Is swap gate equivalent to just exchanging the wire of the two qubits?What happens if two separately entangled qubits are passed through a C-NOT gate?How to implement the “Square root of Swap gate” on the IBM Q (composer)?How to implement a Square root of Swap gate that swaps 2n-qubits on the IBM Q (composer)?Expressing “Square root of Swap” gate in terms of CNOTSWAP gate(s) in the $R(lambda^-1)$ step of the HHL circuit for $4times 4$ systemsWhat happens when I input two qubits starting at state $|00rangle$ into a hadamard gate?Composing the CNOT gate as a tensor product of two level matricesImplement Fredkin gate with square root of swapWhat is the tensorial representation of the quantum swap gate?
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Is swap gate equivalent to just exchanging the wire of the two qubits?
What happens if two separately entangled qubits are passed through a C-NOT gate?How to implement the “Square root of Swap gate” on the IBM Q (composer)?How to implement a Square root of Swap gate that swaps 2n-qubits on the IBM Q (composer)?Expressing “Square root of Swap” gate in terms of CNOTSWAP gate(s) in the $R(lambda^-1)$ step of the HHL circuit for $4times 4$ systemsWhat happens when I input two qubits starting at state $|00rangle$ into a hadamard gate?Composing the CNOT gate as a tensor product of two level matricesImplement Fredkin gate with square root of swapWhat is the tensorial representation of the quantum swap gate?
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Is swap gate equivalent to just exchanging the wire of the two qubits?
if yes why not just switching the wire whenever we want to apply a swap gate?
quantum-gate
edited Jun 14 at 13:41
Sanchayan Dutta
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asked Jun 14 at 10:45
Zzy1130Zzy1130
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add a comment
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$begingroup$
Is swap gate equivalent to just exchanging the wire of the two qubits?
if yes why not just switching the wire whenever we want to apply a swap gate?
quantum-gate
edited Jun 14 at 13:41
Sanchayan Dutta
8,4864 gold badges18 silver badges64 bronze badges
asked Jun 14 at 10:45
Zzy1130Zzy1130
1057 bronze badges
$endgroup$
add a comment
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$begingroup$
Is swap gate equivalent to just exchanging the wire of the two qubits?
if yes why not just switching the wire whenever we want to apply a swap gate?
quantum-gate
edited Jun 14 at 13:41
Sanchayan Dutta
8,4864 gold badges18 silver badges64 bronze badges
asked Jun 14 at 10:45
Zzy1130Zzy1130
1057 bronze badges
$endgroup$
Is swap gate equivalent to just exchanging the wire of the two qubits?
if yes why not just switching the wire whenever we want to apply a swap gate?
quantum-gate
quantum-gate
edited Jun 14 at 13:41
Sanchayan Dutta
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asked Jun 14 at 10:45
Zzy1130Zzy1130
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edited Jun 14 at 13:41
Sanchayan Dutta
8,4864 gold badges18 silver badges64 bronze badges
asked Jun 14 at 10:45
Zzy1130Zzy1130
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edited Jun 14 at 13:41
Sanchayan Dutta
8,4864 gold badges18 silver badges64 bronze badges
edited Jun 14 at 13:41
Sanchayan Dutta
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edited Jun 14 at 13:41
Sanchayan Dutta
8,4864 gold badges18 silver badges64 bronze badges
8,4864 gold badges18 silver badges64 bronze badges
asked Jun 14 at 10:45
Zzy1130Zzy1130
1057 bronze badges
asked Jun 14 at 10:45
Zzy1130Zzy1130
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asked Jun 14 at 10:45
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2 Answers
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Yes, it is equivalent. That's a perfectly valid way to represent it in a quantum circuit diagram.
In practice, it depends what physical implementation you're talking about. If you have 'flying' qubits, then it is possible to just reorder them, or even simply relabel them without actively doing anything. However, many physical implementations of quantum computation use static qubits. That means one particular physical qubit is in a particular place. If you want to make it interact (e.g. perform controlled-not) with another qubit, they usually have to be next to each other. That means performing swap operations along some path, exchanging the states of the qubits so that the state that was on one physical qubit is now on a different one, where you need it to be.
answered Jun 14 at 11:08
DaftWullieDaftWullie
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why not just switching the wire whenever we want to apply a swap gate?
If this is the crux of the question I think the main rationale here is to treat the swap gate as an actual gate.
I only rarely see raw swap gates on their own in diagrams, typically I see them as part of a Fredkin Gate. Basically, by treating it as a gate rather than a crossing of wires, it makes it easier to understand how it can be "controlled" like any other gate.
But yes, an un-controlled swap gate could be implemented on classical bits in electrical wires by swapping the wires' positions.
answered Jun 14 at 19:39
LambdaBetaLambdaBeta
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2 Answers
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2 Answers
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$begingroup$
Yes, it is equivalent. That's a perfectly valid way to represent it in a quantum circuit diagram.
In practice, it depends what physical implementation you're talking about. If you have 'flying' qubits, then it is possible to just reorder them, or even simply relabel them without actively doing anything. However, many physical implementations of quantum computation use static qubits. That means one particular physical qubit is in a particular place. If you want to make it interact (e.g. perform controlled-not) with another qubit, they usually have to be next to each other. That means performing swap operations along some path, exchanging the states of the qubits so that the state that was on one physical qubit is now on a different one, where you need it to be.
answered Jun 14 at 11:08
DaftWullieDaftWullie
21.5k1 gold badge9 silver badges54 bronze badges
$endgroup$
add a comment
|
$begingroup$
Yes, it is equivalent. That's a perfectly valid way to represent it in a quantum circuit diagram.
In practice, it depends what physical implementation you're talking about. If you have 'flying' qubits, then it is possible to just reorder them, or even simply relabel them without actively doing anything. However, many physical implementations of quantum computation use static qubits. That means one particular physical qubit is in a particular place. If you want to make it interact (e.g. perform controlled-not) with another qubit, they usually have to be next to each other. That means performing swap operations along some path, exchanging the states of the qubits so that the state that was on one physical qubit is now on a different one, where you need it to be.
answered Jun 14 at 11:08
DaftWullieDaftWullie
21.5k1 gold badge9 silver badges54 bronze badges
$endgroup$
add a comment
|
$begingroup$
Yes, it is equivalent. That's a perfectly valid way to represent it in a quantum circuit diagram.
In practice, it depends what physical implementation you're talking about. If you have 'flying' qubits, then it is possible to just reorder them, or even simply relabel them without actively doing anything. However, many physical implementations of quantum computation use static qubits. That means one particular physical qubit is in a particular place. If you want to make it interact (e.g. perform controlled-not) with another qubit, they usually have to be next to each other. That means performing swap operations along some path, exchanging the states of the qubits so that the state that was on one physical qubit is now on a different one, where you need it to be.
answered Jun 14 at 11:08
DaftWullieDaftWullie
21.5k1 gold badge9 silver badges54 bronze badges
$endgroup$
Yes, it is equivalent. That's a perfectly valid way to represent it in a quantum circuit diagram.
In practice, it depends what physical implementation you're talking about. If you have 'flying' qubits, then it is possible to just reorder them, or even simply relabel them without actively doing anything. However, many physical implementations of quantum computation use static qubits. That means one particular physical qubit is in a particular place. If you want to make it interact (e.g. perform controlled-not) with another qubit, they usually have to be next to each other. That means performing swap operations along some path, exchanging the states of the qubits so that the state that was on one physical qubit is now on a different one, where you need it to be.
answered Jun 14 at 11:08
DaftWullieDaftWullie
21.5k1 gold badge9 silver badges54 bronze badges
answered Jun 14 at 11:08
DaftWullieDaftWullie
21.5k1 gold badge9 silver badges54 bronze badges
answered Jun 14 at 11:08
DaftWullieDaftWullie
21.5k1 gold badge9 silver badges54 bronze badges
answered Jun 14 at 11:08
DaftWullieDaftWullie
21.5k1 gold badge9 silver badges54 bronze badges
21.5k1 gold badge9 silver badges54 bronze badges
add a comment
|
add a comment
|
$begingroup$
why not just switching the wire whenever we want to apply a swap gate?
If this is the crux of the question I think the main rationale here is to treat the swap gate as an actual gate.
I only rarely see raw swap gates on their own in diagrams, typically I see them as part of a Fredkin Gate. Basically, by treating it as a gate rather than a crossing of wires, it makes it easier to understand how it can be "controlled" like any other gate.
But yes, an un-controlled swap gate could be implemented on classical bits in electrical wires by swapping the wires' positions.
answered Jun 14 at 19:39
LambdaBetaLambdaBeta
1736 bronze badges
$endgroup$
add a comment
|
$begingroup$
why not just switching the wire whenever we want to apply a swap gate?
If this is the crux of the question I think the main rationale here is to treat the swap gate as an actual gate.
I only rarely see raw swap gates on their own in diagrams, typically I see them as part of a Fredkin Gate. Basically, by treating it as a gate rather than a crossing of wires, it makes it easier to understand how it can be "controlled" like any other gate.
But yes, an un-controlled swap gate could be implemented on classical bits in electrical wires by swapping the wires' positions.
answered Jun 14 at 19:39
LambdaBetaLambdaBeta
1736 bronze badges
$endgroup$
add a comment
|
$begingroup$
why not just switching the wire whenever we want to apply a swap gate?
If this is the crux of the question I think the main rationale here is to treat the swap gate as an actual gate.
I only rarely see raw swap gates on their own in diagrams, typically I see them as part of a Fredkin Gate. Basically, by treating it as a gate rather than a crossing of wires, it makes it easier to understand how it can be "controlled" like any other gate.
But yes, an un-controlled swap gate could be implemented on classical bits in electrical wires by swapping the wires' positions.
answered Jun 14 at 19:39
LambdaBetaLambdaBeta
1736 bronze badges
$endgroup$
why not just switching the wire whenever we want to apply a swap gate?
If this is the crux of the question I think the main rationale here is to treat the swap gate as an actual gate.
I only rarely see raw swap gates on their own in diagrams, typically I see them as part of a Fredkin Gate. Basically, by treating it as a gate rather than a crossing of wires, it makes it easier to understand how it can be "controlled" like any other gate.
But yes, an un-controlled swap gate could be implemented on classical bits in electrical wires by swapping the wires' positions.
answered Jun 14 at 19:39
LambdaBetaLambdaBeta
1736 bronze badges
answered Jun 14 at 19:39
LambdaBetaLambdaBeta
1736 bronze badges
answered Jun 14 at 19:39
LambdaBetaLambdaBeta
1736 bronze badges
answered Jun 14 at 19:39
LambdaBetaLambdaBeta
1736 bronze badges
1736 bronze badges
add a comment
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add a comment
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