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this is the question I am referring to

I am sorry to ask such a question, can someone explain how are we able to make the new form of the operator A.

Thank you, any help would be great!

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    $\begingroup$ What do you mean by the new form? Do you mean the second example they provided in which they has a different input and output basis? $\endgroup$ Feb 20 at 17:01
  • $\begingroup$ In the beginning the answer starts with a result, equating the operator A to sum of outer product of basis states, i don't understand how that happened! $\endgroup$
    – Rohit Sahu
    Feb 21 at 2:43
  • $\begingroup$ Yeah I just assumed that was what you meant. Did my answer explain things? $\endgroup$ Feb 21 at 13:59
  • $\begingroup$ I might be wrong, but stating that result is essentially the answer what remains is just tensor or matrix matrix multiplication. Can you explain that in baby steps? :) Thank you! $\endgroup$
    – Rohit Sahu
    Feb 27 at 7:42
  • $\begingroup$ The matrix multiplication or the actioj of A? I'm not sure what part my answer didn't cover. $\endgroup$ Feb 28 at 11:47

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If you mean how does the answer immediately state that $A=|1\rangle \langle 0|+|0\rangle \langle 1|$, the reason is because in the question, it tells us that $A|0\rangle = |1\rangle$ and $A|1\rangle = |0\rangle$

So we can see that we need two operators that perform these mappings:

$|1\rangle \langle 0|0\rangle = |1\rangle$ and $|0\rangle \langle 1|1\rangle=|0\rangle$

A far easier way, however, is simply taking what we know about the behaviour of A, and using the Identity operator along with the inner product function:

$$IAI=\sum_{i}|i\rangle \langle i|A\sum_{j}|j\rangle \langle j|=\sum_{i,j}|i\rangle \langle j|\langle i|A|j\rangle$$

This way, you can calculate the entries of A based on it's actions on the input and output states, along with the associated operator.

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