I Choice of basis in BB84 protocol

[lys04]

In the BB84 protocol, we have two basis, the Z basis which consists of |0> and |1> which represents bit values of 0 and 1 respectively in the Z basis. From this we construct another basis, the X basis which consists of $|+> = \frac{1}{\sqrt(2)} |0> + \frac{1}{\sqrt(2)} |1>$ and $|-> = \frac{1}{\sqrt(2)} |0> - \frac{1}{\sqrt(2)} |1>$ which also represents bit values of 0 and 1 in the X basis.

The purpose of having two basis is to prevent the eavesdropper as the act of measurement in the wrong basis has a chance of changing the state and thus introduce errors which can be detected.
Technically taking any linear combination of the Z basis to form a new basis has this effect but choosing $\frac{1}{\sqrt(2)}$ means the probabilities are symmetric. If the probabilities weren’t symmetric, then this means the probabilities of detecting the Eve for one bit is higher than the other bit, so it’s kind of like a trade off between probabilities?

 

[Nugatory]

so it’s kind of like a trade off between probabilities?

It seems that way at first glance, but no. We detect Eve’s presence only when Alice and Bob find that the keys they have constructed don’t work because at least one bit doesn’t match. And the chances that Eve’s tampering will cause a mismatch are greatest when the axes are 90 degrees apart.
It may be easiest to see this by considering the limiting case in which all the bits are sent using the same axis - as long as Eve knows what that axis is the eavesdropping will be undetectable.

 

[lys04]

It seems that way at first glance, but no.

Why not
Would this be a valid example?
Suppose instead of the X basis was $|+ \rangle = \frac{\sqrt{2}}{\sqrt{10}} |0 \rangle + \frac{\sqrt{8}}{\sqrt{10}} |1 \rangle$
And Alice sends out a 0 encoded in the X basis i.e a $| + \rangle$ and the eve tries to intercept by making a measurement in the rectilinear basis. Then the eve has a probability of $\frac{2}{10}$ of measuring $| 0 \rangle$, i.e 0, and a probability of $\frac{8}{10}$ measuring a $|1 \rangle$.

The eve can be detected if he measures in the wrong basis and this results in the wrong bit value. In my definition of $| + \rangle$, the eve has a probability $\frac{4}{10}$ of being detected. Therefore in n bits of 0's encoded in the X basis, the probability of not detecting the eve is $\left( \frac{6}{10} \right)^n$ and thus the probability of detecting the eve after comparing n qubits is $1- \left( \frac{6}{10} \right) ^n$ which if you compare to having symmetric probabilities which is $1-\left( \frac{3}{4} \right) ^n$, it's higher.