POVM three-qubit circuit for symmetric quantum states
$begingroup$
I have been reading this paper but don't yet understand how to implement a circuit to determine in which state the qubit is not for a cyclic POVM. More specifically, I want to implement a cyclic POVM with $m=3$;
Update: I came across having to implement this unitary matrix:
$$
M=
frac{1}{sqrt{2}}left[ {begin{array}{cc}
1 & 1 \
1 & w \
end{array} } right]
$$
Where $w$ is a third root of unity using rotations, after which I am stuck.
quantum-state quantum-information circuit-construction mathematics quantum-operation
asked Mar 4 at 15:09
xbk365xbk365
213
$endgroup$
add a comment |
$begingroup$
I have been reading this paper but don't yet understand how to implement a circuit to determine in which state the qubit is not for a cyclic POVM. More specifically, I want to implement a cyclic POVM with $m=3$;
Update: I came across having to implement this unitary matrix:
$$
M=
frac{1}{sqrt{2}}left[ {begin{array}{cc}
1 & 1 \
1 & w \
end{array} } right]
$$
Where $w$ is a third root of unity using rotations, after which I am stuck.
quantum-state quantum-information circuit-construction mathematics quantum-operation
asked Mar 4 at 15:09
xbk365xbk365
213
$endgroup$
1
$begingroup$
That's not a unitary matrix unless w=-1..
$endgroup$
– Craig Gidney
Mar 6 at 19:49
add a comment |
$begingroup$
I have been reading this paper but don't yet understand how to implement a circuit to determine in which state the qubit is not for a cyclic POVM. More specifically, I want to implement a cyclic POVM with $m=3$;
Update: I came across having to implement this unitary matrix:
$$
M=
frac{1}{sqrt{2}}left[ {begin{array}{cc}
1 & 1 \
1 & w \
end{array} } right]
$$
Where $w$ is a third root of unity using rotations, after which I am stuck.
quantum-state quantum-information circuit-construction mathematics quantum-operation
asked Mar 4 at 15:09
xbk365xbk365
213
$endgroup$
I have been reading this paper but don't yet understand how to implement a circuit to determine in which state the qubit is not for a cyclic POVM. More specifically, I want to implement a cyclic POVM with $m=3$;
Update: I came across having to implement this unitary matrix:
$$
M=
frac{1}{sqrt{2}}left[ {begin{array}{cc}
1 & 1 \
1 & w \
end{array} } right]
$$
Where $w$ is a third root of unity using rotations, after which I am stuck.
quantum-state quantum-information circuit-construction mathematics quantum-operation
quantum-state quantum-information circuit-construction mathematics quantum-operation
asked Mar 4 at 15:09
xbk365xbk365
213
asked Mar 4 at 15:09
xbk365xbk365
213
edited Mar 4 at 15:51
xbk365
asked Mar 4 at 15:09
xbk365xbk365
213
asked Mar 4 at 15:09
xbk365xbk365
213
asked Mar 4 at 15:09
xbk365xbk365
213
213
1
$begingroup$
That's not a unitary matrix unless w=-1..
$endgroup$
– Craig Gidney
Mar 6 at 19:49
add a comment |
1
$begingroup$
That's not a unitary matrix unless w=-1..
$endgroup$
– Craig Gidney
Mar 6 at 19:49
1
1
$begingroup$
That's not a unitary matrix unless w=-1..
$endgroup$
– Craig Gidney
Mar 6 at 19:49
$begingroup$
That's not a unitary matrix unless w=-1..
$endgroup$
– Craig Gidney
Mar 6 at 19:49
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This is not the unitary that you have to implement: you need a two-qubit unitary
$$
frac{1}{sqrt{3}}left(begin{array}{cccc}
1 & 1 & 1 & 0 \
1 & omega & omega^2 & 0 \
1 & omega^2 & omega & 0 \
0 & 0 & 0 & sqrt{3}
end{array}right),
$$
where $omega=e^{2ipi/3}$, the point being that if you introduce an ancilla qubit in the state 0, apply this unitary, and then measure in the computational basis, the 3 measurement outcomes 00, 01 and 10 correspond to the 3 POVM elements.
I don't (yet) have a circuit implementation for this. You'll see the paper you cite carefully avoids talking about the Fourier transform in non-power of 2 dimensions. You certainly could use the standard constructions based on Givens rotations, but the result is going to be fairly horrible.
Here's an attempt at a circuit. I've made a few tweaks since I last ran it through a computer to check, so it's always possible that a slight error has crept in, but broadly...
Here, I'm using $Z^r$ to denote
$$
left(begin{array}{cc} 1 & 0 \ 0 & e^{ipi r} end{array}right),
$$
and
$$
V=frac{1}{sqrt{3}}left(begin{array}{cc}
1 & sqrt{2} \ -sqrt{2} & 1
end{array}right).
$$
answered Mar 4 at 16:00
DaftWullieDaftWullie
14.9k1541
$endgroup$
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
Mar 4 at 16:09
$begingroup$
@chubakueno You said you wanted $m=3$, which means you need $F_3$, which is a $3times 3$ matrix which we must embed into a $4times 4$ matrix if we're using qubits.
$endgroup$
– DaftWullie
Mar 4 at 16:11
$begingroup$
I understand now. So, is there an example of IQFT for non power of twos there? I would be surprised to find there isn't, but what do I know.
$endgroup$
– chubakueno
Mar 4 at 16:26
$begingroup$
@chubakueno I didn't immediately find one. I've now constructed one for the $m=3$ case, but don't have time right now to write it into a circuit.
$endgroup$
– DaftWullie
Mar 4 at 16:48
1
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
Mar 4 at 18:21
|
show 1 more comment
$begingroup$
Check this
$$frac{1}{sqrt{3}}left(begin{array}{ccc}
1 & 1 & 1 \
1 & omega & omega^2 \
1 & omega^2 & omega \
end{array}right) = left(begin{array}{cc}
H & 0 \
0 & 1 \
end{array}right) cdot frac{1}{sqrt{3}}left(begin{array}{ccc}
sqrt{2} & 0 & 1 \
0 & sqrt{3} & 0 \
1 & 0 & -sqrt{2} \
end{array}right) cdot
M_{3}
$$
$$M_{3} = left(begin{array}{ccc}
1 & 0 & 0 \
0 & frac{1}{sqrt{2}}iomega^2 & frac{1}{sqrt{2}}iomega \
0 & -frac{1}{sqrt{2}}omega^2 & -frac{1}{sqrt{2}}omega \
end{array}right)
$$
$$
frac{1}{sqrt{2}}left(begin{array}{ccc}
iomega^2 & iomega \
-omega^2 & -omega \
end{array}right) = X cdot S cdot X cdot Z cdot H cdot left(begin{array}{ccc}
omega^2 & 0 \
0 & omega \
end{array}right) cdot Z
$$
answered Mar 6 at 18:37
Danylo YDanylo Y
1814
New contributor
Danylo Y is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is not the unitary that you have to implement: you need a two-qubit unitary
$$
frac{1}{sqrt{3}}left(begin{array}{cccc}
1 & 1 & 1 & 0 \
1 & omega & omega^2 & 0 \
1 & omega^2 & omega & 0 \
0 & 0 & 0 & sqrt{3}
end{array}right),
$$
where $omega=e^{2ipi/3}$, the point being that if you introduce an ancilla qubit in the state 0, apply this unitary, and then measure in the computational basis, the 3 measurement outcomes 00, 01 and 10 correspond to the 3 POVM elements.
I don't (yet) have a circuit implementation for this. You'll see the paper you cite carefully avoids talking about the Fourier transform in non-power of 2 dimensions. You certainly could use the standard constructions based on Givens rotations, but the result is going to be fairly horrible.
Here's an attempt at a circuit. I've made a few tweaks since I last ran it through a computer to check, so it's always possible that a slight error has crept in, but broadly...
Here, I'm using $Z^r$ to denote
$$
left(begin{array}{cc} 1 & 0 \ 0 & e^{ipi r} end{array}right),
$$
and
$$
V=frac{1}{sqrt{3}}left(begin{array}{cc}
1 & sqrt{2} \ -sqrt{2} & 1
end{array}right).
$$
answered Mar 4 at 16:00
DaftWullieDaftWullie
14.9k1541
$endgroup$
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
Mar 4 at 16:09
$begingroup$
@chubakueno You said you wanted $m=3$, which means you need $F_3$, which is a $3times 3$ matrix which we must embed into a $4times 4$ matrix if we're using qubits.
$endgroup$
– DaftWullie
Mar 4 at 16:11
$begingroup$
I understand now. So, is there an example of IQFT for non power of twos there? I would be surprised to find there isn't, but what do I know.
$endgroup$
– chubakueno
Mar 4 at 16:26
$begingroup$
@chubakueno I didn't immediately find one. I've now constructed one for the $m=3$ case, but don't have time right now to write it into a circuit.
$endgroup$
– DaftWullie
Mar 4 at 16:48
1
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
Mar 4 at 18:21
|
show 1 more comment
$begingroup$
This is not the unitary that you have to implement: you need a two-qubit unitary
$$
frac{1}{sqrt{3}}left(begin{array}{cccc}
1 & 1 & 1 & 0 \
1 & omega & omega^2 & 0 \
1 & omega^2 & omega & 0 \
0 & 0 & 0 & sqrt{3}
end{array}right),
$$
where $omega=e^{2ipi/3}$, the point being that if you introduce an ancilla qubit in the state 0, apply this unitary, and then measure in the computational basis, the 3 measurement outcomes 00, 01 and 10 correspond to the 3 POVM elements.
I don't (yet) have a circuit implementation for this. You'll see the paper you cite carefully avoids talking about the Fourier transform in non-power of 2 dimensions. You certainly could use the standard constructions based on Givens rotations, but the result is going to be fairly horrible.
Here's an attempt at a circuit. I've made a few tweaks since I last ran it through a computer to check, so it's always possible that a slight error has crept in, but broadly...
Here, I'm using $Z^r$ to denote
$$
left(begin{array}{cc} 1 & 0 \ 0 & e^{ipi r} end{array}right),
$$
and
$$
V=frac{1}{sqrt{3}}left(begin{array}{cc}
1 & sqrt{2} \ -sqrt{2} & 1
end{array}right).
$$
answered Mar 4 at 16:00
DaftWullieDaftWullie
14.9k1541
$endgroup$
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
Mar 4 at 16:09
$begingroup$
@chubakueno You said you wanted $m=3$, which means you need $F_3$, which is a $3times 3$ matrix which we must embed into a $4times 4$ matrix if we're using qubits.
$endgroup$
– DaftWullie
Mar 4 at 16:11
$begingroup$
I understand now. So, is there an example of IQFT for non power of twos there? I would be surprised to find there isn't, but what do I know.
$endgroup$
– chubakueno
Mar 4 at 16:26
$begingroup$
@chubakueno I didn't immediately find one. I've now constructed one for the $m=3$ case, but don't have time right now to write it into a circuit.
$endgroup$
– DaftWullie
Mar 4 at 16:48
1
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
Mar 4 at 18:21
|
show 1 more comment
$begingroup$
This is not the unitary that you have to implement: you need a two-qubit unitary
$$
frac{1}{sqrt{3}}left(begin{array}{cccc}
1 & 1 & 1 & 0 \
1 & omega & omega^2 & 0 \
1 & omega^2 & omega & 0 \
0 & 0 & 0 & sqrt{3}
end{array}right),
$$
where $omega=e^{2ipi/3}$, the point being that if you introduce an ancilla qubit in the state 0, apply this unitary, and then measure in the computational basis, the 3 measurement outcomes 00, 01 and 10 correspond to the 3 POVM elements.
I don't (yet) have a circuit implementation for this. You'll see the paper you cite carefully avoids talking about the Fourier transform in non-power of 2 dimensions. You certainly could use the standard constructions based on Givens rotations, but the result is going to be fairly horrible.
Here's an attempt at a circuit. I've made a few tweaks since I last ran it through a computer to check, so it's always possible that a slight error has crept in, but broadly...
Here, I'm using $Z^r$ to denote
$$
left(begin{array}{cc} 1 & 0 \ 0 & e^{ipi r} end{array}right),
$$
and
$$
V=frac{1}{sqrt{3}}left(begin{array}{cc}
1 & sqrt{2} \ -sqrt{2} & 1
end{array}right).
$$
answered Mar 4 at 16:00
DaftWullieDaftWullie
14.9k1541
$endgroup$
This is not the unitary that you have to implement: you need a two-qubit unitary
$$
frac{1}{sqrt{3}}left(begin{array}{cccc}
1 & 1 & 1 & 0 \
1 & omega & omega^2 & 0 \
1 & omega^2 & omega & 0 \
0 & 0 & 0 & sqrt{3}
end{array}right),
$$
where $omega=e^{2ipi/3}$, the point being that if you introduce an ancilla qubit in the state 0, apply this unitary, and then measure in the computational basis, the 3 measurement outcomes 00, 01 and 10 correspond to the 3 POVM elements.
I don't (yet) have a circuit implementation for this. You'll see the paper you cite carefully avoids talking about the Fourier transform in non-power of 2 dimensions. You certainly could use the standard constructions based on Givens rotations, but the result is going to be fairly horrible.
Here's an attempt at a circuit. I've made a few tweaks since I last ran it through a computer to check, so it's always possible that a slight error has crept in, but broadly...
Here, I'm using $Z^r$ to denote
$$
left(begin{array}{cc} 1 & 0 \ 0 & e^{ipi r} end{array}right),
$$
and
$$
V=frac{1}{sqrt{3}}left(begin{array}{cc}
1 & sqrt{2} \ -sqrt{2} & 1
end{array}right).
$$
answered Mar 4 at 16:00
DaftWullieDaftWullie
14.9k1541
edited Mar 4 at 19:46
answered Mar 4 at 16:00
DaftWullieDaftWullie
14.9k1541
answered Mar 4 at 16:00
DaftWullieDaftWullie
14.9k1541
answered Mar 4 at 16:00
DaftWullieDaftWullie
14.9k1541
14.9k1541
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
Mar 4 at 16:09
$begingroup$
@chubakueno You said you wanted $m=3$, which means you need $F_3$, which is a $3times 3$ matrix which we must embed into a $4times 4$ matrix if we're using qubits.
$endgroup$
– DaftWullie
Mar 4 at 16:11
$begingroup$
I understand now. So, is there an example of IQFT for non power of twos there? I would be surprised to find there isn't, but what do I know.
$endgroup$
– chubakueno
Mar 4 at 16:26
$begingroup$
@chubakueno I didn't immediately find one. I've now constructed one for the $m=3$ case, but don't have time right now to write it into a circuit.
$endgroup$
– DaftWullie
Mar 4 at 16:48
1
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
Mar 4 at 18:21
|
show 1 more comment
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
Mar 4 at 16:09
$begingroup$
@chubakueno You said you wanted $m=3$, which means you need $F_3$, which is a $3times 3$ matrix which we must embed into a $4times 4$ matrix if we're using qubits.
$endgroup$
– DaftWullie
Mar 4 at 16:11
$begingroup$
I understand now. So, is there an example of IQFT for non power of twos there? I would be surprised to find there isn't, but what do I know.
$endgroup$
– chubakueno
Mar 4 at 16:26
$begingroup$
@chubakueno I didn't immediately find one. I've now constructed one for the $m=3$ case, but don't have time right now to write it into a circuit.
$endgroup$
– DaftWullie
Mar 4 at 16:48
1
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
Mar 4 at 18:21
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
Mar 4 at 16:09
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
Mar 4 at 16:09
$begingroup$
@chubakueno You said you wanted $m=3$, which means you need $F_3$, which is a $3times 3$ matrix which we must embed into a $4times 4$ matrix if we're using qubits.
$endgroup$
– DaftWullie
Mar 4 at 16:11
$begingroup$
@chubakueno You said you wanted $m=3$, which means you need $F_3$, which is a $3times 3$ matrix which we must embed into a $4times 4$ matrix if we're using qubits.
$endgroup$
– DaftWullie
Mar 4 at 16:11
$begingroup$
I understand now. So, is there an example of IQFT for non power of twos there? I would be surprised to find there isn't, but what do I know.
$endgroup$
– chubakueno
Mar 4 at 16:26
$begingroup$
I understand now. So, is there an example of IQFT for non power of twos there? I would be surprised to find there isn't, but what do I know.
$endgroup$
– chubakueno
Mar 4 at 16:26
$begingroup$
@chubakueno I didn't immediately find one. I've now constructed one for the $m=3$ case, but don't have time right now to write it into a circuit.
$endgroup$
– DaftWullie
Mar 4 at 16:48
$begingroup$
@chubakueno I didn't immediately find one. I've now constructed one for the $m=3$ case, but don't have time right now to write it into a circuit.
$endgroup$
– DaftWullie
Mar 4 at 16:48
1
1
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
Mar 4 at 18:21
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
Mar 4 at 18:21
|
show 1 more comment
$begingroup$
Check this
$$frac{1}{sqrt{3}}left(begin{array}{ccc}
1 & 1 & 1 \
1 & omega & omega^2 \
1 & omega^2 & omega \
end{array}right) = left(begin{array}{cc}
H & 0 \
0 & 1 \
end{array}right) cdot frac{1}{sqrt{3}}left(begin{array}{ccc}
sqrt{2} & 0 & 1 \
0 & sqrt{3} & 0 \
1 & 0 & -sqrt{2} \
end{array}right) cdot
M_{3}
$$
$$M_{3} = left(begin{array}{ccc}
1 & 0 & 0 \
0 & frac{1}{sqrt{2}}iomega^2 & frac{1}{sqrt{2}}iomega \
0 & -frac{1}{sqrt{2}}omega^2 & -frac{1}{sqrt{2}}omega \
end{array}right)
$$
$$
frac{1}{sqrt{2}}left(begin{array}{ccc}
iomega^2 & iomega \
-omega^2 & -omega \
end{array}right) = X cdot S cdot X cdot Z cdot H cdot left(begin{array}{ccc}
omega^2 & 0 \
0 & omega \
end{array}right) cdot Z
$$
answered Mar 6 at 18:37
Danylo YDanylo Y
1814
New contributor
Danylo Y is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Check this
$$frac{1}{sqrt{3}}left(begin{array}{ccc}
1 & 1 & 1 \
1 & omega & omega^2 \
1 & omega^2 & omega \
end{array}right) = left(begin{array}{cc}
H & 0 \
0 & 1 \
end{array}right) cdot frac{1}{sqrt{3}}left(begin{array}{ccc}
sqrt{2} & 0 & 1 \
0 & sqrt{3} & 0 \
1 & 0 & -sqrt{2} \
end{array}right) cdot
M_{3}
$$
$$M_{3} = left(begin{array}{ccc}
1 & 0 & 0 \
0 & frac{1}{sqrt{2}}iomega^2 & frac{1}{sqrt{2}}iomega \
0 & -frac{1}{sqrt{2}}omega^2 & -frac{1}{sqrt{2}}omega \
end{array}right)
$$
$$
frac{1}{sqrt{2}}left(begin{array}{ccc}
iomega^2 & iomega \
-omega^2 & -omega \
end{array}right) = X cdot S cdot X cdot Z cdot H cdot left(begin{array}{ccc}
omega^2 & 0 \
0 & omega \
end{array}right) cdot Z
$$
answered Mar 6 at 18:37
Danylo YDanylo Y
1814
New contributor
Danylo Y is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Check this
$$frac{1}{sqrt{3}}left(begin{array}{ccc}
1 & 1 & 1 \
1 & omega & omega^2 \
1 & omega^2 & omega \
end{array}right) = left(begin{array}{cc}
H & 0 \
0 & 1 \
end{array}right) cdot frac{1}{sqrt{3}}left(begin{array}{ccc}
sqrt{2} & 0 & 1 \
0 & sqrt{3} & 0 \
1 & 0 & -sqrt{2} \
end{array}right) cdot
M_{3}
$$
$$M_{3} = left(begin{array}{ccc}
1 & 0 & 0 \
0 & frac{1}{sqrt{2}}iomega^2 & frac{1}{sqrt{2}}iomega \
0 & -frac{1}{sqrt{2}}omega^2 & -frac{1}{sqrt{2}}omega \
end{array}right)
$$
$$
frac{1}{sqrt{2}}left(begin{array}{ccc}
iomega^2 & iomega \
-omega^2 & -omega \
end{array}right) = X cdot S cdot X cdot Z cdot H cdot left(begin{array}{ccc}
omega^2 & 0 \
0 & omega \
end{array}right) cdot Z
$$
answered Mar 6 at 18:37
Danylo YDanylo Y
1814
New contributor
Danylo Y is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Check this
$$frac{1}{sqrt{3}}left(begin{array}{ccc}
1 & 1 & 1 \
1 & omega & omega^2 \
1 & omega^2 & omega \
end{array}right) = left(begin{array}{cc}
H & 0 \
0 & 1 \
end{array}right) cdot frac{1}{sqrt{3}}left(begin{array}{ccc}
sqrt{2} & 0 & 1 \
0 & sqrt{3} & 0 \
1 & 0 & -sqrt{2} \
end{array}right) cdot
M_{3}
$$
$$M_{3} = left(begin{array}{ccc}
1 & 0 & 0 \
0 & frac{1}{sqrt{2}}iomega^2 & frac{1}{sqrt{2}}iomega \
0 & -frac{1}{sqrt{2}}omega^2 & -frac{1}{sqrt{2}}omega \
end{array}right)
$$
$$
frac{1}{sqrt{2}}left(begin{array}{ccc}
iomega^2 & iomega \
-omega^2 & -omega \
end{array}right) = X cdot S cdot X cdot Z cdot H cdot left(begin{array}{ccc}
omega^2 & 0 \
0 & omega \
end{array}right) cdot Z
$$
answered Mar 6 at 18:37
Danylo YDanylo Y
1814
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edited Mar 6 at 18:43
answered Mar 6 at 18:37
Danylo YDanylo Y
1814
New contributor
Danylo Y is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered Mar 6 at 18:37
Danylo YDanylo Y
1814
answered Mar 6 at 18:37
Danylo YDanylo Y
1814
1814
New contributor
Danylo Y is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Danylo Y is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Danylo Y 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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1
$begingroup$
That's not a unitary matrix unless w=-1..
$endgroup$
– Craig Gidney
Mar 6 at 19:49