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I'm trying to work out how to do a non-demolition measurement on a qubit using an ancilla photon. I start with an entangled electron and nucleus in the state $$ \frac{1}{\sqrt{2}}(\alpha|\uparrow\Downarrow>+\beta|\downarrow\Uparrow>) $$ and entangle this to a photon in the state $$ \frac{1}{\sqrt{2}}(|R>-|L>) $$ (vertically polarised). The single arrow is the electron spin, the double arrow is the nuclear spin and |R> and |L> are the polarisation states of the photon. I want to use the photon to measure the state of the nuclear spin at this point. To do this I will disentangle the electron with a $\frac{\pi}{2}$ pulse so that I have just the state of the photon and nucleus. Then by measuring the photon in the correct polarisation bases I should be able to determine the state of the nucleus (i.e. the value of $\alpha$ and $\beta$). Is anyone aware of how the maths of this process is generated? So far I have tried this: $$\frac{1}{\sqrt{2}}(|R>-|L>)\otimes\frac{1}{\sqrt{2}}(\alpha|\uparrow\Downarrow>+\beta|\downarrow\Uparrow>) $$ $$=\frac{1}{2}(\alpha|R\uparrow\Downarrow>+\beta|R\downarrow\Uparrow>-\alpha|L\uparrow\Downarrow>-\beta|L\downarrow\Uparrow>). $$ Then a $\frac{\pi}{2}$ pulse on the electron spin has this effect: $$(<\uparrow|+<\downarrow|)\otimes\frac{1}{2}(\alpha|R\uparrow\Downarrow>+\beta|R\downarrow\Uparrow>-\alpha|L\uparrow\Downarrow>-\beta|L\downarrow\Uparrow>) $$ $$=\frac{1}{2}(\alpha|R\Downarrow>+\beta|R\Uparrow>-\alpha|L\Downarrow>-\beta|L\Uparrow>) $$ $$=\frac{1}{2}(\alpha(|R\Downarrow>-|L\Downarrow>)+\beta(|R\Uparrow>-|L\Uparrow>)) $$ but I am unsure how the photon measurement tells me the state of the nuclear spin.

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