If $\mid\phi\rangle=a_0\mid 00\rangle + a_1\mid 01\rangle +a_2\mid 10\rangle +a_3\mid 11\rangle$ and $P_0=\mid 0\rangle\langle 0\mid \otimes I$, how do we show that $\langle\phi \mid P_0\mid \phi\rangle =\langle\phi \mid a_0\mid 00\rangle+\langle \phi\mid a_1\mid 01\rangle =|a_0|^2+|a_1|^2$
I get the following if we only consider the first quibit (assuming that's what $\otimes I$ does,.... I'm still not sure on that either) :
$(a_0 \langle 0\mid +a_1\langle 0\mid)(\mid 0\rangle\langle0\mid)(a_0\mid0\rangle + a_1\mid0\rangle)=$
$(a_0 \langle 0\mid +a_1\langle 0\mid)(a_0\mid 0\rangle\langle0\mid0\rangle+a_1\mid 0\rangle\langle0\mid0\rangle$)$ = a_0^2+2a_0a_1+a_1^2$