2
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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!

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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?" $\endgroup$ – Mark S Oct 9 at 13:55
  • 1
    $\begingroup$ I 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. $\endgroup$ – Markus Heinrich Oct 12 at 9:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.