I understand that a qudit is a quantum $d$-state system. If $d=4$, is this exactly the same as a two-qubit system, which also presents $4$ quantum states? The Hilbert space is the same, right? Are there any theoretical or practical differences?
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$\begingroup$ See also: quantumcomputing.stackexchange.com/questions/2415/… $\endgroup$– user820789Commented Sep 11, 2018 at 20:46
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5 Answers
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For qubits, we usually base all of our operators on the Pauli matrices. Our basic gate set consists of the Pauli matrices themselves, Clifford gates like $H$ and $S$ that map between Pauli matrices, controlled operations like the CNOT that implement a Pauli on one qubit depending on the Pauli eigenstate of another, etc.
For any larger $d$-dimensional quantum system, we have to find the basic set of operators that will play the same role.
One approach is the generalize the Pauli matrices. We choose a group whose order is $d$, and define operators based in that group. This is my go-to text on how to do this, though it is actually focused more on generalizing stabilizer codes.
We could also look to the spin operators for inspiration. The Pauli matrices describe a spin-$1/2$ system. So for higher dimensional systems, we could look at the operators for higher spin. They don’t have the same sort of nice properties, though. So this doesn’t seem to be a popular approach.
Either way, the Hilbert space is the same and universal QC based on them is the same thing. The only difference is our basic gate set. So the numbers of gates required for a given task may have a difference in terms of constants and coefficients. And the maths might be nicer for one than the other. But the complexity will be the same.
answered Apr 22, 2018 at 7:46
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Yes the Hilbert space is the same, but you have to choose the isomorphism $\phi : \; \; (\mathbb{C}^2)^{\otimes 2} \simeq \mathbb{C}^4$. But the different setup will mean some unitaries that will be easy to implement in one setup will be hard in the other. For example, as 2 qubits gates something like $\sigma_z \otimes 1$ will be easy. But if you write that as a 4 by 4 unitary through that isomorphism $\phi$ instead that might not be as easy to implement. You should say both the Hilbert space and the easy operations that you wish to write your program in terms of.
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$\begingroup$ That already seems to help, but would you care to elaborate the examples a little for the non-experts? $\endgroup$ Commented Apr 22, 2018 at 6:33
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A fundamental difference between the two kinds of systems is that a two-qubit system can actually be in an entangled state. On the other hand, a single d=4 dimensional system does not possess entanglement, since entanglement is always defined with respect to more than one party. Consequently, for the purposes of quantum protocols that exploit entanglement as a resource, a two-qubit system and a single 4-dimensional quantum system are very different.
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2$\begingroup$ This is indeed an important point missed by other answers. However, I'd say that it is a fundamental difference, rather than the fundamental difference, since it only applies to cases where you might want to split the state up. $\endgroup$ Commented Apr 24, 2018 at 11:24
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2$\begingroup$ I don't agree with this. Even if you have a $d=4$ qudit, you can still "change the way you look at it", and study the entanglement between two different parts of this system. In other words, given any $\mathcal H_4$, you can always split it up so that $\mathcal H_4=A\otimes B$, and study the entanglement between the subspaces $A$ and $B$ $\endgroup$– glS ♦Commented Jun 21, 2018 at 10:49
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There is also a difference if you consider experiments or implementations. To make a physical qubit, I need to use a two-level quantum system. Qudits than require a more complicated quantum system, e.g., with four levels for a d=4 qudit. The engineering justification for using the more complicated system would be that you than require fewer of the four-level systems.
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The only difference between a "pair of qubits" and a single "four-dimensional qudit" is that when you say you have "two qubits" you are implicitly making some assumptions on the kind of operations you can perform on it.
In particular, it only makes sense to talk of two qubits if they can be treated as two different systems, or, in other words, if it is possible to act locally on them. Similarly, the kinds of operations that one can assume to be able to perform on two qubits are different than those on qudits.
From a practical point of view, the difference is that one tends to consider different operations as "easily available" when talking of sets of qubits rather than (sets of) qudits.
