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?

  • $\begingroup$ Roughly speaking you need as many dimensions as many different observable outcomes you can have. For 1 qubit this is 2 dimensions, for 2 qubits this is 4 dimensions, for $n$ qubits this is $2^n$ dimensions. $\endgroup$
    – Attila Kun
    Commented Sep 4, 2020 at 21:11

2 Answers 2


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...

  • 1
    $\begingroup$ yes indeed. It's been a while since uni and I can tell! I did not consider that the scalar could / should be in $\mathbb{C}$ also. +1 $\endgroup$
    – Dan Ward
    Commented Sep 5, 2020 at 23:59
  • 1
    $\begingroup$ I just thought i'd add a reference which very clearly helped me the compression from 4d to 2d in the Bloch sphere. The wikipedia entry is very clear. Your description is indeed also clear but the additional equations helped me track the collapsing of the freedoms: en.wikipedia.org/wiki/Qubit $\endgroup$
    – Dan Ward
    Commented Sep 13, 2020 at 17:46
  • $\begingroup$ Thanks for the link @DanWard, I didn't consider putting the equations on the answer. Perhaps I could edit my answer to add that? $\endgroup$
    – agiri
    Commented Sep 14, 2020 at 16:51
  • $\begingroup$ please don't feel you must on my account. You're answer was clear, I just needed to see it on paper. $\endgroup$
    – Dan Ward
    Commented Sep 15, 2020 at 17:05

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
  • $\begingroup$ +1, as my comment on the other response I had neglected to allow the scalars to be in $\mathbb{C}$ . I know at least some of what I don't know for Bloch spheres - the length 1 I am aware of, but the term 'Pure' state is yet to be defined so I'm willing to suspend my lack of clarity on that point for now. As of right now anyway this is all just linear algebra shenanigans. I await elucidation on why we're doing any of this, and why we're doing it in $\mathbb{C}^2$.... $\endgroup$
    – Dan Ward
    Commented Sep 6, 2020 at 0:02

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