I was not able to locate any visuals online. The visual I have in my head is a cube w/ bloch spheres as the eight vertices.

I am also curious about a matrix representation, although I am not sure how feasible this is as a qubyte has $2^8$ (256) states.

What is the best way to represent a qubyte?

  • $\begingroup$ Since you created the tag qubyte, please consider writing the tag excerpt for it. Go here. Thank you. $\endgroup$ – Sanchayan Dutta Jul 5 '18 at 8:58

I don't think you'll find a good visual representation. The Bloch sphere for a qubit is a particularly unique coincidence because the number of parameters to represent an arbitrary mixed state is only 1 more than the number of parameters required to represent an arbitrary pure state, and so the pure states can be thought of as the surface to a mixed state's volume.

A cube with Bloch spheres at the vertices is a fair representation if all 8 qubits remain separable at all times. However, you will get in a mess trying to represent entanglement in that picture (you'll need superpositions of several different separable states). As you've said, an arbitrary pure state of a qubyte has $2^8$ states, requiring $2^{16}-2$ real parameters to describe. 8 (pure) Bloch states give you access to only 16 parameters. You're probably best sticking to a complex vector of $2^8$ elements (minus 2 real parameters for global phase and normalisation).

  • $\begingroup$ What about something like a hypercube? $\endgroup$ – meowzz Jun 25 '18 at 8:24
  • 1
    $\begingroup$ You mean an 8-dimensional hypercube? Think about the 1-dimensional hypercube (which, by analogy, should work for a single qubit). The corresponding line already misses out stuff that the Bloch sphere gives you. $\endgroup$ – DaftWullie Jun 25 '18 at 8:34
  • $\begingroup$ An 8-cube! $\endgroup$ – meowzz Jun 25 '18 at 8:51

For a pure state of 8 qubits, the Hilbert space is $2^8$ dimensional. Dropping the normalization and phase information means you are left with the space $\mathbb{CP}^{2^8-1}$.

Unlike a single qubit which give the Bloch sphere $\mathbb{CP}^{2^1-1}$, this is too big to draw directly. Instead one usually draws simpler spaces that capture the essential features. This is thanks to the fact that these spaces are toric varieties. The example of how to draw $\mathbb{CP}^2$ as a triangle with tori above each point is given there. What this amounts to is remembering only the amplitudes first and then realizing that you forgot the phases and fixing that later.

So instead of drawing a line segment (a 1-cube) as you would for a single qubit, you would have to draw a $2^8-1$ simplex. Of course you can't do that, but you can project onto planes that you can draw several 2 dimensional pictures for. So this draws only certain linear combinations of probabilities for the $2^8$ basis states. Do several of these.

A lot of information is lost, because you couldn't draw the full $2^8-1$ complex dimensional ($2^{16}-2$ real dimensional) thing, but by throwing away information about phases and only taking certain linear combinations of probabilities, you get something you can visualize. Also you know what sort of structure you forgot along the way. Like when going from the point on the simplex back to the full complex projective space, you lost all the phases. You can draw those as points on the circle as usual, so you can recover the full information from those two diagrams.

If you want to say that some of the qubits are separable from others, then you get $\mathbb{CP}^{2^n-1} \times \mathbb{CP}^{2^m-1}$ where $n+m=8$. This is also toric and so also has a polytope that replaces the $2^8-1$ simplex. If you want it fully separable this would be $\mathbb{CP}^{2^1-1} \times \cdots \mathbb{CP}^{2^1-1}$ 8 times. The polytope that replaces the $2^8-1$ simplex there is a product of 8 1-cubes as already mentioned. Again above this polytope are some phases you lost along the way, but those are easy to draw in an accompanying diagram.


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