I am relatively new to the Quantum Computing world and was wondering if representing a 3 qubit system using 3 Bloch Spheres feasible and if so what would the correct way to do it?

I understand a Bloch sphere can represent a single qubit.

$$\phi = \alpha |0> + \beta |1>$$ (i.e the $|0> + |1>$ states along with it's amplitudes)

In a 2- qubit system you could represent it using two bloch spheres.

$$\phi = \alpha_1 |00> + \beta_1 |01> + \alpha_2 |10> + \beta_2 |11>$$

So one sphere for the $|00>$ and $|01>$ amplitudes and another sphere for the $|10>$ and $|11>$ amplitudes

But I am a little confused about a 3-qubit system as from what I can gather it is represented by a structure like this (granted I realise different textbooks have different notation for the amplitudes):

$$/phi = \alpha_1 |000> + \beta_1 |001> + \gamma_1 |010> + \delta_1 |011> + \alpha_2 |100> + \beta_2 |101> + \gamma_2|110> + \delta_2|111>$$

How would one divide the amplitudes with their computational basis states in order to represent phi as what I expect should be 3 Bloch spheres (as it is a 3 qubit system) Since Bloch spheres represent a single $|0>+|1>$ ? Or should there be more Bloch spheres?

For example should it actually be represented by 4 Bloch spheres in the manner: $sphere1 = \alpha_1 |000> + \beta_1 |001>$

$sphere2 = \alpha_1 |000> + \beta_1 |001>$

$sphere3 = \alpha_2 |100> + \beta_2 |101>$

$sphere4= \gamma_2|110> + \delta_2|111>$

and if this is the correct manner to represent, why is that?

Apologies if this is a silly questions, I am still very much a beginner in this field!


1 Answer 1


It is not possible two represent an $n$-qubit system with $n$ (3D) Bloch spheres. You would need instead a $(2^{n+1}-1)$ dimensional hypersphere.

That can be simply deduced from the degrees of freedom of the system you want to represent. A $n$-qubit system has $2^{n}$ base vectors. For every base vector there is one complex amplitude, which makes $2^{n+1}$ degrees of freedom. Because of the normalization and the global phase we have to substract two degrees of freedom again, which makes $2^{n+1}-2$ DoF.

$n$ 3D standard bloch spheres on the other hand have just $2n$ free parameters, which is clearly not sufficient to represent the whole system. The linear growth of degrees of freedom would only suffice to represent a classical system.

A simple example, when your proposal would suffice is when the many-body state is just a product state of single particle wavefunctions (qubits), i.e. \begin{align} \left | \Psi\right\rangle = \prod_{n}\left( \cos\theta_n\left |0\right\rangle_n+\exp(-i\phi_n)\sin\theta_n \left |1\right\rangle_n \right) \end{align}

  • $\begingroup$ Thank you for your explanation I realised they I might be wrong but was a little confused as to why. If Bloch spheres aren't the way to go, what would be the best way to represent a 2 or 3 quit systems to do something like compare a before and after of certain operation on a qubit system. E.g 2 qubit system showing before and after using Hadamard gate or the Quantum Fourier Transform etc. $\endgroup$
    – Catherine
    Commented Feb 16, 2017 at 15:25
  • $\begingroup$ Often one plots the density matrix, see e.g. here qutip.org/docs/3.1.0/guide/guide-visualization.html $\endgroup$
    – Jannick
    Commented Feb 16, 2017 at 15:36
  • $\begingroup$ You can ignore global phase to save another DOF. $\endgroup$ Commented Feb 17, 2017 at 0:38
  • $\begingroup$ Yes exactly. I edited my answer. $\endgroup$
    – Jannick
    Commented Feb 17, 2017 at 9:11
  • $\begingroup$ Also all states in this higher dimensional ball don't represent valid density matricies in general. The equivalent of the Bloch sphere in higher dimensions is a complicated convex body. $\endgroup$
    – biryani
    Commented Feb 18, 2017 at 6:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.