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:
$$
M=
frac1sqrt2left[ beginarraycc
1 & 1 \
1 & w \
endarray right]
$$

Where $w$ is a third root of unity using rotations, after which I am stuck.










share|improve this question











$endgroup$







  • 1




    $begingroup$
    That's not a unitary matrix unless w=-1..
    $endgroup$
    – Craig Gidney
    Mar 6 at 19:49















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:
$$
M=
frac1sqrt2left[ beginarraycc
1 & 1 \
1 & w \
endarray right]
$$

Where $w$ is a third root of unity using rotations, after which I am stuck.










share|improve this question











$endgroup$







  • 1




    $begingroup$
    That's not a unitary matrix unless w=-1..
    $endgroup$
    – Craig Gidney
    Mar 6 at 19:49













4












4








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:
$$
M=
frac1sqrt2left[ beginarraycc
1 & 1 \
1 & w \
endarray right]
$$

Where $w$ is a third root of unity using rotations, after which I am stuck.










share|improve this question











$endgroup$




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:
$$
M=
frac1sqrt2left[ beginarraycc
1 & 1 \
1 & w \
endarray right]
$$

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















share|improve this question













share|improve this question




share|improve this question








edited Mar 4 at 15:51







xbk365

















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












  • 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










2 Answers
2






active

oldest

votes


















5












$begingroup$

This is not the unitary that you have to implement: you need a two-qubit unitary
$$
frac1sqrt3left(beginarraycccc
1 & 1 & 1 & 0 \
1 & omega & omega^2 & 0 \
1 & omega^2 & omega & 0 \
0 & 0 & 0 & sqrt3
endarrayright),
$$

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...
enter image description here
Here, I'm using $Z^r$ to denote
$$
left(beginarraycc 1 & 0 \ 0 & e^ipi r endarrayright),
$$

and
$$
V=frac1sqrt3left(beginarraycc
1 & sqrt2 \ -sqrt2 & 1
endarrayright).
$$






share|improve this answer











$endgroup$












  • $begingroup$
    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


















2












$begingroup$

Check this
$$frac1sqrt3left(beginarrayccc
1 & 1 & 1 \
1 & omega & omega^2 \
1 & omega^2 & omega \
endarrayright) = left(beginarraycc
H & 0 \
0 & 1 \
endarrayright) cdot frac1sqrt3left(beginarrayccc
sqrt2 & 0 & 1 \
0 & sqrt3 & 0 \
1 & 0 & -sqrt2 \
endarrayright) cdot
M_3
$$



$$M_3 = left(beginarrayccc
1 & 0 & 0 \
0 & frac1sqrt2iomega^2 & frac1sqrt2iomega \
0 & -frac1sqrt2omega^2 & -frac1sqrt2omega \
endarrayright)
$$



$$
frac1sqrt2left(beginarrayccc
iomega^2 & iomega \
-omega^2 & -omega \
endarrayright) = X cdot S cdot X cdot Z cdot H cdot left(beginarrayccc
omega^2 & 0 \
0 & omega \
endarrayright) cdot Z
$$






share|improve this answer











$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    5












    $begingroup$

    This is not the unitary that you have to implement: you need a two-qubit unitary
    $$
    frac1sqrt3left(beginarraycccc
    1 & 1 & 1 & 0 \
    1 & omega & omega^2 & 0 \
    1 & omega^2 & omega & 0 \
    0 & 0 & 0 & sqrt3
    endarrayright),
    $$

    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...
    enter image description here
    Here, I'm using $Z^r$ to denote
    $$
    left(beginarraycc 1 & 0 \ 0 & e^ipi r endarrayright),
    $$

    and
    $$
    V=frac1sqrt3left(beginarraycc
    1 & sqrt2 \ -sqrt2 & 1
    endarrayright).
    $$






    share|improve this answer











    $endgroup$












    • $begingroup$
      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















    5












    $begingroup$

    This is not the unitary that you have to implement: you need a two-qubit unitary
    $$
    frac1sqrt3left(beginarraycccc
    1 & 1 & 1 & 0 \
    1 & omega & omega^2 & 0 \
    1 & omega^2 & omega & 0 \
    0 & 0 & 0 & sqrt3
    endarrayright),
    $$

    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...
    enter image description here
    Here, I'm using $Z^r$ to denote
    $$
    left(beginarraycc 1 & 0 \ 0 & e^ipi r endarrayright),
    $$

    and
    $$
    V=frac1sqrt3left(beginarraycc
    1 & sqrt2 \ -sqrt2 & 1
    endarrayright).
    $$






    share|improve this answer











    $endgroup$












    • $begingroup$
      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













    5












    5








    5





    $begingroup$

    This is not the unitary that you have to implement: you need a two-qubit unitary
    $$
    frac1sqrt3left(beginarraycccc
    1 & 1 & 1 & 0 \
    1 & omega & omega^2 & 0 \
    1 & omega^2 & omega & 0 \
    0 & 0 & 0 & sqrt3
    endarrayright),
    $$

    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...
    enter image description here
    Here, I'm using $Z^r$ to denote
    $$
    left(beginarraycc 1 & 0 \ 0 & e^ipi r endarrayright),
    $$

    and
    $$
    V=frac1sqrt3left(beginarraycc
    1 & sqrt2 \ -sqrt2 & 1
    endarrayright).
    $$






    share|improve this answer











    $endgroup$



    This is not the unitary that you have to implement: you need a two-qubit unitary
    $$
    frac1sqrt3left(beginarraycccc
    1 & 1 & 1 & 0 \
    1 & omega & omega^2 & 0 \
    1 & omega^2 & omega & 0 \
    0 & 0 & 0 & sqrt3
    endarrayright),
    $$

    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...
    enter image description here
    Here, I'm using $Z^r$ to denote
    $$
    left(beginarraycc 1 & 0 \ 0 & e^ipi r endarrayright),
    $$

    and
    $$
    V=frac1sqrt3left(beginarraycc
    1 & sqrt2 \ -sqrt2 & 1
    endarrayright).
    $$







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Mar 4 at 19:46

























    answered Mar 4 at 16:00









    DaftWullieDaftWullie

    15.2k1542




    15.2k1542











    • $begingroup$
      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










    • $begingroup$
      @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













    2












    $begingroup$

    Check this
    $$frac1sqrt3left(beginarrayccc
    1 & 1 & 1 \
    1 & omega & omega^2 \
    1 & omega^2 & omega \
    endarrayright) = left(beginarraycc
    H & 0 \
    0 & 1 \
    endarrayright) cdot frac1sqrt3left(beginarrayccc
    sqrt2 & 0 & 1 \
    0 & sqrt3 & 0 \
    1 & 0 & -sqrt2 \
    endarrayright) cdot
    M_3
    $$



    $$M_3 = left(beginarrayccc
    1 & 0 & 0 \
    0 & frac1sqrt2iomega^2 & frac1sqrt2iomega \
    0 & -frac1sqrt2omega^2 & -frac1sqrt2omega \
    endarrayright)
    $$



    $$
    frac1sqrt2left(beginarrayccc
    iomega^2 & iomega \
    -omega^2 & -omega \
    endarrayright) = X cdot S cdot X cdot Z cdot H cdot left(beginarrayccc
    omega^2 & 0 \
    0 & omega \
    endarrayright) cdot Z
    $$






    share|improve this answer











    $endgroup$

















      2












      $begingroup$

      Check this
      $$frac1sqrt3left(beginarrayccc
      1 & 1 & 1 \
      1 & omega & omega^2 \
      1 & omega^2 & omega \
      endarrayright) = left(beginarraycc
      H & 0 \
      0 & 1 \
      endarrayright) cdot frac1sqrt3left(beginarrayccc
      sqrt2 & 0 & 1 \
      0 & sqrt3 & 0 \
      1 & 0 & -sqrt2 \
      endarrayright) cdot
      M_3
      $$



      $$M_3 = left(beginarrayccc
      1 & 0 & 0 \
      0 & frac1sqrt2iomega^2 & frac1sqrt2iomega \
      0 & -frac1sqrt2omega^2 & -frac1sqrt2omega \
      endarrayright)
      $$



      $$
      frac1sqrt2left(beginarrayccc
      iomega^2 & iomega \
      -omega^2 & -omega \
      endarrayright) = X cdot S cdot X cdot Z cdot H cdot left(beginarrayccc
      omega^2 & 0 \
      0 & omega \
      endarrayright) cdot Z
      $$






      share|improve this answer











      $endgroup$















        2












        2








        2





        $begingroup$

        Check this
        $$frac1sqrt3left(beginarrayccc
        1 & 1 & 1 \
        1 & omega & omega^2 \
        1 & omega^2 & omega \
        endarrayright) = left(beginarraycc
        H & 0 \
        0 & 1 \
        endarrayright) cdot frac1sqrt3left(beginarrayccc
        sqrt2 & 0 & 1 \
        0 & sqrt3 & 0 \
        1 & 0 & -sqrt2 \
        endarrayright) cdot
        M_3
        $$



        $$M_3 = left(beginarrayccc
        1 & 0 & 0 \
        0 & frac1sqrt2iomega^2 & frac1sqrt2iomega \
        0 & -frac1sqrt2omega^2 & -frac1sqrt2omega \
        endarrayright)
        $$



        $$
        frac1sqrt2left(beginarrayccc
        iomega^2 & iomega \
        -omega^2 & -omega \
        endarrayright) = X cdot S cdot X cdot Z cdot H cdot left(beginarrayccc
        omega^2 & 0 \
        0 & omega \
        endarrayright) cdot Z
        $$






        share|improve this answer











        $endgroup$



        Check this
        $$frac1sqrt3left(beginarrayccc
        1 & 1 & 1 \
        1 & omega & omega^2 \
        1 & omega^2 & omega \
        endarrayright) = left(beginarraycc
        H & 0 \
        0 & 1 \
        endarrayright) cdot frac1sqrt3left(beginarrayccc
        sqrt2 & 0 & 1 \
        0 & sqrt3 & 0 \
        1 & 0 & -sqrt2 \
        endarrayright) cdot
        M_3
        $$



        $$M_3 = left(beginarrayccc
        1 & 0 & 0 \
        0 & frac1sqrt2iomega^2 & frac1sqrt2iomega \
        0 & -frac1sqrt2omega^2 & -frac1sqrt2omega \
        endarrayright)
        $$



        $$
        frac1sqrt2left(beginarrayccc
        iomega^2 & iomega \
        -omega^2 & -omega \
        endarrayright) = X cdot S cdot X cdot Z cdot H cdot left(beginarrayccc
        omega^2 & 0 \
        0 & omega \
        endarrayright) cdot Z
        $$







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Mar 6 at 18:43

























        answered Mar 6 at 18:37









        Danylo YDanylo Y

        29615




        29615



























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