What does an observable in a different basis mean physically?
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So when you have an observable which is the measurement operator acting on the state you get a different result than in a different basis.
What does it mean in physical terms? Does that mean writing your space operation in different style could collapse your state to something different?
Again, we can understand it mathematically but what's in physical terms?
quantum-gate quantum-state quantum-operation
edited 12 hours ago
Blue♦
6,35141455
asked 13 hours ago
bilanushbilanush
1816
$endgroup$
add a comment |
$begingroup$
So when you have an observable which is the measurement operator acting on the state you get a different result than in a different basis.
What does it mean in physical terms? Does that mean writing your space operation in different style could collapse your state to something different?
Again, we can understand it mathematically but what's in physical terms?
quantum-gate quantum-state quantum-operation
edited 12 hours ago
Blue♦
6,35141455
asked 13 hours ago
bilanushbilanush
1816
$endgroup$
2
$begingroup$
Related: What does “measurement in a certain basis” mean?
$endgroup$
– Blue♦
12 hours ago
1
$begingroup$
also related to The meaning of measurements in different bases
$endgroup$
– glS
11 hours ago
add a comment |
$begingroup$
So when you have an observable which is the measurement operator acting on the state you get a different result than in a different basis.
What does it mean in physical terms? Does that mean writing your space operation in different style could collapse your state to something different?
Again, we can understand it mathematically but what's in physical terms?
quantum-gate quantum-state quantum-operation
edited 12 hours ago
Blue♦
6,35141455
asked 13 hours ago
bilanushbilanush
1816
$endgroup$
So when you have an observable which is the measurement operator acting on the state you get a different result than in a different basis.
What does it mean in physical terms? Does that mean writing your space operation in different style could collapse your state to something different?
Again, we can understand it mathematically but what's in physical terms?
quantum-gate quantum-state quantum-operation
quantum-gate quantum-state quantum-operation
edited 12 hours ago
Blue♦
6,35141455
asked 13 hours ago
bilanushbilanush
1816
edited 12 hours ago
Blue♦
6,35141455
asked 13 hours ago
bilanushbilanush
1816
edited 12 hours ago
Blue♦
6,35141455
edited 12 hours ago
Blue♦
6,35141455
edited 12 hours ago
Blue♦
6,35141455
6,35141455
asked 13 hours ago
bilanushbilanush
1816
asked 13 hours ago
bilanushbilanush
1816
asked 13 hours ago
bilanushbilanush
1816
1816
2
$begingroup$
Related: What does “measurement in a certain basis” mean?
$endgroup$
– Blue♦
12 hours ago
1
$begingroup$
also related to The meaning of measurements in different bases
$endgroup$
– glS
11 hours ago
add a comment |
2
$begingroup$
Related: What does “measurement in a certain basis” mean?
$endgroup$
– Blue♦
12 hours ago
1
$begingroup$
also related to The meaning of measurements in different bases
$endgroup$
– glS
11 hours ago
2
2
$begingroup$
Related: What does “measurement in a certain basis” mean?
$endgroup$
– Blue♦
12 hours ago
$begingroup$
Related: What does “measurement in a certain basis” mean?
$endgroup$
– Blue♦
12 hours ago
1
1
$begingroup$
also related to The meaning of measurements in different bases
$endgroup$
– glS
11 hours ago
$begingroup$
also related to The meaning of measurements in different bases
$endgroup$
– glS
11 hours ago
add a comment |
1 Answer
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One way to think about it is that "measuring in a given basis" is how we describe mathematically the act of interacting with the system in different ways.
Taking as an example a qubit, "measuring in the ${lvert0rangle,lvert1rangle}$ basis" is mathematese for "interacting with the system in a way that reveals to the experimenter whether the system is in the physical state which we are thinking of as corresponding to $lvert0rangle$ or $lvert1rangle$". In general, the system will not be in a state that can be directly associated with one of these basis states, in which case the act of measurement "forces" the system to collapse to one of the two states, and this happens with a probability that depends on the actual state of the system, and induces a physical change in the state of the system.
Measuring in a different basis then simply corresponds to another type of physical interaction with the system.
Another, slightly more informal way to think about it is to say that each measurement basis corresponds to a different "question" being asked to the system. "Measuring in ${lvert0rangle,lvert1rangle}$" means to be asking the system whether it's in the $lvert0rangle$ or $lvert1rangle$ state, while "measuring in ${lvert+rangle,lvert-rangle}$" means to be asking the system whether it is in the $lvert+rangle$ or $lvert-rangle$ state. As it happens, in quantum mechanics the act of "asking a question" to the system - that is, of collecting some classical information out of the system - will in general result in a perturbation of the system itself, which in quantum mechanics is captured by means of the noncommutativity of different observables.
answered 11 hours ago
glSglS
4,125639
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1 Answer
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$begingroup$
One way to think about it is that "measuring in a given basis" is how we describe mathematically the act of interacting with the system in different ways.
Taking as an example a qubit, "measuring in the ${lvert0rangle,lvert1rangle}$ basis" is mathematese for "interacting with the system in a way that reveals to the experimenter whether the system is in the physical state which we are thinking of as corresponding to $lvert0rangle$ or $lvert1rangle$". In general, the system will not be in a state that can be directly associated with one of these basis states, in which case the act of measurement "forces" the system to collapse to one of the two states, and this happens with a probability that depends on the actual state of the system, and induces a physical change in the state of the system.
Measuring in a different basis then simply corresponds to another type of physical interaction with the system.
Another, slightly more informal way to think about it is to say that each measurement basis corresponds to a different "question" being asked to the system. "Measuring in ${lvert0rangle,lvert1rangle}$" means to be asking the system whether it's in the $lvert0rangle$ or $lvert1rangle$ state, while "measuring in ${lvert+rangle,lvert-rangle}$" means to be asking the system whether it is in the $lvert+rangle$ or $lvert-rangle$ state. As it happens, in quantum mechanics the act of "asking a question" to the system - that is, of collecting some classical information out of the system - will in general result in a perturbation of the system itself, which in quantum mechanics is captured by means of the noncommutativity of different observables.
answered 11 hours ago
glSglS
4,125639
$endgroup$
add a comment |
$begingroup$
One way to think about it is that "measuring in a given basis" is how we describe mathematically the act of interacting with the system in different ways.
Taking as an example a qubit, "measuring in the ${lvert0rangle,lvert1rangle}$ basis" is mathematese for "interacting with the system in a way that reveals to the experimenter whether the system is in the physical state which we are thinking of as corresponding to $lvert0rangle$ or $lvert1rangle$". In general, the system will not be in a state that can be directly associated with one of these basis states, in which case the act of measurement "forces" the system to collapse to one of the two states, and this happens with a probability that depends on the actual state of the system, and induces a physical change in the state of the system.
Measuring in a different basis then simply corresponds to another type of physical interaction with the system.
Another, slightly more informal way to think about it is to say that each measurement basis corresponds to a different "question" being asked to the system. "Measuring in ${lvert0rangle,lvert1rangle}$" means to be asking the system whether it's in the $lvert0rangle$ or $lvert1rangle$ state, while "measuring in ${lvert+rangle,lvert-rangle}$" means to be asking the system whether it is in the $lvert+rangle$ or $lvert-rangle$ state. As it happens, in quantum mechanics the act of "asking a question" to the system - that is, of collecting some classical information out of the system - will in general result in a perturbation of the system itself, which in quantum mechanics is captured by means of the noncommutativity of different observables.
answered 11 hours ago
glSglS
4,125639
$endgroup$
add a comment |
$begingroup$
One way to think about it is that "measuring in a given basis" is how we describe mathematically the act of interacting with the system in different ways.
Taking as an example a qubit, "measuring in the ${lvert0rangle,lvert1rangle}$ basis" is mathematese for "interacting with the system in a way that reveals to the experimenter whether the system is in the physical state which we are thinking of as corresponding to $lvert0rangle$ or $lvert1rangle$". In general, the system will not be in a state that can be directly associated with one of these basis states, in which case the act of measurement "forces" the system to collapse to one of the two states, and this happens with a probability that depends on the actual state of the system, and induces a physical change in the state of the system.
Measuring in a different basis then simply corresponds to another type of physical interaction with the system.
Another, slightly more informal way to think about it is to say that each measurement basis corresponds to a different "question" being asked to the system. "Measuring in ${lvert0rangle,lvert1rangle}$" means to be asking the system whether it's in the $lvert0rangle$ or $lvert1rangle$ state, while "measuring in ${lvert+rangle,lvert-rangle}$" means to be asking the system whether it is in the $lvert+rangle$ or $lvert-rangle$ state. As it happens, in quantum mechanics the act of "asking a question" to the system - that is, of collecting some classical information out of the system - will in general result in a perturbation of the system itself, which in quantum mechanics is captured by means of the noncommutativity of different observables.
answered 11 hours ago
glSglS
4,125639
$endgroup$
One way to think about it is that "measuring in a given basis" is how we describe mathematically the act of interacting with the system in different ways.
Taking as an example a qubit, "measuring in the ${lvert0rangle,lvert1rangle}$ basis" is mathematese for "interacting with the system in a way that reveals to the experimenter whether the system is in the physical state which we are thinking of as corresponding to $lvert0rangle$ or $lvert1rangle$". In general, the system will not be in a state that can be directly associated with one of these basis states, in which case the act of measurement "forces" the system to collapse to one of the two states, and this happens with a probability that depends on the actual state of the system, and induces a physical change in the state of the system.
Measuring in a different basis then simply corresponds to another type of physical interaction with the system.
Another, slightly more informal way to think about it is to say that each measurement basis corresponds to a different "question" being asked to the system. "Measuring in ${lvert0rangle,lvert1rangle}$" means to be asking the system whether it's in the $lvert0rangle$ or $lvert1rangle$ state, while "measuring in ${lvert+rangle,lvert-rangle}$" means to be asking the system whether it is in the $lvert+rangle$ or $lvert-rangle$ state. As it happens, in quantum mechanics the act of "asking a question" to the system - that is, of collecting some classical information out of the system - will in general result in a perturbation of the system itself, which in quantum mechanics is captured by means of the noncommutativity of different observables.
answered 11 hours ago
glSglS
4,125639
edited 11 hours ago
answered 11 hours ago
glSglS
4,125639
answered 11 hours ago
glSglS
4,125639
answered 11 hours ago
glSglS
4,125639
4,125639
add a comment |
add a comment |
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$begingroup$
Related: What does “measurement in a certain basis” mean?
$endgroup$
– Blue♦
12 hours ago
1
$begingroup$
also related to The meaning of measurements in different bases
$endgroup$
– glS
11 hours ago