Let us consider bipartite system in an entangled mixed state. Since its density matrix can always be diagonalized we can write it in following ways: $$\rho_{AB} = \sum_{j} p_{j} | \psi_j\rangle \langle \psi_j|= \sum_{i,k} C_{i,k} | a_i\rangle \langle a_i|\otimes |b_k \rangle \langle b_k |, \tag1$$

where $\{|\psi_j\rangle\}$ is an orthonormal basis in $H_{AB} = H_{A}\otimes H_{B}$, and orthonoral bases in $H_A$ and $H_B$ are given by $\{|a_i\rangle\}$ and $\{|b_j\rangle\} $.

One may obtain reduced density matrices for subsystems A and B: $$\rho_A = \mathrm{Tr}_B(\rho_{AB})= \sum_{i,k} C_{i,k}| a_i\rangle \langle a_i| \qquad \rho_B =\sum_{i,k} C_{i,k}| b_k\rangle \langle b_k| \tag{2}$$

So my question is: given the spectra (sets of eigenvalues) of $\rho_A$ and $\rho_B$, is it in general possible to recover the spectrum of $\rho_{AB}$? If not, then is there any physical intuition behind this result? Can this result be generalized for any multipartite system?

As a toy model I considered $\mathrm{dim}(H_A) =2;$ and $\mathrm{dim}(H_B) =3$. Once applying (1) and (2) and taking into account the fact that traces of all the density matrices $=1$, I have arrived to system of equations with number of unknowns larger than number of independent equations. So unless I am missing some other constrains this task seems to have no solution in general.


I want to fix points mentioned by Emilio Pisanty and Luzanne.
$\quad$(At least) one thing that I have overlooked is that for arbitrary orthonormal basis $ \{| \psi_j\rangle \}$ expression $\rho_{AB} = \sum_{j} p_{j} | \psi_j\rangle \langle \psi_j|$ doesn't have to be given by diagonal matrix with eigenvalues $p_i$ on its diagonal. Yet (as I believe) there always exists such an orthonormal basis that satisfies aforementioned condition. For instance we can take for $| \psi_k \rangle$ column with only k-th element being nonzero and equal to 1.
$\quad$Now we define orthonormal bases $\{|a_i\rangle\}$ and $\{|b_j\rangle\} $ in exactly the same manner. Then(it seems to me) we do have the following connection: $$| \psi_j \rangle = c^j _{i,k} |a_i\rangle \otimes |b_k\rangle $$ where I don't imply summation over $i$ and $k$. $c^j _{i,k} =1$ for every combination of $i,k $ and $ j$. For this highly specific choice of bases (1) should hold.

Still I imply that my reasoning about recovery of $\rho_{AB}$ from $\rho_{A}$ and $\rho_{B}$ should hold because:

  • set of eigenvalues doesn't depend on choice of basis
  • density matrix is completely(?) defined by set of eigenvalues
  • $\begingroup$ physics.stackexchange.com/questions/1491/… this question seems to be related to mine, yet it seems to me the author discusses some other aspects. $\endgroup$ Commented Oct 9, 2018 at 9:29
  • $\begingroup$ Your expression in $(1)$ is extremely dubious and, at the very least, non-obvious. What makes you think that $\rho_{AB}$ can be decomposed in that way? $\endgroup$ Commented Oct 9, 2018 at 10:33
  • $\begingroup$ I suspect the right hand side of (1) should be $\sum_{i,i',k,k'} C_{i,i',k,k'} |a_i\rangle\langle a_{i'}| \otimes |b_k\rangle\langle b_{k'} |$, as follow from writing $|\psi_j\rangle = \sum_{i,k} c^j_{i,k} |a_i\rangle \otimes |b_k\rangle$ and $\langle\psi_j| = \sum_{i',k'} c^{j,*}_{i',k'} \langle a_{i'}| \otimes \langle b_{k'}|$. And of course such a decomposition is not unique... (we can choose whatever orthonormal bases $(a_i)_i$ and $(b_k)_k$ we want) $\endgroup$
    – Luzanne
    Commented Oct 9, 2018 at 14:22
  • $\begingroup$ @Luzanne Well, it's hard to shut the door completely on the existence of some special basis that will take way some of the complexity from your general expression. (But in this instance I don't think it's doable, or at least not down to Yaroslav's expression.) $\endgroup$ Commented Oct 9, 2018 at 16:05
  • $\begingroup$ @EmilioPisanty If $(a_i)_i$ and $(b_k)_k$ are orthonormal bases as claimed, then (1) as it stands would imply that the $|a_i\rangle \otimes |b_k\rangle$ form a basis of eigenstates for $\rho$. This is demonstrably impossible e.g. for $|\Psi\rangle\langle\Psi|$ (using the notation from your answer), bcs the eigenspace for the eigenvalue $1$ is the 1-dimensional space $\text{Span}\{|\Psi\rangle\}$ which cannot be written as the span of a number of $|a_i\rangle\otimes|b_k\rangle$. $\endgroup$
    – Luzanne
    Commented Oct 9, 2018 at 16:22

2 Answers 2


Leaving aside some dubious aspects of your question (in particular, the fact that your decomposition in $(1)$ is almost certainly not possible in general), there is an easy answer to the core question you pose:

So my question is: given the spectra (sets of eigenvalues) of $\rho_A$ and $\rho_B$, is it in general possible to recover the spectrum of $\rho_{AB}$?

No, this is not possible. This is easy to see by comparing

  • $\rho_{AB}^{(1)} = |\Psi\rangle\langle\Psi|$ where $|\Psi\rangle$ is a maximally entangled state on two qubits, and
  • $\rho_{AB}^{(2)} = \frac14 \mathbb I_4$, the maximally mixed state of two qubits.

Both of these have identical reduced density matrices, $\rho_A=\rho_B = \frac12 \mathbb I_2$ with eigenvalues $(1,1)$, but the spectrum of $\rho_{AB}^{(1)}$ is $(1,0,0,0)$ and the spectrum of $\rho_{AB}^{(2)}$ is $(\frac14,\frac14,\frac14,\frac14)$.


To complement the clean counter-example given by Emilio Pisanty and to try and answer the

If not, then is there any physical intuition behind this result?

part of the question, it is worth noting that the same thing is already true in classical statistical physics.

If I have a probability distribution $p(a,b)$ for the state of a composite classical system, it is not possible to reconstruct this joint probability distribution from the marginal distributions $p_1(a) = \int p(a,b) \,db$ and $p_2(b) = \int p(a,b) \,da$. What the joint distribution $p$ encodes that is not captured by $p_1$ and $p_2$ alone, is knowledge of how the state of system $1$ is correlated with that of system $2$.

In the quantum case, what is lost if we only know the partial density matrices for each system are both the classical and quantum correlations between the two systems.


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