Suppose I had a single particle with a four-dimensional Hilbert space $\mathcal{H}$ spanned by the states $\{ |1\rangle,|2\rangle,|3\rangle,|4\rangle\}$ and suppose I prepared the particle in the superposition state

$$ |\psi\rangle = \frac{1}{\sqrt{2}}\left(|1\rangle + |4\rangle \right). $$

With a trivial relabelling of the basis states as $|1\rangle \rightarrow |00\rangle$, $|2\rangle \rightarrow |01\rangle$, $|3\rangle \rightarrow |10\rangle$ and $|4\rangle \rightarrow |11\rangle$, then it makes it clear that the original Hilbert space $\mathcal{H}$ is isomorphic to the Hilbert space of a pair of spin-$1/2$ particles as $\mathcal{H} \cong \mathcal{H}_{1/2} \otimes \mathcal{H}_{1/2} $, where $\mathcal{H}_{1/2}$ is spanned by the basis $\{ |0\rangle , |1\rangle \}$.

In this basis, the state $|\psi\rangle$ can be trivially rewritten as

$$ |\psi\rangle = \frac{1}{\sqrt{2}}\left(|00\rangle + |11\rangle \right),$$

which is a Bell state - a maximally entangled state of two particles. Well, in the original basis, we have a superposition state of a single particle, whilst in the binary basis we can view it as an entangled state of two particles.

My questions

It would seem as though the original particle is entangled with itself, but it feels like I am cheating as it is simply a result of changing our interpretation of the basis states, so my questions are:

  1. Can a particle be entangled with itself?
  2. If so, does this entanglement have any physical significance? For example, could it be used for quantum computing?
  • 1
    $\begingroup$ Some would call any superposition of states an entangled state, but this is mostly a trivial case from the point of view of practical applications of entanglement. Notation of course doesn't change anything. $\endgroup$
    – Roger V.
    Jan 12, 2023 at 11:07
  • 1
    $\begingroup$ Well, I'd say the meaning of our model (e.g. "choice" of Hilbert space; recall that they are all isomorphic) depends on the physical observables; so a four-dimensional vector space can be the state of e.g. a particle on a lattice with four sites or the Hilbert space of two qubits, for example. The meaning is determined by the physical experiments we want to model. $\endgroup$ Jan 12, 2023 at 11:29
  • $\begingroup$ Particles are entangled when their state cannot be described as the tensor product of states, each in a single-particle Hilbert space. The physical significance of entanglement in, say, secure quantum communication, also rests heavily on the spacial separation of two entangled particles (see Anton Zeilinger's experiments). I find the question if a particle can be entangled with itself of little physical significance. $\endgroup$
    – Kurt G.
    Jan 12, 2023 at 11:40
  • $\begingroup$ There were papers claiming that e.g. the s-wave state of a single electron is an entangled state between left and right half-space, but I always found this rather obscure. As the other comments say, there is a physical meaning to entanglement, which generally involves a tensor product and locality. $\endgroup$ Jan 12, 2023 at 14:11
  • $\begingroup$ there are valid arguments to be made that a single particle in a superposition of different spatial modes ought to be considered "entangled". For example because such superpositions can (in principle) always be converted into entanglement, hence they're "fundamentally the same thing". A standard reference is van Enk's paper: arxiv.org/abs/quant-ph/0507189 $\endgroup$
    – glS
    Jan 16, 2023 at 10:29

1 Answer 1


Your isomorphism is not an isomorphism of the relevant structures.

The state space of a quantum system with many particles is a Hilbert space together with a tensor product decomposition of that Hilbert space. In the case of one particle on a lattice with four sites, the state space is a four-dimensional Hilbert space with the trivial decomposition. In the case of two qubits, the state space is a four-dimensional Hilbert space presented as a tensor product of two two-dimensional Hilbert spaces.

One four-dimensional Hilbert space is isomorphic to another, but not in a way that respects the given decompositions.


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