What's $(\langle 0|\otimes I)(|00\rangle + |11\rangle)$ simplified?

Question

It's a rather simple question. I think I am confused by the fact that using $\langle0|$ on a qubit doesn't result in another qubit. So I'm not sure if I should interpret $\langle 0|$ as the $1\times2$ matrix $(1 \space 0)$.

Answer 1

What's $(\langle 0|\otimes I)(|00\rangle + |11\rangle)$ simplified?

As an alternative to Tristan's very good and explicit answer in terms of the matrix representation, it might be easier/illuminating to work directly with the operator/bra/ket expression.

By definition of the direct product, and remembering, for example, that $|00\rangle$ really means $|0\rangle\otimes|0\rangle$, we have: $$ (\langle 0|\otimes I)(|00\rangle + |11\rangle) =(\langle 0|\otimes I)(|0\rangle\otimes|0\rangle) +(\langle 0|\otimes I)(|1\rangle\otimes|1\rangle) $$ $$ =(\langle 0|0\rangle\otimes I|0\rangle) +(\langle 0|1\rangle\otimes I|1\rangle) $$ $$ =1|0\rangle + 0|1\rangle = |0\rangle $$

Answer 2

By definition, $|0\rangle$ is the vector $\begin{pmatrix}1\\0\end{pmatrix}$, and $\langle0|$ is the conjugate transpose of $|0\rangle$. Thus, $\langle0|$ is the row vector $\begin{pmatrix}1&0\end{pmatrix}$. We can then compute: $$\begin{align} \langle|0|\otimes I &= \begin{pmatrix}1&0\end{pmatrix}\otimes\begin{pmatrix}1&0\\0&1\end{pmatrix}\\ &=\begin{pmatrix}1&0&0&0\\0&1&0&0\end{pmatrix} \end{align}$$ You can then compute the product with $|00\rangle+|11\rangle=\begin{pmatrix}1\\0\\0\\1\end{pmatrix}$: $$(\langle 0|\otimes I)(|00\rangle + |11\rangle)=\begin{pmatrix}1&0&0&0\\0&1&0&0\end{pmatrix}\begin{pmatrix}1\\0\\0\\1\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}=|0\rangle.$$