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Please, help me understand this statement. The outer product notation for matrices also gives an intuitive input-output relation for them. For instance, the matrix |0⟩ ⟨1| + |1⟩ ⟨0| can be read as "output 0 when given a 1 and output 1 when given a 0".

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Note that

$$ |0\rangle\langle 1| = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \end{pmatrix}= \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$ and similarly

$$ |1\rangle\langle 0| = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \end{pmatrix}= \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} $$ and therefore $$ X = |0\rangle\langle 1| + |1\rangle\langle 0| = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} $$

Thus you can see that $X|0\rangle = |1\rangle$ and $X|1\rangle = |0\rangle$. So giving the input is the state $|0\rangle$ then the output is the state $|1\rangle$ and vice versa.

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  • $\begingroup$ Hi Binshumesh, adding to the well-written answer above, you can intuitively think about the 'bra' in the outer product (operator) as the part combining (via the inner product) with the vector input to the operator to give the numeric coefficient of the output vector, and the 'ket' in the operator being the label of the output vector itself. So (|0><1|)|1> =( |0>)(<1|1>) = (<1|1>)(|0>) = (1)(|0>) = |0>. $\endgroup$
    – Dhruv B
    Oct 15 '20 at 18:21
  • $\begingroup$ Thanks for your explanation. $\endgroup$ Oct 19 '20 at 8:23

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