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Understanding the oracle in Deutsch's algorithm


How is the oracle in Grover's search algorithm implemented?How exactly does Simon's algorithm solve the Simon's problem?Grover's algorithm: what to input to Oracle?How exactly is the stated composite state of the two registers being produced using the $R_zz$ controlled rotations?How would I implement the quantum oracle in Deutsch's algorithm?How is the Deutsch-Jozsa algorithm faster than classical for practical implementation?How to create the oracle matrix in Grover's algorithm?How to prove that the query oracle is unitary?How does an oracle function in Grover's algorithm actually work?Grover's algorithm – DES circuit as oracle?






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








4












$begingroup$


I am reading John Watrous' notes from his course CPSC 519 on quantum computing. In a pre-discussion before presenting Deutsch's algorithm to determine whether a function is constant or not, the author presents a function $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, and the diagram:
enter image description here



The inital state is $|0 rangle |1 rangle$, and after the first two Hadamard transforms, will be $$big(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1ranglebig)big(frac 1 sqrt 2 |0rangle-frac 1 sqrt 2 |1ranglebig) .$$



Up to this far I understand. The author then writes: "After performing the $B_f$ operation the state is transformed to:



$$frac 1 2 |0 rangle big(|0 oplus f(0)rangle - |1 oplus f(0)ranglebig) + frac 1 2 |1 rangle big(|0 oplus f(1)rangle) - |1 oplus f(1) ranglebig).$$



I am not sure how this was obtained, from what I understand, the operation should be
$$frac 1 sqrt 2 big( |0rangle + |1ranglebig) otimes big|(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) oplus f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) bigrangle$$ (simply subbing in $x,y$ to $B_f$). Any insights appreciated as this subject is completely new to me, although I have a decent mathematics and computer science background.










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    4












    $begingroup$


    I am reading John Watrous' notes from his course CPSC 519 on quantum computing. In a pre-discussion before presenting Deutsch's algorithm to determine whether a function is constant or not, the author presents a function $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, and the diagram:
    enter image description here



    The inital state is $|0 rangle |1 rangle$, and after the first two Hadamard transforms, will be $$big(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1ranglebig)big(frac 1 sqrt 2 |0rangle-frac 1 sqrt 2 |1ranglebig) .$$



    Up to this far I understand. The author then writes: "After performing the $B_f$ operation the state is transformed to:



    $$frac 1 2 |0 rangle big(|0 oplus f(0)rangle - |1 oplus f(0)ranglebig) + frac 1 2 |1 rangle big(|0 oplus f(1)rangle) - |1 oplus f(1) ranglebig).$$



    I am not sure how this was obtained, from what I understand, the operation should be
    $$frac 1 sqrt 2 big( |0rangle + |1ranglebig) otimes big|(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) oplus f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) bigrangle$$ (simply subbing in $x,y$ to $B_f$). Any insights appreciated as this subject is completely new to me, although I have a decent mathematics and computer science background.










    share|improve this question









    New contributor



    IntegrateThis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$














      4












      4








      4


      1



      $begingroup$


      I am reading John Watrous' notes from his course CPSC 519 on quantum computing. In a pre-discussion before presenting Deutsch's algorithm to determine whether a function is constant or not, the author presents a function $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, and the diagram:
      enter image description here



      The inital state is $|0 rangle |1 rangle$, and after the first two Hadamard transforms, will be $$big(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1ranglebig)big(frac 1 sqrt 2 |0rangle-frac 1 sqrt 2 |1ranglebig) .$$



      Up to this far I understand. The author then writes: "After performing the $B_f$ operation the state is transformed to:



      $$frac 1 2 |0 rangle big(|0 oplus f(0)rangle - |1 oplus f(0)ranglebig) + frac 1 2 |1 rangle big(|0 oplus f(1)rangle) - |1 oplus f(1) ranglebig).$$



      I am not sure how this was obtained, from what I understand, the operation should be
      $$frac 1 sqrt 2 big( |0rangle + |1ranglebig) otimes big|(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) oplus f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) bigrangle$$ (simply subbing in $x,y$ to $B_f$). Any insights appreciated as this subject is completely new to me, although I have a decent mathematics and computer science background.










      share|improve this question









      New contributor



      IntegrateThis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






      $endgroup$




      I am reading John Watrous' notes from his course CPSC 519 on quantum computing. In a pre-discussion before presenting Deutsch's algorithm to determine whether a function is constant or not, the author presents a function $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, and the diagram:
      enter image description here



      The inital state is $|0 rangle |1 rangle$, and after the first two Hadamard transforms, will be $$big(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1ranglebig)big(frac 1 sqrt 2 |0rangle-frac 1 sqrt 2 |1ranglebig) .$$



      Up to this far I understand. The author then writes: "After performing the $B_f$ operation the state is transformed to:



      $$frac 1 2 |0 rangle big(|0 oplus f(0)rangle - |1 oplus f(0)ranglebig) + frac 1 2 |1 rangle big(|0 oplus f(1)rangle) - |1 oplus f(1) ranglebig).$$



      I am not sure how this was obtained, from what I understand, the operation should be
      $$frac 1 sqrt 2 big( |0rangle + |1ranglebig) otimes big|(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) oplus f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle) bigrangle$$ (simply subbing in $x,y$ to $B_f$). Any insights appreciated as this subject is completely new to me, although I have a decent mathematics and computer science background.







      algorithm deutsch-jozsa-algorithm






      share|improve this question









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      IntegrateThis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|improve this question









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      IntegrateThis is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|improve this question




      share|improve this question








      edited May 15 at 9:34









      Sanchayan Dutta

      7,15041658




      7,15041658






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      asked May 14 at 21:25









      IntegrateThisIntegrateThis

      1234




      1234




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          1 Answer
          1






          active

          oldest

          votes


















          8












          $begingroup$

          Remember that when you define the oracle effect as $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, $f(x)$ is a classical function of a classical 1-bit argument, so you do not have a way to compute $f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle)$ (a function of a quantum state).



          The quantum oracles that implement classical functions are defined as follows:



          1. Define the effect of the oracle on all basis states for $|xrangle$ and $|yrangle$: $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $.


          2. This will automatically define the effect of the oracle on all superposition states: the oracle is a quantum operation and has to be linear in the state on which it acts. So if you start with a state $frac12 (|00rangle + |10rangle - |01rangle - |11rangle)$ (which is the state after applying Hadamard gates) and apply the oracle, you need to apply oracle to each basis state separately. You'll get


          $$B_f frac12 (|00rangle + |10rangle - |01rangle - |11rangle) = frac12 (B_f|00rangle + B_f|10rangle - B_f|01rangle - B_f|11rangle) =$$



          $$ = frac12 (|0rangle|0 oplus f(0)rangle + |1rangle|0 oplus f(1)rangle - |0rangle|1 oplus f(0)rangle - |1rangle|1 oplus f(1)rangle)$$



          Which is the same as the expression in the notes, up to a different grouping or terms.




          The part about the oracles being defined by their effect on basis states is implicit in a lot of sources I've seen, and is a frequent source of confusion. If you need more mathematical details on this, we ended up writing it up here.






          share|improve this answer









          $endgroup$








          • 3




            $begingroup$
            Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
            $endgroup$
            – Dr. Sarah Kaiser
            May 14 at 22:00











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            May 14 at 23:02











          Your Answer








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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          8












          $begingroup$

          Remember that when you define the oracle effect as $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, $f(x)$ is a classical function of a classical 1-bit argument, so you do not have a way to compute $f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle)$ (a function of a quantum state).



          The quantum oracles that implement classical functions are defined as follows:



          1. Define the effect of the oracle on all basis states for $|xrangle$ and $|yrangle$: $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $.


          2. This will automatically define the effect of the oracle on all superposition states: the oracle is a quantum operation and has to be linear in the state on which it acts. So if you start with a state $frac12 (|00rangle + |10rangle - |01rangle - |11rangle)$ (which is the state after applying Hadamard gates) and apply the oracle, you need to apply oracle to each basis state separately. You'll get


          $$B_f frac12 (|00rangle + |10rangle - |01rangle - |11rangle) = frac12 (B_f|00rangle + B_f|10rangle - B_f|01rangle - B_f|11rangle) =$$



          $$ = frac12 (|0rangle|0 oplus f(0)rangle + |1rangle|0 oplus f(1)rangle - |0rangle|1 oplus f(0)rangle - |1rangle|1 oplus f(1)rangle)$$



          Which is the same as the expression in the notes, up to a different grouping or terms.




          The part about the oracles being defined by their effect on basis states is implicit in a lot of sources I've seen, and is a frequent source of confusion. If you need more mathematical details on this, we ended up writing it up here.






          share|improve this answer









          $endgroup$








          • 3




            $begingroup$
            Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
            $endgroup$
            – Dr. Sarah Kaiser
            May 14 at 22:00











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            May 14 at 23:02















          8












          $begingroup$

          Remember that when you define the oracle effect as $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, $f(x)$ is a classical function of a classical 1-bit argument, so you do not have a way to compute $f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle)$ (a function of a quantum state).



          The quantum oracles that implement classical functions are defined as follows:



          1. Define the effect of the oracle on all basis states for $|xrangle$ and $|yrangle$: $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $.


          2. This will automatically define the effect of the oracle on all superposition states: the oracle is a quantum operation and has to be linear in the state on which it acts. So if you start with a state $frac12 (|00rangle + |10rangle - |01rangle - |11rangle)$ (which is the state after applying Hadamard gates) and apply the oracle, you need to apply oracle to each basis state separately. You'll get


          $$B_f frac12 (|00rangle + |10rangle - |01rangle - |11rangle) = frac12 (B_f|00rangle + B_f|10rangle - B_f|01rangle - B_f|11rangle) =$$



          $$ = frac12 (|0rangle|0 oplus f(0)rangle + |1rangle|0 oplus f(1)rangle - |0rangle|1 oplus f(0)rangle - |1rangle|1 oplus f(1)rangle)$$



          Which is the same as the expression in the notes, up to a different grouping or terms.




          The part about the oracles being defined by their effect on basis states is implicit in a lot of sources I've seen, and is a frequent source of confusion. If you need more mathematical details on this, we ended up writing it up here.






          share|improve this answer









          $endgroup$








          • 3




            $begingroup$
            Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
            $endgroup$
            – Dr. Sarah Kaiser
            May 14 at 22:00











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            May 14 at 23:02













          8












          8








          8





          $begingroup$

          Remember that when you define the oracle effect as $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, $f(x)$ is a classical function of a classical 1-bit argument, so you do not have a way to compute $f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle)$ (a function of a quantum state).



          The quantum oracles that implement classical functions are defined as follows:



          1. Define the effect of the oracle on all basis states for $|xrangle$ and $|yrangle$: $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $.


          2. This will automatically define the effect of the oracle on all superposition states: the oracle is a quantum operation and has to be linear in the state on which it acts. So if you start with a state $frac12 (|00rangle + |10rangle - |01rangle - |11rangle)$ (which is the state after applying Hadamard gates) and apply the oracle, you need to apply oracle to each basis state separately. You'll get


          $$B_f frac12 (|00rangle + |10rangle - |01rangle - |11rangle) = frac12 (B_f|00rangle + B_f|10rangle - B_f|01rangle - B_f|11rangle) =$$



          $$ = frac12 (|0rangle|0 oplus f(0)rangle + |1rangle|0 oplus f(1)rangle - |0rangle|1 oplus f(0)rangle - |1rangle|1 oplus f(1)rangle)$$



          Which is the same as the expression in the notes, up to a different grouping or terms.




          The part about the oracles being defined by their effect on basis states is implicit in a lot of sources I've seen, and is a frequent source of confusion. If you need more mathematical details on this, we ended up writing it up here.






          share|improve this answer









          $endgroup$



          Remember that when you define the oracle effect as $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $, $f(x)$ is a classical function of a classical 1-bit argument, so you do not have a way to compute $f(frac 1 sqrt 2 |0rangle +frac 1 sqrt 2 |1rangle)$ (a function of a quantum state).



          The quantum oracles that implement classical functions are defined as follows:



          1. Define the effect of the oracle on all basis states for $|xrangle$ and $|yrangle$: $B_f |x rangle |y rangle = |x rangle |y oplus f(x) rangle $.


          2. This will automatically define the effect of the oracle on all superposition states: the oracle is a quantum operation and has to be linear in the state on which it acts. So if you start with a state $frac12 (|00rangle + |10rangle - |01rangle - |11rangle)$ (which is the state after applying Hadamard gates) and apply the oracle, you need to apply oracle to each basis state separately. You'll get


          $$B_f frac12 (|00rangle + |10rangle - |01rangle - |11rangle) = frac12 (B_f|00rangle + B_f|10rangle - B_f|01rangle - B_f|11rangle) =$$



          $$ = frac12 (|0rangle|0 oplus f(0)rangle + |1rangle|0 oplus f(1)rangle - |0rangle|1 oplus f(0)rangle - |1rangle|1 oplus f(1)rangle)$$



          Which is the same as the expression in the notes, up to a different grouping or terms.




          The part about the oracles being defined by their effect on basis states is implicit in a lot of sources I've seen, and is a frequent source of confusion. If you need more mathematical details on this, we ended up writing it up here.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered May 14 at 21:50









          Mariia MykhailovaMariia Mykhailova

          2,1601212




          2,1601212







          • 3




            $begingroup$
            Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
            $endgroup$
            – Dr. Sarah Kaiser
            May 14 at 22:00











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            May 14 at 23:02












          • 3




            $begingroup$
            Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
            $endgroup$
            – Dr. Sarah Kaiser
            May 14 at 22:00











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            May 14 at 23:02







          3




          3




          $begingroup$
          Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
          $endgroup$
          – Dr. Sarah Kaiser
          May 14 at 22:00





          $begingroup$
          Another resource that may be helpful is Learn Quantum Computing with Python and Q# which should have the chapter on Deutsch–Jozsa algorithm up shortly! We work though implementing Deutsch–Jozsa in Q# as well as in Python with QuTiP. <manning.com/books/…>
          $endgroup$
          – Dr. Sarah Kaiser
          May 14 at 22:00













          $begingroup$
          Thanks so much for your help! I am very grateful :)
          $endgroup$
          – IntegrateThis
          May 14 at 23:02




          $begingroup$
          Thanks so much for your help! I am very grateful :)
          $endgroup$
          – IntegrateThis
          May 14 at 23:02










          IntegrateThis is a new contributor. Be nice, and check out our Code of Conduct.









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