Mixing
Decomposing the “sign flip†matrix in terms of Pauli matrices
Clash Royale CLAN TAG#URR8PPP
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3
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Can the matrix:
$A=left[beginmatrix0 & -1 \ -1 & 0endmatrixright]$
be somehow expressed as a product of the $3$ standard Pauli matrices?
I'm being able to diagonalize $A$, but not sure if it can be expressed in terms of Pauli-X, Y, Z.
linear-algebra matrices quantum-computation
asked Jul 27 at 23:20
Blue
15617
add a comment |Â
up vote
3
down vote
favorite
Can the matrix:
$A=left[beginmatrix0 & -1 \ -1 & 0endmatrixright]$
be somehow expressed as a product of the $3$ standard Pauli matrices?
I'm being able to diagonalize $A$, but not sure if it can be expressed in terms of Pauli-X, Y, Z.
linear-algebra matrices quantum-computation
asked Jul 27 at 23:20
Blue
15617
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Can the matrix:
$A=left[beginmatrix0 & -1 \ -1 & 0endmatrixright]$
be somehow expressed as a product of the $3$ standard Pauli matrices?
I'm being able to diagonalize $A$, but not sure if it can be expressed in terms of Pauli-X, Y, Z.
linear-algebra matrices quantum-computation
asked Jul 27 at 23:20
Blue
15617
Can the matrix:
$A=left[beginmatrix0 & -1 \ -1 & 0endmatrixright]$
be somehow expressed as a product of the $3$ standard Pauli matrices?
I'm being able to diagonalize $A$, but not sure if it can be expressed in terms of Pauli-X, Y, Z.
linear-algebra matrices quantum-computation
asked Jul 27 at 23:20
Blue
15617
asked Jul 27 at 23:20
Blue
15617
asked Jul 27 at 23:20
Blue
15617
asked Jul 27 at 23:20
Blue
15617
15617
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
I start by knowing that the flip operator is $F=-sigma_1$.
Then to express as a product of all the sigma matrices we just find a combination which reduces to the negative identity.
The relation
$$sigma_1^2 = sigma_2^2=sigma_3^2=-isigma_1sigma_2sigma_3 = 1$$
lets us construct $(sigma_1sigma_2sigma_3)^2 = -1$ to get
$$F=-sigma_1(sigma_1sigma_2sigma_3)^2$$
$$requirecancel F=-cancelsigma_1sigma_1sigma_2sigma_3sigma_1sigma_2sigma_3$$
We can also recover the solution provided by @euler-is-alive because $sigma_1sigma_2sigma_3=i$ is just a number and so we can freely rearrange as $sigma_1sigma_2sigma_3sigma_2sigma_3$
answered Jul 28 at 1:24
Alexander McFarlane
1,018518
More of a comment but I also thought it's interesting that $F = -sigma_1 = isigma_2sigma_3$
– Alexander McFarlane
Jul 28 at 1:27
add a comment |Â
up vote
3
down vote
$beginbmatrix 0 & -1\ -1 & 0endbmatrix = beginbmatrix 0 & 1\ 1 & 0endbmatrix beginbmatrix 0 & -i\ i & 0endbmatrix beginbmatrix 1 & 0\ 0 & -1endbmatrix beginbmatrix 0 & -i\ i & 0endbmatrix beginbmatrix 1 & 0\ 0 & -1endbmatrix $
answered Jul 27 at 23:40
Euler....IS_ALIVE
2,51511036
Nice! How exactly did you come up with the decomposition?
– Blue
Jul 27 at 23:42
1
You're probably not going to like my answer. I had never heard of the Pauli Matrices before, so I looked them up on Wikipedia, and played around with them for about 5 minutes finding out some properties. There may very well in fact be other decompositions that are more useful for what you are looking for.
– Euler....IS_ALIVE
Jul 27 at 23:53
(Note that since the matrices are involutory, the last 4 terms are simply the multiplicative commutator).
– Euler....IS_ALIVE
Jul 28 at 0:04
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I start by knowing that the flip operator is $F=-sigma_1$.
Then to express as a product of all the sigma matrices we just find a combination which reduces to the negative identity.
The relation
$$sigma_1^2 = sigma_2^2=sigma_3^2=-isigma_1sigma_2sigma_3 = 1$$
lets us construct $(sigma_1sigma_2sigma_3)^2 = -1$ to get
$$F=-sigma_1(sigma_1sigma_2sigma_3)^2$$
$$requirecancel F=-cancelsigma_1sigma_1sigma_2sigma_3sigma_1sigma_2sigma_3$$
We can also recover the solution provided by @euler-is-alive because $sigma_1sigma_2sigma_3=i$ is just a number and so we can freely rearrange as $sigma_1sigma_2sigma_3sigma_2sigma_3$
answered Jul 28 at 1:24
Alexander McFarlane
1,018518
More of a comment but I also thought it's interesting that $F = -sigma_1 = isigma_2sigma_3$
– Alexander McFarlane
Jul 28 at 1:27
add a comment |Â
up vote
2
down vote
accepted
I start by knowing that the flip operator is $F=-sigma_1$.
Then to express as a product of all the sigma matrices we just find a combination which reduces to the negative identity.
The relation
$$sigma_1^2 = sigma_2^2=sigma_3^2=-isigma_1sigma_2sigma_3 = 1$$
lets us construct $(sigma_1sigma_2sigma_3)^2 = -1$ to get
$$F=-sigma_1(sigma_1sigma_2sigma_3)^2$$
$$requirecancel F=-cancelsigma_1sigma_1sigma_2sigma_3sigma_1sigma_2sigma_3$$
We can also recover the solution provided by @euler-is-alive because $sigma_1sigma_2sigma_3=i$ is just a number and so we can freely rearrange as $sigma_1sigma_2sigma_3sigma_2sigma_3$
answered Jul 28 at 1:24
Alexander McFarlane
1,018518
More of a comment but I also thought it's interesting that $F = -sigma_1 = isigma_2sigma_3$
– Alexander McFarlane
Jul 28 at 1:27
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I start by knowing that the flip operator is $F=-sigma_1$.
Then to express as a product of all the sigma matrices we just find a combination which reduces to the negative identity.
The relation
$$sigma_1^2 = sigma_2^2=sigma_3^2=-isigma_1sigma_2sigma_3 = 1$$
lets us construct $(sigma_1sigma_2sigma_3)^2 = -1$ to get
$$F=-sigma_1(sigma_1sigma_2sigma_3)^2$$
$$requirecancel F=-cancelsigma_1sigma_1sigma_2sigma_3sigma_1sigma_2sigma_3$$
We can also recover the solution provided by @euler-is-alive because $sigma_1sigma_2sigma_3=i$ is just a number and so we can freely rearrange as $sigma_1sigma_2sigma_3sigma_2sigma_3$
answered Jul 28 at 1:24
Alexander McFarlane
1,018518
I start by knowing that the flip operator is $F=-sigma_1$.
Then to express as a product of all the sigma matrices we just find a combination which reduces to the negative identity.
The relation
$$sigma_1^2 = sigma_2^2=sigma_3^2=-isigma_1sigma_2sigma_3 = 1$$
lets us construct $(sigma_1sigma_2sigma_3)^2 = -1$ to get
$$F=-sigma_1(sigma_1sigma_2sigma_3)^2$$
$$requirecancel F=-cancelsigma_1sigma_1sigma_2sigma_3sigma_1sigma_2sigma_3$$
We can also recover the solution provided by @euler-is-alive because $sigma_1sigma_2sigma_3=i$ is just a number and so we can freely rearrange as $sigma_1sigma_2sigma_3sigma_2sigma_3$
answered Jul 28 at 1:24
Alexander McFarlane
1,018518
answered Jul 28 at 1:24
Alexander McFarlane
1,018518
answered Jul 28 at 1:24
Alexander McFarlane
1,018518
answered Jul 28 at 1:24
Alexander McFarlane
1,018518
1,018518
More of a comment but I also thought it's interesting that $F = -sigma_1 = isigma_2sigma_3$
– Alexander McFarlane
Jul 28 at 1:27
add a comment |Â
More of a comment but I also thought it's interesting that $F = -sigma_1 = isigma_2sigma_3$
– Alexander McFarlane
Jul 28 at 1:27
More of a comment but I also thought it's interesting that $F = -sigma_1 = isigma_2sigma_3$
– Alexander McFarlane
Jul 28 at 1:27
More of a comment but I also thought it's interesting that $F = -sigma_1 = isigma_2sigma_3$
– Alexander McFarlane
Jul 28 at 1:27
add a comment |Â
up vote
3
down vote
$beginbmatrix 0 & -1\ -1 & 0endbmatrix = beginbmatrix 0 & 1\ 1 & 0endbmatrix beginbmatrix 0 & -i\ i & 0endbmatrix beginbmatrix 1 & 0\ 0 & -1endbmatrix beginbmatrix 0 & -i\ i & 0endbmatrix beginbmatrix 1 & 0\ 0 & -1endbmatrix $
answered Jul 27 at 23:40
Euler....IS_ALIVE
2,51511036
Nice! How exactly did you come up with the decomposition?
– Blue
Jul 27 at 23:42
1
You're probably not going to like my answer. I had never heard of the Pauli Matrices before, so I looked them up on Wikipedia, and played around with them for about 5 minutes finding out some properties. There may very well in fact be other decompositions that are more useful for what you are looking for.
– Euler....IS_ALIVE
Jul 27 at 23:53
(Note that since the matrices are involutory, the last 4 terms are simply the multiplicative commutator).
– Euler....IS_ALIVE
Jul 28 at 0:04
add a comment |Â
up vote
3
down vote
$beginbmatrix 0 & -1\ -1 & 0endbmatrix = beginbmatrix 0 & 1\ 1 & 0endbmatrix beginbmatrix 0 & -i\ i & 0endbmatrix beginbmatrix 1 & 0\ 0 & -1endbmatrix beginbmatrix 0 & -i\ i & 0endbmatrix beginbmatrix 1 & 0\ 0 & -1endbmatrix $
answered Jul 27 at 23:40
Euler....IS_ALIVE
2,51511036
Nice! How exactly did you come up with the decomposition?
– Blue
Jul 27 at 23:42
1
You're probably not going to like my answer. I had never heard of the Pauli Matrices before, so I looked them up on Wikipedia, and played around with them for about 5 minutes finding out some properties. There may very well in fact be other decompositions that are more useful for what you are looking for.
– Euler....IS_ALIVE
Jul 27 at 23:53
(Note that since the matrices are involutory, the last 4 terms are simply the multiplicative commutator).
– Euler....IS_ALIVE
Jul 28 at 0:04
add a comment |Â
up vote
3
down vote
up vote
3
down vote
$beginbmatrix 0 & -1\ -1 & 0endbmatrix = beginbmatrix 0 & 1\ 1 & 0endbmatrix beginbmatrix 0 & -i\ i & 0endbmatrix beginbmatrix 1 & 0\ 0 & -1endbmatrix beginbmatrix 0 & -i\ i & 0endbmatrix beginbmatrix 1 & 0\ 0 & -1endbmatrix $
answered Jul 27 at 23:40
Euler....IS_ALIVE
2,51511036
$beginbmatrix 0 & -1\ -1 & 0endbmatrix = beginbmatrix 0 & 1\ 1 & 0endbmatrix beginbmatrix 0 & -i\ i & 0endbmatrix beginbmatrix 1 & 0\ 0 & -1endbmatrix beginbmatrix 0 & -i\ i & 0endbmatrix beginbmatrix 1 & 0\ 0 & -1endbmatrix $
answered Jul 27 at 23:40
Euler....IS_ALIVE
2,51511036
answered Jul 27 at 23:40
Euler....IS_ALIVE
2,51511036
answered Jul 27 at 23:40
Euler....IS_ALIVE
2,51511036
answered Jul 27 at 23:40
Euler....IS_ALIVE
2,51511036
2,51511036
Nice! How exactly did you come up with the decomposition?
– Blue
Jul 27 at 23:42
1
You're probably not going to like my answer. I had never heard of the Pauli Matrices before, so I looked them up on Wikipedia, and played around with them for about 5 minutes finding out some properties. There may very well in fact be other decompositions that are more useful for what you are looking for.
– Euler....IS_ALIVE
Jul 27 at 23:53
(Note that since the matrices are involutory, the last 4 terms are simply the multiplicative commutator).
– Euler....IS_ALIVE
Jul 28 at 0:04
add a comment |Â
Nice! How exactly did you come up with the decomposition?
– Blue
Jul 27 at 23:42
1
You're probably not going to like my answer. I had never heard of the Pauli Matrices before, so I looked them up on Wikipedia, and played around with them for about 5 minutes finding out some properties. There may very well in fact be other decompositions that are more useful for what you are looking for.
– Euler....IS_ALIVE
Jul 27 at 23:53
(Note that since the matrices are involutory, the last 4 terms are simply the multiplicative commutator).
– Euler....IS_ALIVE
Jul 28 at 0:04
Nice! How exactly did you come up with the decomposition?
– Blue
Jul 27 at 23:42
Nice! How exactly did you come up with the decomposition?
– Blue
Jul 27 at 23:42
1
1
You're probably not going to like my answer. I had never heard of the Pauli Matrices before, so I looked them up on Wikipedia, and played around with them for about 5 minutes finding out some properties. There may very well in fact be other decompositions that are more useful for what you are looking for.
– Euler....IS_ALIVE
Jul 27 at 23:53
You're probably not going to like my answer. I had never heard of the Pauli Matrices before, so I looked them up on Wikipedia, and played around with them for about 5 minutes finding out some properties. There may very well in fact be other decompositions that are more useful for what you are looking for.
– Euler....IS_ALIVE
Jul 27 at 23:53
(Note that since the matrices are involutory, the last 4 terms are simply the multiplicative commutator).
– Euler....IS_ALIVE
Jul 28 at 0:04
(Note that since the matrices are involutory, the last 4 terms are simply the multiplicative commutator).
– Euler....IS_ALIVE
Jul 28 at 0:04
add a comment |Â
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