What does it mean to express a gate in Dirac notation?How does bra-ket notation work?How does Fourier sampling actually work (and solve the parity problem)?What do we mean by the notation $lvert mathbfx, 0rangle$?What does the notation $lvert underlinex rangle$ mean?How to properly write the action of a quantum gate implementing an operator $U$ on the superposition of its eigenvectors?Notation for two entangled registersWhy is correlation in the $X$ basis represented as $Xotimes X = 1$?N&C quantum circuit for Grover's algorithmNotation for two qubit composite product stateWhat is the tensorial representation of the quantum swap gate?
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What does it mean to express a gate in Dirac notation?
How does bra-ket notation work?How does Fourier sampling actually work (and solve the parity problem)?What do we mean by the notation $lvert mathbfx, 0rangle$?What does the notation $lvert underlinex rangle$ mean?How to properly write the action of a quantum gate implementing an operator $U$ on the superposition of its eigenvectors?Notation for two entangled registersWhy is correlation in the $X$ basis represented as $Xotimes X = 1$?N&C quantum circuit for Grover's algorithmNotation for two qubit composite product stateWhat is the tensorial representation of the quantum swap gate?
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When discussing the Dirac notation of an operator, for example, let's just say we have the bit flip gate $X$ if we want to write this in the Dirac notation does that just mean writing it as follows?
$$X|psirangle=X(c_0|0rangle+c_1|1rangle)=c_0|1rangle+c_1|0rangle$$
quantum-gate notation
edited 1 hour ago
Sanchayan Dutta♦
6,72341556
asked 1 hour ago
can'tcauchycan'tcauchy
1945
$endgroup$
add a comment |
$begingroup$
When discussing the Dirac notation of an operator, for example, let's just say we have the bit flip gate $X$ if we want to write this in the Dirac notation does that just mean writing it as follows?
$$X|psirangle=X(c_0|0rangle+c_1|1rangle)=c_0|1rangle+c_1|0rangle$$
quantum-gate notation
edited 1 hour ago
Sanchayan Dutta♦
6,72341556
asked 1 hour ago
can'tcauchycan'tcauchy
1945
$endgroup$
add a comment |
$begingroup$
When discussing the Dirac notation of an operator, for example, let's just say we have the bit flip gate $X$ if we want to write this in the Dirac notation does that just mean writing it as follows?
$$X|psirangle=X(c_0|0rangle+c_1|1rangle)=c_0|1rangle+c_1|0rangle$$
quantum-gate notation
edited 1 hour ago
Sanchayan Dutta♦
6,72341556
asked 1 hour ago
can'tcauchycan'tcauchy
1945
$endgroup$
When discussing the Dirac notation of an operator, for example, let's just say we have the bit flip gate $X$ if we want to write this in the Dirac notation does that just mean writing it as follows?
$$X|psirangle=X(c_0|0rangle+c_1|1rangle)=c_0|1rangle+c_1|0rangle$$
quantum-gate notation
quantum-gate notation
edited 1 hour ago
Sanchayan Dutta♦
6,72341556
asked 1 hour ago
can'tcauchycan'tcauchy
1945
edited 1 hour ago
Sanchayan Dutta♦
6,72341556
asked 1 hour ago
can'tcauchycan'tcauchy
1945
edited 1 hour ago
Sanchayan Dutta♦
6,72341556
edited 1 hour ago
Sanchayan Dutta♦
6,72341556
edited 1 hour ago
Sanchayan Dutta♦
6,72341556
6,72341556
asked 1 hour ago
can'tcauchycan'tcauchy
1945
asked 1 hour ago
can'tcauchycan'tcauchy
1945
asked 1 hour ago
can'tcauchycan'tcauchy
1945
1945
add a comment |
add a comment |
2 Answers
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This might mean using the ketbra notation:
$$X = |1rangle langle0| + |1rangle langle0|$$
This notation describes the effect the operator has on the basis vectors: in this case $X$ converts $|0rangle$ into $|1rangle$ and vice versa.
A couple of other examples:
$$Z = |0rangle langle0| - |1rangle langle1|$$
$$operatornameCNOT = |0 ranglelangle0| otimes I + |1 ranglelangle 1| otimes X = |00rangle langle00| + |01rangle langle01| + |11rangle langle10| + |10rangle langle11|$$
edited 30 mins ago
Sanchayan Dutta♦
6,72341556
answered 1 hour ago
Mariia MykhailovaMariia Mykhailova
1,9401212
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For OP: a quick and intuitive way to derive the CNOT's outer product representation is to see what effect it has on the individual qubits. The CNOT flips the second qubit when the control qubit is $|1rangle$. When the control qubit is $|0rangle$ the second qubit remains intact, which is equivalent to applying the identity gate $I$. So we get $|0ranglelangle 0|otimes I$ for this. However, when the control qubit is $|1rangle$ we need to flip the second qubit's state i.e. we need to apply the $X$ gate to the second qubit. Thus, we get $|1ranglelangle 1|otimes X$ for this.
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– Sanchayan Dutta♦
15 mins ago
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The Dirac notation for the Pauli-$X$ gate is:
$$|1rangle langle0| + |1rangle langle0|.$$
Now you might be wondering where this comes from. The term you're looking for is outer product representation of the $X$ gate. It follows from the spectral decomposition theorem (check Nielsen & Chuang 10th edition, p. 72) which holds for all normal operators. The key point:
In terms of the outer product representation, this means that $M$ can be written as $M=sum_ilambda_i|iranglelangle i|$,where $lambda_i$ are the eigenvalues of $M$,$|irangle$ is an orthonormal basis for $V$, and each $|irangle$ an eigenvector of $M$ with eigenvalue $lambda_i$.
The eigenvectors of the Pauli-$X$ gate are $-|0rangle+|1rangle$ and $|0rangle+|1rangle$, and the corresponding eigenvalues are $-1$ and $+1$ cf. Wolfram Alpha. Normalize the eigenvectors to get an orthonormal basis for $X$ i.e. $frac0rangle+sqrt2,fracsqrt2$. According the spectral decomposition theorem you can represent the $X$ gate as:
$$-1(frac0rangle+sqrt2)(fracsqrt2}) + 1(frac
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For OP: a quick and intuitive way to derive the CNOT's outer product representation is to see what effect it has on the individual qubits. The CNOT flips the second qubit when the control qubit is $) + 1(frac1rangle$ are orthonormal vectors i.e. their inner product $langle psi) + 1(frac1rangle langle0) + 1(frac+) + 1(frac{sqrt2)(fraclangle 0sqrt2)$$
$$=-frac12(|0ranglelangle0|-|0ranglelangle1|-|1ranglelangle0|+|1ranglelangle1|)+frac12(|0ranglelangle0|+|0ranglelangle1|+|1ranglelangle0|+|1ranglelangle1|)$$
$$=|1rangle langle0| + |1rangle langle0|$$To convince you that this result is correct let's apply it on an arbitrary qubit state $c_0|0rangle+c_1 |1rangle$:
$$(|1rangle langle0| + |0rangle langle1|)(c_0|0rangle+c_1|1rangle)$$
$$=c_0|1ranglelangle0|0rangle+c_1|0ranglelangle 1|1rangle$$
$$=c_0 |1rangle + c_1 |0rangle$$So yes, our result is correct and the bits were indeed flipped upon application of $X=|1rangle langle0| + |1rangle langle0|$ to $c_0|0rangle + c_1|1rangle$.The last step followed from the fact that $langle 0|0rangle$ and $langle 1|1rangle$ are both equal to $1$, as $|0rangle$ and $|1rangle$ are orthonormal vectors i.e. their inner product $langle psi|psirangle=1$ by definition.
We're done. As an exercise, find the outer product representation of the Pauli-$Z$ gate by yourself. And definitely, do go through the proof of the spectral theorem in Nielsen and Chung if time permits!
answered 44 mins ago
Sanchayan Dutta♦Sanchayan Dutta
6,72341556
edited 26 mins ago
answered 44 mins ago
Sanchayan Dutta♦Sanchayan Dutta
6,72341556
answered 44 mins ago
Sanchayan Dutta♦Sanchayan Dutta
6,72341556
answered 44 mins ago
Sanchayan Dutta♦Sanchayan Dutta
6,72341556
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