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What does an observable in a different basis mean physically?

What does “measurement in a certain basis” mean?The meaning of measurements in different basesHow does Fourier sampling actually work (and solve the parity problem)?Obtaining gate $e^{-iDelta t Z}$ from elementary gatesWhat does the notation $lvert underline{x} rangle$ mean?Many-Worlds Interpretation and GHZ StatesHow does the stated Pauli decomposition for $operatorname{CPcdot Acdot CP}$ arise?Measurement of a qubit and storage of the information on a bitIs it correct to say that we need controlled gates because unitary matrices are reversible?What is the meaning of the state $|1rangle-|1rangle$?What's a vector in the format of the Bloch Sphere?The meaning of measurements in different bases

2

$begingroup$

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

share|improve this question

edited 8 hours ago

Blue♦

6,35141355

asked 9 hours ago

bilanushbilanush

1816

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Related: What does “measurement in a certain basis” mean?
  $endgroup$
  – Blue♦
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  also related to The meaning of measurements in different bases
  $endgroup$
  – glS
  7 hours ago
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

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

share|improve this question

edited 8 hours ago

Blue♦

6,35141355

asked 9 hours ago

bilanushbilanush

1816

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Related: What does “measurement in a certain basis” mean?
  $endgroup$
  – Blue♦
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  also related to The meaning of measurements in different bases
  $endgroup$
  – glS
  7 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

share|improve this question

edited 8 hours ago

Blue♦

6,35141355

asked 9 hours ago

bilanushbilanush

1816

$endgroup$

share|improve this question

edited 8 hours ago

Blue♦

6,35141355

asked 9 hours ago

bilanushbilanush

1816

share|improve this question

edited 8 hours ago

Blue♦

6,35141355

asked 9 hours ago

bilanushbilanush

1816

edited 8 hours ago

Blue♦

6,35141355

edited 8 hours ago

Blue♦

6,35141355

asked 9 hours ago

bilanushbilanush

1816

asked 9 hours ago

bilanushbilanush

1816

1 Answer
1

active

oldest

votes

2

$begingroup$

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.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

glSglS

4,125639

$endgroup$

add a comment | 

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2

$begingroup$

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.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

glSglS

4,125639

$endgroup$

add a comment | 

2

$begingroup$

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.

share|improve this answer

edited 7 hours ago

answered 7 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.

share|improve this answer

edited 7 hours ago

answered 7 hours ago

glSglS

4,125639

$endgroup$

share|improve this answer

edited 7 hours ago

answered 7 hours ago

glSglS

4,125639

answered 7 hours ago

glSglS

4,125639

answered 7 hours ago

glSglS

4,125639