Asked by Anonymous
Suppose we have a quantum circuit that takes the input |0⟩ and outputs |+⟩, and also takes the input |1⟩ and outputs −|−⟩. If we input √2*i/√3|+⟩ + 1/√3|−⟩, what does the circuit output?
Answers
Answered by
Count Iblis
I'm going to assuming that:
|+> = 1/sqrt(2) [|0> + |1>]
|-> = 1/sqrt(2) [|0> - |1>]
Then there is a unitary operator U that relates output to input as:
|output> = U|input>
You can then calculate the output without having to expand everything in one basis, you can simplify things by invoking the fact that U is unitary.
We want to compute:
|s> = U|k>
with |k> = i sqrt(2/3)|+> + 1/sqrt(3)|->
Note that if:
|b> = U|a>
then
<b| = <a|U-dagger ----->
<b|U = <a|U-dagger U = <a|
So, we have:
<+|s> = <+|U|k> = <0|k> =
i/sqrt(3) + 1/sqrt(6)
<-|s> = <-|U|k> = -<1|k> =
-i/sqrt(3) + 1/sqrt(6)
Therefore:
|s> = [i/sqrt(3) + 1/sqrt(6)]|+>
+ [-i/sqrt(3) + 1/sqrt(6)] |->
|+> = 1/sqrt(2) [|0> + |1>]
|-> = 1/sqrt(2) [|0> - |1>]
Then there is a unitary operator U that relates output to input as:
|output> = U|input>
You can then calculate the output without having to expand everything in one basis, you can simplify things by invoking the fact that U is unitary.
We want to compute:
|s> = U|k>
with |k> = i sqrt(2/3)|+> + 1/sqrt(3)|->
Note that if:
|b> = U|a>
then
<b| = <a|U-dagger ----->
<b|U = <a|U-dagger U = <a|
So, we have:
<+|s> = <+|U|k> = <0|k> =
i/sqrt(3) + 1/sqrt(6)
<-|s> = <-|U|k> = -<1|k> =
-i/sqrt(3) + 1/sqrt(6)
Therefore:
|s> = [i/sqrt(3) + 1/sqrt(6)]|+>
+ [-i/sqrt(3) + 1/sqrt(6)] |->
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