# What is the difference between "maximally entangled" and "entangled" states?

when we talk about bell state we say that these states are maximally entangled. so just wanted to understand is there any difference between just entangled and maximally entangled ?

In the case of bipartite states, this generally means a state that maximises the entanglement entropy, that is, the von Neumann entropy of the reduced states. Or if you prefer, the Shannon entropy of the vector of eigenvalues of $$\operatorname{Tr}_B(\rho)$$.
So, for a bipartite two-qubit state, a state $$\rho$$ is maximally entangled if $$\rho_A\equiv\operatorname{Tr}_B(\rho)$$ has eigenvalues $$(\frac12,\frac12)$$. For a pure bipartite state $$|\psi\rangle$$, this is equivalent to asking its Schmidt coefficients to be $$(\frac1{\sqrt2},\frac1{\sqrt2})$$, meaning that the state must have the form $$|\psi\rangle = \frac1{\sqrt2}(|u_1,v_1\rangle+|u_2,v_2\rangle)$$ for any set of states $$|u_i\rangle,|v_i\rangle$$ such that $$\langle u_1,u_2\rangle=\langle v_1,v_2\rangle=0$$.
Now we have a notion of which states are entangled and are also able, in some cases, to assert that one state is more entangled than another. This naturally raises the question whether there is a maximally entangled state, i.e. one that is more entangled than all others. Indeed, at least in two-party systems consisting of two fixed ddimensional sub-systems (sometimes called qudits), such states exist. It turns out that any pure state that is local unitarily equivalent to $$|\psi_d^+\rangle = \frac{|0,0\rangle+\cdots + |d-1,d-1\rangle}{\sqrt d}$$ is maximally entangled. This is well justified, because as we shall see in the next subsection, any pure or mixed state of two d-dimensional systems can be prepared from such states with certainty using only LOCC operations. We shall later also see that the non-existence of an equivalent statement in multi-particle systems is one of the reasons for the difficulty in establishing a theory of multi-particle entanglement.