### Archive

Archive for July, 2013

## The Spectrum of the Partial Transpose of a Density Matrix

July 3rd, 2013

It is a simple fact that, given any density matrix (i.e., quantum state) $\rho\in M_n$, the eigenvalues of $\rho$ are the same as the eigenvalues of $\rho^T$ (the transpose of $\rho$). However, strange things can happen if we instead only apply the transpose to one half of a quantum state. That is, if $\rho\in M_m \otimes M_n$ then its eigenvalues in general will be very different from the eigenvalues of $(id_m\otimes T)(\rho)$, where $id_m$ is the identity map on $M_m$ and $T$ is the transpose map on $M_n$ (the map $id_m\otimes T$ is called the partial transpose).

In fact, even though $\rho$ is positive semidefinite (since it is a density matrix), the matrix $(id_m\otimes T)(\rho)$ in general can have negative eigenvalues. To see this, define $p:={\rm min}\{m,n\}$ and let $\rho=|\psi\rangle\langle\psi|$, where

$|\psi\rangle=\displaystyle\frac{1}{\sqrt{p}}\sum_{j=1}^{p}|j\rangle\otimes|j\rangle$

is the standard maximally-entangled pure state. It then follows that

$(id_m\otimes T)(\rho)=\displaystyle\frac{1}{p}\sum_{i,j=1}^{p}|i\rangle\langle j|\otimes|j\rangle\langle i|$,

which has $p(p+1)/2$ eigenvalues equal to $1/p$$p(p-1)/2$ eigenvalues equal to $-1/p$, and $p|m-n|$ eigenvalues equal to $0$.

The fact that $(id_m\otimes T)(\rho)$ can have negative eigenvalues is another way of saying that the transpose map is positive but not completely positive, and thus plays a big role in entanglement theory. In this post we consider the question of how exactly the partial transpose map can transform the eigenvalues of $\rho$:

Question. For which ordered lists $\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_{mn}\in\mathbb{R}$ does there exist a density matrix $\rho$ such that $(id_m\otimes T)(\rho)$ has eigenvalues $\lambda_1,\lambda_2,\ldots,\lambda_{mn}$?

### The Answer for Pure States

In the case when $\rho$ is a pure state (i.e., has rank 1), we can completely characterize the eigenvalues of $(id_m\otimes T)(\rho)$ by making use of the Schmidt decomposition. In particular, we have the following:

Theorem 1. Let $|\phi\rangle$ have Schmidt rank $r$ and Schmidt coefficients $\alpha_1\geq\alpha_2\geq\cdots\geq\alpha_r>0$. Then the spectrum of $(id_m\otimes T)(|\phi\rangle\langle\phi|)$ is

$\{\alpha_i^2 : 1\leq i\leq r\}\cup\{\pm\alpha_i\alpha_j:1\leq i,

together with the eigenvalue $0$ with multiplicity $p|n-m|+p^2-r^2$.

Proof. If $|\phi\rangle$ has Schmidt decomposition

$\displaystyle|\phi\rangle=\sum_{i=1}^r\alpha_i|a_i\rangle\otimes|b_i\rangle$

then

$\displaystyle(id_m\otimes T)(|\phi\rangle\langle\phi|)=\sum_{i,j=1}^r\alpha_i\alpha_j|a_i\rangle\langle a_j|\otimes|b_j\rangle\langle b_i|.$

It is then straightforward to verify, for all $1\leq i, that:

• $|a_i\rangle\otimes|b_i\rangle$ is an eigenvector with eigenvalue $\alpha_i^2$;
• $|a_i\rangle\otimes|b_j\rangle\pm|a_j\rangle\otimes|b_i\rangle$ is an eigenvector with eigenvalue $\pm\alpha_i\alpha_j$; and
• ${\rm rank}\big((id_m\otimes T)(|\phi\rangle\langle\phi|)\big)= r^2$, from which it follows that the remaining $p|n-m|+p^2-r^2$ eigenvalues are $0$.

Despite such a simple characterization in the case of rank-1 density matrices, there is no known characterization for general density matrices, since eigenvalues aren’t well-behaved under convex combinations.

### The Number of Negative Eigenvalues

Instead of asking for a complete characterization of the possible spectra of $(id_m\otimes T)(\rho)$, for now we focus on the simpler question that asks how many of the eigenvalues of $(id_m\otimes T)(\rho)$ can be negative. Theorem 1 answers this question when $\rho=|\phi\rangle\langle\phi|$ is a pure state: the number of negative eigenvalues is $r(r-1)/2$, where r is the Schmidt rank of $|\phi\rangle$. Since $r\leq p$, it follows that $(id_m\otimes T)(\rho)$ has at most $p(p-1)/2$ negative eigenvalues when $\rho$ is a pure state.

It was conjectured in [1] that a similar fact holds for general (not necessarily pure) density matrices $\rho$ as well. In particular, they conjectured that if $\rho\in M_n\otimes M_n$ then $(id_n\otimes T)(\rho)$ has at most $n(n-1)/2$ negative eigenvalues. However, this conjecture is easily shown to be false just by randomly-generating many density matrices $\rho$ and then counting the number of negative eigenvalues of $(id_n\otimes T)(\rho)$; density matrices whose partial transposes have more than $n(n-1)/2$ negative eigenvalues are very common.

In [2,3], it was shown that if $\rho\in M_m\otimes M_n$ then $(id_m\otimes T)(\rho)$ can not have more than $(m-1)(n-1)$ negative eigenvalues. In [4], this bound was shown to be tight when ${\rm min}\{m,n\}=2$ by explicitly constructing density matrices $\rho\in M_2\otimes M_n$ such that $(id_2\otimes T)(\rho)$ has $n-1$ negative eigenvalues. Similarly, this bound was shown to be tight via explicit construction when $m=n=3$ in [3]. Finally, it was shown in [5] that this bound is tight in general. That is, we have the following result:

Theorem 2. The maximum number of negative eigenvalues that $(id_m\otimes T)(\rho)$ can have when $\rho\in M_m\otimes M_n$ is $(m-1)(n-1)$.

It is worth pointing out that the method used in [5] to prove that this bound is tight is not completely analytic. Instead, a numerical method was presented that is proved to always generate a density matrix $\rho\in M_m\otimes M_n$ such that $(id_m\otimes T)(\rho)$ has $(m-1)(n-1)$ negative eigenvalues. Code that implements this numerical procedure in MATLAB is available here, but no general analytic form for such density matrices is known.

### Other Bounds on the Spectrum

Unfortunately, not a whole lot more is known about the spectrum of $(id_m\otimes T)(\rho)$. Here are some miscellaneous other results that impose certain restrictions on its maximal and minimal eigenvalues (which we denote by $\lambda_\textup{max}$ and $\lambda_\textup{min}$, respectively):

Theorem 3 [3]. $1\geq\lambda_\textup{max}\geq\lambda_\textup{min}\geq -1/2$.

Theorem 4 [2]. $\lambda_\textup{min}\geq\lambda_\textup{max}(1-{\rm min}\{m,n\})$.

Theorem 5 [6]. If $(id_m\otimes T)(\rho)$ has $q$ negative eigenvalues then

$\displaystyle\lambda_\textup{min}\geq\lambda_\textup{max}\Big(1-\big\lceil\tfrac{1}{2}\big(m+n-\sqrt{(m-n)^2+4q-4}\big)\big\rceil\Big)$ and

$\displaystyle\lambda_\textup{min}\geq\lambda_\textup{max}\Big(1-\frac{mn\sqrt{mn-1}}{q\sqrt{mn-1}+\sqrt{mnq-q^2}}\Big)$.

However, these bounds in general are fairly weak and the question of what the possible spectra of $(id_m\otimes T)(\rho)$ are is still far beyond our grasp.