when we specify quantum states in $\mathbb C^2$, why do we only have $2$ basis states?

Question

I am just starting to get up to speed with quantum computing via the Quiskit learning path: online tutorial

Here they explain the Dirac notation and use it to describe quantum states as elements in $\mathbb{C}^2$ (square could be for the initial lesson but my question holds in n.

i.e a quantum state $| a \rangle = \begin{pmatrix}a_1 \\ a_2 \end{pmatrix}$ where $a_1, a_2 \in \mathbb{C}$

Shortly after this the concept of the basis is introduced, with the x, y and z examples given. Their orthonormality is stressed. In all three cases this 'basis' is two dimensional.

So as a concrete example $ \{| 0 \rangle , | 1 \rangle \}$ is given as an orthonormal basis for the space to describe and measure quantum states.

Clearly this pair of basis is not an orthonormal basis of $\mathbb{C}^2$ so I understand there may be additional constraints on the space of possible valid quantum states. But then I have not yet seen why it is necessary to embed this apparently two dimensional space into teh 4-dimension $\mathbb{C}^2$.

We have touched on the Bloch Sphere which is a two dimensional representation of pure states but that is derived from the two dimensional orthonormal basis rather than the other way around - but perhaps this is a more profound representation of the space than it seems at this point.

What is the reason we both need a 4-dimensional space to describe our possible quantum states and can work with a two dimensional basis?

Answer 1

Clearly this pair of basis is not an orthonormal basis of $C^2$...

$\{\lvert0\rangle,\lvert1\rangle\}$ is an orthonormal basis of $C^2$. $C^2$ is a 2-dimensional complex vector space, which means that every element of the space is essentially a vector of 2 complex numbers.
$\lvert0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}$, and $\lvert1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix}$ are both vectors in this space because 0 and 1 are also complex numbers. The vectors are also orthogonal: $$ \begin{aligned} \langle0\rvert1\rangle & = \begin{pmatrix} 1\:\:0 \end{pmatrix}\begin{pmatrix} 0\\1 \end{pmatrix}\\ & = 1(0) + 0(1)\\ & = 0 \end{aligned} $$

About the 4-dimensional space: yes, representing 2 complex numbers requires 4 real numbers. However, keep in mind that every quantum state must be normalized. When you specify 3 of these 4 real numbers, you lose your degree of freedom in choosing the 4th one.
Further, the global phase of a quantum state is typically ignored because it has no impact on the probability distributions associated with measurement. This additional redundancy brings down your total degrees of freedom to 2. These 2 parameters correspond to the angles $\theta$ and $\phi$ which specify a location on the surface of the Bloch Sphere.

I hope that answers your question...

Answer 2

Any complex vector $v \in \mathbb{C}^n$ can be written as a linear combination $$ v = \sum_{i=1}^n \alpha_ib_i, $$ where $b_i$ are basis vectors and $\alpha_i \in \mathbb{C}$ are coefficients. Since real numbers are subset of complex ones and $\alpha_i$ are complex numbers, basis vectors can be real.

In your particular example, it is not problem to use basis composed of real vectors $|0\rangle$ and $|1\rangle$ since complex amplitudes are "hidden" in complex coefficients of the linear combination.

Regarding ortogonality, a dot product on space $\mathbb{C}^n$ for vectors $v$ and $w$, both from that space, is defined as $$ v \cdot w = \sum_{i=1}^n v_i w_i^*, $$ where $w_i^*$ is complex conjugated number to $w_i$. Since both members of vectors $|0\rangle$ and $|1\rangle$ are real, $w_i^*=w_i$ and hence these vectors are orthogonal in $\mathbb{C}^2$.

On the Bloch sphere. In the end you need only two parameters - angles $\theta$ and $\varphi$ since you have only two degrees of freedom. This is given by constraints imposed on qubits:

• lenght of a vector describing qubit is 1
• global phase can be neglected as two qubits differing in the global phase only are physically indistinguisable