Can we say, in general, that the qubits of a system of n-qubits are entangled if at least one basis vector is missing from the probability distribution? For example, in a system of 3-qubits, after a certain circuitry, I measure all the qubits. I found that

c1% probability of |000>, 
c2% of |001>,
c3% of |010>,
c4% of |101>,
c5% of |111>

so that c1+c2+c3+c4+c5 = 100.

Can I then say that the system of 3-qubits is an entangled one?


3 Answers 3


There's two sides to this. First, the question you actually ask: No. Consider the state $$ |00\rangle $$ This is clearly not entangled, but misses many basis states. Of course, for the specific example you give, once you know what the values of the ci coefficients are, you might be able to say something about the entanglement. (You might need to know more, such as the amplitudes rather than just the probabilities).

On the other side (I know you didn't ask), but there are entangled states which contain all basis states. For example, $$ \frac12(|00\rangle+|01\rangle+|10\rangle-|11\rangle). $$You can see that this is entangled by writing it as $$ \frac{1}{\sqrt{2}}(|0+\rangle+|1-\rangle), $$ which is just one of the standard Bell states + a Hadamard rotation on one qubit (which therefore cannot affect the entanglement).


In general this is not correct. Consider two systems: one in state $\frac{1}{\sqrt2}(|00\rangle + |01\rangle)$ and another in state $\frac{1}{\sqrt2}(|00\rangle + |11\rangle)$. Both of them have exactly two basis vectors missing from the measurement results, but the first one can be represented as $|0\rangle\otimes\frac{1}{\sqrt2}(|0\rangle + |1\rangle)$ and thus is not entangled, while the second one cannot be represented as a tensor product of 1-qubit states and is entangled.

  • $\begingroup$ That means $\frac{1}{\sqrt{2}(|100\rangle + |101\rangle)$ is not entangled, but $\frac{1}{\sqrt{2}(|100\rangle + |001\rangle)$ is entangled, is it? (Not sure why the latex is not rendering) $\endgroup$ Jun 15, 2020 at 6:17
  • 1
    $\begingroup$ (You missed a '}' after \sqrt{2}; that's why your latex rendering is inverted). But yeah, $\frac{1}{\sqrt{2}}(|100\rangle + |101\rangle) = \frac{1}{\sqrt{2}}(|10\rangle \otimes (|0\rangle + |1\rangle)) = |10+\rangle$, so that is not entangled. However, the other state can be written as $\frac{1}{\sqrt{2}}((|10\rangle\rangle_{13} + |01\rangle_{13})\otimes |0\rangle_{2})$, so the first and third qubits are entangled. $\endgroup$
    – JSdJ
    Jun 15, 2020 at 7:07

Recall that for a tripartite quantum state $| \psi_{ABC} \rangle$, one needs to specify the partition along which we're studying entanglement. For example, consider the $W$-state $| \mathrm{W} \rangle = \frac{1}{\sqrt{3}} \left( | 100 \rangle + | 010 \rangle + | 001 \rangle \right)$, which is entangled across all partitions. However, a state of the form, $\frac{1}{\sqrt{2}}\left( | 00 \rangle + | 11 \rangle \right) \otimes | \alpha \rangle$, where $| \alpha \rangle$ is any single-qubit state, is entangled across the $A:BC$ partition but is not entangled on the $AB:C$ partition.

Note that, in general, unless you specify the partitions (for a $n$-qubit system with $n>2$) and the coefficients (in your case the set $\{ c_{1}, \cdots, c_{5} \}$), one cannot say if the state is entangled or not. Moreover, the question becomes even more complex if the state in consideration is a mixed state and not a pure state. Here's a (classic) review paper on entanglement: An introduction to entanglement measures by Plenio and Virmani. If this is a bit advanced for you then I'd recommend studying the standard textbook: Nielsen and Chuang.

Also, if you're looking for numerical tools to check if a given state with some coefficients is entangled or not, consider using tools like QETLAB (or others, an exhaustive list can be found on Quantiki). For example, here is a numerical function to check if a given pure state is entangled or not. But I'd highly recommend learning some more foundational ideas before using the numerical tool.


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