Measuring but not looking at the result
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Once a state is measured, but we don't look at the result, is the state now written as a density matrix, that is, the probability that it could land on a measurement operator multiplied by the operator applied on the state, this summed up for every measurement operator that it could land on contained in the measurement?
measurement density-matrix decoherence
asked 2 hours ago
Tinkidinki
2346
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up vote
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down vote
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Once a state is measured, but we don't look at the result, is the state now written as a density matrix, that is, the probability that it could land on a measurement operator multiplied by the operator applied on the state, this summed up for every measurement operator that it could land on contained in the measurement?
measurement density-matrix decoherence
asked 2 hours ago
Tinkidinki
2346
1
This answer is highly relevant.
– DaftWullie
1 hour ago
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Once a state is measured, but we don't look at the result, is the state now written as a density matrix, that is, the probability that it could land on a measurement operator multiplied by the operator applied on the state, this summed up for every measurement operator that it could land on contained in the measurement?
measurement density-matrix decoherence
asked 2 hours ago
Tinkidinki
2346
Once a state is measured, but we don't look at the result, is the state now written as a density matrix, that is, the probability that it could land on a measurement operator multiplied by the operator applied on the state, this summed up for every measurement operator that it could land on contained in the measurement?
measurement density-matrix decoherence
measurement density-matrix decoherence
asked 2 hours ago
Tinkidinki
2346
asked 2 hours ago
Tinkidinki
2346
asked 2 hours ago
Tinkidinki
2346
asked 2 hours ago
Tinkidinki
2346
asked 2 hours ago
Tinkidinki
2346
2346
1
This answer is highly relevant.
– DaftWullie
1 hour ago
add a comment |
1
This answer is highly relevant.
– DaftWullie
1 hour ago
1
1
This answer is highly relevant.
– DaftWullie
1 hour ago
This answer is highly relevant.
– DaftWullie
1 hour ago
add a comment |
2 Answers
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Suppose you have a state $rho$, and a random process that changes this to a state $rho_j$ with probability $p_j$. If you know what the value of $j$ is, your knowledge of the resulting state will be described by the corresponding $rho_j$. If you have no information regarding $j$, your knowledge will be described by
$$sum_j ~ p_j ~ rho_j$$
This is a general statement that holds for any random process. For the case you describe, which is measurement, the possible outcomes can often be described by a set of projectors ${P_j}$. For these
$$ p_j = {rm tr}~(~P_j~rho~), ~~~ rho_j = frac{P_j rho P_j}{p_j}.$$
Probabilities for more general measurements can be calculated by more general operators, but figuring out the post-measurement states for these is not always as easy.
edited 54 mins ago
Blue♦
5,63511352
answered 57 mins ago
James Wootton
5,9101942
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1
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In the Copenhagen interpretation, there are only two kinds of things that one can do, one is evolution and other is the measurement. Measuring but not looking is equivalent to measuring the system and hence projecting it to one of the possible eigenstates. (Or maybe you can clarify more what you meant by not looking?)
And after the system is probed in the measurement, it is no longer in a superposition and no longer in the statistical mixture anymore. It just becomes a decohered density matrix with a single element in the measurement (projector) basis. Density matrix representation then becomes trivial.
(I think in the latter part of your question you are pointing the completeness of probability in the measurement which summed over all the measurement projectors will be unity. But this has to do with the act of measurement, once that is performed, there is only one deterministic state.)
answered 2 hours ago
Siddhānt Singh
404114
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Suppose you have a state $rho$, and a random process that changes this to a state $rho_j$ with probability $p_j$. If you know what the value of $j$ is, your knowledge of the resulting state will be described by the corresponding $rho_j$. If you have no information regarding $j$, your knowledge will be described by
$$sum_j ~ p_j ~ rho_j$$
This is a general statement that holds for any random process. For the case you describe, which is measurement, the possible outcomes can often be described by a set of projectors ${P_j}$. For these
$$ p_j = {rm tr}~(~P_j~rho~), ~~~ rho_j = frac{P_j rho P_j}{p_j}.$$
Probabilities for more general measurements can be calculated by more general operators, but figuring out the post-measurement states for these is not always as easy.
edited 54 mins ago
Blue♦
5,63511352
answered 57 mins ago
James Wootton
5,9101942
add a comment |
up vote
2
down vote
Suppose you have a state $rho$, and a random process that changes this to a state $rho_j$ with probability $p_j$. If you know what the value of $j$ is, your knowledge of the resulting state will be described by the corresponding $rho_j$. If you have no information regarding $j$, your knowledge will be described by
$$sum_j ~ p_j ~ rho_j$$
This is a general statement that holds for any random process. For the case you describe, which is measurement, the possible outcomes can often be described by a set of projectors ${P_j}$. For these
$$ p_j = {rm tr}~(~P_j~rho~), ~~~ rho_j = frac{P_j rho P_j}{p_j}.$$
Probabilities for more general measurements can be calculated by more general operators, but figuring out the post-measurement states for these is not always as easy.
edited 54 mins ago
Blue♦
5,63511352
answered 57 mins ago
James Wootton
5,9101942
add a comment |
up vote
2
down vote
up vote
2
down vote
Suppose you have a state $rho$, and a random process that changes this to a state $rho_j$ with probability $p_j$. If you know what the value of $j$ is, your knowledge of the resulting state will be described by the corresponding $rho_j$. If you have no information regarding $j$, your knowledge will be described by
$$sum_j ~ p_j ~ rho_j$$
This is a general statement that holds for any random process. For the case you describe, which is measurement, the possible outcomes can often be described by a set of projectors ${P_j}$. For these
$$ p_j = {rm tr}~(~P_j~rho~), ~~~ rho_j = frac{P_j rho P_j}{p_j}.$$
Probabilities for more general measurements can be calculated by more general operators, but figuring out the post-measurement states for these is not always as easy.
edited 54 mins ago
Blue♦
5,63511352
answered 57 mins ago
James Wootton
5,9101942
Suppose you have a state $rho$, and a random process that changes this to a state $rho_j$ with probability $p_j$. If you know what the value of $j$ is, your knowledge of the resulting state will be described by the corresponding $rho_j$. If you have no information regarding $j$, your knowledge will be described by
$$sum_j ~ p_j ~ rho_j$$
This is a general statement that holds for any random process. For the case you describe, which is measurement, the possible outcomes can often be described by a set of projectors ${P_j}$. For these
$$ p_j = {rm tr}~(~P_j~rho~), ~~~ rho_j = frac{P_j rho P_j}{p_j}.$$
Probabilities for more general measurements can be calculated by more general operators, but figuring out the post-measurement states for these is not always as easy.
edited 54 mins ago
Blue♦
5,63511352
answered 57 mins ago
James Wootton
5,9101942
edited 54 mins ago
Blue♦
5,63511352
edited 54 mins ago
Blue♦
5,63511352
edited 54 mins ago
Blue♦
5,63511352
5,63511352
answered 57 mins ago
James Wootton
5,9101942
answered 57 mins ago
James Wootton
5,9101942
answered 57 mins ago
James Wootton
5,9101942
5,9101942
add a comment |
add a comment |
up vote
1
down vote
In the Copenhagen interpretation, there are only two kinds of things that one can do, one is evolution and other is the measurement. Measuring but not looking is equivalent to measuring the system and hence projecting it to one of the possible eigenstates. (Or maybe you can clarify more what you meant by not looking?)
And after the system is probed in the measurement, it is no longer in a superposition and no longer in the statistical mixture anymore. It just becomes a decohered density matrix with a single element in the measurement (projector) basis. Density matrix representation then becomes trivial.
(I think in the latter part of your question you are pointing the completeness of probability in the measurement which summed over all the measurement projectors will be unity. But this has to do with the act of measurement, once that is performed, there is only one deterministic state.)
answered 2 hours ago
Siddhānt Singh
404114
add a comment |
up vote
1
down vote
In the Copenhagen interpretation, there are only two kinds of things that one can do, one is evolution and other is the measurement. Measuring but not looking is equivalent to measuring the system and hence projecting it to one of the possible eigenstates. (Or maybe you can clarify more what you meant by not looking?)
And after the system is probed in the measurement, it is no longer in a superposition and no longer in the statistical mixture anymore. It just becomes a decohered density matrix with a single element in the measurement (projector) basis. Density matrix representation then becomes trivial.
(I think in the latter part of your question you are pointing the completeness of probability in the measurement which summed over all the measurement projectors will be unity. But this has to do with the act of measurement, once that is performed, there is only one deterministic state.)
answered 2 hours ago
Siddhānt Singh
404114
add a comment |
up vote
1
down vote
up vote
1
down vote
In the Copenhagen interpretation, there are only two kinds of things that one can do, one is evolution and other is the measurement. Measuring but not looking is equivalent to measuring the system and hence projecting it to one of the possible eigenstates. (Or maybe you can clarify more what you meant by not looking?)
And after the system is probed in the measurement, it is no longer in a superposition and no longer in the statistical mixture anymore. It just becomes a decohered density matrix with a single element in the measurement (projector) basis. Density matrix representation then becomes trivial.
(I think in the latter part of your question you are pointing the completeness of probability in the measurement which summed over all the measurement projectors will be unity. But this has to do with the act of measurement, once that is performed, there is only one deterministic state.)
answered 2 hours ago
Siddhānt Singh
404114
In the Copenhagen interpretation, there are only two kinds of things that one can do, one is evolution and other is the measurement. Measuring but not looking is equivalent to measuring the system and hence projecting it to one of the possible eigenstates. (Or maybe you can clarify more what you meant by not looking?)
And after the system is probed in the measurement, it is no longer in a superposition and no longer in the statistical mixture anymore. It just becomes a decohered density matrix with a single element in the measurement (projector) basis. Density matrix representation then becomes trivial.
(I think in the latter part of your question you are pointing the completeness of probability in the measurement which summed over all the measurement projectors will be unity. But this has to do with the act of measurement, once that is performed, there is only one deterministic state.)
answered 2 hours ago
Siddhānt Singh
404114
answered 2 hours ago
Siddhānt Singh
404114
answered 2 hours ago
Siddhānt Singh
404114
answered 2 hours ago
Siddhānt Singh
404114
404114
add a comment |
add a comment |
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This answer is highly relevant.
– DaftWullie
1 hour ago