# $n$ qubit vs. a $d=2^n$ qudit states and measurements

The pure states of a qudit inhabit the $$\mathbb{CP}(d-1)$$ manifold.

1. Is it true that the pure states of $$n$$ qubits live on the $$\mathbb{CP}(2^n-1)$$ manifold?

2. If the answer to the first question is yes, then how do the sets of measurements on $$n$$ qubits and an $$d=2^n$$ qudit compare? Intuitively, I feel like the latter should be bigger due to the locality issues. However, if we stick to a particular bases, there seem to be a one-to-one correspondence.

• What is a set of measurement? Jun 18, 2019 at 1:24
• All the possible measurement operators. Jun 18, 2019 at 10:59
• Why would locality be an issue for the n-qubit system? Since we can measure the system in an entangled basis as well? Jun 18, 2019 at 11:08

Without additional assumptions or context, there is no fundamental difference between an "$$2^n$$-dimensional qudit" and "$$n$$ qubits". Any "qudit system" over $$2^n$$ modes for some integer $$n$$ can be thought of as a system of $$n$$ qubits. Equivalently, an $$n$$-qubit system is nothing but a $$2^n$$-dimensional qudit system.

The difference is in the fact that if you talk of qubit systems, there is an implicit assumption that you are going to "prefer" operations that act on few qubits at the same time (the precise meaning of this will strongly depend on the context). In other words, if you are studying an $$n$$-qubit system, you are likely to be going to use an operatorial basis for the space of operators that is built out of products of single-qubit operations.

For example, say you have a "four-dimensional qudit system". The "computational basis" for this system will usually be written as $$\{\lvert 1\rangle,\lvert 2\rangle,\lvert 3\rangle,\lvert 4\rangle\}$$. Nonetheless, you can simply "reinterpret" these by writing them in binary notation: $$\lvert 1\rangle\to\lvert00\rangle, \quad\lvert 2\rangle\to\lvert01\rangle, \quad\lvert 3\rangle\to\lvert10\rangle, \quad\lvert 4\rangle\to\lvert11\rangle.$$ Then, an operation that only makes first and second modes interfere corresponds to a "local operation on the second qubit", and similarly for the other possible operations. In a situation in which these kinds of "local" operations are natural, it is then also natural to introduce the notion of entanglement, which needs a notion of "partiteness" to make sense.

Interestingly, note that if you think of the possible states as those of a single $$n$$-dimensional qudit, there is no notion of entanglement, only of correlations. On the other hand, if you reinterpret the states as those of a number of qubits, then you can meaningfully talk of entanglement between the different modes.

Finally, it's worth nothing that for practical purposes there is a significant difference between "$$2^n$$-dimensional qudit" and "$$n$$ qubits". You could describe a system of $$n$$ interacting physical systems (say, interacting atoms) as possible states of a single qudit, but that would result in a very awkward formalism that would obscure the physically interesting aspects of the problem.

Addressing now more directly the raised points:

Is it true that the pure states of $$n$$ qubits live on the $$\mathbb{CP}(2^n−1)$$ manifold?

As of above, yes. At a purely abstract level there is absolutely no difference between the two things.

If the answer to the first question is yes, then how do the sets of measurements on $$n$$ qubits and an $$d=2^n$$ qudit compare? Intuitively, I feel like the latter should be bigger due to the locality issues. However, if we stick to a particular bases, there seem to be a one-to-one correspondence.

Again, what measurements you are going to use fully depends on what you are trying to model. If you think of the system as comprised of $$n$$ qubits, it is likely (though not necessary) that you are going to consider local measurements.

To talk of how "big" a set of measurements is, you should first specify what kind of measurements you are considering. If you stick to projective measurements, then the maximum number of measurement outcomes will be $$2^n$$ in both cases, the difference will be in how the projectors are built. For the qudit system you might consider a set of projectors of the form $$\mathbb P_k\equiv\lvert k\rangle\!\langle k\rvert$$, while for $$n$$ qubits the projectors might be written as products of two-outcome projective measurements $$\mathbb P_\ell^\pm$$, projecting on one of the two possible states of the $$\ell$$-th qubit.

• I find the sentence "On the other hand, if you reinterpret the states as those of a number of qubits, then you can meaningfully talk of entanglement between the different modes" to be quite profound. If $n$ is a power of two can we think of an $n$-mode qudit as a tensor-product of $\log_2 n$ "virtual" qubits? As long as $n$ is composite I guess, we can always factor $n$ to its constituent qudits. Then, a superposition of modes of the $n$-mode qudit would correspond to entanglement among the virtual qubits. Dec 29, 2022 at 19:33
• yes, I'd say so. I'm not sure how "profound" this actually is though. I think of it as the quantum counterpart of the fact that for any (classical) probability vector over $pq$ outcomes, we can talk about "correlations between two outcomes" (making a choice on how to assign the $pq$ outcomes to the two parties), one with $p$ and the other with $q$ outcomes.
– glS
Dec 30, 2022 at 0:32
• Well, a superposition of a single qudit can be isomorphic to an entanglement of smaller subsystems. In that sense, there’s no classical counterpart. A harmonic oscillator with four modes on a single particle can be in a superposition and be isomorphic to a Bell pair - but there’s no spatial locality to confuse things with a four mode qudit in superposition. Could we play the CHSH game with such a qudit? Does that even make sense? Dec 30, 2022 at 2:22
• @MarkS a harmonic oscillator is an infinite-dimensional system so that somewhat complicates things, though the same principle will apply. But I mean the classical analogy is superposition : entanglement = single probability distribution : correlations between outcomes. Of course in the quantum case the correlations are at the "deeper" level of amplitudes, but the principle that you can in principle arbitrarily partition the outcomes into different parties is identical. You can discuss nonlocality and LHV for these things but w/o an actual notion of locality it's probably not very useful
– glS
Dec 30, 2022 at 7:37