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Going from a circuit to the quantum state output of the circuit
How to construct matrix of regular and “flipped” 2-qubit CNOT?Why is it important to eliminate the garbage qubits?Why do we use ancilla qubits for error syndrome measurements?How exactly is the stated composite state of the two registers being produced using the $R_zz$ controlled rotations?SWAP gate(s) in the $R(lambda^-1)$ step of the HHL circuit for $4times 4$ systemsDecomposition of arbitrary 2 qubit operatorCalculating measurement result of quantum swap circuitN&C quantum circuit for Grover's algorithmHow to derive a circuit from given equations?Approximating unitary matrices — restricted gatesetIs the intuition of quantum parallelism always correct?
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$begingroup$
I'm looking at the following lecture notes where we start with the circuit below for some state $vertpsirangle_L$ that picks up an error to become $Evertpsirangle_L$
It is later claimed in the notes that the syndrome extraction part of the circuit can be represented by the following operation on $Evertpsirangle_L$.
$$E|psirangle_L|0rangle_A rightarrow frac12left(I_1 I_2+Z_1 Z_2right) E|psirangle_L|0rangle_A+frac12left(I_1I_2-Z_1 Z_2right) E|psirangle_L|1rangle_A$$
How does one see this? I can write the Hadamard and the control $Z_1Z_2$ gates as 8x8 matrices but this seems like a tedious way to do it. The alternative is to express the control $Z_1Z_2$ gates using something like this answer. However, I was unable to do it this way either.
So the question is - how do I see that the following line is true just by looking at the circuit?
$$E|psirangle_L|0rangle_A rightarrow frac12left(I_1 I_2+Z_1 Z_2right) E|psirangle_L|0rangle_A+frac12left(I_1I_2-Z_1 Z_2right) E|psirangle_L|1rangle_A$$
quantum-gate error-correction
asked Jul 29 at 16:09
user1936752user1936752
3086 bronze badges
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add a comment |
$begingroup$
I'm looking at the following lecture notes where we start with the circuit below for some state $vertpsirangle_L$ that picks up an error to become $Evertpsirangle_L$
It is later claimed in the notes that the syndrome extraction part of the circuit can be represented by the following operation on $Evertpsirangle_L$.
$$E|psirangle_L|0rangle_A rightarrow frac12left(I_1 I_2+Z_1 Z_2right) E|psirangle_L|0rangle_A+frac12left(I_1I_2-Z_1 Z_2right) E|psirangle_L|1rangle_A$$
How does one see this? I can write the Hadamard and the control $Z_1Z_2$ gates as 8x8 matrices but this seems like a tedious way to do it. The alternative is to express the control $Z_1Z_2$ gates using something like this answer. However, I was unable to do it this way either.
So the question is - how do I see that the following line is true just by looking at the circuit?
$$E|psirangle_L|0rangle_A rightarrow frac12left(I_1 I_2+Z_1 Z_2right) E|psirangle_L|0rangle_A+frac12left(I_1I_2-Z_1 Z_2right) E|psirangle_L|1rangle_A$$
quantum-gate error-correction
asked Jul 29 at 16:09
user1936752user1936752
3086 bronze badges
$endgroup$
add a comment |
$begingroup$
I'm looking at the following lecture notes where we start with the circuit below for some state $vertpsirangle_L$ that picks up an error to become $Evertpsirangle_L$
It is later claimed in the notes that the syndrome extraction part of the circuit can be represented by the following operation on $Evertpsirangle_L$.
$$E|psirangle_L|0rangle_A rightarrow frac12left(I_1 I_2+Z_1 Z_2right) E|psirangle_L|0rangle_A+frac12left(I_1I_2-Z_1 Z_2right) E|psirangle_L|1rangle_A$$
How does one see this? I can write the Hadamard and the control $Z_1Z_2$ gates as 8x8 matrices but this seems like a tedious way to do it. The alternative is to express the control $Z_1Z_2$ gates using something like this answer. However, I was unable to do it this way either.
So the question is - how do I see that the following line is true just by looking at the circuit?
$$E|psirangle_L|0rangle_A rightarrow frac12left(I_1 I_2+Z_1 Z_2right) E|psirangle_L|0rangle_A+frac12left(I_1I_2-Z_1 Z_2right) E|psirangle_L|1rangle_A$$
quantum-gate error-correction
asked Jul 29 at 16:09
user1936752user1936752
3086 bronze badges
$endgroup$
I'm looking at the following lecture notes where we start with the circuit below for some state $vertpsirangle_L$ that picks up an error to become $Evertpsirangle_L$
It is later claimed in the notes that the syndrome extraction part of the circuit can be represented by the following operation on $Evertpsirangle_L$.
$$E|psirangle_L|0rangle_A rightarrow frac12left(I_1 I_2+Z_1 Z_2right) E|psirangle_L|0rangle_A+frac12left(I_1I_2-Z_1 Z_2right) E|psirangle_L|1rangle_A$$
How does one see this? I can write the Hadamard and the control $Z_1Z_2$ gates as 8x8 matrices but this seems like a tedious way to do it. The alternative is to express the control $Z_1Z_2$ gates using something like this answer. However, I was unable to do it this way either.
So the question is - how do I see that the following line is true just by looking at the circuit?
$$E|psirangle_L|0rangle_A rightarrow frac12left(I_1 I_2+Z_1 Z_2right) E|psirangle_L|0rangle_A+frac12left(I_1I_2-Z_1 Z_2right) E|psirangle_L|1rangle_A$$
quantum-gate error-correction
quantum-gate error-correction
asked Jul 29 at 16:09
user1936752user1936752
3086 bronze badges
asked Jul 29 at 16:09
user1936752user1936752
3086 bronze badges
asked Jul 29 at 16:09
user1936752user1936752
3086 bronze badges
asked Jul 29 at 16:09
user1936752user1936752
3086 bronze badges
asked Jul 29 at 16:09
user1936752user1936752
3086 bronze badges
3086 bronze badges
add a comment |
add a comment |
1 Answer
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$begingroup$
Let's represent controlled $Z_1Z_2$ gate in the projector formalism, as described in this answer:
$$C_AZ_1Z_2 = |0ranglelangle0|_A I_1I_2 + |1ranglelangle1|_A Z_1Z_2 $$
This just tells you to apply identity gates to qubits 1 and 2 if the ancilla is in the $|0rangle$ state and to apply Z gates to qubits 1 and 2 if the ancilla is in the $|1rangle$ state - which is the definition of the controlled gate.
Now let's apply this to the state $colorblue_AE|psirangle_L$ (this is the state of the system after the first Hadamard gate of syndrome extraction):
$$C_AZ_1Z_2 big( colorblue_AE|psirangle_L big) = big( colorblue_A I_1I_2 + colorblue1ranglelangle1_A Z_1Z_2 big) bigg( frac1sqrt2colorblue0rangle + _AE|psirangle_L bigg) = $$
$$= frac1sqrt2 big( colorblue0rangle_A otimes I_1I_2 E|psirangle_L + colorblue1rangle_A otimes Z_1Z_2 E|psirangle_L big)$$
Finally, apply the last Hadamard gate to the ancilla; after that the state of the system becomes
$$frac1sqrt2 big( colorblue_A otimes I_1I_2 E|psirangle_L + colorblue-rangle_A otimes Z_1Z_2 E|psirangle_L big) = $$
$$= frac12 big( colorblue0rangle + _A otimes I_1I_2 E|psirangle_L + colorblue0rangle - _A otimes Z_1Z_2 E|psirangle_L big) = $$
(after reordering the terms and grouping same ancilla states together)
$$= frac12 colorblue0rangle_A otimes left(I_1 I_2+Z_1 Z_2right) E|psirangle_L + frac12 colorblue1rangle_A otimes left(I_1I_2-Z_1 Z_2right) E|psirangle_L$$
which is exactly the state you need to get.
answered Jul 29 at 18:53
Mariia MykhailovaMariia Mykhailova
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1 Answer
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1 Answer
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$begingroup$
Let's represent controlled $Z_1Z_2$ gate in the projector formalism, as described in this answer:
$$C_AZ_1Z_2 = |0ranglelangle0|_A I_1I_2 + |1ranglelangle1|_A Z_1Z_2 $$
This just tells you to apply identity gates to qubits 1 and 2 if the ancilla is in the $|0rangle$ state and to apply Z gates to qubits 1 and 2 if the ancilla is in the $|1rangle$ state - which is the definition of the controlled gate.
Now let's apply this to the state $colorblue_AE|psirangle_L$ (this is the state of the system after the first Hadamard gate of syndrome extraction):
$$C_AZ_1Z_2 big( colorblue_AE|psirangle_L big) = big( colorblue_A I_1I_2 + colorblue1ranglelangle1_A Z_1Z_2 big) bigg( frac1sqrt2colorblue0rangle + _AE|psirangle_L bigg) = $$
$$= frac1sqrt2 big( colorblue0rangle_A otimes I_1I_2 E|psirangle_L + colorblue1rangle_A otimes Z_1Z_2 E|psirangle_L big)$$
Finally, apply the last Hadamard gate to the ancilla; after that the state of the system becomes
$$frac1sqrt2 big( colorblue_A otimes I_1I_2 E|psirangle_L + colorblue-rangle_A otimes Z_1Z_2 E|psirangle_L big) = $$
$$= frac12 big( colorblue0rangle + _A otimes I_1I_2 E|psirangle_L + colorblue0rangle - _A otimes Z_1Z_2 E|psirangle_L big) = $$
(after reordering the terms and grouping same ancilla states together)
$$= frac12 colorblue0rangle_A otimes left(I_1 I_2+Z_1 Z_2right) E|psirangle_L + frac12 colorblue1rangle_A otimes left(I_1I_2-Z_1 Z_2right) E|psirangle_L$$
which is exactly the state you need to get.
answered Jul 29 at 18:53
Mariia MykhailovaMariia Mykhailova
3,2751 gold badge3 silver badges20 bronze badges
$endgroup$
add a comment |
$begingroup$
Let's represent controlled $Z_1Z_2$ gate in the projector formalism, as described in this answer:
$$C_AZ_1Z_2 = |0ranglelangle0|_A I_1I_2 + |1ranglelangle1|_A Z_1Z_2 $$
This just tells you to apply identity gates to qubits 1 and 2 if the ancilla is in the $|0rangle$ state and to apply Z gates to qubits 1 and 2 if the ancilla is in the $|1rangle$ state - which is the definition of the controlled gate.
Now let's apply this to the state $colorblue_AE|psirangle_L$ (this is the state of the system after the first Hadamard gate of syndrome extraction):
$$C_AZ_1Z_2 big( colorblue_AE|psirangle_L big) = big( colorblue_A I_1I_2 + colorblue1ranglelangle1_A Z_1Z_2 big) bigg( frac1sqrt2colorblue0rangle + _AE|psirangle_L bigg) = $$
$$= frac1sqrt2 big( colorblue0rangle_A otimes I_1I_2 E|psirangle_L + colorblue1rangle_A otimes Z_1Z_2 E|psirangle_L big)$$
Finally, apply the last Hadamard gate to the ancilla; after that the state of the system becomes
$$frac1sqrt2 big( colorblue_A otimes I_1I_2 E|psirangle_L + colorblue-rangle_A otimes Z_1Z_2 E|psirangle_L big) = $$
$$= frac12 big( colorblue0rangle + _A otimes I_1I_2 E|psirangle_L + colorblue0rangle - _A otimes Z_1Z_2 E|psirangle_L big) = $$
(after reordering the terms and grouping same ancilla states together)
$$= frac12 colorblue0rangle_A otimes left(I_1 I_2+Z_1 Z_2right) E|psirangle_L + frac12 colorblue1rangle_A otimes left(I_1I_2-Z_1 Z_2right) E|psirangle_L$$
which is exactly the state you need to get.
answered Jul 29 at 18:53
Mariia MykhailovaMariia Mykhailova
3,2751 gold badge3 silver badges20 bronze badges
$endgroup$
add a comment |
$begingroup$
Let's represent controlled $Z_1Z_2$ gate in the projector formalism, as described in this answer:
$$C_AZ_1Z_2 = |0ranglelangle0|_A I_1I_2 + |1ranglelangle1|_A Z_1Z_2 $$
This just tells you to apply identity gates to qubits 1 and 2 if the ancilla is in the $|0rangle$ state and to apply Z gates to qubits 1 and 2 if the ancilla is in the $|1rangle$ state - which is the definition of the controlled gate.
Now let's apply this to the state $colorblue_AE|psirangle_L$ (this is the state of the system after the first Hadamard gate of syndrome extraction):
$$C_AZ_1Z_2 big( colorblue_AE|psirangle_L big) = big( colorblue_A I_1I_2 + colorblue1ranglelangle1_A Z_1Z_2 big) bigg( frac1sqrt2colorblue0rangle + _AE|psirangle_L bigg) = $$
$$= frac1sqrt2 big( colorblue0rangle_A otimes I_1I_2 E|psirangle_L + colorblue1rangle_A otimes Z_1Z_2 E|psirangle_L big)$$
Finally, apply the last Hadamard gate to the ancilla; after that the state of the system becomes
$$frac1sqrt2 big( colorblue_A otimes I_1I_2 E|psirangle_L + colorblue-rangle_A otimes Z_1Z_2 E|psirangle_L big) = $$
$$= frac12 big( colorblue0rangle + _A otimes I_1I_2 E|psirangle_L + colorblue0rangle - _A otimes Z_1Z_2 E|psirangle_L big) = $$
(after reordering the terms and grouping same ancilla states together)
$$= frac12 colorblue0rangle_A otimes left(I_1 I_2+Z_1 Z_2right) E|psirangle_L + frac12 colorblue1rangle_A otimes left(I_1I_2-Z_1 Z_2right) E|psirangle_L$$
which is exactly the state you need to get.
answered Jul 29 at 18:53
Mariia MykhailovaMariia Mykhailova
3,2751 gold badge3 silver badges20 bronze badges
$endgroup$
Let's represent controlled $Z_1Z_2$ gate in the projector formalism, as described in this answer:
$$C_AZ_1Z_2 = |0ranglelangle0|_A I_1I_2 + |1ranglelangle1|_A Z_1Z_2 $$
This just tells you to apply identity gates to qubits 1 and 2 if the ancilla is in the $|0rangle$ state and to apply Z gates to qubits 1 and 2 if the ancilla is in the $|1rangle$ state - which is the definition of the controlled gate.
Now let's apply this to the state $colorblue_AE|psirangle_L$ (this is the state of the system after the first Hadamard gate of syndrome extraction):
$$C_AZ_1Z_2 big( colorblue_AE|psirangle_L big) = big( colorblue_A I_1I_2 + colorblue1ranglelangle1_A Z_1Z_2 big) bigg( frac1sqrt2colorblue0rangle + _AE|psirangle_L bigg) = $$
$$= frac1sqrt2 big( colorblue0rangle_A otimes I_1I_2 E|psirangle_L + colorblue1rangle_A otimes Z_1Z_2 E|psirangle_L big)$$
Finally, apply the last Hadamard gate to the ancilla; after that the state of the system becomes
$$frac1sqrt2 big( colorblue_A otimes I_1I_2 E|psirangle_L + colorblue-rangle_A otimes Z_1Z_2 E|psirangle_L big) = $$
$$= frac12 big( colorblue0rangle + _A otimes I_1I_2 E|psirangle_L + colorblue0rangle - _A otimes Z_1Z_2 E|psirangle_L big) = $$
(after reordering the terms and grouping same ancilla states together)
$$= frac12 colorblue0rangle_A otimes left(I_1 I_2+Z_1 Z_2right) E|psirangle_L + frac12 colorblue1rangle_A otimes left(I_1I_2-Z_1 Z_2right) E|psirangle_L$$
which is exactly the state you need to get.
answered Jul 29 at 18:53
Mariia MykhailovaMariia Mykhailova
3,2751 gold badge3 silver badges20 bronze badges
answered Jul 29 at 18:53
Mariia MykhailovaMariia Mykhailova
3,2751 gold badge3 silver badges20 bronze badges
answered Jul 29 at 18:53
Mariia MykhailovaMariia Mykhailova
3,2751 gold badge3 silver badges20 bronze badges
answered Jul 29 at 18:53
Mariia MykhailovaMariia Mykhailova
3,2751 gold badge3 silver badges20 bronze badges
3,2751 gold badge3 silver badges20 bronze badges
add a comment |
add a comment |
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