Preparing odd integers using quantum computationHow do I add 1+1 using a quantum computer?Embedding classical information into norm of a quantum stateImplementation of the oracle of Grover's algorithm on IBM Q using three qubitsSWAP gate(s) in the $R(lambda^-1)$ step of the HHL circuit for $4times 4$ systemsHow to prevent future loops using a control qubit?Quantum addition and modulo operation using gatesQuantum secret Sharing using GHZ state paperPerform quantum gate operations using state vectors and matrices
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Preparing odd integers using quantum computation
How do I add 1+1 using a quantum computer?Embedding classical information into norm of a quantum stateImplementation of the oracle of Grover's algorithm on IBM Q using three qubitsSWAP gate(s) in the $R(lambda^-1)$ step of the HHL circuit for $4times 4$ systemsHow to prevent future loops using a control qubit?Quantum addition and modulo operation using gatesQuantum secret Sharing using GHZ state paperPerform quantum gate operations using state vectors and matrices
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This is just a basic question. I need to output odd integers till $15$, i.e $1,3,5,7,9,11,13,15$. Since $15$ requires $4$ bits, I prepare initial state by using Hadamard gate on initial $|0rangle$, i.e $$H|0rangle otimes H|0rangle otimes H|0rangle$$ $$=|000rangle+|001rangle+|010rangle+|011rangle+|100rangle+|101rangle+|110rangle+|111rangle.$$ Then take the tensor product with $|1rangle$, i.e $$(H|0rangle otimes H|0rangle otimes H|0rangle)|1rangle$$ $$=|0001rangle+|0011rangle+|0101rangle+|0111rangle+|1001rangle+|1011rangle+|1101rangle+|1111rangle.$$ Is this right?
Actually, I know the ideas but don't know how to write it in the quantum computing terms, that involves using terms as registers, qubits
quantum-gate quantum-state
edited Apr 15 at 10:57
Sanchayan Dutta
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asked Apr 15 at 7:13
UpstartUpstart
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$begingroup$
This is just a basic question. I need to output odd integers till $15$, i.e $1,3,5,7,9,11,13,15$. Since $15$ requires $4$ bits, I prepare initial state by using Hadamard gate on initial $|0rangle$, i.e $$H|0rangle otimes H|0rangle otimes H|0rangle$$ $$=|000rangle+|001rangle+|010rangle+|011rangle+|100rangle+|101rangle+|110rangle+|111rangle.$$ Then take the tensor product with $|1rangle$, i.e $$(H|0rangle otimes H|0rangle otimes H|0rangle)|1rangle$$ $$=|0001rangle+|0011rangle+|0101rangle+|0111rangle+|1001rangle+|1011rangle+|1101rangle+|1111rangle.$$ Is this right?
Actually, I know the ideas but don't know how to write it in the quantum computing terms, that involves using terms as registers, qubits
quantum-gate quantum-state
edited Apr 15 at 10:57
Sanchayan Dutta
8,1224 gold badges18 silver badges63 bronze badges
asked Apr 15 at 7:13
UpstartUpstart
4131 silver badge9 bronze badges
$endgroup$
add a comment
|
$begingroup$
This is just a basic question. I need to output odd integers till $15$, i.e $1,3,5,7,9,11,13,15$. Since $15$ requires $4$ bits, I prepare initial state by using Hadamard gate on initial $|0rangle$, i.e $$H|0rangle otimes H|0rangle otimes H|0rangle$$ $$=|000rangle+|001rangle+|010rangle+|011rangle+|100rangle+|101rangle+|110rangle+|111rangle.$$ Then take the tensor product with $|1rangle$, i.e $$(H|0rangle otimes H|0rangle otimes H|0rangle)|1rangle$$ $$=|0001rangle+|0011rangle+|0101rangle+|0111rangle+|1001rangle+|1011rangle+|1101rangle+|1111rangle.$$ Is this right?
Actually, I know the ideas but don't know how to write it in the quantum computing terms, that involves using terms as registers, qubits
quantum-gate quantum-state
edited Apr 15 at 10:57
Sanchayan Dutta
8,1224 gold badges18 silver badges63 bronze badges
asked Apr 15 at 7:13
UpstartUpstart
4131 silver badge9 bronze badges
$endgroup$
This is just a basic question. I need to output odd integers till $15$, i.e $1,3,5,7,9,11,13,15$. Since $15$ requires $4$ bits, I prepare initial state by using Hadamard gate on initial $|0rangle$, i.e $$H|0rangle otimes H|0rangle otimes H|0rangle$$ $$=|000rangle+|001rangle+|010rangle+|011rangle+|100rangle+|101rangle+|110rangle+|111rangle.$$ Then take the tensor product with $|1rangle$, i.e $$(H|0rangle otimes H|0rangle otimes H|0rangle)|1rangle$$ $$=|0001rangle+|0011rangle+|0101rangle+|0111rangle+|1001rangle+|1011rangle+|1101rangle+|1111rangle.$$ Is this right?
Actually, I know the ideas but don't know how to write it in the quantum computing terms, that involves using terms as registers, qubits
quantum-gate quantum-state
quantum-gate quantum-state
edited Apr 15 at 10:57
Sanchayan Dutta
8,1224 gold badges18 silver badges63 bronze badges
asked Apr 15 at 7:13
UpstartUpstart
4131 silver badge9 bronze badges
edited Apr 15 at 10:57
Sanchayan Dutta
8,1224 gold badges18 silver badges63 bronze badges
asked Apr 15 at 7:13
UpstartUpstart
4131 silver badge9 bronze badges
edited Apr 15 at 10:57
Sanchayan Dutta
8,1224 gold badges18 silver badges63 bronze badges
edited Apr 15 at 10:57
Sanchayan Dutta
8,1224 gold badges18 silver badges63 bronze badges
edited Apr 15 at 10:57
Sanchayan Dutta
8,1224 gold badges18 silver badges63 bronze badges
8,1224 gold badges18 silver badges63 bronze badges
asked Apr 15 at 7:13
UpstartUpstart
4131 silver badge9 bronze badges
asked Apr 15 at 7:13
UpstartUpstart
4131 silver badge9 bronze badges
asked Apr 15 at 7:13
UpstartUpstart
4131 silver badge9 bronze badges
4131 silver badge9 bronze badges
add a comment
|
add a comment
|
1 Answer
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It depends what you mean by "output odd integers till 15". If you actually mean "prepare a uniform superposition of all odd, positive integers up to and including 15", then yes, you are essentially correct. The only thing to say is that your initial preparation using Hadamards is only over 3 qubits, not 4 (you've got 4 qubits on the left-hand side and 3 on the right-hand side). Similarly for your second equation, you've got 5 qubits on the left-hand side and 4 on the right. You would also be safer to include the normalisation factor.
answered Apr 15 at 7:18
DaftWullieDaftWullie
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i have edited it
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– Upstart
Apr 15 at 7:21
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@Upstart better :)
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– DaftWullie
Apr 15 at 7:38
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1 Answer
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$begingroup$
It depends what you mean by "output odd integers till 15". If you actually mean "prepare a uniform superposition of all odd, positive integers up to and including 15", then yes, you are essentially correct. The only thing to say is that your initial preparation using Hadamards is only over 3 qubits, not 4 (you've got 4 qubits on the left-hand side and 3 on the right-hand side). Similarly for your second equation, you've got 5 qubits on the left-hand side and 4 on the right. You would also be safer to include the normalisation factor.
answered Apr 15 at 7:18
DaftWullieDaftWullie
20.4k1 gold badge9 silver badges51 bronze badges
$endgroup$
$begingroup$
i have edited it
$endgroup$
– Upstart
Apr 15 at 7:21
$begingroup$
@Upstart better :)
$endgroup$
– DaftWullie
Apr 15 at 7:38
add a comment
|
$begingroup$
It depends what you mean by "output odd integers till 15". If you actually mean "prepare a uniform superposition of all odd, positive integers up to and including 15", then yes, you are essentially correct. The only thing to say is that your initial preparation using Hadamards is only over 3 qubits, not 4 (you've got 4 qubits on the left-hand side and 3 on the right-hand side). Similarly for your second equation, you've got 5 qubits on the left-hand side and 4 on the right. You would also be safer to include the normalisation factor.
answered Apr 15 at 7:18
DaftWullieDaftWullie
20.4k1 gold badge9 silver badges51 bronze badges
$endgroup$
$begingroup$
i have edited it
$endgroup$
– Upstart
Apr 15 at 7:21
$begingroup$
@Upstart better :)
$endgroup$
– DaftWullie
Apr 15 at 7:38
add a comment
|
$begingroup$
It depends what you mean by "output odd integers till 15". If you actually mean "prepare a uniform superposition of all odd, positive integers up to and including 15", then yes, you are essentially correct. The only thing to say is that your initial preparation using Hadamards is only over 3 qubits, not 4 (you've got 4 qubits on the left-hand side and 3 on the right-hand side). Similarly for your second equation, you've got 5 qubits on the left-hand side and 4 on the right. You would also be safer to include the normalisation factor.
answered Apr 15 at 7:18
DaftWullieDaftWullie
20.4k1 gold badge9 silver badges51 bronze badges
$endgroup$
It depends what you mean by "output odd integers till 15". If you actually mean "prepare a uniform superposition of all odd, positive integers up to and including 15", then yes, you are essentially correct. The only thing to say is that your initial preparation using Hadamards is only over 3 qubits, not 4 (you've got 4 qubits on the left-hand side and 3 on the right-hand side). Similarly for your second equation, you've got 5 qubits on the left-hand side and 4 on the right. You would also be safer to include the normalisation factor.
answered Apr 15 at 7:18
DaftWullieDaftWullie
20.4k1 gold badge9 silver badges51 bronze badges
answered Apr 15 at 7:18
DaftWullieDaftWullie
20.4k1 gold badge9 silver badges51 bronze badges
answered Apr 15 at 7:18
DaftWullieDaftWullie
20.4k1 gold badge9 silver badges51 bronze badges
answered Apr 15 at 7:18
DaftWullieDaftWullie
20.4k1 gold badge9 silver badges51 bronze badges
20.4k1 gold badge9 silver badges51 bronze badges
$begingroup$
i have edited it
$endgroup$
– Upstart
Apr 15 at 7:21
$begingroup$
@Upstart better :)
$endgroup$
– DaftWullie
Apr 15 at 7:38
add a comment
|
$begingroup$
i have edited it
$endgroup$
– Upstart
Apr 15 at 7:21
$begingroup$
@Upstart better :)
$endgroup$
– DaftWullie
Apr 15 at 7:38
$begingroup$
i have edited it
$endgroup$
– Upstart
Apr 15 at 7:21
$begingroup$
i have edited it
$endgroup$
– Upstart
Apr 15 at 7:21
$begingroup$
@Upstart better :)
$endgroup$
– DaftWullie
Apr 15 at 7:38
$begingroup$
@Upstart better :)
$endgroup$
– DaftWullie
Apr 15 at 7:38
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