3.2 Two quantum bits

The way we can describe two quantum bits is very similar to how we described two probabilistic bits. The main difference is that we have amplitudes instead of probabilities, meaning that they can be negative and are normalized in a different way (see Section 2.1.1). A general two-qubit quantum state looks as follows:

|ψ⟩=ψ00⁢|00⟩+ψ01⁢|01⟩+ψ10⁢|10⟩+ψ11⁢|11⟩\left|\psi\right\rangle=\psi_{00}\left|00\right\rangle+\psi_{01}\left|01\right% \rangle+\psi_{10}\left|10\right\rangle+\psi_{11}\left|11\right\rangle (3.31)

where ψi⁢j∈[−1,1]\psi_{ij}\in[-1,1] and

ψ002+ψ012+ψ102+ψ112=1.\psi_{00}^{2}+\psi_{01}^{2}+\psi_{10}^{2}+\psi_{11}^{2}=1.

We write |a,b⟩\left|a,b\right\rangle instead of [a,b][a,b] to make it clear that we are now dealing with quantum bits and not probabilistic bits. Just like we did in Eq. 3.3 for probabilistic bits, we can identify the four basis states |a,b⟩\left|a,b\right\rangle with the four basis vectors

|00⟩=(1000),|01⟩=(0100),|10⟩=(0010),|11⟩=(0001).\displaystyle\left|00\right\rangle=\begin{pmatrix}1\\ 0\\ 0\\ 0\end{pmatrix},\qquad\left|01\right\rangle=\begin{pmatrix}0\\ 1\\ 0\\ 0\end{pmatrix},\qquad\left|10\right\rangle=\begin{pmatrix}0\\ 0\\ 1\\ 0\end{pmatrix},\qquad\left|11\right\rangle=\begin{pmatrix}0\\ 0\\ 0\\ 1\end{pmatrix}. (3.48)

Similarly to Eq. 3.1, the general two-qubit state in Eq. 3.31 can be represented by a 4-vector

(ψ00ψ01ψ10ψ11).\begin{pmatrix}\psi_{00}\\ \psi_{01}\\ \psi_{10}\\ \psi_{11}\end{pmatrix}.

However, it quickly becomes cumbersome to manipulate such vectors. They are also not so easy to visualize anymore, so we will mostly work with the |a,b⟩\left|a,b\right\rangle notation in Eq. 3.31.

Another advantage of this notation is that it is much easier to combine quantum systems. Recall from Section 3.1.7 that we used the tensor product operation “⊗\otimes” to combine two independent probabilistic bits into a joint two-bit system. The same operation (except with [a][a] replaced by |a⟩\left|a\right\rangle) works also for qubits. In particular, just like we combined the basis vectors of probabilistic bits in Eq. 3.25, we can also do this for qubits:

|0⟩⊗|0⟩=|00⟩,|0⟩⊗|1⟩=|01⟩,|1⟩⊗|0⟩=|10⟩,|1⟩⊗|1⟩=|11⟩.\left|0\right\rangle\otimes\left|0\right\rangle=\left|00\right\rangle,\qquad% \left|0\right\rangle\otimes\left|1\right\rangle=\left|01\right\rangle,\qquad% \left|1\right\rangle\otimes\left|0\right\rangle=\left|10\right\rangle,\qquad% \left|1\right\rangle\otimes\left|1\right\rangle=\left|11\right\rangle. (3.49)

Notice that this corresponds to simply concatenating the bit strings. This gives you an alternative way to view the four two-qubit basis vectors in Eq. 3.48.

We can extend the tensor product operation to combine two arbitrary one-qubit states |α⟩=α0⁢|0⟩+α1⁢|1⟩\left|\alpha\right\rangle=\alpha_{0}\left|0\right\rangle+\alpha_{1}\left|1\right\rangle and |β⟩=β0⁢|0⟩+β1⁢|1⟩\left|\beta\right\rangle=\beta_{0}\left|0\right\rangle+\beta_{1}\left|1\right\rangle in the following way:

|α⟩⊗|β⟩\displaystyle\left|\alpha\right\rangle\otimes\left|\beta\right\rangle =(α0⁢|0⟩+α1⁢|1⟩)⊗(β0⁢|0⟩+β1⁢|1⟩)\displaystyle=\lparen\alpha_{0}\left|0\right\rangle+\alpha_{1}\left|1\right% \rangle\rparen\otimes\lparen\beta_{0}\left|0\right\rangle+\beta_{1}\left|1% \right\rangle\rparen (3.50)
=α0⁢β0⁢|00⟩+α0⁢β1⁢|01⟩+α1⁢β0⁢|10⟩+α1⁢β1⁢|11⟩.\displaystyle=\alpha_{0}\beta_{0}\left|00\right\rangle+\alpha_{0}\beta_{1}% \left|01\right\rangle+\alpha_{1}\beta_{0}\left|10\right\rangle+\alpha_{1}\beta% _{1}\left|11\right\rangle.

Two-qubit states of this form are called product states. In Section 3.2.3, we will discuss how to construct these states using quantum operations. Importantly, just like for two probabilistic bits, not all two-qubit states are product states.

Exercise 3.7 ( Tensor product and product states ).

Recall the states |+⟩\left|+\right\rangle and |−⟩\left|-\right\rangle from 2.1.

  1. 1.
    ​

    Write |+⟩⊗|−⟩\left|+\right\rangle\otimes\left|-\right\rangle in the same form as Eq. 3.31.

  2. 2.
    ​

    Is the state 12⁢(|00⟩+|01⟩+|10⟩+|11⟩)\frac{1}{2}\left(\left|00\right\rangle+\left|01\right\rangle+\left|10\right% \rangle+\left|11\right\rangle\right) a product state?

Solution.
  1. 1.
    |+⟩⊗|−⟩=\displaystyle\left|+\right\rangle\otimes\left|-\right\rangle= (12⁢|0⟩+12⁢|1⟩)⊗(12⁢|0⟩−12⁢|1⟩)\displaystyle\left(\frac{1}{\sqrt{2}}\left|0\right\rangle+\frac{1}{\sqrt{2}}% \left|1\right\rangle\right)\otimes\left(\frac{1}{\sqrt{2}}\left|0\right\rangle% -\frac{1}{\sqrt{2}}\left|1\right\rangle\right)
    =\displaystyle= 12⁢|00⟩−12⁢|01⟩+12⁢|10⟩−12⁢|11⟩.\displaystyle\frac{1}{2}\left|00\right\rangle-\frac{1}{2}\left|01\right\rangle% +\frac{1}{2}\left|10\right\rangle-\frac{1}{2}\left|11\right\rangle.
  2. 2.
    12⁢(|00⟩+|01⟩+|10⟩+|11⟩)=12⁢(|0⟩+|1⟩)⊗12⁢(|0⟩+|1⟩)=|+⟩⊗|+⟩.\displaystyle\frac{1}{2}(\left|00\right\rangle+\left|01\right\rangle+\left|10% \right\rangle+\left|11\right\rangle)=\frac{1}{\sqrt{2}}(\left|0\right\rangle+% \left|1\right\rangle)\otimes\frac{1}{\sqrt{2}}(\left|0\right\rangle+\left|1% \right\rangle)=\left|+\right\rangle\otimes\left|+\right\rangle.
    ​

    So this is indeed a product state.

In the following, we will discuss the rules for measuring and manipulating two quantum bits. While these rules will follow in complete analogy to the case of two probabilistic bits, we will find some new and surprising phenomena along the way. Happily, this week Quirky is also able to help us explore the world of two quantum bits! To begin, go to:

https://www.quantum-quest.org/quirky

and click on “Quest 3” and then on “Two Qubits”. Your web browser will look similarly to Fig. 3.4. As compared to last week’s Quirky, we now have two quantum bits, which are initialized in state |00⟩\left|00\right\rangle. In addition, there are three new boxes: ZZ, HH, and ∙\bullet (and again the mystery box is gone). We will discuss those in the remainder of this chapter.

Refer to caption

Figure 3.4: Quirky for Quest 3.