Understanding Deutch's AlgorithmHow would I implement the quantum oracle in Deutsch's algorithm?Deutsch Algorithm on a Quantum Turing MachineHow is the Deutsch-Jozsa algorithm faster than classical for practical implementation?Deutsch algorithm with equal input bitsWhy doesn't Deutsch-Jozsa Algorithm show that P ≠ BQP?How to understand Deutsch-Jozsa algorithm from an adiabatic perspective?Deutsch–Jozsa algorithm: why is $f$ constant?How does an oracle function in Grover's algorithm actually work?What can be a mini research project based on Grover's algorithm or the Deutsch Jozsa algorithm?Deutsch-Jozsa algorithm as a generalization of Bernstein-Vazirani

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Understanding Deutch's Algorithm


How would I implement the quantum oracle in Deutsch's algorithm?Deutsch Algorithm on a Quantum Turing MachineHow is the Deutsch-Jozsa algorithm faster than classical for practical implementation?Deutsch algorithm with equal input bitsWhy doesn't Deutsch-Jozsa Algorithm show that P ≠ BQP?How to understand Deutsch-Jozsa algorithm from an adiabatic perspective?Deutsch–Jozsa algorithm: why is $f$ constant?How does an oracle function in Grover's algorithm actually work?What can be a mini research project based on Grover's algorithm or the Deutsch Jozsa algorithm?Deutsch-Jozsa algorithm as a generalization of Bernstein-Vazirani






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








1












$begingroup$


I am reading John Waltrous 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









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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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    1












    $begingroup$


    I am reading John Waltrous 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.






    $endgroup$














      1












      1








      1





      $begingroup$


      I am reading John Waltrous 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 Waltrous 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.







      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 3 hours ago









      Mariia Mykhailova

      2,0751212




      2,0751212






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      asked 4 hours ago









      IntegrateThisIntegrateThis

      1084




      1084




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          1 Answer
          1






          active

          oldest

          votes


















          4












          $begingroup$

          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









          $endgroup$








          • 2




            $begingroup$
            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
            3 hours ago











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            2 hours ago











          Your Answer








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          1 Answer
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          1 Answer
          1






          active

          oldest

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          active

          oldest

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          active

          oldest

          votes









          4












          $begingroup$

          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









          $endgroup$








          • 2




            $begingroup$
            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
            3 hours ago











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            2 hours ago















          4












          $begingroup$

          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









          $endgroup$








          • 2




            $begingroup$
            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
            3 hours ago











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            2 hours ago













          4












          4








          4





          $begingroup$

          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 3 hours ago









          Mariia MykhailovaMariia Mykhailova

          2,0751212




          2,0751212







          • 2




            $begingroup$
            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
            3 hours ago











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            2 hours ago












          • 2




            $begingroup$
            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
            3 hours ago











          • $begingroup$
            Thanks so much for your help! I am very grateful :)
            $endgroup$
            – IntegrateThis
            2 hours ago







          2




          2




          $begingroup$
          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
          3 hours ago





          $begingroup$
          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
          3 hours ago













          $begingroup$
          Thanks so much for your help! I am very grateful :)
          $endgroup$
          – IntegrateThis
          2 hours ago




          $begingroup$
          Thanks so much for your help! I am very grateful :)
          $endgroup$
          – IntegrateThis
          2 hours ago










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









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