This is a screenshot from my lecture:

enter image description here

Why does the $\text{CNOT}$ here output $a ⊕ b$? Does the "$⊕$" operation even make sense for a quantum state?


1 Answer 1


Consider action of the $\text{CNOT}$ on the basis states $\{ |00\rangle,|01\rangle,|10\rangle,|11\rangle \}\,.$

$$|00\rangle \to |00\rangle \,, $$ $$|01\rangle \to |01\rangle \,, $$ $$|10\rangle \to |11\rangle \,, $$ $$|11\rangle \to |10\rangle \,. $$

Now, comparing this with $|ab\rangle$, You can observe that $$|a,b\rangle \to |a, a\oplus b\rangle \,. $$

Hence, the notation makes sense. $\text{CNOT}$ keeps the control qubit as it is and performs the $\text{XOR}$ operation of two states and outputs the result in the second, the target qubit.

  • $\begingroup$ Thanks @FDGod ! This makes sense for basis states, but how would it look like if both inputs were in a superposition? Like: $$\alpha_1|0\rangle + \beta_1|1\rangle $$ $$\alpha_2|0\rangle + \beta_2|1\rangle $$ $\endgroup$
    – Siddarth
    Feb 1 at 15:36
  • $\begingroup$ Does this help? $\endgroup$
    – FDGod
    Feb 1 at 16:09

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