This will sound like a philosophical question (i.e. a question with no actual theoretical content), but I dont intend it as such. Admittedly, my question is based on a very vague intuition.

Also note, I'm a beginner at QM.

When two particles X, Y are perfectly entangled in quantum mechanics, then there is no measurement that we can perform on one of the particles such that we know what the outcome of the measurement will be.

We can write the quantum state of the combined system Z in terms of the tensor product of the separate states of X and Y. I.e. if we denote by u the up state and by d the down state, then the combined system Z can be written as a linear combination of uu, ud, du, dd.

However, these four states are only one possible basis for expressing the combined quantum state of Z. Moreover, the above fact about perfectly entangled states leads me intuitively to the following conjecture: Is it possible to write the combined quantum state in a basis such that none of the basis vectors can be written as tensor products of the individual qubits?

If that is possible, then intuitively I would guess that there is nothing inherently "correct" about seeing the combined quantum system Z as consisting of X and Y (even though these were originally two separate particles and were subsequently entangled to each other). Instead, shouldn't it be possible to "decompose" Z differently, into a particle W and V, whose separate wavefunctions are different from those of X and Y.

Please forgive me if this question is too vague to answer.

  • $\begingroup$ "Fully entangled" doesn't mean anything. Perhaps you're trying to capture the meaning of the term maximally entangled instead. $\endgroup$ Mar 16, 2018 at 20:28
  • $\begingroup$ @EmilioPisanty, yes I meant maximally entangled. $\endgroup$
    – user56834
    Mar 17, 2018 at 4:47

2 Answers 2


The answer to your highlighted question,

Is it possible to write the combined quantum state in a basis such that none of the basis vectors can be written as tensor products of the individual qubits?

is yes. The Bell-state basis is the canonical such example but there's obviously an infinity of possible choices. It's also important to note that this is a statement about the Hilbert space of the particles, i.e. it's a statement about the system, not the state of the system. It remains true regardless of whether the system is in a separable state or in an entangled one.

As to whether this mathematical fact suddenly means that the particles suddenly "meld" into a "single particle" ─ no. That's just plain nonsense. A particles is not just the Hilbert space that describes it; it also carries with it the core observables (operators within that Hilbert space) that describe its properties. If you tensor together two spin-1/2 particles and then manipulate their shared state into a maximally entangled state, then their spin operators $\vec \sigma^{(1)}$ and $\vec \sigma^{(2)}$ remain perfectly concrete per-particle properties (once tensored with the identity operator on the other half); they don't suddenly "vanish" into nothingness.

What does happen, on the other hand, is that each individual system ceases to have a (pure) state of its own: there is one global pure state, but there is no guarantee that each individual particle will be describable by a pure quantum state on its individual Hilbert space. However, this is not a statement about the particle itself, but (in practice) about our knowledge of its current configuration.

  • $\begingroup$ Thank you for the answer. Let me clarify though: Perhaps the question in the title is not formulated so well. My question is not whether they "meld" into a "single particle". My question is more something like: is it possible to decompose the combined system into two particles that are not the same particles as the two original ones? To make a very very crude analogy, if we imagine each particle's wave function is a piece of clay, and then combine the two clays into one, we can then decompose the big piece of clay into two parts that are different than the original two. $\endgroup$
    – user56834
    Mar 17, 2018 at 4:55
  • $\begingroup$ Even thought the observables in the hilbert space remain intact, we could also measure the state of the system with completely different observables. For example, we could have an observable for each basis vector in the Bell-state basis. My question is basically: is there a fundamental distinction between a state vector in the bell-state basis and its corresponding observable, versus a state vector of one of the original particles and its corresponding observable? $\endgroup$
    – user56834
    Mar 17, 2018 at 5:01
  • $\begingroup$ I've already addressed that aspect; I have nothing to add at this time. $\endgroup$ Mar 17, 2018 at 12:42

Though I think your question is misguided, I think there is actually something there. Here's something that sounds rather similar to the idea you seem to have in mind. I think it's related to entanglement, but it is mostly about indistinguishability.

In QM, there is the possibility for particles to be indistinguishable in principle. If two such particles are brought close together -- one from the left, one from the right -- and then separated again, it is impossible to tell which one is which: impossible, not because we lack the technology to find the answer but because the question does not even have an answer. Indistinguishable particles come in two flavors, bosons and fermions (and also anyons). I will only talk about fermions here, but something similar happens for bosons.

Electrons are fermions, and this means they are not allowed to share the same quantum state (this is the Pauli exclusion principle). When two electrons occupy the same spatial state (as they do in atomic orbitals, for example), it is necessary that they occupy different spin states: note that this condition must be satisfied regardless of the measurement basis an observer might choose. It turns out that there is exactly one spin state of two electrons that does satisfy the condition, the entangled singlet state $\frac{1}{\sqrt{2}}(\lvert \uparrow,\downarrow \rangle-\lvert \downarrow,\uparrow \rangle)$. On the other hand, if these two electrons had been clearly distinguishable in position and neither one was entangled to anything else, their spins could take on any state in the four-dimensional space spanned by $\{\lvert \uparrow \uparrow \rangle, \lvert \uparrow \downarrow \rangle, \lvert \downarrow \uparrow \rangle, \lvert \downarrow \downarrow \rangle\}$. Placing the electrons in the same orbital reduces the accessible region of the Hilbert space to a single state; moreover, it is an entangled state.

Now that does sound reminiscent of your question (in bold), but I would caution that this is really about indistinguishability, and the entanglement arises as a by-product. Specifically, note that when two distinguishable particles are entangled, these considerations don't apply and the full tensor product Hilbert space is accessible; you certainly could work in an entangled basis (as Emilio Pisanti said), but it is not forced upon you the way it was in the above examples.


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