Suppose we initially had n qubits in the state |ψ⟩=∑x∈{0,1}nαx|x⟩, on which we were going to perform Fourier sampling (i.e. perform Hadamard transform H⊗n and sample). Unfortunately, a clumsy lab assistant mistakenly applied X to one of the qubits and forgot which one it was.
3a. Suppose we initially had n qubits in the state |ψ⟩=∑x∈{0,1}nαx|x⟩, on which we were going to perform Fourier sampling (i.e. perform Hadamard transform H⊗n and sample). Unfortunately, a clumsy lab assistant mistakenly applied X to one of the qubits and forgot which one it was.
.βx′=βx
.βx′=−βx
.βx′=(−1)xiβx
.βx′=(−1)x⋅xβx
βx′=βx⊕ei
Hi Elviria, do you have any answer for this exam?
ƒÀx�Œ=(−1)xiƒÀx
what is that?
Another way to view it is to say that |ψ′⟩=Xi|ψ⟩ where Xi means X applied on the i-th qubit. We would like to claim that the application of Xi to the original state is effectively the same as applying some A to the i-th qubit after the Hadamard transform, i.e, H⊗n|ψ′⟩=H⊗nXi|ψ⟩=AiH⊗n|ψ⟩. What is A?
I
X
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H
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3A .βx′=(−1)xiβx
To answer this question, we need to consider the effect of applying the X gate (also known as a Pauli-X gate or a bit-flip gate) on a single qubit in the state |ψ⟩.
When the X gate is applied to a qubit, it flips the value of that qubit. In other words, if the qubit was in the state |0⟩, it will be changed to |1⟩, and if it was in the state |1⟩, it will be changed to |0⟩.
Since we don't know which qubit the clumsy lab assistant applied the X gate to, we need to consider all possible scenarios.
Let's consider the case where the X gate was applied to the i-th qubit. In this case, the state of the qubits after the X gate is applied would be:
|ψ'⟩ = X⊗I⊗(n-i-1)|ψ⟩
Where X⊗I⊗(n-i-1) represents the tensor product of the X gate applied to the i-th qubit and identity gates (I) applied to the rest of the qubits.
To find the new coefficients βx' in the state |ψ'⟩, we can apply the X⊗I⊗(n-i-1) gate to the original coefficients βx.
The effect of applying the X gate on a single qubit is that it multiplies the coefficient βx by (-1) if the corresponding qubit in x is 1. In other words:
βx' = (-1)^(xi) * βx
Where xi is the value of the i-th qubit in x.
So, for each coefficient βx, we need to compute xi and multiply it by (-1)^xi to get βx'. The final state |ψ'⟩ would be the same sum over x, but with the updated coefficients βx'.
Therefore, the correct answer for βx' in the case when the X gate was applied to the i-th qubit is βx' = (-1)^(xi) * βx.
In the case where we don't know which qubit the X gate was applied to, we need to consider all possible qubits and update the coefficients accordingly. This means we need to consider all possible values of xi for each qubit and apply the corresponding multiplication factor (-1)^(xi) to the coefficients βx.
To summarize, the updated coefficients βx' after applying the X gate to one of the qubits, but not knowing which one, would be:
βx' = ∏(i=1 to n) ((-1)^(xi) * βx)
Where xi is the value of the i-th qubit in x, and ∏ represents the product over all qubits.