7

This question is inspired by "What is the difference between a qudit system with d=4 and a two-qubit system?", as an experimental follow-up.

Consider for illustration these two particular cases:

In general I'm referring to experimental cases where in practice there is an always-on-but-sometimes-weak coupling between two two-state systems, producing a ground quadruplet.

My question is: in experiments such as these, are 2·qubit and d=4 qudit (a) strictly distinguishable beasts, or (b) theoretical idealizations which are more or less adequate depending on practical considerations?

João Bravo
  • 145
  • 11
agaitaarino
  • 3,907
  • 2
  • 13
  • 42

1 Answers1

2

For theoretical purposes, I would say that describing two qubits either as exactly that, two qubits ($\mathbb{C}^2\otimes\mathbb{C}^2$), or as a single $d=4$ spin, ($\mathbb{C}^4$) are essentially equivalent, assuming you have universal control over the whole Hilbert space, because it means you can do whatever you want. The distinction is usually most applicable when you separate the two qubits over some distance, and cannot easily implement a key gate in the universal set (i.e. the two-qubit interaction). But here, you're explicitly stating that that interaction is present. So, the theory claims it makes no difference; you can always do anything you want.

I expect that in practice (although I'm not an experimentalist), the difference comes down to the error mechanisms, which will be different between what you might actually describe as 3 different settings: two qubits $\mathbb{C}^2\otimes\mathbb{C}^2$, a single spin $\mathbb{C}^4$, or the Hilbert space structure implied by the two-qubit interaction you mentioned, $\mathbb{C}\oplus\mathbb{C}^3$. The energy levels in each case are quite different, which will affect the relaxation properties, and presumably more general interactions with the environment as well.

DaftWullie
  • 62,671
  • 4
  • 55
  • 140