Can a single qutrit in superposition be considered entangled?

2

Often in quantum computing the idea of quantum superposition is introduced well before the concept of entanglement. I suspect this may be because our conception of (classical) computing privileges bits, and hence we also privilege qubits in a Hilbert space of dimension $d=2$. It's easy enough to consider a single qubit in superposition, but transitioning to entanglement requires a plurality of such particles.

Or does it?

For example, suppose we lived in a world that privileged qudits, with $d=4$; e.g. four-level quantum systems as opposed to two-level qubits. We can think of our system (say, a particle-in-a-box or a harmonic-oscillator or what-have-you); our qudit could be in any superposition of $\{\vert 0\rangle,\vert 1\rangle,\vert 2\rangle,\vert 3\rangle\}$.

We can think of a particle in a superposition of $\vert \Psi\rangle=\frac{1}{\sqrt{2}}\vert 0\rangle\pm\vert 3\rangle$, or $\vert\Phi\rangle=\frac{1}{\sqrt{2}}\vert 1\rangle\pm\vert 2\rangle$.

Now if we envision our (single) qudit instead as two virtual qubits, with a mapping/isomorphism such as:

$$\vert 0\rangle_{qudit}=\vert 00\rangle_{qubit}$$ $$\vert 1\rangle_{qudit}=\vert 01\rangle_{qubit}$$ $$\vert 2\rangle_{qudit}=\vert 10\rangle_{qubit}$$ $$\vert 3\rangle_{qudit}=\vert 11\rangle_{qubit},$$

then we can see that both $\vert \Psi\rangle$ and $\vert \Phi\rangle$ are the Bell states, e.g. are entangled.

This works nicely for $d=4$ or any other power of $2$. But would it work for any other dimension, such as $d=3$?

Can we decompose a qutrit that is in superposition into smaller components, and ask whether the qutrit thusly is in some sense entangled?

Mark S

Posted 2020-10-07T21:47:47.023

Reputation: 4 273

Answers

5

To talk about entanglement, you have to first identify subsystems. In your $d=4$ example, you defined an isomorphism $\mathbb{C}^4\simeq \mathbb{C}^2\otimes\mathbb{C}^2$ via the identification of basis states. Whether this is meaningful, depends on the context/the physical scenario you have in mind. But it definitely can be.

For $d=3$, this is never possible. Why? Because you have to single out subsystems i.e. you have to define a tensor product structure. But necessarily, if your Hilbert space is $\mathcal H \simeq \mathcal H_1 \otimes \mathcal H_2$, then $\dim\mathcal H = \dim \mathcal H_1 \times \dim\mathcal H_2$. So if $\mathcal H$ has prime dimension, it cannot be factored (non-trivially). The trivial factorisation is of course always possible, this is $\mathcal H \simeq \mathcal H \otimes \mathbb C$. But you can easily see that in this case, no entanglement is possible.

(Maybe unrelated) note: I have observed multiple times that people confuse subsystems with subspaces. Subspace give rise to a direct sum decomposition, most commonly $\mathcal H = U\oplus U^\perp$. This is vastly different from a tensor product structure!

Markus Heinrich

Posted 2020-10-07T21:47:47.023

Reputation: 928

Thanks! There's another question that I'll have to consider how to formalize, along the lines of "what are the requirements for such a factorization to be meaningful?". If we had a harmonic oscillator that can be in a superposition of one of $n$ eigenstates for some non-prime $n=p\times q$, then this can be factored non-trivially into into the tensor product of two different Hilbert spaces of dimension $p$ and $q$. Can we then think of these two separate Hilbert spaces as somehow separate "particles" or separate "qudits?" – Mark S – 2020-10-09T13:55:41.500

1I guess the question is what IS a separate "particle" in the first place? Maybe some kind of system which you can consider to be isolated in a certain limit? In this case, your example would not really meet the criteria. But one could maybe think about these "subsystems" as "logical" qudits which your big system is simulating. – Markus Heinrich – 2020-10-12T09:09:21.200