Suppose that a qubit is in the state |φ〉 = a|0〉 + 1 − a2 |1〉 where a ∈ [−1, 1]. If we �rst
perform a standard basis measurement on this qubit and then perform a |u〉, |u⊥ 〉-basis mea-
√
surement where |u〉 = b|0〉 + 1 − b2 |1〉 for some b ∈ [−1, 1], what is the probability that the
outcome of the second measurement is u, in terms of a and b?
To find the probability of the outcome of the second measurement being u, we need to find the projection of the state |φ〉 onto the |u〉 basis and square the result.
Let's start by calculating the projection of the state |φ〉 onto the |u〉 basis. The projection can be obtained by taking the inner product of the state |φ〉 and the basis state |u〉:
< u | φ〉 = < u | (a|0〉 + (1 - a^2)|1〉)
= a< u | 0〉 + (1 - a^2)< u | 1〉
= a<b|0〉 + (1 - a^2)<b|1〉
Now, let's expand |u〉 in terms of basis states:
|u〉 = b|0〉 + (1 - b^2)|1〉
Substituting this into the previous equation:
< u | φ〉 = a<b|0〉 + (1 - a^2)<b|1〉
= ab<b|0〉 + (1 - a^2)(1 - b^2)<b|1〉
Next, let's find the probabilities associated with the outcome of the second measurement. The probability of obtaining the outcome |u〉 is given by the square of the projection:
P(u) = |< u | φ〉|^2
= |ab<b|0〉 + (1 - a^2)(1 - b^2)<b|1〉|^2
Since the basis states |0〉 and |1〉 are orthogonal, <b|0〉 and <b|1〉 are also orthogonal. Thus, their inner product is 0:
<b|0〉<b|1〉 = 0
Therefore, the probability P(u) simplifies to:
P(u) = |ab<b|0〉|^2
= |ab|^2
Finally, remembering that a and b are both within the range [-1, 1], the probability of the outcome of the second measurement being u in terms of a and b is:
P(u) = |ab|^2