Composing the CNOT gate as a tensor product of two level matrices












3












$begingroup$


I don't understand, why they all the time use the control not gate. As far as I understand it, if you apply two 2 level operations on two qubits then you get a 4 x 4 matrix by the tensor product. So how would you express the CNOT gate as a product of the one matrix acting upon the first qubit and the second acting on the second qubit?



Another thing which I have difficulty understanding is why we use so many single qubits and CNOT gates for implementing the Toffoli gate? Can't we implement it using just an AND between the first two qubits and then applying a controlled NOT for the third one?










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  • $begingroup$
    Refer to this question: How to construct matrix of regular and “flipped” 2-qubit CNOT?
    $endgroup$
    – Josu Etxezarreta Martinez
    9 hours ago


















3












$begingroup$


I don't understand, why they all the time use the control not gate. As far as I understand it, if you apply two 2 level operations on two qubits then you get a 4 x 4 matrix by the tensor product. So how would you express the CNOT gate as a product of the one matrix acting upon the first qubit and the second acting on the second qubit?



Another thing which I have difficulty understanding is why we use so many single qubits and CNOT gates for implementing the Toffoli gate? Can't we implement it using just an AND between the first two qubits and then applying a controlled NOT for the third one?










share|improve this question









New contributor




bilanush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Refer to this question: How to construct matrix of regular and “flipped” 2-qubit CNOT?
    $endgroup$
    – Josu Etxezarreta Martinez
    9 hours ago
















3












3








3





$begingroup$


I don't understand, why they all the time use the control not gate. As far as I understand it, if you apply two 2 level operations on two qubits then you get a 4 x 4 matrix by the tensor product. So how would you express the CNOT gate as a product of the one matrix acting upon the first qubit and the second acting on the second qubit?



Another thing which I have difficulty understanding is why we use so many single qubits and CNOT gates for implementing the Toffoli gate? Can't we implement it using just an AND between the first two qubits and then applying a controlled NOT for the third one?










share|improve this question









New contributor




bilanush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I don't understand, why they all the time use the control not gate. As far as I understand it, if you apply two 2 level operations on two qubits then you get a 4 x 4 matrix by the tensor product. So how would you express the CNOT gate as a product of the one matrix acting upon the first qubit and the second acting on the second qubit?



Another thing which I have difficulty understanding is why we use so many single qubits and CNOT gates for implementing the Toffoli gate? Can't we implement it using just an AND between the first two qubits and then applying a controlled NOT for the third one?







quantum-gate gate-synthesis tensor-product






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bilanush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question









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bilanush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









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share|improve this question








edited 9 hours ago









Blue

6,12331354




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asked 9 hours ago









bilanushbilanush

192




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bilanush is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.












  • $begingroup$
    Refer to this question: How to construct matrix of regular and “flipped” 2-qubit CNOT?
    $endgroup$
    – Josu Etxezarreta Martinez
    9 hours ago




















  • $begingroup$
    Refer to this question: How to construct matrix of regular and “flipped” 2-qubit CNOT?
    $endgroup$
    – Josu Etxezarreta Martinez
    9 hours ago


















$begingroup$
Refer to this question: How to construct matrix of regular and “flipped” 2-qubit CNOT?
$endgroup$
– Josu Etxezarreta Martinez
9 hours ago






$begingroup$
Refer to this question: How to construct matrix of regular and “flipped” 2-qubit CNOT?
$endgroup$
– Josu Etxezarreta Martinez
9 hours ago












2 Answers
2






active

oldest

votes


















3












$begingroup$

The whole point is that CNOT cannot be written in the form $Aotimes B$. This is absolutely essential because if we only ever had operators of the form $Aotimes B$, states of the form $|psirangleotimes|phirangle$ would always remain separable. There's be no entanglement, and all quantum circuits would be easy to simulate on a classical computer.



As for the Toffoli construction you describe, we require that in a quantum circuit, all gates are reversible. The AND gate is not reversible as you can see from the way the number of outputs is smaller than the number of inputs (so it's impossible to know what the inputs were just by looking at the outputs).






share|improve this answer









$endgroup$













  • $begingroup$
    As to the first part, so you are saying CNOT is for entanglement? I thought it's just an operator that we need for reversability. But either way, isn't it a theorem which states that every 4 level matrix can be expressed as a multiple of 2 level matrices? As for second part, Toffoli gate
    $endgroup$
    – bilanush
    6 hours ago












  • $begingroup$
    As for second part on Toffoli gate . I understand your point. But can't we just perform another output which would specify the input . I mean , perform some operation/s that would reveal the input. My point is , that it can be in the end much easier. No?
    $endgroup$
    – bilanush
    5 hours ago










  • $begingroup$
    Also, I am reading a book. And it looks strange but it says there, that for a generalization of Toffoli gate, we can perform AND operations on all n controll-qubits and then apply a U operation on the k qubits you have for action conditioned on the control ones. So why here do they allow for an and operation if I have general n qubits for control and k for operation.
    $endgroup$
    – bilanush
    4 hours ago










  • $begingroup$
    1) CNOT 'can' be for entanglement or otherwise, it just depends upon the requirement. Not anything restrictive. Any 4-level can be decomposed in CNOT and single qubit gates, this always follows from the universality. This is nowhere violating anything. 2) No, you cannot measure the system to perform a conditional operation, that won't be a true conditional operation as I pointed out in my answer. 3) Can you clarify what you mean by "AND" operation on qubits? Because there is no notion of an "AND" gate in quantum literature.
    $endgroup$
    – Siddhānt Singh
    3 hours ago






  • 1




    $begingroup$
    "can't we just perform another output which would specify the input". Actually, that's exactly what the Toffoli gate is. So you'd be going around in circles.
    $endgroup$
    – DaftWullie
    2 hours ago



















1












$begingroup$

Any two qubit Controlled-U gate is not decomposable as $Aotimes B$ in general. The whole point is that a conditional operator has a non-local nature and it gives rise to non-trivial correlations when operated on a state.



The Physics behind is that; imagine a Controlled operation which can be factored as $Aotimes B$, and the sub-systems (or the two qubits) are separated non-locally in space. then the only way to perform a controlled conditional operation is to measure the state of a first qubit, convey the outcome classically to the second qubit and apply the desired unitary on the second qubit, and finally re-initialize the first qubit. But now note that this is actually a 'true' controlled gate because it cannot work for an arbitrary control qubit.



Hence, the actual nature of a conditional gate is non-local, which essentially implies that it cannot be factored as $Aotimes B$ over the sub-spaces. The intuition can be taken as the operation reads system $A$ and simultaneously changes system $B$ based on the state of system $A$. This is strictly non-local in nature. Physically how it is implemented in experiments is most often by a single exchange of quanta between the states, which is triggered by the state of one sub-system. Hence, this gives rise to entanglement between the sub-systems. It is most often called the 'entanglement generator'.



Also, you can think of $operatorname{CNOT}$ gate to be the most simplified two-qubit gate which can generate the class of two-qubit non-local gates when combined with few other gates. This is already proved by the Universality of single-qubit gates and the $operatorname{CNOT}$ gate.



About the Toffoli gate, the most obvious way to implement it (after taking in mind the universality) is to decompose it in parts of $operatorname{CNOT}$ gates and single qubit gates. This is what is done. And because reversibility is a must requirement for a quantum operation, you cannot use any analogous irreversible gate like the $operatorname{AND}$ gate in quantum circuits. To perform the $operatorname{AND}$ analogous operation as you pointed out, this is further broken up in parts of the single qubit and $operatorname{CNOT}$ gates (to ensure the reversibility and unitarity).



PS: One essential reason why this decomposition is done because in experiments all we can implement are these single qubit and $operatorname{CNOT}$ gates only. Any such protocol must be decomposed in these gates to cope up with some experimental implementation. And these are intuitive as well, rather than using multi-qubit gates which are nothing but a series of the single qubit gates and $operatorname{CNOT}$s.






share|improve this answer











$endgroup$













  • $begingroup$
    Thanks. So CNOT is basically an entanglement operator? Are all entanglement operators are two qubits operations ? Or even a regular 4x4 matrix is considered a two qubits operation?
    $endgroup$
    – bilanush
    3 hours ago










  • $begingroup$
    And please, in my book it proves that all NxN matrices can be decomposed to multiple of two level matrices so... So is a CNOT gate? Or is a CNOT gate an exception?
    $endgroup$
    – bilanush
    3 hours ago










  • $begingroup$
    By that, I meant it is a fundamental operation to give rise to an entangled state. If you talk about the entanglement between two qubits only, then yes. But any highly entangled state with any $n$ qubits can be decomposed with CNOTs in a complicated manner. A $4times 4$ is a valid two-qubit operation but it does not necessarily mean it will be non-local.
    $endgroup$
    – Siddhānt Singh
    3 hours ago










  • $begingroup$
    Please read the answer, it is already made clear that a CNOT cannot be decomposed further, more essentially if it is written in the basis of the states being operated.
    $endgroup$
    – Siddhānt Singh
    3 hours ago










  • $begingroup$
    So if I understand you properly. Then each operation acting on two qubits is called two qubits operation even if it's a tensor product of two single qubits matrcies. But among them, only matrices which can't be composed by single qubit matrices are defined as performing entanglement. As to your second comment, the book quantum computation chapter 4 proves that any NxN can be decomposed to a multiple of 2 level matrices he even proves it, so why do you say a control-not gate cannot? Isn't the proof a general proof regarding any N×N matrix?
    $endgroup$
    – bilanush
    2 hours ago











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

The whole point is that CNOT cannot be written in the form $Aotimes B$. This is absolutely essential because if we only ever had operators of the form $Aotimes B$, states of the form $|psirangleotimes|phirangle$ would always remain separable. There's be no entanglement, and all quantum circuits would be easy to simulate on a classical computer.



As for the Toffoli construction you describe, we require that in a quantum circuit, all gates are reversible. The AND gate is not reversible as you can see from the way the number of outputs is smaller than the number of inputs (so it's impossible to know what the inputs were just by looking at the outputs).






share|improve this answer









$endgroup$













  • $begingroup$
    As to the first part, so you are saying CNOT is for entanglement? I thought it's just an operator that we need for reversability. But either way, isn't it a theorem which states that every 4 level matrix can be expressed as a multiple of 2 level matrices? As for second part, Toffoli gate
    $endgroup$
    – bilanush
    6 hours ago












  • $begingroup$
    As for second part on Toffoli gate . I understand your point. But can't we just perform another output which would specify the input . I mean , perform some operation/s that would reveal the input. My point is , that it can be in the end much easier. No?
    $endgroup$
    – bilanush
    5 hours ago










  • $begingroup$
    Also, I am reading a book. And it looks strange but it says there, that for a generalization of Toffoli gate, we can perform AND operations on all n controll-qubits and then apply a U operation on the k qubits you have for action conditioned on the control ones. So why here do they allow for an and operation if I have general n qubits for control and k for operation.
    $endgroup$
    – bilanush
    4 hours ago










  • $begingroup$
    1) CNOT 'can' be for entanglement or otherwise, it just depends upon the requirement. Not anything restrictive. Any 4-level can be decomposed in CNOT and single qubit gates, this always follows from the universality. This is nowhere violating anything. 2) No, you cannot measure the system to perform a conditional operation, that won't be a true conditional operation as I pointed out in my answer. 3) Can you clarify what you mean by "AND" operation on qubits? Because there is no notion of an "AND" gate in quantum literature.
    $endgroup$
    – Siddhānt Singh
    3 hours ago






  • 1




    $begingroup$
    "can't we just perform another output which would specify the input". Actually, that's exactly what the Toffoli gate is. So you'd be going around in circles.
    $endgroup$
    – DaftWullie
    2 hours ago
















3












$begingroup$

The whole point is that CNOT cannot be written in the form $Aotimes B$. This is absolutely essential because if we only ever had operators of the form $Aotimes B$, states of the form $|psirangleotimes|phirangle$ would always remain separable. There's be no entanglement, and all quantum circuits would be easy to simulate on a classical computer.



As for the Toffoli construction you describe, we require that in a quantum circuit, all gates are reversible. The AND gate is not reversible as you can see from the way the number of outputs is smaller than the number of inputs (so it's impossible to know what the inputs were just by looking at the outputs).






share|improve this answer









$endgroup$













  • $begingroup$
    As to the first part, so you are saying CNOT is for entanglement? I thought it's just an operator that we need for reversability. But either way, isn't it a theorem which states that every 4 level matrix can be expressed as a multiple of 2 level matrices? As for second part, Toffoli gate
    $endgroup$
    – bilanush
    6 hours ago












  • $begingroup$
    As for second part on Toffoli gate . I understand your point. But can't we just perform another output which would specify the input . I mean , perform some operation/s that would reveal the input. My point is , that it can be in the end much easier. No?
    $endgroup$
    – bilanush
    5 hours ago










  • $begingroup$
    Also, I am reading a book. And it looks strange but it says there, that for a generalization of Toffoli gate, we can perform AND operations on all n controll-qubits and then apply a U operation on the k qubits you have for action conditioned on the control ones. So why here do they allow for an and operation if I have general n qubits for control and k for operation.
    $endgroup$
    – bilanush
    4 hours ago










  • $begingroup$
    1) CNOT 'can' be for entanglement or otherwise, it just depends upon the requirement. Not anything restrictive. Any 4-level can be decomposed in CNOT and single qubit gates, this always follows from the universality. This is nowhere violating anything. 2) No, you cannot measure the system to perform a conditional operation, that won't be a true conditional operation as I pointed out in my answer. 3) Can you clarify what you mean by "AND" operation on qubits? Because there is no notion of an "AND" gate in quantum literature.
    $endgroup$
    – Siddhānt Singh
    3 hours ago






  • 1




    $begingroup$
    "can't we just perform another output which would specify the input". Actually, that's exactly what the Toffoli gate is. So you'd be going around in circles.
    $endgroup$
    – DaftWullie
    2 hours ago














3












3








3





$begingroup$

The whole point is that CNOT cannot be written in the form $Aotimes B$. This is absolutely essential because if we only ever had operators of the form $Aotimes B$, states of the form $|psirangleotimes|phirangle$ would always remain separable. There's be no entanglement, and all quantum circuits would be easy to simulate on a classical computer.



As for the Toffoli construction you describe, we require that in a quantum circuit, all gates are reversible. The AND gate is not reversible as you can see from the way the number of outputs is smaller than the number of inputs (so it's impossible to know what the inputs were just by looking at the outputs).






share|improve this answer









$endgroup$



The whole point is that CNOT cannot be written in the form $Aotimes B$. This is absolutely essential because if we only ever had operators of the form $Aotimes B$, states of the form $|psirangleotimes|phirangle$ would always remain separable. There's be no entanglement, and all quantum circuits would be easy to simulate on a classical computer.



As for the Toffoli construction you describe, we require that in a quantum circuit, all gates are reversible. The AND gate is not reversible as you can see from the way the number of outputs is smaller than the number of inputs (so it's impossible to know what the inputs were just by looking at the outputs).







share|improve this answer












share|improve this answer



share|improve this answer










answered 9 hours ago









DaftWullieDaftWullie

13.6k1539




13.6k1539












  • $begingroup$
    As to the first part, so you are saying CNOT is for entanglement? I thought it's just an operator that we need for reversability. But either way, isn't it a theorem which states that every 4 level matrix can be expressed as a multiple of 2 level matrices? As for second part, Toffoli gate
    $endgroup$
    – bilanush
    6 hours ago












  • $begingroup$
    As for second part on Toffoli gate . I understand your point. But can't we just perform another output which would specify the input . I mean , perform some operation/s that would reveal the input. My point is , that it can be in the end much easier. No?
    $endgroup$
    – bilanush
    5 hours ago










  • $begingroup$
    Also, I am reading a book. And it looks strange but it says there, that for a generalization of Toffoli gate, we can perform AND operations on all n controll-qubits and then apply a U operation on the k qubits you have for action conditioned on the control ones. So why here do they allow for an and operation if I have general n qubits for control and k for operation.
    $endgroup$
    – bilanush
    4 hours ago










  • $begingroup$
    1) CNOT 'can' be for entanglement or otherwise, it just depends upon the requirement. Not anything restrictive. Any 4-level can be decomposed in CNOT and single qubit gates, this always follows from the universality. This is nowhere violating anything. 2) No, you cannot measure the system to perform a conditional operation, that won't be a true conditional operation as I pointed out in my answer. 3) Can you clarify what you mean by "AND" operation on qubits? Because there is no notion of an "AND" gate in quantum literature.
    $endgroup$
    – Siddhānt Singh
    3 hours ago






  • 1




    $begingroup$
    "can't we just perform another output which would specify the input". Actually, that's exactly what the Toffoli gate is. So you'd be going around in circles.
    $endgroup$
    – DaftWullie
    2 hours ago


















  • $begingroup$
    As to the first part, so you are saying CNOT is for entanglement? I thought it's just an operator that we need for reversability. But either way, isn't it a theorem which states that every 4 level matrix can be expressed as a multiple of 2 level matrices? As for second part, Toffoli gate
    $endgroup$
    – bilanush
    6 hours ago












  • $begingroup$
    As for second part on Toffoli gate . I understand your point. But can't we just perform another output which would specify the input . I mean , perform some operation/s that would reveal the input. My point is , that it can be in the end much easier. No?
    $endgroup$
    – bilanush
    5 hours ago










  • $begingroup$
    Also, I am reading a book. And it looks strange but it says there, that for a generalization of Toffoli gate, we can perform AND operations on all n controll-qubits and then apply a U operation on the k qubits you have for action conditioned on the control ones. So why here do they allow for an and operation if I have general n qubits for control and k for operation.
    $endgroup$
    – bilanush
    4 hours ago










  • $begingroup$
    1) CNOT 'can' be for entanglement or otherwise, it just depends upon the requirement. Not anything restrictive. Any 4-level can be decomposed in CNOT and single qubit gates, this always follows from the universality. This is nowhere violating anything. 2) No, you cannot measure the system to perform a conditional operation, that won't be a true conditional operation as I pointed out in my answer. 3) Can you clarify what you mean by "AND" operation on qubits? Because there is no notion of an "AND" gate in quantum literature.
    $endgroup$
    – Siddhānt Singh
    3 hours ago






  • 1




    $begingroup$
    "can't we just perform another output which would specify the input". Actually, that's exactly what the Toffoli gate is. So you'd be going around in circles.
    $endgroup$
    – DaftWullie
    2 hours ago
















$begingroup$
As to the first part, so you are saying CNOT is for entanglement? I thought it's just an operator that we need for reversability. But either way, isn't it a theorem which states that every 4 level matrix can be expressed as a multiple of 2 level matrices? As for second part, Toffoli gate
$endgroup$
– bilanush
6 hours ago






$begingroup$
As to the first part, so you are saying CNOT is for entanglement? I thought it's just an operator that we need for reversability. But either way, isn't it a theorem which states that every 4 level matrix can be expressed as a multiple of 2 level matrices? As for second part, Toffoli gate
$endgroup$
– bilanush
6 hours ago














$begingroup$
As for second part on Toffoli gate . I understand your point. But can't we just perform another output which would specify the input . I mean , perform some operation/s that would reveal the input. My point is , that it can be in the end much easier. No?
$endgroup$
– bilanush
5 hours ago




$begingroup$
As for second part on Toffoli gate . I understand your point. But can't we just perform another output which would specify the input . I mean , perform some operation/s that would reveal the input. My point is , that it can be in the end much easier. No?
$endgroup$
– bilanush
5 hours ago












$begingroup$
Also, I am reading a book. And it looks strange but it says there, that for a generalization of Toffoli gate, we can perform AND operations on all n controll-qubits and then apply a U operation on the k qubits you have for action conditioned on the control ones. So why here do they allow for an and operation if I have general n qubits for control and k for operation.
$endgroup$
– bilanush
4 hours ago




$begingroup$
Also, I am reading a book. And it looks strange but it says there, that for a generalization of Toffoli gate, we can perform AND operations on all n controll-qubits and then apply a U operation on the k qubits you have for action conditioned on the control ones. So why here do they allow for an and operation if I have general n qubits for control and k for operation.
$endgroup$
– bilanush
4 hours ago












$begingroup$
1) CNOT 'can' be for entanglement or otherwise, it just depends upon the requirement. Not anything restrictive. Any 4-level can be decomposed in CNOT and single qubit gates, this always follows from the universality. This is nowhere violating anything. 2) No, you cannot measure the system to perform a conditional operation, that won't be a true conditional operation as I pointed out in my answer. 3) Can you clarify what you mean by "AND" operation on qubits? Because there is no notion of an "AND" gate in quantum literature.
$endgroup$
– Siddhānt Singh
3 hours ago




$begingroup$
1) CNOT 'can' be for entanglement or otherwise, it just depends upon the requirement. Not anything restrictive. Any 4-level can be decomposed in CNOT and single qubit gates, this always follows from the universality. This is nowhere violating anything. 2) No, you cannot measure the system to perform a conditional operation, that won't be a true conditional operation as I pointed out in my answer. 3) Can you clarify what you mean by "AND" operation on qubits? Because there is no notion of an "AND" gate in quantum literature.
$endgroup$
– Siddhānt Singh
3 hours ago




1




1




$begingroup$
"can't we just perform another output which would specify the input". Actually, that's exactly what the Toffoli gate is. So you'd be going around in circles.
$endgroup$
– DaftWullie
2 hours ago




$begingroup$
"can't we just perform another output which would specify the input". Actually, that's exactly what the Toffoli gate is. So you'd be going around in circles.
$endgroup$
– DaftWullie
2 hours ago













1












$begingroup$

Any two qubit Controlled-U gate is not decomposable as $Aotimes B$ in general. The whole point is that a conditional operator has a non-local nature and it gives rise to non-trivial correlations when operated on a state.



The Physics behind is that; imagine a Controlled operation which can be factored as $Aotimes B$, and the sub-systems (or the two qubits) are separated non-locally in space. then the only way to perform a controlled conditional operation is to measure the state of a first qubit, convey the outcome classically to the second qubit and apply the desired unitary on the second qubit, and finally re-initialize the first qubit. But now note that this is actually a 'true' controlled gate because it cannot work for an arbitrary control qubit.



Hence, the actual nature of a conditional gate is non-local, which essentially implies that it cannot be factored as $Aotimes B$ over the sub-spaces. The intuition can be taken as the operation reads system $A$ and simultaneously changes system $B$ based on the state of system $A$. This is strictly non-local in nature. Physically how it is implemented in experiments is most often by a single exchange of quanta between the states, which is triggered by the state of one sub-system. Hence, this gives rise to entanglement between the sub-systems. It is most often called the 'entanglement generator'.



Also, you can think of $operatorname{CNOT}$ gate to be the most simplified two-qubit gate which can generate the class of two-qubit non-local gates when combined with few other gates. This is already proved by the Universality of single-qubit gates and the $operatorname{CNOT}$ gate.



About the Toffoli gate, the most obvious way to implement it (after taking in mind the universality) is to decompose it in parts of $operatorname{CNOT}$ gates and single qubit gates. This is what is done. And because reversibility is a must requirement for a quantum operation, you cannot use any analogous irreversible gate like the $operatorname{AND}$ gate in quantum circuits. To perform the $operatorname{AND}$ analogous operation as you pointed out, this is further broken up in parts of the single qubit and $operatorname{CNOT}$ gates (to ensure the reversibility and unitarity).



PS: One essential reason why this decomposition is done because in experiments all we can implement are these single qubit and $operatorname{CNOT}$ gates only. Any such protocol must be decomposed in these gates to cope up with some experimental implementation. And these are intuitive as well, rather than using multi-qubit gates which are nothing but a series of the single qubit gates and $operatorname{CNOT}$s.






share|improve this answer











$endgroup$













  • $begingroup$
    Thanks. So CNOT is basically an entanglement operator? Are all entanglement operators are two qubits operations ? Or even a regular 4x4 matrix is considered a two qubits operation?
    $endgroup$
    – bilanush
    3 hours ago










  • $begingroup$
    And please, in my book it proves that all NxN matrices can be decomposed to multiple of two level matrices so... So is a CNOT gate? Or is a CNOT gate an exception?
    $endgroup$
    – bilanush
    3 hours ago










  • $begingroup$
    By that, I meant it is a fundamental operation to give rise to an entangled state. If you talk about the entanglement between two qubits only, then yes. But any highly entangled state with any $n$ qubits can be decomposed with CNOTs in a complicated manner. A $4times 4$ is a valid two-qubit operation but it does not necessarily mean it will be non-local.
    $endgroup$
    – Siddhānt Singh
    3 hours ago










  • $begingroup$
    Please read the answer, it is already made clear that a CNOT cannot be decomposed further, more essentially if it is written in the basis of the states being operated.
    $endgroup$
    – Siddhānt Singh
    3 hours ago










  • $begingroup$
    So if I understand you properly. Then each operation acting on two qubits is called two qubits operation even if it's a tensor product of two single qubits matrcies. But among them, only matrices which can't be composed by single qubit matrices are defined as performing entanglement. As to your second comment, the book quantum computation chapter 4 proves that any NxN can be decomposed to a multiple of 2 level matrices he even proves it, so why do you say a control-not gate cannot? Isn't the proof a general proof regarding any N×N matrix?
    $endgroup$
    – bilanush
    2 hours ago
















1












$begingroup$

Any two qubit Controlled-U gate is not decomposable as $Aotimes B$ in general. The whole point is that a conditional operator has a non-local nature and it gives rise to non-trivial correlations when operated on a state.



The Physics behind is that; imagine a Controlled operation which can be factored as $Aotimes B$, and the sub-systems (or the two qubits) are separated non-locally in space. then the only way to perform a controlled conditional operation is to measure the state of a first qubit, convey the outcome classically to the second qubit and apply the desired unitary on the second qubit, and finally re-initialize the first qubit. But now note that this is actually a 'true' controlled gate because it cannot work for an arbitrary control qubit.



Hence, the actual nature of a conditional gate is non-local, which essentially implies that it cannot be factored as $Aotimes B$ over the sub-spaces. The intuition can be taken as the operation reads system $A$ and simultaneously changes system $B$ based on the state of system $A$. This is strictly non-local in nature. Physically how it is implemented in experiments is most often by a single exchange of quanta between the states, which is triggered by the state of one sub-system. Hence, this gives rise to entanglement between the sub-systems. It is most often called the 'entanglement generator'.



Also, you can think of $operatorname{CNOT}$ gate to be the most simplified two-qubit gate which can generate the class of two-qubit non-local gates when combined with few other gates. This is already proved by the Universality of single-qubit gates and the $operatorname{CNOT}$ gate.



About the Toffoli gate, the most obvious way to implement it (after taking in mind the universality) is to decompose it in parts of $operatorname{CNOT}$ gates and single qubit gates. This is what is done. And because reversibility is a must requirement for a quantum operation, you cannot use any analogous irreversible gate like the $operatorname{AND}$ gate in quantum circuits. To perform the $operatorname{AND}$ analogous operation as you pointed out, this is further broken up in parts of the single qubit and $operatorname{CNOT}$ gates (to ensure the reversibility and unitarity).



PS: One essential reason why this decomposition is done because in experiments all we can implement are these single qubit and $operatorname{CNOT}$ gates only. Any such protocol must be decomposed in these gates to cope up with some experimental implementation. And these are intuitive as well, rather than using multi-qubit gates which are nothing but a series of the single qubit gates and $operatorname{CNOT}$s.






share|improve this answer











$endgroup$













  • $begingroup$
    Thanks. So CNOT is basically an entanglement operator? Are all entanglement operators are two qubits operations ? Or even a regular 4x4 matrix is considered a two qubits operation?
    $endgroup$
    – bilanush
    3 hours ago










  • $begingroup$
    And please, in my book it proves that all NxN matrices can be decomposed to multiple of two level matrices so... So is a CNOT gate? Or is a CNOT gate an exception?
    $endgroup$
    – bilanush
    3 hours ago










  • $begingroup$
    By that, I meant it is a fundamental operation to give rise to an entangled state. If you talk about the entanglement between two qubits only, then yes. But any highly entangled state with any $n$ qubits can be decomposed with CNOTs in a complicated manner. A $4times 4$ is a valid two-qubit operation but it does not necessarily mean it will be non-local.
    $endgroup$
    – Siddhānt Singh
    3 hours ago










  • $begingroup$
    Please read the answer, it is already made clear that a CNOT cannot be decomposed further, more essentially if it is written in the basis of the states being operated.
    $endgroup$
    – Siddhānt Singh
    3 hours ago










  • $begingroup$
    So if I understand you properly. Then each operation acting on two qubits is called two qubits operation even if it's a tensor product of two single qubits matrcies. But among them, only matrices which can't be composed by single qubit matrices are defined as performing entanglement. As to your second comment, the book quantum computation chapter 4 proves that any NxN can be decomposed to a multiple of 2 level matrices he even proves it, so why do you say a control-not gate cannot? Isn't the proof a general proof regarding any N×N matrix?
    $endgroup$
    – bilanush
    2 hours ago














1












1








1





$begingroup$

Any two qubit Controlled-U gate is not decomposable as $Aotimes B$ in general. The whole point is that a conditional operator has a non-local nature and it gives rise to non-trivial correlations when operated on a state.



The Physics behind is that; imagine a Controlled operation which can be factored as $Aotimes B$, and the sub-systems (or the two qubits) are separated non-locally in space. then the only way to perform a controlled conditional operation is to measure the state of a first qubit, convey the outcome classically to the second qubit and apply the desired unitary on the second qubit, and finally re-initialize the first qubit. But now note that this is actually a 'true' controlled gate because it cannot work for an arbitrary control qubit.



Hence, the actual nature of a conditional gate is non-local, which essentially implies that it cannot be factored as $Aotimes B$ over the sub-spaces. The intuition can be taken as the operation reads system $A$ and simultaneously changes system $B$ based on the state of system $A$. This is strictly non-local in nature. Physically how it is implemented in experiments is most often by a single exchange of quanta between the states, which is triggered by the state of one sub-system. Hence, this gives rise to entanglement between the sub-systems. It is most often called the 'entanglement generator'.



Also, you can think of $operatorname{CNOT}$ gate to be the most simplified two-qubit gate which can generate the class of two-qubit non-local gates when combined with few other gates. This is already proved by the Universality of single-qubit gates and the $operatorname{CNOT}$ gate.



About the Toffoli gate, the most obvious way to implement it (after taking in mind the universality) is to decompose it in parts of $operatorname{CNOT}$ gates and single qubit gates. This is what is done. And because reversibility is a must requirement for a quantum operation, you cannot use any analogous irreversible gate like the $operatorname{AND}$ gate in quantum circuits. To perform the $operatorname{AND}$ analogous operation as you pointed out, this is further broken up in parts of the single qubit and $operatorname{CNOT}$ gates (to ensure the reversibility and unitarity).



PS: One essential reason why this decomposition is done because in experiments all we can implement are these single qubit and $operatorname{CNOT}$ gates only. Any such protocol must be decomposed in these gates to cope up with some experimental implementation. And these are intuitive as well, rather than using multi-qubit gates which are nothing but a series of the single qubit gates and $operatorname{CNOT}$s.






share|improve this answer











$endgroup$



Any two qubit Controlled-U gate is not decomposable as $Aotimes B$ in general. The whole point is that a conditional operator has a non-local nature and it gives rise to non-trivial correlations when operated on a state.



The Physics behind is that; imagine a Controlled operation which can be factored as $Aotimes B$, and the sub-systems (or the two qubits) are separated non-locally in space. then the only way to perform a controlled conditional operation is to measure the state of a first qubit, convey the outcome classically to the second qubit and apply the desired unitary on the second qubit, and finally re-initialize the first qubit. But now note that this is actually a 'true' controlled gate because it cannot work for an arbitrary control qubit.



Hence, the actual nature of a conditional gate is non-local, which essentially implies that it cannot be factored as $Aotimes B$ over the sub-spaces. The intuition can be taken as the operation reads system $A$ and simultaneously changes system $B$ based on the state of system $A$. This is strictly non-local in nature. Physically how it is implemented in experiments is most often by a single exchange of quanta between the states, which is triggered by the state of one sub-system. Hence, this gives rise to entanglement between the sub-systems. It is most often called the 'entanglement generator'.



Also, you can think of $operatorname{CNOT}$ gate to be the most simplified two-qubit gate which can generate the class of two-qubit non-local gates when combined with few other gates. This is already proved by the Universality of single-qubit gates and the $operatorname{CNOT}$ gate.



About the Toffoli gate, the most obvious way to implement it (after taking in mind the universality) is to decompose it in parts of $operatorname{CNOT}$ gates and single qubit gates. This is what is done. And because reversibility is a must requirement for a quantum operation, you cannot use any analogous irreversible gate like the $operatorname{AND}$ gate in quantum circuits. To perform the $operatorname{AND}$ analogous operation as you pointed out, this is further broken up in parts of the single qubit and $operatorname{CNOT}$ gates (to ensure the reversibility and unitarity).



PS: One essential reason why this decomposition is done because in experiments all we can implement are these single qubit and $operatorname{CNOT}$ gates only. Any such protocol must be decomposed in these gates to cope up with some experimental implementation. And these are intuitive as well, rather than using multi-qubit gates which are nothing but a series of the single qubit gates and $operatorname{CNOT}$s.







share|improve this answer














share|improve this answer



share|improve this answer








edited 3 hours ago









Blue

6,12331354




6,12331354










answered 3 hours ago









Siddhānt SinghSiddhānt Singh

720115




720115












  • $begingroup$
    Thanks. So CNOT is basically an entanglement operator? Are all entanglement operators are two qubits operations ? Or even a regular 4x4 matrix is considered a two qubits operation?
    $endgroup$
    – bilanush
    3 hours ago










  • $begingroup$
    And please, in my book it proves that all NxN matrices can be decomposed to multiple of two level matrices so... So is a CNOT gate? Or is a CNOT gate an exception?
    $endgroup$
    – bilanush
    3 hours ago










  • $begingroup$
    By that, I meant it is a fundamental operation to give rise to an entangled state. If you talk about the entanglement between two qubits only, then yes. But any highly entangled state with any $n$ qubits can be decomposed with CNOTs in a complicated manner. A $4times 4$ is a valid two-qubit operation but it does not necessarily mean it will be non-local.
    $endgroup$
    – Siddhānt Singh
    3 hours ago










  • $begingroup$
    Please read the answer, it is already made clear that a CNOT cannot be decomposed further, more essentially if it is written in the basis of the states being operated.
    $endgroup$
    – Siddhānt Singh
    3 hours ago










  • $begingroup$
    So if I understand you properly. Then each operation acting on two qubits is called two qubits operation even if it's a tensor product of two single qubits matrcies. But among them, only matrices which can't be composed by single qubit matrices are defined as performing entanglement. As to your second comment, the book quantum computation chapter 4 proves that any NxN can be decomposed to a multiple of 2 level matrices he even proves it, so why do you say a control-not gate cannot? Isn't the proof a general proof regarding any N×N matrix?
    $endgroup$
    – bilanush
    2 hours ago


















  • $begingroup$
    Thanks. So CNOT is basically an entanglement operator? Are all entanglement operators are two qubits operations ? Or even a regular 4x4 matrix is considered a two qubits operation?
    $endgroup$
    – bilanush
    3 hours ago










  • $begingroup$
    And please, in my book it proves that all NxN matrices can be decomposed to multiple of two level matrices so... So is a CNOT gate? Or is a CNOT gate an exception?
    $endgroup$
    – bilanush
    3 hours ago










  • $begingroup$
    By that, I meant it is a fundamental operation to give rise to an entangled state. If you talk about the entanglement between two qubits only, then yes. But any highly entangled state with any $n$ qubits can be decomposed with CNOTs in a complicated manner. A $4times 4$ is a valid two-qubit operation but it does not necessarily mean it will be non-local.
    $endgroup$
    – Siddhānt Singh
    3 hours ago










  • $begingroup$
    Please read the answer, it is already made clear that a CNOT cannot be decomposed further, more essentially if it is written in the basis of the states being operated.
    $endgroup$
    – Siddhānt Singh
    3 hours ago










  • $begingroup$
    So if I understand you properly. Then each operation acting on two qubits is called two qubits operation even if it's a tensor product of two single qubits matrcies. But among them, only matrices which can't be composed by single qubit matrices are defined as performing entanglement. As to your second comment, the book quantum computation chapter 4 proves that any NxN can be decomposed to a multiple of 2 level matrices he even proves it, so why do you say a control-not gate cannot? Isn't the proof a general proof regarding any N×N matrix?
    $endgroup$
    – bilanush
    2 hours ago
















$begingroup$
Thanks. So CNOT is basically an entanglement operator? Are all entanglement operators are two qubits operations ? Or even a regular 4x4 matrix is considered a two qubits operation?
$endgroup$
– bilanush
3 hours ago




$begingroup$
Thanks. So CNOT is basically an entanglement operator? Are all entanglement operators are two qubits operations ? Or even a regular 4x4 matrix is considered a two qubits operation?
$endgroup$
– bilanush
3 hours ago












$begingroup$
And please, in my book it proves that all NxN matrices can be decomposed to multiple of two level matrices so... So is a CNOT gate? Or is a CNOT gate an exception?
$endgroup$
– bilanush
3 hours ago




$begingroup$
And please, in my book it proves that all NxN matrices can be decomposed to multiple of two level matrices so... So is a CNOT gate? Or is a CNOT gate an exception?
$endgroup$
– bilanush
3 hours ago












$begingroup$
By that, I meant it is a fundamental operation to give rise to an entangled state. If you talk about the entanglement between two qubits only, then yes. But any highly entangled state with any $n$ qubits can be decomposed with CNOTs in a complicated manner. A $4times 4$ is a valid two-qubit operation but it does not necessarily mean it will be non-local.
$endgroup$
– Siddhānt Singh
3 hours ago




$begingroup$
By that, I meant it is a fundamental operation to give rise to an entangled state. If you talk about the entanglement between two qubits only, then yes. But any highly entangled state with any $n$ qubits can be decomposed with CNOTs in a complicated manner. A $4times 4$ is a valid two-qubit operation but it does not necessarily mean it will be non-local.
$endgroup$
– Siddhānt Singh
3 hours ago












$begingroup$
Please read the answer, it is already made clear that a CNOT cannot be decomposed further, more essentially if it is written in the basis of the states being operated.
$endgroup$
– Siddhānt Singh
3 hours ago




$begingroup$
Please read the answer, it is already made clear that a CNOT cannot be decomposed further, more essentially if it is written in the basis of the states being operated.
$endgroup$
– Siddhānt Singh
3 hours ago












$begingroup$
So if I understand you properly. Then each operation acting on two qubits is called two qubits operation even if it's a tensor product of two single qubits matrcies. But among them, only matrices which can't be composed by single qubit matrices are defined as performing entanglement. As to your second comment, the book quantum computation chapter 4 proves that any NxN can be decomposed to a multiple of 2 level matrices he even proves it, so why do you say a control-not gate cannot? Isn't the proof a general proof regarding any N×N matrix?
$endgroup$
– bilanush
2 hours ago




$begingroup$
So if I understand you properly. Then each operation acting on two qubits is called two qubits operation even if it's a tensor product of two single qubits matrcies. But among them, only matrices which can't be composed by single qubit matrices are defined as performing entanglement. As to your second comment, the book quantum computation chapter 4 proves that any NxN can be decomposed to a multiple of 2 level matrices he even proves it, so why do you say a control-not gate cannot? Isn't the proof a general proof regarding any N×N matrix?
$endgroup$
– bilanush
2 hours ago










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