I am currently working with an exercise set about discriminating between two different quantum states. We consider a 2-dimensional Hilbert space expanded by $|\phi_1\rangle$, $|\phi_2\rangle$. The system is secretly prepared into one of the following states at random:
$|\psi_1\rangle=|{\phi_1}\rangle \tag{1}\\$ $|\psi_2\rangle=\text{cos}\theta|\phi_1\rangle+\text{sin}\theta |\phi_2\rangle \tag{2}$
Where angle $\theta$ is a real number between 0 and $\frac{\pi}{2}$
We are handed one of these states and want to indentify if we receive $|\psi_1\rangle$ or $|\psi_2\rangle$.
To start with we consider another basis $|e_1\rangle$, $|e_2\rangle$ which is rotated with respect to $|\phi_1\rangle$ and $|\phi_2\rangle$:
$|e_1\rangle=\text{cos}\alpha|\phi_1\rangle+\text{sin}\alpha|\phi_2\rangle\\$
$|e_2\rangle=-\text{sin}\alpha|\phi_1\rangle+\text{cos}\alpha|\phi_2\rangle$
Where angle $\alpha$ originates from the fact that basis $|e_1\rangle$, $|e_2\rangle$ is rotated relative to $|\phi_1\rangle$, $|\phi_2\rangle$.
I have derived the following expressions using $|\phi_1\rangle$, $|\phi_2\rangle$ for the projectors corresponding to a measurement in basis $|e_1\rangle$, $|e_2\rangle$:
$|\phi_1\rangle=-\text{sin}\alpha |e_2\rangle \tag{3}$
$|\phi_2\rangle=\text{sin}\alpha |e_1\rangle+\text{cos} \alpha |e_2\rangle \tag{4}$
Now; We want to use basis $|e_1\rangle$, $|e_2\rangle$ to discriminate between the states $|\psi_1\rangle$, $|\psi_2\rangle$. The average probability of picking the wrong state is:
$P_e=P(\psi_1|\psi_2)P(\psi_1)+P(\psi_2|\psi_1)P(\psi_2)$
We want to derive an explicit expression for $P_e$ using (1), (2), (3) and (4).
Contemporary solution:
I insert (3) and (4) into (1) and (2) to get:
$|\psi_1\rangle=-\text{sin}\alpha|e_2\rangle\\$
$|\psi_2\rangle=\text{sin}\theta\text{sin}\alpha |e_1\rangle+\text{sin}(\theta-\alpha)|e_2\rangle$
I remember that the probability of a quantum state is given as $\text{P}(a,b)= \langle a|b \rangle ^2$. Using this approach, I get:
$P(\psi_1)=\langle\psi_1|\psi_1\rangle^2=\text{sin}^2\alpha$
$P(\psi_2)=\langle\psi_2|\psi_2\rangle^2=\text{sin}^2\theta\text{sin}^2\alpha + \text{sin}^2(\theta-\alpha)$
$P(\psi_1,\psi_2)=\langle\psi_1|\psi_2\rangle^2=-\text{sin}\alpha\text{sin}(\theta-\alpha)$
$P(\psi_2,\psi_1)=\langle\psi_2|\psi_1\rangle^2=-\text{sin}\alpha\text{sin}(\theta-\alpha)$
$P_e=-\text{sin}\alpha\text{sin}(\theta-\alpha)[\text{sin}^2\theta\text{sin}^2\alpha + \text{sin}^2(\theta-\alpha)]-\text{sin}^3\alpha\text{sin}(\theta-\alpha)$
I do not have the solutions to the exercise available, so I have cannot determine whether this approach is correct. I would greatly appreciate any help to explain if this approach is correct or if another approach is better.