Do qubits need to be in superposition to be entangled?

Put another way, can qubits be entangled but not in superposition?


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    $\begingroup$ Can a number be even but not a sum? $\endgroup$ – WillO Feb 9 at 17:22

The problem that this question is that for any given system "being in a superposition" isn't actually a statement that makes sense in isolation.

This is because, from a first-principles perspective, the state space $\mathcal H$ of any quantum system is just a complex vector space which is completely 'homogeneous', i.e. there's nothing that really distinguishes any state from any other state. Thus:

  • Any state is just a state, i.e. a vector in $\mathcal H$, which means that it can be expressed as a linear combination of only one vector in $\mathcal H$, which isn't really what you want to count as a 'superposition' state.
  • On the other hand, for every state $\psi$ in the state space $\mathcal H$, there exists a pair of states $\phi$ and $\chi$ (actually, an infinite set of such pairs), such that $$ \psi = \phi + \chi.$$ So, in any meaningful technical sense, all states are superpositions.

In practice, though, you've generally got more information about your system than just the fully abstract perspective, and in particular you've usually got a preferred set of states (usually a basis) in mind. And with that set of states $\{\varphi_j\}$ in hand, it does make sense to ask

is $\psi$ one of the $\varphi_j$, or is it a superposition of several of them,

and to call that second category the "superposition states" of your system.

However, as above, if you change the set of states that you're holding as 'special', then the separation of which states are and are not "superposition states" will move.

As Chiral Anomaly points out in his (great) answer, when we're talking about entanglement between two qubits, it does make sense to pick a certain basis for each of the qubits and then hold the tensor product of the states in those bases as 'special'; as Chiral points out, with that set in hand, the answer is yes - every entangled state must be a superposition of states in that separable basis. (And as Dan and ZeroTheHero point out, that condition is necessary but it is not sufficient: there are states which are not superpositions of those chosen basis states, but which are still separable and therefore not entangled.)

However, in general, the fact that "$\psi$ is a superposition state" is such a fragile statement, which makes no sense without some very clearly delineated context, means that it is very rarely used in professional practice without a suitable qualifier that specifies the set of states that's being held as special.


The fact that one qubit might be in a superposition does not imply it is entangled with another qubit.

For instance, let $\vert 1\rangle=\frac{1}{\sqrt{2}}\left(\vert +\rangle+\vert -\rangle\right)$ and $\vert 2\rangle=\vert +\rangle$. Then $\vert 1\rangle$ is in a superposition but the two-qubit state $\vert 1\rangle\vert 2\rangle$ is not entangled as it is obviously separable.

Conversely, consider the state $$ \vert +\rangle\vert +\rangle + \vert -\rangle\vert -\rangle\, . $$ It is certainly entangled but none of the invididual qubits are in a superposition.

  • $\begingroup$ This addresses the contrapositive of the question. OP asks "does A implies B?" (does entanglement imply superposition), and this answer provides an example where $¬$A but still B (not entangled, but still in superposition) - which has no bearing on the question itself. $\endgroup$ – Emilio Pisanty Feb 9 at 18:30
  • $\begingroup$ @EmilioPisanty point taken but superposition of what? $\endgroup$ – ZeroTheHero Feb 9 at 19:32
  • $\begingroup$ Well, that's kind of the problem with the question as posed. "the state $\psi$ is a superposition" is not a statement that has any meaning in isolation. $\endgroup$ – Emilio Pisanty Feb 9 at 20:16
  • $\begingroup$ @EmilioPisanty sorry mon... didn’t want to come off as snappy. $\endgroup$ – ZeroTheHero Feb 9 at 20:50
  • $\begingroup$ @ Zero No snappiness even detected from my side =). $\endgroup$ – Emilio Pisanty Feb 9 at 21:38

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