Nathaniel Johnston » 2013

The Spectrum of the Partial Transpose of a Density Matrix

July 3rd, 2013

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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

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

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 $1\leq i<j\leq r$ , 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.

Update [August 21, 2017]: Everett Patterson and I have now written a paper about this topic.

References

R. Xi-Jun, H. Yong-Jian, W. Yu-Chun, and G. Guang-Can. Partial transposition on bipartite system. Chinese Phys. Lett., 25:35, 2008.
N. Johnston and D. W. Kribs. A family of norms with applications in quantum information theory. J. Math. Phys., 51:082202, 2010. E-print: arXiv:0909.3907 [quant-ph]
S. Rana. Negative eigenvalues of partial transposition of arbitrary bipartite states. Phys. Rev. A, 87:054301, 2013. E-print: arXiv:1304.6775 [quant-ph]
L. Chen, D. Z. Djokovic. Qubit-qudit states with positive partial transpose. Phys. Rev. A, 86:062332, 2012. E-print: arXiv:1210.0111 [quant-ph]
N. Johnston. Non-positive-partial-transpose subspaces can be as large as any entangled subspace. Phys. Rev. A, 87:064302, 2013. E-print: arXiv:1305.0257 [quant-ph]
N. Johnston. Norms and Cones in the Theory of Quantum Entanglement. PhD thesis, University of Guelph, 2012.

Tags: Quantum Entanglement, Quantum Information Theory, Research

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The Spectrum of the Partial Transpose of a Density Matrix

The Answer for Pure States

The Number of Negative Eigenvalues

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