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I have a CS background and am studying quantum computing by myself. Struggling at the moment with the Dirac notation for the teleportation circuit. Here we go:

The circuit starts with the EPR pair between Alice and Bob (qubits entangled):

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Then the $CNOT$ is applied (so far so good):

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Then the Hadamard (still ok):

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But then, the terms are regrouped in order to get the equation below, but I couldn't figure out how that was done.

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For instance, I can't figure out how $\beta|1\rangle$ can possibly be multiplying $|00\rangle$ as $\beta$ is only multiplying the terms $|01\rangle$ and $|10\rangle$ part of the equation.

I checked Thomas Wong's book as well as Nielsen and Chuang, but they don't really detail this part at all.

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2 Answers 2

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Consider the following equivalency:

$$ |xy\rangle = |x\rangle|y\rangle = |x\rangle \otimes |y\rangle $$

While $\otimes$ is a notation for the Kronecker product. Take a look there at the properties of the Kronecker product, but in particular and regarding your question, it's important to understand that the Kronecker product is bilinear, associative and non-commutative.

In light of these properties we can write:

$$ |\psi\rangle = \frac{1}{2} \big[\alpha(|0\rangle + |1\rangle)(|00\rangle + |11\rangle) + \beta(|0\rangle - |1\rangle)(|10\rangle + |01\rangle) \big] = \\ \ \frac{1}{2}(\alpha|000\rangle + \alpha|011\rangle + \alpha|100\rangle + \alpha|111\rangle + \beta|010\rangle + \beta|001\rangle - \beta|110\rangle - \beta|101\rangle) = \\ \ \frac{1}{2}\big[ |00\rangle(\alpha|0\rangle + \beta|1\rangle) + |01\rangle(\alpha|1\rangle + \beta|0\rangle) + |10\rangle(\alpha|0\rangle - \beta|1\rangle) + |11\rangle(\alpha|1\rangle - \beta|0\rangle) \big] $$

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  • $\begingroup$ thank you for the detailed answer and for fixing the issues in the question as well. This is really helpful and makes this community quite welcoming to newbies like me. $\endgroup$
    – neilson
    Sep 11 at 8:50
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In their notation, $|ab\rangle|c \rangle$ is the same thing as $|a \rangle |bc \rangle $.

They’re not moving the first qubit around, they’re regrouping the qubits.

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