Do qubits need to be in superposition to be entangled?

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


closed as unclear what you're asking by Norbert Schuch, Kyle Oman, Wolphram jonny, John Rennie, Jon Custer Feb 10 at 17:54

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Can a number be even but not a sum? $\endgroup$ – WillO Feb 9 at 17:22

By definition, a pure state of two qubits is separable if and only if it can be written as a product of two single-qubit states. In other words, if the two-qubit state can be written $|a\rangle\otimes|b\rangle$ where $|a\rangle$ and $|b\rangle$ are single-qubit states, then the two-qubit state is separable.

A two-qubit state is called entangled if it is not separable.

The single-qubit state $|a\rangle$ may itself be a superposition, and in fact it is always a superposition of single-qubit basis vectors in most bases. The same comment applies to the other single-qubit state $|b\rangle$.

However, in order for the two qubits to be entangled with each other, the state is necessarily a superposition of separable two-qubit states. (This condition is not sufficient, but it is necessary.) If this is what "superposition" means in the OP, then the answer is yes: two qubits do need to be in a superposition (of separable two-qubit states) in order to be entangled with each other.


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 Dan Yand 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 Dan 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

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