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What this question is about:

Given the situation, that it is only known that a system is in a state $|\psi_i\rangle$ with probability $p_i$, one defines the density operator $$\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|\tag{1},$$ which describes the state of the system. If a system $\mathcal{AB}$, composed of two subsystems $\mathcal{A}$ and $\mathcal{B}$, is described by a density operator $\rho$, one defines the reduced density matrix $$\rho_{\mathcal{A}}:=\text{Tr}_{\mathcal{B}}\left(\rho\right)\tag{2}$$ as a description of the state of the subsystem $\mathcal{A}$. This is a useful description of the state of the subsystem because

one is able to calculate the expected value of $O_{\mathcal{A}}\otimes I$, which is $\text{Tr}(\rho(O_{\mathcal{A}}\otimes I))$, by calculating the expecte value of $O_{\mathcal{A}}$, which is $\text{Tr}(\rho_{\mathcal{A}}O_{\mathcal{A}})$ (which can be shown using (2)).

The question:

Is the reduced density matrix itself a density matrix? E.g. can it be treated as a density operator of a system just like density operators from the definition (1)? In my opinion this doesn't follow from (2) necessarily, because it's not clear, that the reduced density matrix is (mathematically) an object in the shape of (1).

Since no book explicitly answers this question, I am tempted to believe that the reduced density matrix is a density matrix because of (quote) - so because the calculation of an expected value is analog to the case of density matrices, which implies they are objects of the same "shape".

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  • $\begingroup$ Have you tried to prove the defining properties of a density matrix for the reduced density matrix? $\endgroup$ Jun 19 at 14:15
  • $\begingroup$ Thank you. Yes I have, although I was not sure if that works, since density matrices are (in the books I use) defined by (1) and afterwards awarded the properties you mention in your answer - yet not defined by them. $\endgroup$
    – manuel459
    Jun 19 at 14:45
  • $\begingroup$ Please see the edit in my answer. The upshot of this is that every positive semi-definite operator of unit trace on a finite-dimensional complex Hilbert space is a density matrix and a density matrix is positive semi-definite and of unit trace. $\endgroup$ Jun 19 at 14:56

1 Answer 1

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Density matrices are, by definition, positive semi-definite trace-class operators of unit trace, see for example this. In the following we will only discuss the finite-dimensional case, so we have to show that the reduced density matrix is positive semi-definite and of unit trace.$^\dagger$


To start, consider a complex bipartite Hilbert space $H=H_A\otimes H_B$ of finite dimension. Let $\rho$ be a density operator on $H$. In particular it holds that $$ \langle \psi|\rho|\psi\rangle \geq 0 \tag{1}$$ for all $|\psi\rangle \in H$ and $$\mathrm{Tr} \rho =1 \tag{2} \quad .$$

We define the reduced density operator $\rho_A$ of $\rho$ as $$ \rho_A:=\mathrm{Tr}_B\,\rho\tag{3} := \sum\limits_{k \in K} \left( \mathbb I_A\otimes\langle \psi_k|\right) \,\rho\, \left(\mathbb I_A\otimes |\psi_k\rangle\right) \quad , $$ for some orthonormal basis $\{|\psi_k\rangle\}_{k\in K}$ in $H_B$, cf. here. Note that $\rho_A$ is a linear operator on $H_A$. We compute its trace:

$$\mathrm{Tr}^{(A)} \rho_A = \sum\limits_{j \in J}\sum\limits_{k \in K} \langle \varphi_j|\otimes \langle \psi_k| \, \rho\, |\varphi_j\rangle \otimes |\psi_k\rangle = \mathrm{Tr} \rho \overset{(2)}{=} 1 \quad , \tag{4} $$

where $\{|\varphi_j\rangle\}_{j\in J}$ and $\{|\varphi_j\rangle \otimes |\psi_k\rangle\}_{j\in J,k\in K}$ are orthonormal bases in $H_A$ and $H$, respectively. We are left to show that $\rho_A$ is positive semi-definite. To do so, define $|\phi_k\rangle := |\varphi\rangle \otimes |\psi_k\rangle$ for an arbitrary $|\varphi\rangle \in H_A$. Then

$$0 \overset{(1)}{\leq}\sum\limits_{k\in K}\langle \phi_k|\rho|\phi_k\rangle = \sum\limits_{k \in K} \langle \varphi| \otimes \langle \psi_k|\,\rho\,|\varphi\rangle \otimes |\psi_k\rangle\overset{(3)}{=}\langle \varphi| \rho_A |\varphi\rangle \tag{5} \quad . $$

In conclusion, since $\rho_A$ is a (linear) positive semi-definite operator of unit trace on $H_A$, it is a density operator.


$^\dagger$ That this generalizes your definition can be seen as follows (again, working with on a finite-dimensional complex Hilbert space): If an operator $\sigma$ is positive semi-definite, it is hermitian (cf. the last part of this answer) and thus admits a spectral decomposition: $$\sigma = \sum\limits_i \lambda_i\, |\lambda_i\rangle \langle \lambda_i| \quad , \tag{6}$$ with $\lambda_i \geq 0$. The trace condition now ensures that $ \sum\limits_i \lambda_i = 1$, so it is of the form given by your equation $(1)$ and hence a density matrix. Conversely, for an operator of the form $$\sigma := \sum\limits_i p_i\, |\psi_i\rangle \langle \psi_i| \tag{7}$$ with $ \sum\limits_i p_i=1$, $p_i \geq 0$ and $|\psi_i\rangle$ normalized (but not necessarily orthogonal), i.e. a density matrix, it is not hard to prove that $\sigma \geq 0$ and $\mathrm{Tr}\, \sigma =1$.

To summarize: Proving that $\rho_A$ is positive semi-definite and of unit trace shows that it admits a representation of the form $(6)$ and thus proves that it is a density operator in terms of your definition.

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  • $\begingroup$ Is there a certain (maybe mathematical and the definitions regarding) reason some authors (e.g. Marinescu and Marinescu ISBN: 978-0123838742 or Nakahara and Ohmi ISBN: 978-0750309837) say that the density operator is a positive-semidefinite hermitian matrix with trace 1? Why do so when its automatically hermitian when it is positive-semidefinite? $\endgroup$
    – manuel459
    Jun 20 at 12:16
  • $\begingroup$ @manuel459 Hmhm, good question, I don't know. As I said: Positive semi-definiteness implies hermiticity for a complex Hilbert space, cf. Wikipedia. Most references I know don't mention hermiticity. One reason I could imagine is that they don't want to prove/ require the knowledge that this implication holds. $\endgroup$ Jun 20 at 12:25
  • $\begingroup$ Thank you very much! $\endgroup$
    – manuel459
    Jun 20 at 12:29
  • $\begingroup$ @manuel459 Glad I could help. As usual: If you think this answers your question(s), consider to accept the answer. $\endgroup$ Jun 20 at 12:32

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