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Say, qubit $\left|a\right\rangle = \alpha_1|0\rangle + \beta_1|1\rangle$ and $|b\rangle = \alpha_2|0\rangle + \beta_2|1\rangle$.

After $\sqrt{\text{SWAP}}$(a,b) what are new probability amplitudes of $a$ and $b$ in terms of $\alpha_1,\, \alpha_2,\, \beta_1,\, \beta_2$?

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The overall state of the input is $|a\rangle|b\rangle$, which we can represent as:

$$ \left(\begin{array}{c} \alpha_1\alpha_2 \\ \alpha_1\beta_2 \\ \beta_1\alpha_2 \\ \beta_1\beta_2 \end{array}\right) $$ We apply the square root of swap gate (note that there are different ways that this matrix could be written ), $$ \sqrt{\text{SWAP}}=\left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & e^{i\pi/4}/\sqrt{2} & e^{-i\pi/4}/\sqrt{2} & 0 \\ 0 & e^{-i\pi/4}/\sqrt{2} & e^{i\pi/4}/\sqrt{2} & 0 \\ 0 & 0 & 0 & 1 \end{array}\right). $$ Hence, we're after the calculation $$ \left(\begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & e^{i\pi/4}/\sqrt{2} & e^{-i\pi/4}/\sqrt{2} & 0 \\ 0 & e^{-i\pi/4}/\sqrt{2} & e^{i\pi/4}/\sqrt{2} & 0 \\ 0 & 0 & 0 & 1 \end{array}\right)\left(\begin{array}{c} \alpha_1\alpha_2 \\ \alpha_1\beta_2 \\ \beta_1\alpha_2 \\ \beta_2\beta_2 \end{array}\right)=\left(\begin{array}{c} \alpha_1\alpha_2 \\ e^{i\pi/4}(\alpha_1\beta_2-i\beta_1\alpha_2)/\sqrt{2} \\ e^{-i\pi/4}(\alpha_1\beta_2+i\beta_1\alpha_2)/\sqrt{2} \\ \beta_1\beta_2 \end{array}\right). $$ Converting this back to Dirac notation gives us the final answer $$\alpha_1\alpha_2|00\rangle+\frac{e^{i\pi/4}}{\sqrt{2}}(\alpha_1\beta_2-i\beta_1\alpha_2)|01\rangle+ \frac{e^{-i\pi/4}}{\sqrt{2}}(\alpha_1\beta_2+i\beta_1\alpha_2)|10\rangle+\beta_1\beta_2|11\rangle. $$ Note that the probability amplitudes of each term are joint probability amplitudes of the whole system. As a general rule, you cannot separate this state into (state of a)$\otimes$(state of b), and therefore cannot identify individual probability amplitudes for each system.

Perhaps I could also suggest a bit of a notational change? If you want to refer to the two qubits as $a$ and $b$, do not also refer to the states of the two qubits as $|a\rangle$ and $|b\rangle$, as that's only likely to lead to confusion when the states change.

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Just apply corresponding matrix of this gate on the vector $$ |a\rangle |b\rangle = \alpha_1\alpha_2|00\rangle + \alpha_1\beta_2|01\rangle + \beta_1\alpha_2|10\rangle + \beta_1\beta_2|11\rangle $$

Note that in general the resulting state will be entangled. Hence every qubit individually will be in the mixed state.

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