If I have either of the following states:

$$\left|\psi_0\right\rangle = \frac{1}{2}\left|0000\right\rangle + \frac{1}{2}\left|0011\right\rangle + \frac{1}{2}\left|1100\right\rangle + \frac{1}{2}\left|1111\right\rangle$$

$$\left|\psi_1\right\rangle = \frac{1}{\sqrt{3}}\left|001\right\rangle -\frac{1}{\sqrt{3}} \left|010\right\rangle + \frac{1}{\sqrt{3}}\left|100\right\rangle$$

How can I check to see whether they are entangled or not? I have seen people say that the state $\left|\psi_1\right\rangle$ is entangled because "there are correlations between the three qubits", but I can't find what this really means.

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    $\begingroup$ Does this answer your question? How do I show that a two-qubit state is an entangled state? $\endgroup$ Commented Jan 2, 2020 at 21:10
  • $\begingroup$ @MartinVesely Thanks, I had seen that answer, but was looking for hopefully another method (even if less formal and complete) which I could use on the fly to check if an n-qubit superposition was entangled or not. $\endgroup$
    – Cascades
    Commented Jan 2, 2020 at 23:39
  • $\begingroup$ @Cascades Oh hahaha, in that case, my answer probably isn't very helpful! $\endgroup$ Commented Jan 2, 2020 at 23:54
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    $\begingroup$ @JackCeroni To be fair I did find your clearer to read than the previously mentioned one... so please don't take it down! It just gets a bit cumbersome in an exam setting having to expand out 4+ qubit systems. $\endgroup$
    – Cascades
    Commented Jan 3, 2020 at 0:08
  • $\begingroup$ Glad to hear that! It definitely is cumbersome haha $\endgroup$ Commented Jan 3, 2020 at 1:12

2 Answers 2


The definition of an entangled state is one that cannot be decomposed into an arbitrary Kronecker product of individual state vectors corresponding to individual quantum systems. Mathematically speaking, an un-entangled state $|\psi\rangle$ is one that can be written in the following way:

$$|\psi\rangle \ = \ \displaystyle\bigotimes_n |\psi_n\rangle$$

Where $|\psi_n\rangle$ is the state of the $n$-th qudit, whatever you want. In the case of an $D$-level discrete quantum system, we can write each $|\psi_n\rangle$ as a normalized linear combination of $D$ orthogonal basis vectors:

$$|\psi_n\rangle \ = \ \displaystyle\sum_{d} c^n_d |d\rangle \ \ \ \text{with} \ \ \ \displaystyle\sum_d |c^n_d|^2 \ = \ 1$$

In the case of a qubit, $D \ = \ 2$ and we have $|0\rangle$ and $|1\rangle$ as our basis states. We thus expand $|\psi\rangle$:

$$|\psi\rangle \ = \ \displaystyle\bigotimes_n \displaystyle\sum_{d} c^n_d |d\rangle$$

When we expand our tensor product, we will get a linear combination of $D^N$ different state vectors, each with some unknown coefficient. In general, for the $j$-th term of the sum, our coefficient $c_j$ is given by a product of $D^N$ different values of $c^n_d$ (I'm not going to do the explicit expansion for the $j$-th term in the case of a $d$-level quantum system, I don't even want to do the binomial expansion for a $2$-level system):

$$c_j \ = \ c^{n_{j_1}}_{d_{j_1}} \ c^{n_{j_2}}_{d_{j_2}} \ ... \ c^{n_{j_{D^N}}}_{d_{j_{D^N}}}$$

But we already know $c_j$, as we know $|\psi\rangle$, and $c_j$ is given by:

$$c_j \ = \ \langle j | \psi \rangle$$

Where $|j\rangle$ is the $j$-th basis vector in the expansion. Thus, all we have to do is solve the system of $D^N$ equations for $D^N$ unknowns. If we can in fact do this, then we know that we have an un-coupled (un-entangled) state. If we can't we know our state is entangled!

If this was slightly confusing an abstract, let's work through a basic example, and prove that the maximally entangled qubit Bell state:

$$|\psi\rangle \ = \ \frac{|00\rangle \ + \ |11\rangle}{\sqrt{2}}$$

is in fact entangled using the outlined method. Our goal is to write $|\psi\rangle$ as a tensor product of two individual state vectors:

$$|\psi\rangle \ = \ \frac{|00\rangle \ + \ |11\rangle}{\sqrt{2}} \ = \ (a|0\rangle \ + \ b|1\rangle) \ \otimes \ (c|0\rangle \ + \ d|1\rangle)$$ $$\Rightarrow \ (a|0\rangle \ + \ b|1\rangle) \ \otimes \ (c|0\rangle \ + \ d|1\rangle) \ = \ ac |00\rangle \ + \ ad |01\rangle \ + \ bc|10\rangle \ + \ bd|11\rangle$$

Now, since each of our basis vectors are orthogonal, we can equate the components to get a system of equations:

$$ac \ = \ \frac{1}{\sqrt{2}}$$ $$ad \ = \ 0$$ $$bc \ = \ 0$$ $$bd \ = \ \frac{1}{\sqrt{2}}$$

Rearranging, we get:

$$d \ = \ \frac{1}{\sqrt{2} b} \ \ \ \text{and} \ \ \ a \ = \ \frac{1}{\sqrt{2} c}$$

So we get:

$$ad \ = \ \frac{1}{2bc} \ = \ 0$$

Which is a contradiction (this is not possible). Thus, we cannot express the Bell state as a tensor product of individual state vectors. It follows that the Bell state is in fact entangled.

If you are into some really hardcore stuff, we can do something similar with a continuous-variable quantum system (qumodes, for instance). In this case, we no longer expand out our state $|\psi_n\rangle$ as a discrete sum of basis states, but as a continuous integral over a continuum:

$$|\psi_n\rangle \ = \ \displaystyle\int \psi_n(x) |x\rangle dx$$

We then have:

$$|\psi\rangle \ = \ \displaystyle\bigotimes_n \displaystyle\int \psi_n(x) |x\rangle dx \ = \ \displaystyle\int \ ... \ \displaystyle\int \displaystyle\bigotimes_n \psi_n(x_n) |x_n\rangle \ dx_1 \ ... \ dx_n$$

In general, $|\psi\rangle$ is given by:

$$|\psi\rangle \ = \ \displaystyle\int \ ... \ \displaystyle\int \psi(x_1, \ ..., \ x_n) \displaystyle\bigotimes_n |x_n\rangle \ dx_1 \ ... \ dx_n$$

And, analogous to the discrete case, we know the function $\psi(x_1, \ ..., \ x_n)$. Thus, in the un-entangled case, we have:

$$\displaystyle\prod_n \psi_n(x_n) \ = \ \psi(x_1, \ ..., \ x_n)$$

So instead of finding discrete components, our goal is to find a collection of functions $\psi_j(x_j)$, such that the product of all these functions gives us the general $\psi$. If we can't do this, then our state is entangled. As an example, consider the function:

$$\psi(x, \ y) \ = \ x \ + \ y \ + \ xy$$

We want to find $\psi_1$ and $\psi_2$ such that:

$$\psi_1(x) \psi_2(y) \ = \ x \ + \ y \ + \ xy$$

We have:

$$\psi_2(y) \ = \ \frac{x \ + \ y \ + \ xy}{\psi_1(x)} \ = \ \frac{x}{\psi_1(x)} \ + \ \frac{y}{\psi_1(x)} \ + \ \frac{xy}{\psi_1(x)}$$

Now, $\psi_2(y)$ is a function of only $y$, we can't have $x$ present in any of the terms. We quickly realize that this is impossible, as the only way to get rid of the $x$ in the first term is if $\psi_1(x) \ = \ ax$, but this would mean that the second term contains $x$. This same logic can be applied to all of the terms. In addition, cancellation is obviously of no help here. Thus, the state given as:

$$|\psi\rangle \ = \ \displaystyle\int \displaystyle\int (x \ + \ y \ + \ xy) |x, \ y\rangle dx dy$$

must be entangled.


After writing this answer, I came across this question, which is very similar, and already has answers. Despite this, I don't think this question should be closed as my answer discusses a more general approach to this problem.


Your comments talk about an exam setting. Probably the most reliable thing to do is calculate the partial trace of each qubit. If any of them is mixed, there is entanglement in the system. (I’m not sure that a yes/no question about the existence of entanglement is a good question in a system of more than 2 qubits, but...) In your particular examples, you can gain a lot of efficiency by noticing that there’s a permutation symmetry between the qubits, so that all the reduced density matrices are basically the same. You only have to check one of them.


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