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 14 hours ago
xbk365xbk365
213
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add a comment |
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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 14 hours ago
xbk365xbk365
213
New contributor
xbk365 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 |
$begingroup$
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 14 hours ago
xbk365xbk365
213
New contributor
xbk365 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
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 14 hours ago
xbk365xbk365
213
New contributor
xbk365 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 14 hours ago
xbk365xbk365
213
New contributor
xbk365 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 13 hours ago
xbk365
asked 14 hours ago
xbk365xbk365
213
New contributor
xbk365 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 14 hours ago
xbk365xbk365
213
asked 14 hours ago
xbk365xbk365
213
213
New contributor
xbk365 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
xbk365 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
xbk365 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
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 13 hours ago
DaftWullieDaftWullie
14.5k1541
$endgroup$
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
13 hours ago
$begingroup$
@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
13 hours ago
$begingroup$
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
13 hours ago
$begingroup$
@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
12 hours ago
1
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
11 hours ago
|
show 1 more comment
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1 Answer
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active
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1 Answer
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oldest
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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 13 hours ago
DaftWullieDaftWullie
14.5k1541
$endgroup$
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
13 hours ago
$begingroup$
@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
13 hours ago
$begingroup$
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
13 hours ago
$begingroup$
@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
12 hours ago
1
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
11 hours ago
|
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 13 hours ago
DaftWullieDaftWullie
14.5k1541
$endgroup$
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
13 hours ago
$begingroup$
@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
13 hours ago
$begingroup$
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
13 hours ago
$begingroup$
@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
12 hours ago
1
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
11 hours ago
|
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 13 hours ago
DaftWullieDaftWullie
14.5k1541
$endgroup$
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 13 hours ago
DaftWullieDaftWullie
14.5k1541
edited 9 hours ago
answered 13 hours ago
DaftWullieDaftWullie
14.5k1541
answered 13 hours ago
DaftWullieDaftWullie
14.5k1541
answered 13 hours ago
DaftWullieDaftWullie
14.5k1541
14.5k1541
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
13 hours ago
$begingroup$
@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
13 hours ago
$begingroup$
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
13 hours ago
$begingroup$
@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
12 hours ago
1
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
11 hours ago
|
show 1 more comment
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
13 hours ago
$begingroup$
@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
13 hours ago
$begingroup$
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
13 hours ago
$begingroup$
@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
12 hours ago
1
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
11 hours ago
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
13 hours ago
$begingroup$
Hmm, I was actually talking about F2, not the whole unitary. Or, I am missing something?
$endgroup$
– chubakueno
13 hours ago
$begingroup$
@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
13 hours ago
$begingroup$
@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
13 hours ago
$begingroup$
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
13 hours ago
$begingroup$
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
13 hours ago
$begingroup$
@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
12 hours ago
$begingroup$
@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
12 hours ago
1
1
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
11 hours ago
$begingroup$
@xbk365 once i’m done with the evening’s childcare responsibilities...
$endgroup$
– DaftWullie
11 hours ago
|
show 1 more comment
xbk365 is a new contributor. Be nice, and check out our Code of Conduct.
xbk365 is a new contributor. Be nice, and check out our Code of Conduct.
xbk365 is a new contributor. Be nice, and check out our Code of Conduct.
xbk365 is a new contributor. Be nice, and check out our Code of Conduct.
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