The Two Qubit States and Their Transformation

Two qubit states

Let’s say we have two qubits. We can call them $|a\rangle$ and $|b\rangle$. Each of the two qubits has its own probability amplitudes: $|a\rangle=a_0|0\rangle+a_1|1\rangle=\begin{bmatrix}a_0 \\ a_1\end{bmatrix}$ and $|b\rangle=b_0|0\rangle+b_1|1\rangle=\begin{bmatrix}b_0 \\ b_1\end{bmatrix}$. When we look at these two qubits concurrently, there are four different combinations of the basis states. Each of these combinations has its probability amplitude. These are the products of the probability amplitudes of the two corresponding states.

• $a_0|0\rangle b_0 |0\rangle$
• $a_0|0\rangle b_1 |1\rangle$
• $a_1|1\rangle b_0 |0\rangle$
• $a_1|1\rangle b_1 |1\rangle$

These four states form a quantum system on their own. Therefore, we can represent them in a single equation. While we are free to choose an arbitrary name for the state, we use $|ab\rangle$ because this state is the collective quantum state of $|a\rangle$ and $|b\rangle$.

$$|ab\rangle=|a\rangle\otimes|b\rangle=a_0 b_0|0\rangle|0\rangle+a_0 b_1|0\rangle|1\rangle+a_1 b_0|1\rangle|0\rangle+a_1 b_1|1\rangle|1\rangle$$

In this equation, $|ab\rangle$ is an arbitrary name. The last term is the four combinations reordered to have the amplitudes at the beginning. But what does $|a\rangle\otimes|b\rangle$ mean?

The term $|a\rangle\otimes|b\rangle$ is the tensor product of the two vectors $|a\rangle$ and $|b\rangle$.

The tensor product (denoted by the symbol $\otimes$) is the mathematical way of calculating the amplitudes. In general, the tensor product of two vectors $v$ and $w$ is a vector of all combinations. Like this:

With $v=\begin{bmatrix}v_0 \\ v_1 \\ \vdots \\ v_n \end{bmatrix}$and $w=\begin{bmatrix}w_0 \\ w_1 \\ \vdots \\ w_n \end{bmatrix}$then $v\otimes w=\begin{bmatrix}v_0w_0\\v_0w_1\\\vdots\\v_0w_n\\v_1w_0\\v_1w_1\\\vdots\\v_1w_n\\\vdots\\v_nw_n\end{bmatrix}$

For our system of two qubits, it is $|a\rangle\otimes|b\rangle=\begin{bmatrix}a_0\cdot\begin{bmatrix}b_0\\b_1\end{bmatrix}\\ a_1\cdot\begin{bmatrix}b_0\\b_1\end{bmatrix}\end{bmatrix}=\begin{bmatrix}a_0 b_0 \\ a_0 b_1 \\ a_1 b_0 \\ a_1 b_1\end{bmatrix}$.

The tensor product $|a\rangle\otimes|b\rangle$ is the explicit notation of $|a\rangle|b\rangle$. Both terms mean the same thing.

We can represent a qubit system in a column vector or the sum of the states and their amplitudes.

$$|ab\rangle=|a\rangle\otimes|b\rangle=a_0 b_0|0\rangle|0\rangle+a_0 b_1|0\rangle|1\rangle+a_1 b_0|1\rangle|0\rangle+a_1 b_1|1\rangle|1\rangle=\begin{bmatrix}a_0 b_0 \\ a_0 b_1 \\ a_1 b_0 \\ a_1 b_1\end{bmatrix}$$

This representation of the qubit state is similar to the single-qubit state $|\psi\rangle$. The only difference is the larger number of dimensions the two-qubit system has. It has four basis state vectors instead of two.

All the rules that govern a single qubit apply to a system that consists of two qubits. It works similarly. Accordingly, the sum of all probabilities—remember here that the probability of a state is the amplitude square—must be 1:

$$|a_0b_0|^2+|a_0b_1|^2+|a_1b_0|^2+|a_1b_1|^2=1$$

Unsurprisingly, working with a two-qubit system works similarly to working with a one-qubit system, too. The only difference is, again, that the larger number of dimensions the vectors and matrices have.