Number of conditions for a two-particle state to be decomposable

Question

Suppose we have a general two-particle state $ \Phi (x_1, x_2 ) = \sum_{n_1,n_2} \phi_{n_1,n_2}(x_1,x_2)|n_1,n_2> $, where $n_1$ can be any of $n$ possible states, and $n_2$ can be any of $m$ states. If the state is decomposable then the coefficients $\phi_{1,2}$ can be decomposed into $\phi_{n_1} * \phi_{n_2} $.

It seems to me that for the decomposable state there are $n + m$ independent coefficients ($n$ coefficients describing the state of partice 1, and $m$ coefficients describing the state of particle 2), and if the system is not decomposable, i.e., if there is entanglement between the two particles, then the number of independent coefficients is $n * m$ ($n$ possibilities for $n_1$ times $ m$ possibilities for $n_2$ in $\phi_{n_1,n_2}$). If this logic is correct, then the number of conditions to be fulfilled by a decomposable state is $nm - n - m$. However, according to the book I am studying the number of conditions is $nm - n - m +1 $. I wonder why there is an extra condition.

I am not considering normalization, because the states are seen as rays in projective space, and furthermore both the decomposable and entangled states would have to fulfill the same normalization requirements so I guess there would be no difference.

Answer 1

If you divide out normalization & overall phase, the two states have $n-1$ and $m-1$ independent (complex) degrees of freedom, respectively. On the other hand, the joint two-particle state has $nm-1$ independent degrees of freedom. The difference is $nm-n-m+1$.

Differently speaking, if you leave in phase and normalization, the two states (in a tensor product) have only $n+m-1$ degrees of freedom, since phase + normalization is a joint property.

---

EDIT: Alternative derivation:

If we write $|\phi\rangle = \sum_{ij} M_{ij} |i\rangle |j\rangle$, we are asking for the number of conditions such that $M_{ij} = a_i b_j$. Clearly, this means that $M$ is determined by its first row and column (which can be choosen freely). The remaining $(n-1)(m-1)=nm-n-m+1$ elements cannot be chosen and enumerate the constraints. Again, the first row and column together only have $n+m-1$ independent variables.