\section{Two Halves of a Qubit}


\definition{}
Just as before, we'll represent multi-quibit states as linear combinations of multi-qubit basis states. \par
For example, a two-qubit state $\ket{ab}$ is the four-dimensional unit vector
\begin{equation}
	\begin{bmatrix}
		a \\ b \\ c \\ d
	\end{bmatrix}
	= a \ket{00} + b\ket{01} + c\ket{10} + d\ket{11}
\end{equation}

As always, multi-qubit states are unit vectors. \par
Thus, $a^2 + b^2 + c^2 + d^2 = 1$ in the two-bit case above.


\problem{}
Say we have two qubits $\ket{\psi}$ and $\ket{\varphi}$. \par
Show that $\ket{\psi} \otimes \ket{\varphi}$ is always a unit vector (and is thus a valid quantum state).

\vfill


\definition{Measurement II}<measureii>
Measurement of a two-qubit state works just like measurement of a one-qubit state: \par
If we measure $a\ket{00} + b\ket{01} + c\ket{10} + d\ket{11}$, \par
we get one of the four basis states with the following probabilities:

\begin{itemize}
	\item $\mathcal{P}(\ket{00}) = a^2$
	\item $\mathcal{P}(\ket{01}) = b^2$
	\item $\mathcal{P}(\ket{10}) = c^2$
	\item $\mathcal{P}(\ket{11}) = d^2$
\end{itemize}
Of course, the sum of all the above probabilities is $1$.


\problem{}
Consider the two-qubit state
$\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3}}{4} \ket{10} + \frac{1}{4} \ket{11}$

\begin{itemize}[itemsep=2mm]
	\item If we measure both bits of $\ket{\psi}$ simultaneously, \par
	what is the probability of getting each of $\ket{00}$, $\ket{01}$, $\ket{10}$, and $\ket{11}$?

	\item If we measure the ONLY the first qubit, what is the probability we get $\ket{0}$? How about $\ket{1}$? \par
	\hint{There are two basis states in which the first qubit is $\ket{0}$.}

	\item Say we measured the second bit and read $\ket{1}$. \par
	If we now measure the first bit, what is the probability of getting $\ket{0}$?
\end{itemize}



\vfill
\pagebreak

\problem{}
Again, consider the two-qubit state
$\ket{\psi} = \frac{1}{\sqrt{2}} \ket{00} + \frac{1}{2} \ket{01} + \frac{\sqrt{3}}{4} \ket{10} + \frac{1}{4} \ket{11}$ \par
If we measure the first qubit of $\ket{\psi}$ and get $\ket{0}$, what is the resulting state of $\ket{\phi}$? \par
What would the state be if we'd measured $\ket{1}$ instead?

\vfill

\problem{}
Consider the three-qubit state $\ket{\psi} = c_0\ket{000} + c_1\ket{001} + ... + c_7 \ket{111}$. \par
Say we measure the first two qubits and get $\ket{00}$. What is the resulting state of $\ket{\psi}$?

\begin{solution}
	We measure $\ket{00}$ with probability $c_0^2 + c_1^2$, and $\ket{\psi}$ collapses to
	\begin{equation*}
		\frac{c_0\ket{000} + c_1\ket{001}}{\sqrt{c_0^2 + c_1^2}}
	\end{equation*}
\end{solution}





\vfill
\pagebreak



\definition{Entanglement}
Some product states can be factored into a tensor product of individual qubit states. For example,
\begin{equation*}
	\frac{1}{2} \bigl(\ket{00} + \ket{01} + \ket{10} + \ket{11}\bigr)
	= \frac{1}{\sqrt{2}}\bigl( \ket{0} + \ket{1} \bigr) \otimes
		\frac{1}{\sqrt{2}}\bigl( \ket{0} - \ket{1} \bigr)
\end{equation*}
Such states are called \textit{product states.} States that aren't product states are called \textit{entangled} states.

\problem{}
Factor the following product state:
\begin{equation*}
	\frac{1}{2\sqrt{2}} \bigl(\sqrt{3}\ket{00} - \sqrt{3}\ket{01} + \ket{10} - \ket{11}\bigr)
\end{equation*}

\begin{solution}
	\begin{equation*}
		\frac{1}{2\sqrt{2}} \biggl(\sqrt{3}\ket{00} - \sqrt{3}\ket{01} + \ket{10} - \ket{11}\biggr)
		= \biggl( \frac{\sqrt{3}}{2}\ket{0} + \frac{1}{2}\ket{1} \biggr) \otimes
		\biggl(\frac{1}{\sqrt{2}}\ket{0} - \frac{1}{\sqrt{2}}\ket{1} \biggr)
	\end{equation*}
\end{solution}


\vfill

\problem{}
Show that the following is an entangled state.
\begin{equation*}
	\frac{1}{\sqrt{2}}\ket{00} + \frac{1}{\sqrt{2}}\ket{11}
\end{equation*}

\begin{solution}
	$
	\left[
	\begin{smallmatrix}
		a_0 \\ a_1
	\end{smallmatrix}
	\right]
	\otimes
	\left[
		\begin{smallmatrix}
			b_0 \\ b_1
		\end{smallmatrix}
	\right]
	=
	a_0b_0\ket{00} + a_0b_1\ket{01} + a_1b_0\ket{10} + a_1b_1\ket{11}
	$

	\vspace{2mm}

	So, we have that $a_1b_1 = a_0b_0 = \sqrt{2}^{-1}$ \par
	But $a_0b_1 = a_1b_0 = 0$, so one of $a_0$ and $b_1$ must be zero. \par
	We thus have a contradiction.
\end{solution}


\vfill
\pagebreak