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Take the Bell state $$\frac{1}{\sqrt{2}}\left(|00\rangle +|11\rangle\right) $$ and measure the first qubit with respect to some axis $\vec{n}_1$ on the Bloch sphere, i.e. measure the observable $$\vec{n}_1\cdot \vec{\sigma}$$ where $\vec{n}_1$ is normalized and $\vec{\sigma}$ is the vector of Pauli matrices.

Now, measure the second qubit with respect to a different axis $\vec{n}_2$.

What is the probability of equal measurement outcomes as a function of the two axes $\vec{n}_1,\vec{n}_2$, or, say, as a function of the angle between the axes?

I have read somewhere that the right answer is $p(\text{same outcome})=\cos^2\left(\frac{\theta}{2}\right)$ where $\theta$ is the angle between the axes.

I am looking for an easy way to obtain this result, or any kind of intuition. I believe I am capable of just carrying out the full-blown calculation myself. In fact, I started but it is very tedious and does not seem to provide me with any intuition about the geometry that's happening here.

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You can calculate the probability of a given answer $\pm 1$ to each measurement by evaluating $$ \langle B|\frac{I+(-1)^{\eta_1}\vec{n}_1\cdot\vec{\sigma}}{2}\otimes\frac{I+(-1)^{\eta_2}\vec{n}_2\cdot\vec{\sigma}}{2}|B\rangle $$ Thus, the probability of equal measurement outcomes is $$ \langle B|\left(\frac{I+\vec{n}_1\cdot\vec{\sigma}}{2}\otimes\frac{I+\vec{n}_2\cdot\vec{\sigma}}{2}+\frac{I-\vec{n}_1\cdot\vec{\sigma}}{2}\otimes\frac{I-\vec{n}_2\cdot\vec{\sigma}}{2}\right)|B\rangle. $$ This simplifies to $$ \langle B|\left(\frac{I+\vec{n}_1\cdot\vec{\sigma}\otimes\vec{n}_2\cdot\vec{\sigma}}{2}\right)|B\rangle. $$ I can also write this as $$ \frac12+\frac12\text{Tr}\left(|B\rangle\langle B|\vec{n}_1\cdot\vec{\sigma}\otimes\vec{n}_2\cdot\vec{\sigma}\right), $$ with $$ |B\rangle\langle B|=\frac14(X\otimes X-Y\otimes Y+I+Z\otimes Z). $$ So, the calculation that we want is all products of Pauli matrices. They are traceless unless the product is identity. That gives us an easy way to pick out terms: $$ \frac12+\frac12(n_1^Xn_2^X-n_1^Yn_2^Y+n_1^Zn_2^Z). $$ This is not quite the result you were expecting because of a negative sign (unless your bases have no $Y$-component, as would be the typical case of a Bell test, for example), so you cannot write it in terms of the angle between the two vectors. If you want to do that, then you should use the Bell state $|B_{11}\rangle=(|01\rangle-|10\rangle)/\sqrt{2}$ because $|B_{11}\rangle\langle B_{11}|=\frac12(I-Z\otimes Z-X\otimes X-Y\otimes Y)$, and hence we find that the probability of getting equal results is $$ \frac12\left(1-\vec{n}_1\cdot\vec{n}_2\right). $$ If the angle between the two vectors is $\theta$, then this is $\frac12(1-\cos\theta)=\sin^2\frac{\theta}{2}$.

If you want some sort of intuition, think of the Bell state as performing a teleportation. (Mathematically, $\text{Tr}_1(|\psi\rangle\langle\psi|\otimes I \cdot|B\rangle\langle B|)=\frac12|\psi\rangle\langle\psi|^T$ on the second qubit.) So, it's like the first state being teleported through to the second state, where the inner product is calculated.

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  • $\begingroup$ Cool, thx for the answer. I really like the calculation, it is easy to follow and short enough. I dont understand how you think about this in terms of teleportation, maybe you can elaborate more on this. I am familiar with the basic teleportation protocol but still. Finally, the form of your result prohibits me to write the result in terms of the angle between the axes or do you see an easy way to fix this? $\endgroup$ – Marsl Oct 14 at 15:16
  • $\begingroup$ @Marsl I don't think that inside this question is the best place for elaborating about the teleportation intutition - the point of an intuitive explanation is to try and explain something to someone they already understand so that they can avoid doing the maths. Giving an in-depth explanation entirely defeats the purpose! $\endgroup$ – DaftWullie Oct 15 at 6:09
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Intuitively, you can rotate $\vec{n}_1$ to $Z$. As $Z$ axis has two antipodal points $|0\rangle$ and $|1\rangle$, let $\vec{n}_1$ have two antipodal points $|b_0\rangle$ and $|b_1\rangle$. Now the Bell state can be rewritten as $\frac{1}{\sqrt{2}}(|b_0b_0\rangle+|b_1b_1\rangle)$. Now in this new basis, the calculation shall be much easier.

To be precise, please clarify your notation first. What is your $\vec{n}_1$? Is it an axis or unitary operator? In Quantum Computation and Quantum Information, section 4.2, an axis $\hat{n}$ is defined as three dimensional real unit vector. Only a rotation around the axis, not the axis itself, becomes an unitary operator. Apparently, the two antipodal points are eigenvectors for any such unitary operator since these two points do not move under rotation. Again, with this understanding of axis, you can always rotate $\hat{n}_1$ to the $Z$ axis by first rotating it around $Z$ to the $X-Y$ plane and then around $Y$ to $Z$.

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  • $\begingroup$ Are your $|b_0\rangle, |b_1\rangle$ the eigenvectors to the operator $\vec{n}_1\cdot \vec{\sigma}$? Can you explain why you can rotate $\vec{n}_1$ to $\sigma_z$? And finally, the calculation is almost as tedious as before, I believe... (unless you can convince me otherwise, I dont see an easy way to calculate the overlaps other than a lot of trigonometric relations etc.) $\endgroup$ – Marsl Oct 13 at 19:06
  • $\begingroup$ I dont know whether your rewriting is correct. As far as I am aware, the Bell state (say as a density operator) is only invariant under $U\otimes U^*$ ? $\endgroup$ – Marsl Oct 15 at 20:21

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