145 lines
4.2 KiB
TeX
145 lines
4.2 KiB
TeX
\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 |