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As I asked in this question: https://quantumcomputing.stackexchange.com/questions/36998/how-can-i-calculate-the-measuring-probabilities-of-a-two-qubit-state-along-a-cer/37000#37000 From here I know how to calculate the probability of measuring a general state at an arbitrary angle. I now deal with an exercise with photons instead of spins and for simplicity just with one qubit. Now again, I wanted to calculate the measurement result with the projection operator.

$$P_+ = \frac{1}{2} ( I + \sin \phi \sigma_x + \cos \phi \sigma_z )$$

$$\langle z | P_+| z \rangle = \frac{1}{2} (1 + \cos \phi)$$

But this is the official solution I received:

Eve projects the state along $|\psi\rangle = \cos(\phi)|z\rangle + \sin(\phi)|-z\rangle$. For a photon $\mid\uparrow\rangle$, Eve gets $\mid\uparrow\rangle$ with probability:

$$P(\uparrow) = |\langle \phi|z \rangle|^2 = \cos^2(\phi).$$

Now the probabilities don’t match up, and I am wondering, where my mistake lies. If I could get any hints on why my approach is wrong here, and when to use it, I would be very grateful.

Extra information:

$$| z \rangle $$ is the basis that is used in this question, together with $$| x \rangle $$ Also, yes, $$\phi$$ is scalar. Here is the question attached: The vertical polarisation is defined to be equal to|z>. The final probability that is asked in the answer is the sum over the probability of the vertical and the diagonal polarisation. I am currently still stuck with calculating the first of these. enter image description here

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$|\langle\psi|z\rangle|^2=\cos^2(\phi/2)$, which matches the probability predicted using the projector. Somebody has mixed up $\phi$ and $\phi/2$.

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  • $\begingroup$ Thanks for your answer! But unfortunately I made an error transporting the equations of the solution over here, the $$|\psi> $$ state has angles of $$\theta $$ and not $$0.5 * \theta$$, as I corrected now. With this, which are also the angles used for the measuring operator in my approach, there exists a difference from this approach to mine. So sadly I must ask, if someone possibly has an idea where my mistake is. $\endgroup$
    – Alex1111
    Commented Mar 11 at 13:43
  • $\begingroup$ I suggest that you look up multiple angle formulae. $\endgroup$
    – alanf
    Commented Mar 11 at 14:36

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