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!
$\begingroup$
$\endgroup$
5
-
1$\begingroup$ 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? $\endgroup$– GaussStrifeFeb 20, 2022 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 SahuFeb 21, 2022 at 2:43
-
$\begingroup$ Yeah I just assumed that was what you meant. Did my answer explain things? $\endgroup$– GaussStrifeFeb 21, 2022 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 SahuFeb 27, 2022 at 7:42
-
$\begingroup$ The matrix multiplication or the actioj of A? I'm not sure what part my answer didn't cover. $\endgroup$– GaussStrifeFeb 28, 2022 at 11:47
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
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.
