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I'm reading this paper (Link to pdf) about a test of entanglement with three particles.

I wanted to ask if there is any mathematical shortcut to express one quantum state on another basis like the author has done it from H and V basis to H', V' , L and R? Or for example writing a superposition of computational basis in terms of + and - states

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  • $\begingroup$ The authors of the paper you have linked are just using more "physically" motivated labels for the elements of their 2-dimensional Hilbert space. You could have as easily picked the labels "0" instead of "H" and "1" instead of "V" for the basis of the Hilbert space and then you would have $|H'\rangle$ is the same as $|+\rangle$ and $|V'\rangle$ the same as $|-\rangle$ and so on; they are not changing the basis they are just picking different labels for the basis of the space. $\endgroup$
    – Condo
    Jun 11 '21 at 18:38
  • $\begingroup$ Okay how do they represent that state on a different basis? Arbitrarily? How do you express for example a superposition of computational states 0 and 1 on Bell basis? $\endgroup$ Jun 12 '21 at 8:34
  • $\begingroup$ Starting from a state written in some basis $B$ you can determine what the state looks like on another basis $B'$ by projecting the state from $B$ to $B'$. To do this you form the projection operators corresponding to the basis $B'$ and then apply these to your state to obtain the coefficients of the new basis elements. $\endgroup$
    – Condo
    Jun 14 '21 at 14:54
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From this paper, suppose you have three bases $S_x$, $S_y$ and $S_z$. And suppose you define the states of the $S_x$ and $S_y$ bases on the $S_z$ basis as follows:

$$ |+x\rangle = \frac{1}{\sqrt{2}}(|+z\rangle + |-z\rangle) \\ |-x\rangle = \frac{1}{\sqrt{2}}(|+z\rangle - |-z\rangle) \\ |+y\rangle = \frac{1}{\sqrt{2}}(|+z\rangle + i|-z\rangle) \\ |-y\rangle = \frac{1}{\sqrt{2}}(|+z\rangle - i|-z\rangle) $$

Now, imagine you want to write $|+x\rangle$ on the $S_y$ basis. We can define $|+x\rangle$ on the $S_y$ basis as

$$ \begin{pmatrix} \langle+y |+x\rangle \\ \langle-y |+x\rangle \end{pmatrix}_y $$

We do this to get the probability that $|+x\rangle$ will be found on state $|+y\rangle$ and $|-y\rangle$, respectively. (For the next step, remember that $\langle\pm y|$ is the complex conjugate of $|\pm y\rangle$; thus, the sign of $i$ will change.) Solving for the first element we get

$$ \begin{align} \langle+y |+x\rangle &= \frac{1}{2}\left(\langle + z| - i \langle - z|\right)\left(|+z\rangle + |-z\rangle \right) \\ &= \frac{1}{2}\left( \langle+z|+z\rangle + \langle+z|-z\rangle - i \langle-z|+z\rangle - i \langle-z|-z\rangle \right) \\ &= \frac{1}{2}(1-i) \end{align} $$

For the above, remember that $\langle\pm z | \pm z\rangle = 1$ and $\langle\mp z | \pm z\rangle = 0$. Doing something similar for the second entry of the vector we get that $\langle -y | +x \rangle = \frac{1}{2}(1+i)$. Therefore, $|+x\rangle$ in the $S_y$ basis is

$$ \frac{1}{2}\begin{pmatrix} 1-i \\ 1+i \end{pmatrix}_y $$

Now, let's do this between the computational basis and the Hadamard basis. Suppose you have an arbitrary state $|\psi\rangle=\alpha |0\rangle + \beta|1\rangle$ that you want to express in the Hadamard basis. Following the procedure above, you get

$$ \begin{pmatrix} \langle + | \psi \rangle \\ \langle - | \psi \rangle \end{pmatrix}_H = \frac{1}{\sqrt{2}} \begin{pmatrix} \alpha + \beta \\ \alpha - \beta \end{pmatrix}_H $$

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Given a quantum state $|\psi\rangle$, you can performa a basis change on this state using the $I$, the Identity operator, which of course can be expanded as $\sum_{i}|i\rangle\langle i|$, where $|i\rangle$ is simply a particular basis from a complete set of basis states. Applying $I$ to $|\psi\rangle$, we get $$I|\psi\rangle=\sum_{i}|i\rangle\langle i|\psi\rangle=\sum_{i}\langle i|\psi\rangle|i\rangle$$ where $\langle i|\psi\rangle$ is the square root of the probability of observing $|\psi\rangle$ in the state $|i\rangle$.

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  • $\begingroup$ But I still don't know how to make use of equations 2 and 3 (on the paper) to transform the equation 1 and get the equation 4 $\endgroup$ Jun 19 '21 at 11:44

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