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I am reading a book on quantum computing. The author is constructing an arbitrary 2-Qbit state from unitary transformations. I need help understanding on step in his logic.

He starts by noting that the general 2-Qbit state has the form

$$ |\Psi\rangle = a_{00}|00\rangle + a_{01}|01\rangle + a_{10}|10\rangle + a_{11}|11\rangle $$

This can also be written as:

$$ |\Psi\rangle = |0\rangle \otimes |\psi\rangle + |1\rangle \otimes |\phi\rangle $$ $$ |\psi\rangle = a_{00}|0\rangle + a_{01}|1\rangle $$ $$ |\phi\rangle = a_{10}|0\rangle + a_{11}|1\rangle $$

So far so good. Next the author says apply $ \textbf{u} \otimes \textbf{1} $ to $ |\Psi\rangle $, where $ \textbf{u} $ is a linear transformation, whose action on the computational basis is of the form:

$$ \textbf{u}|0\rangle = a|0\rangle + b|1\rangle, \quad \textbf{u}|1\rangle = -b^*|0\rangle + a^*|1\rangle; \quad |a|^2 + |b|^2 = 1 $$

The author doesn't state this, but I assume $ a,b \in \mathbb{C} $ and $ a^* $ and $ b^* $ are the complex conjugates of $ a $ and $ b $ respectively. Also unstated but assumed by me is that $ \textbf{u} $ is a $ 2x2 $ matrix and $ \textbf{1} $ is the $ 2x2 $ identity.

Now comes the part I don't understand. The author states:

$$ (\textbf{u} \otimes \textbf{1})|\Psi\rangle = (a|0\rangle + b|1\rangle) \otimes |\psi\rangle + (-b^*|0\rangle + a^*|1\rangle) \otimes |\phi\rangle $$

In order for the above to be true, it would seem that

$$ (\textbf{u} \otimes \textbf{1})|\Psi\rangle = \textbf{u} |0\rangle \otimes |\psi\rangle + \textbf{u}|1\rangle \otimes |\phi\rangle $$

But I don't understand why.

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    $\begingroup$ It's the distributive law, along with the tensor product identity $(\mathbf{u}\otimes\mathbf{1})(|0\rangle \otimes |\psi\rangle) = \mathbf{u} |0\rangle \otimes \mathbf{1} |\psi\rangle.$ $\endgroup$ – Peter Shor Oct 9 '17 at 2:11
  • $\begingroup$ Thanks! I was not aware that the definition of tensor products on linear maps was the distributive law you mentioned. $\endgroup$ – Max Oct 9 '17 at 2:30
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In general, $(A\otimes B)(C \otimes D)=(AC) \otimes (BD)$, in quantum computing this is interpreted as applying a transformation to only one state of a composite system.

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