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POVM three-qubit circuit for symmetric quantum states

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4

$begingroup$

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

share|improve this question

edited 5 hours ago

xbk365

asked 6 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 | 

4

$begingroup$

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

share|improve this question

edited 5 hours ago

xbk365

asked 6 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

share|improve this question

edited 5 hours ago

xbk365

asked 6 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.

$endgroup$

share|improve this question

edited 5 hours ago

xbk365

asked 6 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.

share|improve this question

edited 5 hours ago

xbk365

asked 6 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 6 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 6 hours ago

xbk365xbk365

213

1 Answer
1

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4

$begingroup$

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
$$
V=frac{1}{sqrt{3}}left(begin{array}{cc}
1 & sqrt{2} \ -sqrt{2} & 1
end{array}right).
$$

share|improve this answer

edited 1 hour ago

answered 5 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
  5 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
  4 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
  4 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
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @xbk365 once i’m done with the evening’s childcare responsibilities...
  $endgroup$
  – DaftWullie
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

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4

$begingroup$

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
$$
V=frac{1}{sqrt{3}}left(begin{array}{cc}
1 & sqrt{2} \ -sqrt{2} & 1
end{array}right).
$$

share|improve this answer

edited 1 hour ago

answered 5 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
  5 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
  4 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
  4 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
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @xbk365 once i’m done with the evening’s childcare responsibilities...
  $endgroup$
  – DaftWullie
  2 hours ago
  

  
  
  
  

 | 
show 1 more comment

4

$begingroup$

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
$$
V=frac{1}{sqrt{3}}left(begin{array}{cc}
1 & sqrt{2} \ -sqrt{2} & 1
end{array}right).
$$

share|improve this answer

edited 1 hour ago

answered 5 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
  5 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
  4 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
  4 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
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @xbk365 once i’m done with the evening’s childcare responsibilities...
  $endgroup$
  – DaftWullie
  2 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
$$
V=frac{1}{sqrt{3}}left(begin{array}{cc}
1 & sqrt{2} \ -sqrt{2} & 1
end{array}right).
$$

share|improve this answer

edited 1 hour ago

answered 5 hours ago

DaftWullieDaftWullie

14.5k1541

$endgroup$

share|improve this answer

edited 1 hour ago

answered 5 hours ago

DaftWullieDaftWullie

14.5k1541

answered 5 hours ago

DaftWullieDaftWullie

14.5k1541

answered 5 hours ago

DaftWullieDaftWullie

14.5k1541