# When is a state entangled?

I have read from What's the difference between an entangled state, a superposed state and a cat state? that an entangled state is one that cannot be expressed as product state. Suppose we have the following state

$$|\psi\rangle = |\alpha\rangle_1 |\beta\rangle_2$$

which can be seen as a product of two separable single-mode coherent states. Thus it is not entangled. But if we were given a sum of number states like the following, how could we ascertain whether it can be algebraically expressed as a product of two separable single-mode coherent states?

$$\sum_{k=0}^{N} \binom{N}{k}|k\rangle_1 |N-k\rangle_2$$

• Your first formula is a Product state, so by definition it is not entangled. Did you intend to say it was? Commented Jun 17 at 13:25
• Is the title coherent with the body? I mean like in the context of this question. "When is a state entangled?" and "whether it can be algebraically expressed as a product" seem opposite. Commented Jun 18 at 13:01

Write your state as $$\sum_{i,j=0}^n \alpha_{ij}|k_i\rangle |k_j\rangle$$ and check whether the matrix $$M=(\alpha_{ij})$$ has rank one.

• If we were to be able to diagonalize it, we would obtain a sum of states such that there's an equal number of photons in mode 1 and mode 2. Why does that make it not entangled? Commented Jun 17 at 15:49
• @DanielJanjani : If the matrix is both diagonal and rank one then your state is a product state $k_i\rangle\otimes k_i\rangle$ for some $i$. Commented Jun 17 at 17:49
• @WillO Can you expand on this a little more? I know you answered the question, and I think you are correct, but it is not really clear why. Or maybe link a reference?
– hft
Commented Jun 17 at 22:43
• @DanielJanjani "If we were to be able to diagonalize it..." Diagonalize what? The matrix of $\alpha_{ij}$? This will not change the rank... If you have a direct-product state matrix that is not diagonal, you are going to find that all columns are the same up to a constant factor or are all zeros (i.e., only one linearly independent column, i.e., rank 1). If you diagonalize it, you are going to find that every columns is all zeros except for one, which is all zeros except for a single one... OTOH, if you have an entangled state matrix you are going to find that the rank is greater than one.
– hft
Commented Jun 17 at 22:50
• @hft : 1) A matrix is rank one if and only if it is the product of a column vector and a row vector. 2) A state $\sum_{ij} \alpha_{ij} k_i\rangle$ is a product state if and only if it is of the form $\sum_i\beta_i k_i\rangle\otimes \sum_j\gamma_j k_j\rangle$, if and only if the matrix of $\alpha$'s is the product of a column of $\beta's$ times a row of $\gamma$'s, if and only if the matrix of $\alpha$'s is a matrix of rank $1$ Commented Jun 17 at 23:49

...an entangled state is one that cannot be expressed as product state.

This is correct.

But if we were given a sum of number states like the following, how could we ascertain whether it can be algebraically expressed as a product of two separable single-mode coherent states?

$$\sum_{k=0}^{N} \binom{N}{k}|k\rangle_1 |N-k\rangle_2$$

You have already received an answer that you can apply generally, namely that if the rank of the coefficient matrix is equal to one then the state is unentangled.

But, it might also be instructive to look at an example for, say, $$N=2$$.

In the $$N=2$$ case your (unnormalized) example state is: $$|\psi\rangle = |0\rangle\otimes|2\rangle + 2|1\rangle\otimes|1\rangle + |2\rangle\otimes|0\rangle\;,\tag{A}$$

On the other hand, the most general $$N=2$$ direct product state, has the form: $$\left(\alpha |0\rangle + \beta |1\rangle + \gamma|2\rangle\right) \otimes \left(\delta|0\rangle + \epsilon|1\rangle + \zeta|2\rangle\right)\tag{B}$$

By comparing Eq. (B) to Eq. (A) we see that, if Eq. (A) were a direct product state, the direct product coefficients would have to satisfy $$\begin{matrix} \alpha\delta = 0\;, &\alpha\epsilon = 0\;, &\alpha\zeta=1\;,\\ \beta\delta = 0\;, &\beta\epsilon = 2\;, &\beta\zeta=0\;,\\ \gamma\delta = 1\;, &\gamma\epsilon = 0\;, &\gamma\zeta=0\;, \end{matrix}$$ which is not possible. For example, the first equation above tells us that either $$\alpha=0$$ or $$\delta=0$$, but the third equation above tells us that $$\alpha$$ cannot be zero and the seventh equation above tells us that $$\delta$$ cannot be zero.

This argument generalizes to any $$N$$ (for your example state of a direct product of two kets).

You can write the general direct product state as: $$\left(a_0|0\rangle + \ldots + a_N|N\rangle\right)\otimes\left(b_0|0\rangle+\ldots +b_N|N\rangle\right)\;,\tag{C}$$ but for this general direct product state to be your example state, you must have $$a_0b_0=0$$ and thus either $$a_0=0$$ or $$b_0=0$$. But you must also have $$a_0 b_N=1\neq 0$$ and thus $$a_0\neq 0$$ and you must also have $$a_N b_0=1\neq 0$$ and thus $$b_0\neq 0$$. Thus if you assume that your example state is a direct product state then you arrive at a contradiction. Therefore by reductio ad absurdum, your example state must be entangled.

More generally, when dealing with the direct product of two kets, the general form of a direct product state is given in Eq. (C), which can be re-written as $$\sum_{ij}a_i b_j |i\rangle_1|j\rangle_2\;,$$ whereas a general, possibly entangled, state can be written in the same direct product basis as $$\sum_{ij}\alpha_{ij} |i\rangle_1|j\rangle_2\;,$$ and therefore, when the state is a direct product we have the matrix of $$\alpha_{ij}$$ being: $$(\alpha) = \left( \begin{matrix} \vec a b_0 & \vec a b_1 &\ldots & \vec a b_N \end{matrix}\;, \right)\tag{D}$$ which clearly has rank one, since the rank of a matrix is the dimension of the vector space spanned by its columns, and all the columns are multiples of $$\vec a$$. The converse also holds, since if a matrix has rank one then it can be written in the form of Eq. (D).

Returning to OP's example of $$|\Psi\rangle = \sum_{k=0}^{N} \binom{N}{k}|k\rangle_1 |N-k\rangle_2\;,$$ the matrix of coefficients is $$(\alpha) = \overbrace{\left( \begin{matrix} 0 & 0 &\ldots & 1 \\ 0 & \ldots & N & 0 \\ \ldots \\ 0 & N & 0 & 0 \\ 1 & 0 & 0 & 0 \end{matrix}\;, \right)}^{N+1}$$ which has rank N+1 and thus is entangled unless N=0.

• If $N=0$ wouldn't the rank be 2? Commented Jun 19 at 18:25
• right... the notation is a little confusing. Commented Jun 19 at 19:12
• nevermind! I was just looking at your matrix and setting N=0 without minding the dimension. My bad totally. I wrote up an answer but it basically duplicated yours albeit in a different notation so I was reading your answer in my mindset. Commented Jun 20 at 1:17