Two Quantum Bits

Some states of two quantum bits can be symbolized by the juxtaposition (or multiplication) of states of each quantum bit. In particular, the four logical states $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{0}}\mbox{$\rangl... ...$}{\mathfrak{0}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$},$ and $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{1}}\mbox{$\rangl... ...t$}{\mathfrak{1}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ are acceptable pure states for two quantum bits. In these expressions, we have distinguished the qubits by position (first or second). It is easier to manipulate state expressions if we explicitly name the qubits, say $\mathsf {A}$ and $\mathsf {B}$. We can then distinguish the kets by writing, for example, $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {A}}}}$ for a state of qubit $\mathsf {A}$. Now the state $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfrak{0}}\mbox{$\rangl... ...t$}{\mathfrak{1}}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ can be written with explicit qubit names (or ``labels'') as

$$\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\mathfr... ...pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {BA}}}}.$$ (21)

Having explicit labels allows us to unambiguously reorder the states in a product of states belonging to different qubits. We say that kets for different qubits ``commute''.

So far we have seen four states of two qubits, which are the logical states that correspond to the states of two bits. As in the case of one qubit, the superposition principle can be used to get all the other pure states. Each state of two qubits is therefore of the form

$$\alpha\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\... ...pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {AB}}}},$$ (22)

where $\alpha,\beta,\gamma,$ and $\delta$ are complex numbers. Again, there is a column vector form for the state:

$$\left(\begin{array}{c}\alpha\\ \beta\\ \gamma\\ \delta\end{array}\right),$$ (23)

and this vector has to be of unit length, that is $\vert\alpha\vert^2+\vert\beta\vert^2+\vert\gamma\vert^2+\vert\delta\vert^2=1$. When using the vector form for qubit states, one has to be careful about the convention used for ordering the coefficients.

Other examples of two-qubit states in ket notation are the following:

$$\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi_1}\mbox... ...angle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {AB}}}} = {1\over\sqrt{2}}\Big(\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt... ...angle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {B}}}},$$

$$\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi_2}\mbox... ...angle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {AB}}}} = {1\over\sqrt{2}}\Big(\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt... ...e\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {B}}}}\Big)$$

$$= {1\over 2}\Big(\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert... ...\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {AB}}}}\Big)$$

$$\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi_3}\mbox... ...angle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {AB}}}} = {1\over \sqrt{2}}\Big(\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3p... ...hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {AB}}}}\Big),$$

$$\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi_4}\mbox... ...angle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {AB}}}} = {1\over \sqrt{2}}\Big(\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3p... ...hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {AB}}}}\Big).$$ (24)

The first two of these states have the special property that they can be written as a product $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\phi_1}\mbox{$\rangle\hspa... ...rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {B}}}}$ of a state of qubit $A$ and a state of qubit $B$. The second expression for $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi_2}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}$ shows that the product decomposition is not always easy to see. Such states are called ``product'' states. The last two states, $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi_3}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {AB}}}}$ and $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi_4}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {AB}}}}$ are two of the famous Bell states. They have no such representation as a product of independent states of each qubit. They are said to be ``entangled'' because they contain a uniquely quantum correlation between the two qubits. Pbits can also have correlations that cannot be decomposed into product states, but the entangled states have additional properties that make them very useful. For example, if $\mathsf {A}$lice and $\mathsf {B}$ob each have one of the qubits that together are in the state $\mbox{$\vert\hspace*{-3pt}\vert\hspace*{-3pt}\vert$}{\psi_3}\mbox{$\rangle\hspace*{-4.3pt}\rangle\hspace*{-4.3pt}\rangle$}_{{}_{\!\!{\mathsf {AB}}}}$, they can use them to create a secret bit for encrypting their digital communications.