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As I asked in this question: How can I calculate the measuring probabilities of a two qubit state along a certain axis? 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.

Edit to answer the comment: $$| z \rangle $$ 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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  • $\begingroup$ What is $|z\rangle$? Also, what is $|\phi\rangle$ from the last equation? Isn't $\phi$ a scalar? assuming the measurement axis is $$\vec{n} = \begin{bmatrix}\sin\phi \\ 0 \\cos\phi \end{bmatrix}\,.$$ Can you add the complete question statement as-it-is? $\endgroup$
    – FDGod
    Commented Mar 10 at 4:34
  • $\begingroup$ I edited the question. $\endgroup$
    – Alex1111
    Commented Mar 10 at 8:42

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