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Writing the notation when gates act on non successive registers

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3

$begingroup$

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

share|improve this question

edited May 28 at 14:10

Sanchayan Dutta

7,44841660

asked May 28 at 10:11

UpstartUpstart

35019

$endgroup$

add a comment | 

3

$begingroup$

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

share|improve this question

edited May 28 at 14:10

Sanchayan Dutta

7,44841660

asked May 28 at 10:11

UpstartUpstart

35019

$endgroup$

add a comment | 

$begingroup$

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

share|improve this question

edited May 28 at 14:10

Sanchayan Dutta

7,44841660

asked May 28 at 10:11

UpstartUpstart

35019

$endgroup$

share|improve this question

edited May 28 at 14:10

Sanchayan Dutta

7,44841660

asked May 28 at 10:11

UpstartUpstart

35019

share|improve this question

edited May 28 at 14:10

Sanchayan Dutta

7,44841660

asked May 28 at 10:11

UpstartUpstart

35019

edited May 28 at 14:10

Sanchayan Dutta

7,44841660

edited May 28 at 14:10

Sanchayan Dutta

7,44841660

asked May 28 at 10:11

UpstartUpstart

35019

asked May 28 at 10:11

UpstartUpstart

35019

2 Answers
2

active

oldest

votes

3

$begingroup$

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.

share|improve this answer

answered May 28 at 15:52

DaftWullieDaftWullie

17k1644

$endgroup$

add a comment | 

2

$begingroup$

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
$$
U = R_a otimes I otimes R_c.
$$

share|improve this answer

edited May 28 at 17:36

answered May 28 at 11:48

NelimeeNelimee

1,860430

$endgroup$

add a comment | 

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3

$begingroup$

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.

share|improve this answer

answered May 28 at 15:52

DaftWullieDaftWullie

17k1644

$endgroup$

add a comment | 

3

$begingroup$

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.

share|improve this answer

answered May 28 at 15:52

DaftWullieDaftWullie

17k1644

$endgroup$

add a comment | 

$begingroup$

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.

share|improve this answer

answered May 28 at 15:52

DaftWullieDaftWullie

17k1644

$endgroup$

share|improve this answer

answered May 28 at 15:52

DaftWullieDaftWullie

17k1644

answered May 28 at 15:52

DaftWullieDaftWullie

17k1644

answered May 28 at 15:52

DaftWullieDaftWullie

17k1644

2

$begingroup$

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
$$
U = R_a otimes I otimes R_c.
$$

share|improve this answer

edited May 28 at 17:36

answered May 28 at 11:48

NelimeeNelimee

1,860430

$endgroup$

add a comment | 

2

$begingroup$

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
$$
U = R_a otimes I otimes R_c.
$$

share|improve this answer

edited May 28 at 17:36

answered May 28 at 11:48

NelimeeNelimee

1,860430

$endgroup$

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
$$
U = R_a otimes I otimes R_c.
$$

share|improve this answer

edited May 28 at 17:36

answered May 28 at 11:48

NelimeeNelimee

1,860430

$endgroup$

share|improve this answer

edited May 28 at 17:36

answered May 28 at 11:48

NelimeeNelimee

1,860430

answered May 28 at 11:48

NelimeeNelimee

1,860430

answered May 28 at 11:48

NelimeeNelimee

1,860430