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.


1 Answer 1


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*}

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.