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This must be trivial but I can't find a clear explanation of the notation of $| a, a \oplus b \rangle$.

It is the resulting state after applying a CNOT gate to $|a,b\rangle$: $\rm{CNOT}|a,b\rangle = | a, a \oplus b \rangle$.

Is it correct that $| a, a \oplus b \rangle = | a\rangle \otimes |a \oplus b \rangle$?

Suppose that $|a\rangle=a_1 |0\rangle + a_2 |1\rangle$, and $|b\rangle=b_1 |0\rangle + b_2 |1\rangle$, then $|a \oplus b \rangle = ?$

I know that $\rm{CNOT}|a,b\rangle=a_1 |0\rangle \otimes |b\rangle + a_2 |1\rangle \otimes (b_2 |0\rangle + b_1 |1\rangle)$. Then how to go from here to $| a, a \oplus b \rangle$?

Thank you in advance.

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1 Answer 1

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As you correctly wrote $|a,a\oplus b\rangle:=|a\rangle\otimes|a\oplus b\rangle$ where $a$ as well as $a\oplus b$ are either $0$ or $1$. The reason for the latter is that the logic symbol $\oplus$ stands for "exclusive or" so it acts on $\{0,1\}^2$ via $0\oplus 0:=0=:1\oplus 1$ and $0\oplus 1:=1=:1\oplus 0$. Thus given any two vectors $\psi,\phi\in\mathbb C^2$ the expression $|\psi\oplus\phi\rangle$ has no inherent meaning.

The reason one defines ${\rm CNOT}|a,b\rangle:=|a,a\oplus b\rangle$ for $a,b\in\{0,1\}$ first is of course that it can be extended from basis states $|0,0\rangle,|0,1\rangle,\ldots$ to all of $\mathbb C^2\otimes\mathbb C^2$ via linearity: \begin{align*} {\rm CNOT}(|\psi\rangle\otimes|\phi\rangle)&=\langle 0|\psi\rangle \langle 0|\phi\rangle{\rm CNOT}|0,0\rangle+\langle 0|\psi\rangle \langle 1|\phi\rangle{\rm CNOT}|0,1\rangle+\ldots\\ &=\langle 0|\psi\rangle \langle 0|\phi\rangle|0,0\rangle+\langle 0|\psi\rangle \langle 1|\phi\rangle|0,1\rangle+ \langle 1|\psi\rangle \langle 0|\phi\rangle|1,1\rangle+\langle 1|\psi\rangle \langle 1|\phi\rangle|1,0\rangle \end{align*}

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  • $\begingroup$ Thanks a lot. Now it is clear. $\endgroup$
    – fishjojo
    Commented Apr 11 at 22:06

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