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;
$begingroup$
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:
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
edited 3 hours ago
Mariia Mykhailova
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asked 4 hours ago
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add a comment |
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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:
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
edited 3 hours ago
Mariia Mykhailova
2,0751212
asked 4 hours ago
IntegrateThisIntegrateThis
1084
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.
$endgroup$
add a comment |
$begingroup$
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:
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
edited 3 hours ago
Mariia Mykhailova
2,0751212
asked 4 hours ago
IntegrateThisIntegrateThis
1084
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:
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
deutsch-jozsa-algorithm
edited 3 hours ago
Mariia Mykhailova
2,0751212
asked 4 hours ago
IntegrateThisIntegrateThis
1084
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.
edited 3 hours ago
Mariia Mykhailova
2,0751212
asked 4 hours ago
IntegrateThisIntegrateThis
1084
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.
edited 3 hours ago
Mariia Mykhailova
2,0751212
edited 3 hours ago
Mariia Mykhailova
2,0751212
edited 3 hours ago
Mariia Mykhailova
2,0751212
2,0751212
asked 4 hours ago
IntegrateThisIntegrateThis
1084
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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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asked 4 hours ago
IntegrateThisIntegrateThis
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asked 4 hours ago
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1 Answer
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$begingroup$
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:
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 $.
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.
answered 3 hours ago
Mariia MykhailovaMariia Mykhailova
2,0751212
$endgroup$
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
add a comment |
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1 Answer
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1 Answer
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$begingroup$
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:
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 $.
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.
answered 3 hours ago
Mariia MykhailovaMariia Mykhailova
2,0751212
$endgroup$
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
add a comment |
$begingroup$
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:
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 $.
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.
answered 3 hours ago
Mariia MykhailovaMariia Mykhailova
2,0751212
$endgroup$
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
add a comment |
$begingroup$
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:
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 $.
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.
answered 3 hours ago
Mariia MykhailovaMariia Mykhailova
2,0751212
$endgroup$
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:
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 $.
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.
answered 3 hours ago
Mariia MykhailovaMariia Mykhailova
2,0751212
answered 3 hours ago
Mariia MykhailovaMariia Mykhailova
2,0751212
answered 3 hours ago
Mariia MykhailovaMariia Mykhailova
2,0751212
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
add a comment |
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
add a comment |
IntegrateThis is a new contributor. Be nice, and check out our Code of Conduct.
IntegrateThis is a new contributor. Be nice, and check out our Code of Conduct.
IntegrateThis is a new contributor. Be nice, and check out our Code of Conduct.
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