# What are the individual probabilities after √SWAP gate?

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 $$\alpha_1,\, \alpha_2,\, \beta_1,\, \beta_2$$?

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