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Is a ccH, ccX and ccH equivalent to a cH, ccX and cH sequence?
If all quantum gates must be unitary, what about measurement?Processing density capabilities in a quantum processorWhat is the quantum circuit equivalent of a (delayed choice) quantum eraser?Why are quantum gates unitary and not special unitary?Is the Kraus representation of a quantum channel equivalent to a unitary evolution in an enlarged space?Explicit Conversion Between Universal Gate SetsWhat quantum gate is XNOR equivalent to?How are multi-qubit gates extended into larger registers?How to decompose a controlled unitary $C(U)$ operation where $U$ is a 2-qubit gate?Bug in IBM backend?
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It appears to me that a ccH, ccX and ccH sequence is exactly equivalent to a cH, ccX and cH gate sequence. Is there any quick way to see/verify this?
quantum-gate
asked Jun 30 at 7:45
Sanchayan DuttaSanchayan Dutta
7,9724 gold badges16 silver badges62 bronze badges
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add a comment |
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It appears to me that a ccH, ccX and ccH sequence is exactly equivalent to a cH, ccX and cH gate sequence. Is there any quick way to see/verify this?
quantum-gate
asked Jun 30 at 7:45
Sanchayan DuttaSanchayan Dutta
7,9724 gold badges16 silver badges62 bronze badges
$endgroup$
add a comment |
$begingroup$
It appears to me that a ccH, ccX and ccH sequence is exactly equivalent to a cH, ccX and cH gate sequence. Is there any quick way to see/verify this?
quantum-gate
asked Jun 30 at 7:45
Sanchayan DuttaSanchayan Dutta
7,9724 gold badges16 silver badges62 bronze badges
$endgroup$
It appears to me that a ccH, ccX and ccH sequence is exactly equivalent to a cH, ccX and cH gate sequence. Is there any quick way to see/verify this?
quantum-gate
quantum-gate
asked Jun 30 at 7:45
Sanchayan DuttaSanchayan Dutta
7,9724 gold badges16 silver badges62 bronze badges
asked Jun 30 at 7:45
Sanchayan DuttaSanchayan Dutta
7,9724 gold badges16 silver badges62 bronze badges
asked Jun 30 at 7:45
Sanchayan DuttaSanchayan Dutta
7,9724 gold badges16 silver badges62 bronze badges
asked Jun 30 at 7:45
Sanchayan DuttaSanchayan Dutta
7,9724 gold badges16 silver badges62 bronze badges
asked Jun 30 at 7:45
Sanchayan DuttaSanchayan Dutta
7,9724 gold badges16 silver badges62 bronze badges
7,9724 gold badges16 silver badges62 bronze badges
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2 Answers
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No control equals each control $forall U : U = C(U) cdot barC(U)$
Opposite controls commute $forall U, V : [C(U), barC(V)] = 0$
No control equals each control $forall U : U = C(U) cdot barC(U)$
Self-inverse operations self-cancel
Done
More generally, for any "V conjugated by U" operation of the form $U_a cdot C_b(V_a) cdot U_a^-1$, the $U$ operation can gain or lose any controls that $V$ has without changing the circuit's effect.
answered Jun 30 at 11:42
Craig GidneyCraig Gidney
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There is a way to simplify it down slightly - for some controlled unitary $Cleft(Uright)=Ioplus U$ and some arbitrary unitary $V$ (of the same dimension as $U$), $$left(Iotimes Vright).Cleft(Uright).left(Iotimes V^daggerright) = beginpmatrixV&0\0&VendpmatrixbeginpmatrixI&0\0&UendpmatrixbeginpmatrixV^dagger&0\0&V^daggerendpmatrix = Cleft(VUV^daggerright),$$ so it's now clear that $left[Iotimes Cleft(Hright)right].Cleft(Cleft(Xright)right).left[Iotimes Cleft(Hright)right] = Cleft(Cleft(Hright).Cleft(Xright).Cleft(Hright)right)$.
Funnily enough, if instead of using $Iotimes V$ in the above and instead used $Cleft(Vright)$, we'd find that $$Cleft(Vright).Cleft(Uright).Cleft(V^daggerright) = beginpmatrixI&0\0&VendpmatrixbeginpmatrixI&0\0&UendpmatrixbeginpmatrixI&0\0&V^daggerendpmatrix = Cleft(VUV^daggerright)$$ and so it's not just clear that beginalign*Cleft(Cleft(Hright)right).Cleft(Cleft(Xright)right).Cleft(Cleft(Hright)right) &= Cleft(Cleft(Hright).Cleft(Xright).Cleft(Hright)right) \ &=left[Iotimes Cleft(Hright)right].Cleft(Cleft(Xright)right).left[Iotimes Cleft(Hright)right]endalign* but the more general version $$left(Iotimes Vright).Cleft(Uright).left(Iotimes V^daggerright) = Cleft(Vright).Cleft(Uright).Cleft(V^daggerright)$$ is also true, for any unitaries $U$ and $V$
My use of brackets in the control operation is maybe slightly more unusual, but hopefully it makes it more obvious what the operation is
answered Jun 30 at 11:31
Mithrandir24601♦Mithrandir24601
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2 Answers
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2 Answers
2
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active
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$begingroup$
Start
No control equals each control $forall U : U = C(U) cdot barC(U)$
Opposite controls commute $forall U, V : [C(U), barC(V)] = 0$
No control equals each control $forall U : U = C(U) cdot barC(U)$
Self-inverse operations self-cancel
Done
More generally, for any "V conjugated by U" operation of the form $U_a cdot C_b(V_a) cdot U_a^-1$, the $U$ operation can gain or lose any controls that $V$ has without changing the circuit's effect.
answered Jun 30 at 11:42
Craig GidneyCraig Gidney
5,2783 silver badges23 bronze badges
$endgroup$
add a comment |
$begingroup$
Start
No control equals each control $forall U : U = C(U) cdot barC(U)$
Opposite controls commute $forall U, V : [C(U), barC(V)] = 0$
No control equals each control $forall U : U = C(U) cdot barC(U)$
Self-inverse operations self-cancel
Done
More generally, for any "V conjugated by U" operation of the form $U_a cdot C_b(V_a) cdot U_a^-1$, the $U$ operation can gain or lose any controls that $V$ has without changing the circuit's effect.
answered Jun 30 at 11:42
Craig GidneyCraig Gidney
5,2783 silver badges23 bronze badges
$endgroup$
add a comment |
$begingroup$
Start
No control equals each control $forall U : U = C(U) cdot barC(U)$
Opposite controls commute $forall U, V : [C(U), barC(V)] = 0$
No control equals each control $forall U : U = C(U) cdot barC(U)$
Self-inverse operations self-cancel
Done
More generally, for any "V conjugated by U" operation of the form $U_a cdot C_b(V_a) cdot U_a^-1$, the $U$ operation can gain or lose any controls that $V$ has without changing the circuit's effect.
answered Jun 30 at 11:42
Craig GidneyCraig Gidney
5,2783 silver badges23 bronze badges
$endgroup$
Start
No control equals each control $forall U : U = C(U) cdot barC(U)$
Opposite controls commute $forall U, V : [C(U), barC(V)] = 0$
No control equals each control $forall U : U = C(U) cdot barC(U)$
Self-inverse operations self-cancel
Done
More generally, for any "V conjugated by U" operation of the form $U_a cdot C_b(V_a) cdot U_a^-1$, the $U$ operation can gain or lose any controls that $V$ has without changing the circuit's effect.
answered Jun 30 at 11:42
Craig GidneyCraig Gidney
5,2783 silver badges23 bronze badges
edited Jun 30 at 11:47
answered Jun 30 at 11:42
Craig GidneyCraig Gidney
5,2783 silver badges23 bronze badges
answered Jun 30 at 11:42
Craig GidneyCraig Gidney
5,2783 silver badges23 bronze badges
answered Jun 30 at 11:42
Craig GidneyCraig Gidney
5,2783 silver badges23 bronze badges
5,2783 silver badges23 bronze badges
add a comment |
add a comment |
$begingroup$
There is a way to simplify it down slightly - for some controlled unitary $Cleft(Uright)=Ioplus U$ and some arbitrary unitary $V$ (of the same dimension as $U$), $$left(Iotimes Vright).Cleft(Uright).left(Iotimes V^daggerright) = beginpmatrixV&0\0&VendpmatrixbeginpmatrixI&0\0&UendpmatrixbeginpmatrixV^dagger&0\0&V^daggerendpmatrix = Cleft(VUV^daggerright),$$ so it's now clear that $left[Iotimes Cleft(Hright)right].Cleft(Cleft(Xright)right).left[Iotimes Cleft(Hright)right] = Cleft(Cleft(Hright).Cleft(Xright).Cleft(Hright)right)$.
Funnily enough, if instead of using $Iotimes V$ in the above and instead used $Cleft(Vright)$, we'd find that $$Cleft(Vright).Cleft(Uright).Cleft(V^daggerright) = beginpmatrixI&0\0&VendpmatrixbeginpmatrixI&0\0&UendpmatrixbeginpmatrixI&0\0&V^daggerendpmatrix = Cleft(VUV^daggerright)$$ and so it's not just clear that beginalign*Cleft(Cleft(Hright)right).Cleft(Cleft(Xright)right).Cleft(Cleft(Hright)right) &= Cleft(Cleft(Hright).Cleft(Xright).Cleft(Hright)right) \ &=left[Iotimes Cleft(Hright)right].Cleft(Cleft(Xright)right).left[Iotimes Cleft(Hright)right]endalign* but the more general version $$left(Iotimes Vright).Cleft(Uright).left(Iotimes V^daggerright) = Cleft(Vright).Cleft(Uright).Cleft(V^daggerright)$$ is also true, for any unitaries $U$ and $V$
My use of brackets in the control operation is maybe slightly more unusual, but hopefully it makes it more obvious what the operation is
answered Jun 30 at 11:31
Mithrandir24601♦Mithrandir24601
2,6762 gold badges9 silver badges35 bronze badges
$endgroup$
add a comment |
$begingroup$
There is a way to simplify it down slightly - for some controlled unitary $Cleft(Uright)=Ioplus U$ and some arbitrary unitary $V$ (of the same dimension as $U$), $$left(Iotimes Vright).Cleft(Uright).left(Iotimes V^daggerright) = beginpmatrixV&0\0&VendpmatrixbeginpmatrixI&0\0&UendpmatrixbeginpmatrixV^dagger&0\0&V^daggerendpmatrix = Cleft(VUV^daggerright),$$ so it's now clear that $left[Iotimes Cleft(Hright)right].Cleft(Cleft(Xright)right).left[Iotimes Cleft(Hright)right] = Cleft(Cleft(Hright).Cleft(Xright).Cleft(Hright)right)$.
Funnily enough, if instead of using $Iotimes V$ in the above and instead used $Cleft(Vright)$, we'd find that $$Cleft(Vright).Cleft(Uright).Cleft(V^daggerright) = beginpmatrixI&0\0&VendpmatrixbeginpmatrixI&0\0&UendpmatrixbeginpmatrixI&0\0&V^daggerendpmatrix = Cleft(VUV^daggerright)$$ and so it's not just clear that beginalign*Cleft(Cleft(Hright)right).Cleft(Cleft(Xright)right).Cleft(Cleft(Hright)right) &= Cleft(Cleft(Hright).Cleft(Xright).Cleft(Hright)right) \ &=left[Iotimes Cleft(Hright)right].Cleft(Cleft(Xright)right).left[Iotimes Cleft(Hright)right]endalign* but the more general version $$left(Iotimes Vright).Cleft(Uright).left(Iotimes V^daggerright) = Cleft(Vright).Cleft(Uright).Cleft(V^daggerright)$$ is also true, for any unitaries $U$ and $V$
My use of brackets in the control operation is maybe slightly more unusual, but hopefully it makes it more obvious what the operation is
answered Jun 30 at 11:31
Mithrandir24601♦Mithrandir24601
2,6762 gold badges9 silver badges35 bronze badges
$endgroup$
add a comment |
$begingroup$
There is a way to simplify it down slightly - for some controlled unitary $Cleft(Uright)=Ioplus U$ and some arbitrary unitary $V$ (of the same dimension as $U$), $$left(Iotimes Vright).Cleft(Uright).left(Iotimes V^daggerright) = beginpmatrixV&0\0&VendpmatrixbeginpmatrixI&0\0&UendpmatrixbeginpmatrixV^dagger&0\0&V^daggerendpmatrix = Cleft(VUV^daggerright),$$ so it's now clear that $left[Iotimes Cleft(Hright)right].Cleft(Cleft(Xright)right).left[Iotimes Cleft(Hright)right] = Cleft(Cleft(Hright).Cleft(Xright).Cleft(Hright)right)$.
Funnily enough, if instead of using $Iotimes V$ in the above and instead used $Cleft(Vright)$, we'd find that $$Cleft(Vright).Cleft(Uright).Cleft(V^daggerright) = beginpmatrixI&0\0&VendpmatrixbeginpmatrixI&0\0&UendpmatrixbeginpmatrixI&0\0&V^daggerendpmatrix = Cleft(VUV^daggerright)$$ and so it's not just clear that beginalign*Cleft(Cleft(Hright)right).Cleft(Cleft(Xright)right).Cleft(Cleft(Hright)right) &= Cleft(Cleft(Hright).Cleft(Xright).Cleft(Hright)right) \ &=left[Iotimes Cleft(Hright)right].Cleft(Cleft(Xright)right).left[Iotimes Cleft(Hright)right]endalign* but the more general version $$left(Iotimes Vright).Cleft(Uright).left(Iotimes V^daggerright) = Cleft(Vright).Cleft(Uright).Cleft(V^daggerright)$$ is also true, for any unitaries $U$ and $V$
My use of brackets in the control operation is maybe slightly more unusual, but hopefully it makes it more obvious what the operation is
answered Jun 30 at 11:31
Mithrandir24601♦Mithrandir24601
2,6762 gold badges9 silver badges35 bronze badges
$endgroup$
There is a way to simplify it down slightly - for some controlled unitary $Cleft(Uright)=Ioplus U$ and some arbitrary unitary $V$ (of the same dimension as $U$), $$left(Iotimes Vright).Cleft(Uright).left(Iotimes V^daggerright) = beginpmatrixV&0\0&VendpmatrixbeginpmatrixI&0\0&UendpmatrixbeginpmatrixV^dagger&0\0&V^daggerendpmatrix = Cleft(VUV^daggerright),$$ so it's now clear that $left[Iotimes Cleft(Hright)right].Cleft(Cleft(Xright)right).left[Iotimes Cleft(Hright)right] = Cleft(Cleft(Hright).Cleft(Xright).Cleft(Hright)right)$.
Funnily enough, if instead of using $Iotimes V$ in the above and instead used $Cleft(Vright)$, we'd find that $$Cleft(Vright).Cleft(Uright).Cleft(V^daggerright) = beginpmatrixI&0\0&VendpmatrixbeginpmatrixI&0\0&UendpmatrixbeginpmatrixI&0\0&V^daggerendpmatrix = Cleft(VUV^daggerright)$$ and so it's not just clear that beginalign*Cleft(Cleft(Hright)right).Cleft(Cleft(Xright)right).Cleft(Cleft(Hright)right) &= Cleft(Cleft(Hright).Cleft(Xright).Cleft(Hright)right) \ &=left[Iotimes Cleft(Hright)right].Cleft(Cleft(Xright)right).left[Iotimes Cleft(Hright)right]endalign* but the more general version $$left(Iotimes Vright).Cleft(Uright).left(Iotimes V^daggerright) = Cleft(Vright).Cleft(Uright).Cleft(V^daggerright)$$ is also true, for any unitaries $U$ and $V$
My use of brackets in the control operation is maybe slightly more unusual, but hopefully it makes it more obvious what the operation is
answered Jun 30 at 11:31
Mithrandir24601♦Mithrandir24601
2,6762 gold badges9 silver badges35 bronze badges
answered Jun 30 at 11:31
Mithrandir24601♦Mithrandir24601
2,6762 gold badges9 silver badges35 bronze badges
answered Jun 30 at 11:31
Mithrandir24601♦Mithrandir24601
2,6762 gold badges9 silver badges35 bronze badges
answered Jun 30 at 11:31
Mithrandir24601♦Mithrandir24601
2,6762 gold badges9 silver badges35 bronze badges
2,6762 gold badges9 silver badges35 bronze badges
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