I came over this unclear claim which i wondered someone could clarify:
"The partial transpose of an entangled state has at most one negative eigenvalue."
I wondered if this holds for all states or just some, searched around but couldn’t find some clear explanation/or proof of the claim.
I was wondering what the conditions for this claim are and if someone has a proof?
Adding to @Frederik's answer, there is a more general statement proven in this paper that
Partial transposition of any $m \otimes n$ state can not have more than $(m − 1)(n − 1)$ number of negative eigenvalues.
which of course reduces to "at most one" in the case $m=n=2$.
I believe it was first shown in the 1998 paper "Local description of quantum inseparability" by Sanpera et al. (direct link) that the partial transpose of a two-qubit state has at most one negative eigenvalue. This statement is true for all two-qubit states, entangled or separable, but for separable states taking the partial transpose does not violate positivity so negative eigenvalues can only arise this way for entangled states in the first place.
The reason for this is rather simple, if we accept the following intermediary result (Theorem 2 in the paper by Sanpera et al.)
Lemma. Given any two state vectors $\psi_1,\psi_2\in\mathbb C^4\simeq\mathbb C^2\otimes\mathbb C^2$ define the plane $\mathcal S:={\rm span}\{\psi_1,\psi_2\}$. Then $\mathcal S$ contains at least one product vector, that is, there exist states $v,w\in\mathbb C^2$ such that $v\otimes w\in\mathcal S$.
With this in mind, let us prove that the partial transpose of a two-qubit state has at most one negative eigenvalue: Assume by way of contradiction that there exists a state $\rho\in\mathbb C^{4\times 4}$ such that $\rho^{T_B}$ had two negative eigenvalues. Then $\rho^{T_B}=\sum_{j=1}^4 r_j|\psi_j\rangle\langle\psi_j|$ for some orthonormal basis $\{\psi_j\}_{j=1}^4\subset\mathbb C^4$ and $r_1,r_2<0$, $r_3,r_4\in\mathbb R$. By the previous lemma, the span of $\psi_1,\psi_2$ contains some product state $v\otimes w$, i.e., there exist $(\alpha_1,\alpha_2)\in\mathbb C^2\setminus\{(0,0)\}$ such that $\alpha_1\psi_1+\alpha_2\psi_2=v\otimes w$ is separable. Let us compute the expectation value of $\rho^{T_B}$ under $v\otimes w$: \begin{align*} \langle v\otimes w|\rho^{T_B}|v\otimes w\rangle&=\sum_{a,b=1}^2\sum_{j=1}^4\big\langle \alpha_a\psi_a \big| \,r_j|\psi_j\rangle\langle\psi_j|\, \big| \alpha_b\psi_b \big\rangle\\ &= \sum_{a,b=1}^2\sum_{j=1}^4\overline{\alpha_a}\alpha_br_j\langle\psi_a|\psi_j\rangle\langle\psi_j|\psi_b\rangle\\ &= \sum_{a,b=1}^2\sum_{j=1}^4\overline{\alpha_a}\alpha_br_j\delta_{aj}\delta_{jb}\\ &= \sum_{a=1}^2\overline{\alpha_a}\alpha_ar_a=\sum_{a=1}^2|\alpha_a|^2r_a \end{align*} Recalling that $r_1,r_2<0$ and at least one of the $\alpha_a$ is non-zero, this shows $\langle v\otimes w|\rho^{T_B}|v\otimes w\rangle<0$. This is a problem because this violates positivity of $\rho$ for the following reason: using the basis expansion for the computational basis $|0\rangle,|1\rangle$ \begin{align*} 0>\langle v\otimes w|\rho^{T_B}|v\otimes w\rangle&=\sum_{a,b=1}^2\big\langle v\otimes \langle a|w\rangle a\big|\rho^{T_B}|v\otimes\langle b|w\rangle b\big\rangle\\ &=\sum_{a,b=1}^2 \overline{\langle a|w\rangle}\langle b|w\rangle\langle v\otimes a|\rho^{T_B}|v\otimes b\rangle\\ &=\sum_{a,b=1}^2 \overline{\langle a|w\rangle}\langle b|w\rangle\langle v\otimes b|\rho|v\otimes a\rangle\\ &=\Big\langle v\otimes\Big(\sum_{b=1}^2\overline{\langle b|w\rangle}|b\rangle\Big)\Big|\rho\Big|v\otimes\Big(\sum_{a=1}^2\overline{\langle a|w\rangle}|a\rangle\Big)\Big\rangle\\ &=\langle v\otimes\overline w|\rho|v\otimes\overline w\rangle\,. \end{align*} Here we defined $\overline w:=\sum_j\overline{\langle j|w\rangle}|j\rangle\in\mathbb C^2$ (which is still a state by Parseval's identity), and in the third step we used the definition of the partial transpose. So, in conclusion, if there existed a qubit-qubit state $\rho$ for which the partial transpose had two negative eigenvalues, then there would exist a qubit-qubit pure state ($v\otimes\overline w$) which had a negative probability of occurring ($\langle v\otimes\overline w|\rho|v\otimes\overline w\rangle<0$), thus violating $\rho\geq 0$. $\square$
Note that the qubit assumption in this result is necessary, else the statement is false: Consider the (maximally entangled) qutrit pure state $|\psi\rangle:=\frac1{\sqrt3}(|00\rangle+|11\rangle+|22\rangle)$, i.e., $$ |\psi\rangle\langle\psi|=\frac13\begin{pmatrix}1&0&0&0&1&0&0&0&1\\ 0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0\\ 1&0&0&0&1&0&0&0&1\\ 0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0\\ 0&0&0&0&0&0&0&0&0\\ 1&0&0&0&1&0&0&0&1\\ \end{pmatrix}. $$ Its partial transpose reads \begin{align*} |\psi\rangle\langle\psi|^{T_2}&=\frac13\begin{pmatrix} \begin{pmatrix}1&0&0\\0&0&0\\0&0&0\end{pmatrix}^T&\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}^T&\begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}^T\\ \begin{pmatrix}0&0&0\\1&0&0\\0&0&0\end{pmatrix}^T&\begin{pmatrix}0&0&0\\0&1&0\\0&0&0\end{pmatrix}^T&\begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix}^T\\ \begin{pmatrix}0&0&0\\0&0&0\\1&0&0\end{pmatrix}^T&\begin{pmatrix}0&0&0\\0&0&0\\0&1&0\end{pmatrix}^T&\begin{pmatrix}0&0&0\\0&0&0\\0&0&1\end{pmatrix}^T \end{pmatrix}\\ &=\frac13\begin{pmatrix}1&0&0&0&0&0&0&0&0\\ 0&0&0&1&0&0&0&0&0\\ 0&0&0&0&0&0&1&0&0\\ 0&1&0&0&0&0&0&0&0\\ 0&0&0&0&1&0&0&0&0\\ 0&0&0&0&0&0&0&1&0\\ 0&0&1&0&0&0&0&0&0\\ 0&0&0&0&0&1&0&0&0\\ 0&0&0&0&0&0&0&0&1\\ \end{pmatrix}; \end{align*} this matrix has six positive eigenvalues (all $\frac13$) and three negative eigenvalues (all $-\frac13$).
