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Can someone shows me how to proof this equality:

$\frac{1}{\sqrt2}(\alpha|000⟩+\alpha|011⟩ + \beta|100⟩ + \beta|111⟩ )$ =

$ \frac{1}{2\sqrt2}[(|00⟩+|11⟩) \otimes (\alpha|0⟩+\beta|1⟩) + (|01⟩+|10⟩) \otimes (\alpha|1⟩+\beta|0⟩) $

$+ (|00⟩-|11⟩) \otimes (\alpha|0⟩-\beta|1⟩) + (|01⟩-|10⟩) \otimes (\alpha|1⟩-\beta|0⟩) ]$

It is from this video .

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    $\begingroup$ Welcome to QCSE. This sounds like homework/coursework which, although not disallowed on this site, usually requires that you put in some more effort to explain what you've done and where you've gotten stuck. Can you edit your question to include such detail? $\endgroup$
    – Mark S
    Jul 22 at 20:01
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    $\begingroup$ it is not a homework, it is from qiskit video youtu.be/… $\endgroup$
    – cometta
    Jul 22 at 23:56
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A few words on normalization constants:

Let us firstly assume that $|\alpha|^2+|\beta|^2 = 1$. This allow us to assume that states $\alpha|0\rangle + \beta|1\rangle$ is properly normalized.

Now let's look at the left side. A sum of probabilities of all possible outcomes of the state is $|\frac{1}{\sqrt{2}}\alpha|^2 + |\frac{1}{\sqrt{2}}\alpha|^2 + |\frac{1}{\sqrt{2}}\beta|^2 + |\frac{1}{\sqrt{2}}\beta|^2 = 2|\frac{1}{\sqrt{2}}\alpha|^2+2|\frac{1}{\sqrt{2}}\beta|^2=2\frac{1}{2}|\alpha|^2+2\frac{1}{2}|\beta|^2=|\alpha|^2+|\beta|^2=1$. This means that the state on the left side is properly normalized.

Finally turn our attention to the right side. Bell states on this sides, i.e. $|00\rangle\pm|11\rangle$ and $|01\rangle\pm|10\rangle$ are properly normalized if the normalization constant is $\frac{1}{\sqrt{2}}$ - this value is factored out in front of the parenthes. Moreover, we assumed that $|\alpha|^2+|\beta|^2=1$, so each member on the right side is properly normalized state. The right side is a superpositon of four such states. Assuming that each state can occur with the same probability, i.e. $1/4$ because we have four states, a probability amplitude of each right-side member is $1/2 = \sqrt{1/4}$. One half can be factored out in front of the parenthses.

To sum up, states on both sides are properly normalized.

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Before getting into your problem, remember that $|\psi\rangle \otimes |\phi\rangle=|\psi\phi\rangle$, so we can use the latter as a short hand. Also, remember the following:

$$ (|a\rangle + |b\rangle) \otimes (|c\rangle + |d\rangle) = |ac\rangle + |ad\rangle + |bc\rangle + |bd\rangle $$

So that works just as normal multiplication of two binomials.

Now, let's focus on each term of the sum you gave in your question. First one:

$$ (|00\rangle + |11\rangle) \otimes (\alpha|0\rangle + \beta|1\rangle) = \alpha|000\rangle + \beta|001\rangle + \alpha|110\rangle + \beta|111\rangle $$

Since $\alpha, \beta$ are just scalars, we can write them at the beggining just as we did above. Doing the same for the other three terms, you get:

$$ \begin{align} (|01\rangle + |10\rangle) \otimes (\alpha|1\rangle + \beta|0\rangle) &= \alpha|011\rangle + \beta|010\rangle + \alpha|101\rangle + \beta|100\rangle \\ (|00\rangle - |11\rangle) \otimes (\alpha|0\rangle - \beta|1\rangle) &= \alpha|000\rangle - \beta|001\rangle - \alpha|110\rangle +\beta|111\rangle \\ (|01\rangle - |10\rangle) \otimes (\alpha|1\rangle - \beta|0\rangle) &= \alpha|011\rangle -\beta|010\rangle-\alpha|101\rangle + \beta|100\rangle \end{align} $$

Finally, just add these four terms and you'll see that you get the expression you're looking for. (Note that I didn't pay much attention to the normalizaiton factors here, but that's just a scalar, so it works anyways.)

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  • $\begingroup$ you seem to derive it in reversed. My question is how one term can become 4 terms. Also how to get the scalar value 1/(2 sqrt(2)) $\endgroup$
    – cometta
    Jul 23 at 1:31
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    $\begingroup$ @cometta Well, it just works as you would normally factorize an algebraic expression. And about the scalar value, I don't think it's $1/2 \sqrt{2}$ at any point during quantum teleportation. So the lecturer might have made a mistake. I recommend you look here. $\endgroup$
    – epelaaez
    Jul 23 at 2:53
  • $\begingroup$ @cometta: you seem to derive it in reversed - this is not a problem, just switch sides of the formula and you have a general approach how to solve problems like yours. $\endgroup$ Jul 23 at 8:06
  • $\begingroup$ @MartinVesely is the normalized term 1/(2 sqrt(2)) correct or wrong? sorry i'm confuse , should I change 1/(2 sqrt(2)) to 1/2 ? $\endgroup$
    – cometta
    Jul 23 at 8:40
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    $\begingroup$ @cometta: Please see my answer below. The constants seems to be right. $\endgroup$ Jul 23 at 11:24

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