Let $H_A$ and $H_B$ be two finite dimensional Hilbert spaces and $\rho_{AB}$ a density operator acting on $H_A\otimes H_B$. I am to show that $\rho_{A} = \operatorname{tr}_B\rho_{AB}$ is also a density operator.

I've been able to show $\operatorname{tr}\rho_A = 1$, but am struggling to show $\rho_A$ is a positive semi-definite operator. I feel like it should be simple, but just don't know where to take it. Here's an attempt:

The positive semi-definite-ness of $\rho_{AB}$ shows us that $$\sum_{ijk\ell}\overline{\psi}_{ij}p_{ijk\ell}\psi_{k\ell} \geq 0$$ for all $|\psi\rangle = \sum_{ij}\psi_{ij}|a_i\rangle\otimes|b_j\rangle$ and $p_{ijk\ell}$ are the matrix elements in the tensor product basis. That said I have no idea how to extrapolate any information from this in order to apply it to the elements of $\rho_A$.

  • $\begingroup$ When you want to prove something, start from the result and work your way backwards. So write the positivity condition (as the one you give) but for the matrix elements of the reduced density operator! (Then try to construct a suitable $\psi_{ij}$.) $\endgroup$ Feb 20, 2021 at 21:15
  • $\begingroup$ @Jakob that's exactly what I've written above. Indeed I can then calculate the partial trace and get $\langle\phi|\rho_A|\phi\rangle = \sum_{ijn}\overline{\phi}_ip_{injn}\phi_j$. I'm still lost. $\endgroup$ Feb 20, 2021 at 21:30
  • $\begingroup$ Try taking $\psi = \phi \otimes \sum_j b_j$. $\endgroup$
    – Krup'a
    Feb 20, 2021 at 21:55
  • $\begingroup$ @NateStemen : Are you familiar with the Schmidt decomposition? $\endgroup$ Feb 20, 2021 at 21:59
  • $\begingroup$ @Jakob yes. perhaps I should write $\rho_{AB}$ in the schmidt basis? $\endgroup$ Feb 20, 2021 at 22:09

1 Answer 1


Let $\mathscr{H}\equiv \mathscr{H}_{\mathrm{A}} \otimes \mathscr{H}_\mathrm{B}$. Then consider a density operator $\rho$ on $\mathscr{H}$ in its spectral decomposition: $$\rho = \sum\limits_k \lambda_k \, |\lambda_k\rangle \langle \lambda_k| \quad , $$

with $ \langle \lambda_k|\lambda_q\rangle = \delta_{kq}$, $\lambda_k \geq 0$ and $\sum\limits_k \lambda_k = 1$. We calculate the reduced density matrix of subsystem $\mathrm{A}$ as

$$ \rho_{\mathrm{A}} \equiv \mathrm{Tr}_{\mathrm{B}}(\rho) = \sum\limits_k \lambda_k \, \mathrm{Tr}_{\mathrm{B}}(|\lambda_k\rangle \langle \lambda_k|) \quad , $$ where the second equality follows from the linearity of the partial trace. We then note that we can express each rank-one projection $ |\lambda_k\rangle \langle \lambda_k|$ in terms of the Schmidt decomposition of $|\lambda_k\rangle \in \mathscr{H}$. By doing so, we find that

$$\mathrm{Tr}_{\mathrm{B}}(|\lambda_k\rangle \langle \lambda_k|) = \sum\limits_j |\alpha^k_j|^2\, |a^k_j\rangle \langle a^k_j| \quad . $$

Here, $\{ |a^k_j\rangle \}_j$ denotes a complete and orthonormal basis of $\mathscr{H}_{\mathrm{A}}$. It is easy to see that $\mathrm{Tr}_{\mathrm{B}}(|\lambda_k\rangle \langle \lambda_k|)$ is positive semi-definite for all $k$ and since $\lambda_k \geq 0$ we find that $\rho_{\mathrm{A}} \geq 0$. Additionally, the normalization of $\rho_{\mathrm{A}}$ follows from the normalization of the Schmidt coefficients.

  • $\begingroup$ I'm not sure I follow. What is $|a_m^k\rangle$? It seems you've only defined $\{|a^k\rangle\}$ $\endgroup$ Feb 21, 2021 at 2:30

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