Writing the notation when gates act on non successive registers
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Suppose I have registers $|arangle^{l}|brangle^{l} |crangle^{l}$ and want an adder mod $l$ gate between the $a$ and $c$ registers. Let $R$ be the adder mod $l$ gate. So is this the correct notation for an operator $U$ that implements this $$ U=Rotimes I_b^{otimes l}.$$ But how do I convey that $R$ is between $a$ and $c$ and $I$ is for the register $b$?
quantum-gate quantum-state notation tensor-product
edited May 28 at 14:10
Sanchayan Dutta
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asked May 28 at 10:11
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Suppose I have registers $|arangle^{l}|brangle^{l} |crangle^{l}$ and want an adder mod $l$ gate between the $a$ and $c$ registers. Let $R$ be the adder mod $l$ gate. So is this the correct notation for an operator $U$ that implements this $$ U=Rotimes I_b^{otimes l}.$$ But how do I convey that $R$ is between $a$ and $c$ and $I$ is for the register $b$?
quantum-gate quantum-state notation tensor-product
edited May 28 at 14:10
Sanchayan Dutta
8,2694 gold badges18 silver badges64 bronze badges
asked May 28 at 10:11
UpstartUpstart
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Suppose I have registers $|arangle^{l}|brangle^{l} |crangle^{l}$ and want an adder mod $l$ gate between the $a$ and $c$ registers. Let $R$ be the adder mod $l$ gate. So is this the correct notation for an operator $U$ that implements this $$ U=Rotimes I_b^{otimes l}.$$ But how do I convey that $R$ is between $a$ and $c$ and $I$ is for the register $b$?
quantum-gate quantum-state notation tensor-product
edited May 28 at 14:10
Sanchayan Dutta
8,2694 gold badges18 silver badges64 bronze badges
asked May 28 at 10:11
UpstartUpstart
4171 silver badge9 bronze badges
$endgroup$
Suppose I have registers $|arangle^{l}|brangle^{l} |crangle^{l}$ and want an adder mod $l$ gate between the $a$ and $c$ registers. Let $R$ be the adder mod $l$ gate. So is this the correct notation for an operator $U$ that implements this $$ U=Rotimes I_b^{otimes l}.$$ But how do I convey that $R$ is between $a$ and $c$ and $I$ is for the register $b$?
quantum-gate quantum-state notation tensor-product
quantum-gate quantum-state notation tensor-product
edited May 28 at 14:10
Sanchayan Dutta
8,2694 gold badges18 silver badges64 bronze badges
asked May 28 at 10:11
UpstartUpstart
4171 silver badge9 bronze badges
edited May 28 at 14:10
Sanchayan Dutta
8,2694 gold badges18 silver badges64 bronze badges
asked May 28 at 10:11
UpstartUpstart
4171 silver badge9 bronze badges
edited May 28 at 14:10
Sanchayan Dutta
8,2694 gold badges18 silver badges64 bronze badges
edited May 28 at 14:10
Sanchayan Dutta
8,2694 gold badges18 silver badges64 bronze badges
edited May 28 at 14:10
Sanchayan Dutta
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8,2694 gold badges18 silver badges64 bronze badges
asked May 28 at 10:11
UpstartUpstart
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asked May 28 at 10:11
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asked May 28 at 10:11
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Personally, I would just define $R_{ac}$ to be the unitary that acts $R$ between registers $a$ and $c$, and acts as identity everywhere else.
answered May 28 at 15:52
DaftWullieDaftWullie
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As always with notation there is not a "correct" way of doing things: it's just arbitrary conventions.
The most readable notation I see for your example involves separating the unitary $R$ into 2 virtual unitary matrices:
$R_a$ the portion that acts on $vert a rangle^l$.
$R_c$ the portion that acts on $vert c rangle^l$.
and "defining" $R$ as
$$
R = R_a otimes R_c.
$$
I called the matrices $R_a$ and $R_c$ "virtual unitaries because it is likely that they do not exist: the decomposition $R = R_a otimes R_c$ will probably be impossible to compute because the matrix $R$ cannot be split as 2 separate transformations on $vert a rangle^l$ and $vert c rangle^l$.
Warning with this kind of non-standard notation: as the matrices involved are not really matrices (they are introduced just for the notation and might not exist), it may add more complexity/confusion than it helps.
In the end, your operation on $vert a rangle^lvert b rangle^lvert c rangle^l$ may be written as
$$
U = R_a otimes I otimes R_c.
$$
answered May 28 at 11:48
NelimeeNelimee
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2 Answers
2
active
oldest
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2 Answers
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$begingroup$
Personally, I would just define $R_{ac}$ to be the unitary that acts $R$ between registers $a$ and $c$, and acts as identity everywhere else.
answered May 28 at 15:52
DaftWullieDaftWullie
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$begingroup$
Personally, I would just define $R_{ac}$ to be the unitary that acts $R$ between registers $a$ and $c$, and acts as identity everywhere else.
answered May 28 at 15:52
DaftWullieDaftWullie
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$endgroup$
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$begingroup$
Personally, I would just define $R_{ac}$ to be the unitary that acts $R$ between registers $a$ and $c$, and acts as identity everywhere else.
answered May 28 at 15:52
DaftWullieDaftWullie
21k1 gold badge9 silver badges53 bronze badges
$endgroup$
Personally, I would just define $R_{ac}$ to be the unitary that acts $R$ between registers $a$ and $c$, and acts as identity everywhere else.
answered May 28 at 15:52
DaftWullieDaftWullie
21k1 gold badge9 silver badges53 bronze badges
answered May 28 at 15:52
DaftWullieDaftWullie
21k1 gold badge9 silver badges53 bronze badges
answered May 28 at 15:52
DaftWullieDaftWullie
21k1 gold badge9 silver badges53 bronze badges
answered May 28 at 15:52
DaftWullieDaftWullie
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As always with notation there is not a "correct" way of doing things: it's just arbitrary conventions.
The most readable notation I see for your example involves separating the unitary $R$ into 2 virtual unitary matrices:
$R_a$ the portion that acts on $vert a rangle^l$.
$R_c$ the portion that acts on $vert c rangle^l$.
and "defining" $R$ as
$$
R = R_a otimes R_c.
$$
I called the matrices $R_a$ and $R_c$ "virtual unitaries because it is likely that they do not exist: the decomposition $R = R_a otimes R_c$ will probably be impossible to compute because the matrix $R$ cannot be split as 2 separate transformations on $vert a rangle^l$ and $vert c rangle^l$.
Warning with this kind of non-standard notation: as the matrices involved are not really matrices (they are introduced just for the notation and might not exist), it may add more complexity/confusion than it helps.
In the end, your operation on $vert a rangle^lvert b rangle^lvert c rangle^l$ may be written as
$$
U = R_a otimes I otimes R_c.
$$
answered May 28 at 11:48
NelimeeNelimee
2,0995 silver badges35 bronze badges
$endgroup$
add a comment
|
$begingroup$
As always with notation there is not a "correct" way of doing things: it's just arbitrary conventions.
The most readable notation I see for your example involves separating the unitary $R$ into 2 virtual unitary matrices:
$R_a$ the portion that acts on $vert a rangle^l$.
$R_c$ the portion that acts on $vert c rangle^l$.
and "defining" $R$ as
$$
R = R_a otimes R_c.
$$
I called the matrices $R_a$ and $R_c$ "virtual unitaries because it is likely that they do not exist: the decomposition $R = R_a otimes R_c$ will probably be impossible to compute because the matrix $R$ cannot be split as 2 separate transformations on $vert a rangle^l$ and $vert c rangle^l$.
Warning with this kind of non-standard notation: as the matrices involved are not really matrices (they are introduced just for the notation and might not exist), it may add more complexity/confusion than it helps.
In the end, your operation on $vert a rangle^lvert b rangle^lvert c rangle^l$ may be written as
$$
U = R_a otimes I otimes R_c.
$$
answered May 28 at 11:48
NelimeeNelimee
2,0995 silver badges35 bronze badges
$endgroup$
add a comment
|
$begingroup$
As always with notation there is not a "correct" way of doing things: it's just arbitrary conventions.
The most readable notation I see for your example involves separating the unitary $R$ into 2 virtual unitary matrices:
$R_a$ the portion that acts on $vert a rangle^l$.
$R_c$ the portion that acts on $vert c rangle^l$.
and "defining" $R$ as
$$
R = R_a otimes R_c.
$$
I called the matrices $R_a$ and $R_c$ "virtual unitaries because it is likely that they do not exist: the decomposition $R = R_a otimes R_c$ will probably be impossible to compute because the matrix $R$ cannot be split as 2 separate transformations on $vert a rangle^l$ and $vert c rangle^l$.
Warning with this kind of non-standard notation: as the matrices involved are not really matrices (they are introduced just for the notation and might not exist), it may add more complexity/confusion than it helps.
In the end, your operation on $vert a rangle^lvert b rangle^lvert c rangle^l$ may be written as
$$
U = R_a otimes I otimes R_c.
$$
answered May 28 at 11:48
NelimeeNelimee
2,0995 silver badges35 bronze badges
$endgroup$
As always with notation there is not a "correct" way of doing things: it's just arbitrary conventions.
The most readable notation I see for your example involves separating the unitary $R$ into 2 virtual unitary matrices:
$R_a$ the portion that acts on $vert a rangle^l$.
$R_c$ the portion that acts on $vert c rangle^l$.
and "defining" $R$ as
$$
R = R_a otimes R_c.
$$
I called the matrices $R_a$ and $R_c$ "virtual unitaries because it is likely that they do not exist: the decomposition $R = R_a otimes R_c$ will probably be impossible to compute because the matrix $R$ cannot be split as 2 separate transformations on $vert a rangle^l$ and $vert c rangle^l$.
Warning with this kind of non-standard notation: as the matrices involved are not really matrices (they are introduced just for the notation and might not exist), it may add more complexity/confusion than it helps.
In the end, your operation on $vert a rangle^lvert b rangle^lvert c rangle^l$ may be written as
$$
U = R_a otimes I otimes R_c.
$$
answered May 28 at 11:48
NelimeeNelimee
2,0995 silver badges35 bronze badges
edited May 28 at 17:36
answered May 28 at 11:48
NelimeeNelimee
2,0995 silver badges35 bronze badges
answered May 28 at 11:48
NelimeeNelimee
2,0995 silver badges35 bronze badges
answered May 28 at 11:48
NelimeeNelimee
2,0995 silver badges35 bronze badges
2,0995 silver badges35 bronze badges
add a comment
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