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?

  • $\begingroup$ If people here consider that this is a duplicate, I can happily delete it and instead suggest an edit to the original question in order to expand its scope and include the details I'm interested in. $\endgroup$ Apr 22, 2018 at 9:16
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    $\begingroup$ I don't believe this is a duplicate question. I asked on Meta to get clarity on what questions should and shouldn't be separated. $\endgroup$ Apr 22, 2018 at 18:08
  • $\begingroup$ My take: I think we should be thinking in future users and in search engines (and also in present users, caring to not alienate them/us/each other). Not as a rule but as a gut-feeling: what do we want people to find, in the most typical case? In these terms, I feel we should go for questions&answers that are complete, but not complex (as in: not reducible to simpler-yet-complete questions&answers). $\endgroup$ Apr 22, 2018 at 18:24

1 Answer 1


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.

  • $\begingroup$ For clarification, when you write $\mathbb{C}\oplus\mathbb{C}^3$, is this related to what I would call -talking about the result of the interaction between two spins S=1/2- "a singlet and a triplet"? $\endgroup$ May 6, 2018 at 22:04
  • $\begingroup$ @agaitaarino yes, exactly. When you add the interaction, you get the singlet ground state and a 3-fold degenerate excited state, the triplet. $\endgroup$
    – DaftWullie
    May 7, 2018 at 6:25

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