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


1 Answer 1


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

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.