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

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.

-

Is there a reason for the x-dependence in the state? – Norbert Schuch Nov 26 '11 at 2:57
$x$ is supposed to represent a three dimensional vector. Sorry if that was unclear, maybe I should have written it in bold. – Raphael R. Nov 26 '11 at 4:14
No, it's just that x being or not being there seems to have nothing to do with your question, as you are only taking about the "spin" degree of freedom. – Norbert Schuch Nov 26 '11 at 5:52

## 1 Answer

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.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 29, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9488779306411743, "perplexity_flag": "head"}