Condition that a tripartite/multipartite qubit state does/does not admit a Schmidt decomposition?

Generalization of Schmidt decomposition

The Schmidt decomposition $|psi_{AB}rangle = sum_ilambda_i|i_Arangle|i_Brangle$, with $lambda_i$ positive real numbers and $|i_Arangle$ and $|i_Brangle$ (possibly incomplete) orthonormal bases, of a bipartite state $|psi_{AB}rangle$ can be thought of as a type of singular-value decomposition. Specifically, if the amplitudes $psi_{ij} = langle i|langle j|psi_{AB}rangle$ are arranged into a $2times 2$ matrix rather than the usual $4$-vector, then $lambda_i$ are precisely the non-zero singular values of the matrix $psi_{ij}$. This explains why $lambda_i$ are positive real numbers. See the proof of theorem 2.7 on page 109 in Nielsen & Chuang for more details.

The decomposition can be generalized to $n$-partite state as

$$ |psi_{A_1A_2 dots A_n}rangle = sum_i lambda_i |i_{A_1}rangle |i_{A_2}rangledots |i_{A_n}rangletag1 $$

where as before $lambda_i$ are positive real numbers and $|i_{A_k}rangle$ are (possibly incomplete) orthonormal bases. If $n=2$, then decomposition $(1)$ exists for every pure state $|psi_{A_1A_2}rangle$. If $n>2$, then there are pure states without decomposition $(1)$.

Physical characterization

Physically, the $n$-partite states that admit decomposition $(1)$ can be characterized as the states that are free of $m$-partite entanglement for every $1<m<n$. More precisely, $|psi_{A_1A_2 dots A_n}rangle$ admits decomposition $(1)$ if and only if for every $k = 1,dots n-1$ and subsystems $A_{i_1}dots A_{i_k}$, the state

$$ rho_{A_{i_1}dots A_{i_k}} = mathrm{tr}_{A_{i_1}dots A_{i_k}}left(|psi_{A_1A_2 dots A_n}ranglelanglepsi_{A_1A_2 dots A_n}|right) $$

is separable. One way of thinking about these states is that all (if any) entanglement that they have lives between all $n$ subsystems. Conversely, states that do not admit $(1)$ necessarily contain some entanglement between fewer than $n$ subsystems.

This characterization is not very useful to computationally check whether a given $n$-partite state admits decomposition $(1)$. Nevertheless, it helps to intuitively understand the class of Schmidt-decomposable states. In particular, it explains why a generic $n$-partite state for $n>2$ does not admit $(1)$. It also explains why all bipartite states admit $(1)$ $-$ it is impossible to entangle fewer than $2$ subsystems.

Necessary and sufficient conditions in the tripartite case

The necessary and sufficient conditions for a state to admit decomposition $(1)$ are given in the paper cited in a comment above by @Rammus. We reproduce the results of the paper in the case when the Schmidt decomposition

$$ |psi_{ABC}rangle = sum_i lambda_i |i_Arangle|i_{BC}rangletag2 $$

of a tripartite state $|psi_{ABC}rangle$ associated with the partitioning of $ABC$ into $A$ and $BC$ has distinct coefficients, i.e. $lambda_i ne lambda_j$ for $i ne j$. Define the matrices $Omega_i$ as

$$ Omega_{i,jk} = langle j_B|langle k_C|i_{BC}rangle. $$

In other words, $Omega_i$ is the matrix of amplitudes of the state $|i_{BC}rangle$. The necessary conditions for $|psi_{ABC}rangle$ to admit decomposition $(1)$ is for all $Omega_i$ to be rank one and $Omega_i^daggerOmega_{i'} = 0$ and $Omega_iOmega_{i'}^dagger = 0$ for $ine i'$.

If $lambda_i$ are not all distinct, then the necessary and sufficient conditions become more complicated, because $(2)$ and thus $Omega_i$ are no longer unique. See the paper for more details.

Example: $W$ state

The $W$ state is

$$ |Wrangle = frac{|001rangle + |010rangle + |100rangle}{sqrt{3}} = frac{sqrt{2}}{sqrt{3}}|0rangleotimesfrac{|01rangle + |10rangle}{sqrt{2}} + frac{1}{sqrt{3}}|1rangleotimes |00rangle. $$

Thus, $lambda_0=frac{sqrt{2}}{sqrt{3}}$, $lambda_1 = frac{1}{sqrt{3}}$, $|0_{BC}rangle = (|01rangle+|10rangle)/sqrt{2}$ and $|1_{BC}rangle = |00rangle$ so

$$ Omega_0 = frac{1}{sqrt{2}}begin{pmatrix} 0 & 1 1 & 0 end{pmatrix} quad Omega_1 = begin{pmatrix} 1 & 0 0 & 0 end{pmatrix} $$

and we see that $Omega_1$ is rank one, but $Omega_0$ is not. Therefore, $W$ does not admit decomposition $(1)$.