The Spectrum of the Partial Transpose of a Density Matrix
It is a simple fact that, given any density matrix (i.e., quantum state) , the eigenvalues of are the same as the eigenvalues of (the transpose of ). However, strange things can happen if we instead only apply the transpose to one half of a quantum state. That is, if then its eigenvalues in general will be very different from the eigenvalues of , where is the identity map on and is the transpose map on (the map is called the partial transpose).
In fact, even though is positive semidefinite (since it is a density matrix), the matrix in general can have negative eigenvalues. To see this, define and let , where
is the standard maximally-entangled pure state. It then follows that
which has eigenvalues equal to , eigenvalues equal to , and eigenvalues equal to .
The fact that 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 :
Question. For which ordered lists does there exist a density matrix such that has eigenvalues ?
The Answer for Pure States
In the case when is a pure state (i.e., has rank 1), we can completely characterize the eigenvalues of by making use of the Schmidt decomposition. In particular, we have the following:
Theorem 1. Let have Schmidt rank and Schmidt coefficients . Then the spectrum of is
together with the eigenvalue with multiplicity .
Proof. If has Schmidt decomposition
It is then straightforward to verify, for all , that:
- is an eigenvector with eigenvalue ;
- is an eigenvector with eigenvalue ; and
- , from which it follows that the remaining eigenvalues are .
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 , for now we focus on the simpler question that asks how many of the eigenvalues of can be negative. Theorem 1 answers this question when is a pure state: the number of negative eigenvalues is , where r is the Schmidt rank of . Since , it follows that has at most negative eigenvalues when is a pure state.
It was conjectured in  that a similar fact holds for general (not necessarily pure) density matrices as well. In particular, they conjectured that if then has at most negative eigenvalues. However, this conjecture is easily shown to be false just by randomly-generating many density matrices and then counting the number of negative eigenvalues of ; density matrices whose partial transposes have more than negative eigenvalues are very common.
In [2,3], it was shown that if then can not have more than negative eigenvalues. In , this bound was shown to be tight when by explicitly constructing density matrices such that has negative eigenvalues. Similarly, this bound was shown to be tight via explicit construction when in . Finally, it was shown in  that this bound is tight in general. That is, we have the following result:
Theorem 2. The maximum number of negative eigenvalues that can have when is .
It is worth pointing out that the method used in  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 such that has 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 . Here are some miscellaneous other results that impose certain restrictions on its maximal and minimal eigenvalues (which we denote by and , respectively):
Theorem 3 . .
Theorem 4 . .
Theorem 5 . If has negative eigenvalues then
However, these bounds in general are fairly weak and the question of what the possible spectra of are is still far beyond our grasp.
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