sure what you have measured $q_r$, but your measurement apparatus does not transfer the information correctly (what Norbert Schuch called "classical scrambling" in his comment), then your state is $$ \rho_{q_{\text{readout}}} = \int dq \, P(q_{\text{readout}}|q)\,\rho_q. $$$$ \rho_{q_r} = \int dq \, P(q|q_r)\,\rho_q. $$ (or a sum if your probability distribution is discrete.) In this caseAs Noiralef commented, if$P(q|q_r)$ has to be calculated from $P(q_r|q)$ given above using Bayes' theorem. Your initial guess could be a uniform distribution for $P(q)$, though this does not necessarily imply a uniform distribution for $P(q_r)$.

If you ignore your measurement record entirely, the the state of your system is $$\rho_{\text{ignore}} = \int dq\,\rho_q.$$$$\rho_{\text{ignore}} = \int dq\, P(q)\rho_q = \int dq\,\hat\Pi_q\rho(t)\hat\Pi_q.$$ This is the unconditional state. In this case (and in the former) all coherences between different $q$ subsectors are gone (if you write the density matrix in a basis where different blocks correspond to different $q$, then all off-diagonal blocks are zero).

Finally, you could consider the case in which the measurement is weak. This is the case in which we really should know more about the measurement. Assuming everything is continuous, you can write down a family of Kraus operators $$\hat{\Upsilon}_{q_r}=P(q_{\text{r}}|\hat Q), $$ where you just replaced the number $q$ with the operator $\hat Q$ in your expression above. In this case the state of your system post-measurement is $$ \rho_{q_r} = \frac{\hat\Upsilon_{q_r}\rho(t)\hat\Upsilon_{q_r}}{\text{Tr}[\rho(t)\hat\Upsilon_{q_r}^\dagger\Upsilon_{q_r}]}. $$$$ \rho_{q_r} = \frac{\hat\Upsilon_{q_r}\rho(t)\hat\Upsilon_{q_r}^\dagger}{\text{Tr}[\rho(t)\hat\Upsilon_{q_r}^\dagger\Upsilon_{q_r}]}. $$ When you now look at the unconditional state by integrating over $q_r$, you'll find it's not block diagonal!

Edit: Thanks to Noiralef for pointing out typos and Bayes' theorem.

Assuming the system and probe are initially uncorrelated, the initial density matrix is $$ \rho(0) = \rho_S(0) \otimes \rho_P(0). $$

After interaction for a time $t$, the system and probe are entangled $$ \rho(t) = e^{-i H_{\text{full}}t}\rho(0)e^{iH_{\text{full}}t} \equiv \rho_{SP}.$$

Then the observable $\hat Q$ is measured, which can be written $$ \hat Q = \sum_q q\hat\Pi_q,\qquad\text{with}\quad\hat\Pi_q=\sum_j|{q,j}\rangle\langle{q,j}|,$$ i.e., $\hat Q$ has a discrete spectrum (continuous involves replacing the sum with an integral), where the eigenvalue $q$ has an eigenspace spanned by $|q,j\rangle$ (it is degenerate if there is more than on $j$, which is the typical situation).

In a projective measurement of $\hat Q$, you obtain the eigenvalue $q$ with a probability $\text{Tr}[\rho(t)\hat Q]$. The state of your system is then $$\rho_q=\frac{\hat\Pi_q\rho(t)\hat\Pi_q}{\text{Tr}[\rho(t)\hat\Pi_q]}. $$

If you're not sure what you have measured (what Norbert Schuch called "classical scrambling" in his comment), then your state is $$ \rho_{q_{\text{readout}}} = \int dq \, P(q_{\text{readout}}|q)\,\rho_q. $$ (or a sum if your probability distribution is discrete.) In this case, if you ignore your measurement record, the state of your system is $$\rho_{\text{ignore}} = \int dq\,\rho_q.$$ This is the unconditional state. In this case (and in the former) all coherences between different $q$ subsectors are gone (if you write the density matrix in a basis where different blocks correspond to different $q$, then all off-diagonal blocks are zero).

Finally, you could consider the case in which the measurement is weak. This is the case in which we really should know more about the measurement. Assuming everything is continuous, you can write down a family of Kraus operators $$\hat{\Upsilon}_{q_r}=P(q_{\text{r}}|\hat Q), $$ where you just replaced the number $q$ with the operator $\hat Q$ in your expression above. In this case the state of your system post-measurement is $$ \rho_{q_r} = \frac{\hat\Upsilon_{q_r}\rho(t)\hat\Upsilon_{q_r}}{\text{Tr}[\rho(t)\hat\Upsilon_{q_r}^\dagger\Upsilon_{q_r}]}. $$ When you now look at the unconditional state by integrating over $q_r$, you'll find it's not block diagonal!

Assuming the system and probe are initially uncorrelated, the initial density matrix is $$ \rho(0) = \rho_S(0) \otimes \rho_P(0). $$

After interaction for a time $t$, the system and probe are entangled $$ \rho(t) = e^{-i H_{\text{full}}t}\rho(0)e^{iH_{\text{full}}t} \equiv \rho_{SP}.$$

Then the observable $\hat Q$ is measured, which can be written $$ \hat Q = \sum_q q\hat\Pi_q,\qquad\text{with}\quad\hat\Pi_q=\sum_j|{q,j}\rangle\langle{q,j}|,$$ i.e., $\hat Q$ has a discrete spectrum (continuous involves replacing the sum with an integral), where the eigenvalue $q$ has an eigenspace spanned by $|q,j\rangle$ (it is degenerate if there is more than on $j$, which is the typical situation).

In a projective measurement of $\hat Q$, you obtain the eigenvalue $q$ with a probability $P(q)=\text{Tr}[\rho(t)\hat Q]$. The state of your system is then $$\rho_q=\frac{\hat\Pi_q\rho(t)\hat\Pi_q}{P(q)}. $$

If you have measured $q_r$, but your measurement apparatus does not transfer the information correctly (what Norbert Schuch called "classical scrambling" in his comment), then your state is $$ \rho_{q_r} = \int dq \, P(q|q_r)\,\rho_q. $$ (or a sum if your probability distribution is discrete.) As Noiralef commented, $P(q|q_r)$ has to be calculated from $P(q_r|q)$ given above using Bayes' theorem. Your initial guess could be a uniform distribution for $P(q)$, though this does not necessarily imply a uniform distribution for $P(q_r)$.

If you ignore your measurement record entirely, the state of your system is $$\rho_{\text{ignore}} = \int dq\, P(q)\rho_q = \int dq\,\hat\Pi_q\rho(t)\hat\Pi_q.$$ This is the unconditional state. In this case (and in the former) all coherences between different $q$ subsectors are gone (if you write the density matrix in a basis where different blocks correspond to different $q$, then all off-diagonal blocks are zero).

Finally, you could consider the case in which the measurement is weak. This is the case in which we really should know more about the measurement. Assuming everything is continuous, you can write down a family of Kraus operators $$\hat{\Upsilon}_{q_r}=P(q_{\text{r}}|\hat Q), $$ where you just replaced the number $q$ with the operator $\hat Q$ in your expression above. In this case the state of your system post-measurement is $$ \rho_{q_r} = \frac{\hat\Upsilon_{q_r}\rho(t)\hat\Upsilon_{q_r}^\dagger}{\text{Tr}[\rho(t)\hat\Upsilon_{q_r}^\dagger\Upsilon_{q_r}]}. $$ When you now look at the unconditional state by integrating over $q_r$, you'll find it's not block diagonal!

Edit: Thanks to Noiralef for pointing out typos and Bayes' theorem.

Source Link
Daniel
Daniel
  • 725
  • 5
  • 18

Assuming the system and probe are initially uncorrelated, the initial density matrix is $$ \rho(0) = \rho_S(0) \otimes \rho_P(0). $$

After interaction for a time $t$, the system and probe are entangled $$ \rho(t) = e^{-i H_{\text{full}}t}\rho(0)e^{iH_{\text{full}}t} \equiv \rho_{SP}.$$

Then the observable $\hat Q$ is measured, which can be written $$ \hat Q = \sum_q q\hat\Pi_q,\qquad\text{with}\quad\hat\Pi_q=\sum_j|{q,j}\rangle\langle{q,j}|,$$ i.e., $\hat Q$ has a discrete spectrum (continuous involves replacing the sum with an integral), where the eigenvalue $q$ has an eigenspace spanned by $|q,j\rangle$ (it is degenerate if there is more than on $j$, which is the typical situation).

In a projective measurement of $\hat Q$, you obtain the eigenvalue $q$ with a probability $\text{Tr}[\rho(t)\hat Q]$. The state of your system is then $$\rho_q=\frac{\hat\Pi_q\rho(t)\hat\Pi_q}{\text{Tr}[\rho(t)\hat\Pi_q]}. $$

If you're not sure what you have measured (what Norbert Schuch called "classical scrambling" in his comment), then your state is $$ \rho_{q_{\text{readout}}} = \int dq \, P(q_{\text{readout}}|q)\,\rho_q. $$ (or a sum if your probability distribution is discrete.) In this case, if you ignore your measurement record, the state of your system is $$\rho_{\text{ignore}} = \int dq\,\rho_q.$$ This is the unconditional state. In this case (and in the former) all coherences between different $q$ subsectors are gone (if you write the density matrix in a basis where different blocks correspond to different $q$, then all off-diagonal blocks are zero).

Finally, you could consider the case in which the measurement is weak. This is the case in which we really should know more about the measurement. Assuming everything is continuous, you can write down a family of Kraus operators $$\hat{\Upsilon}_{q_r}=P(q_{\text{r}}|\hat Q), $$ where you just replaced the number $q$ with the operator $\hat Q$ in your expression above. In this case the state of your system post-measurement is $$ \rho_{q_r} = \frac{\hat\Upsilon_{q_r}\rho(t)\hat\Upsilon_{q_r}}{\text{Tr}[\rho(t)\hat\Upsilon_{q_r}^\dagger\Upsilon_{q_r}]}. $$ When you now look at the unconditional state by integrating over $q_r$, you'll find it's not block diagonal!