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Suppose we have Hilbert space factorisable in to K subsystems $$ \mathcal{H} = \mathcal{H}_1\otimes...\otimes\mathcal{H}_K $$ in which we can express a pure state as the Schmidt decomposition $$ |\psi\rangle=\displaystyle\sum_{i_1...i_K} c_{i_1...i_K}|i_1\rangle^{(1)}\otimes...\otimes|i_K\rangle^{(K)} = \displaystyle\sum_{i_1...i_K} c_{i_1...i_K}|i_1,...,i_K\rangle = \sum_I c_I |I\rangle $$

The density operator for a pure state is then

$$ \rho = \sum_{I,J} c_{I}c_J^* |I\rangle\langle J| = \sum_{I,J} a_{IJ} |I\rangle\langle J| $$

Suppose we then use our Schmidt decomposition to form the eigenstates

$$ |\psi_n\rangle = \sum_{I} c_{I,n}|I\rangle $$

From the spectral theorem we can then form a mixed density operator as

$$ \rho = \sum_n \lambda_n |\psi_n\rangle\langle\psi_n| = \sum_{n,I} \lambda_n c_{I,n}c^*_{I,n} |I\rangle\langle I| = \sum_I b_I |I\rangle\langle I| $$ Suppose now I consider the following composition in to a bipartite system $$ \mathcal{H} = (\mathcal{H}_1\otimes...\otimes\mathcal{H}_{J-1})\otimes(\mathcal{H}_J\otimes...\otimes\mathcal{H}_K) = \mathcal{H}_A\otimes\mathcal{H}_B $$ and I wish to consider a reduced density operator $\rho_A$, so I wish to write a general density operator as $$ \rho_{AB} = \sum_{n,m} p_{nm} (\rho_A^n \otimes \rho_B^m) $$ I would therefore like as general a choice of $\rho_A^n$,$\rho_B^m$ as possible and my intuition tells me to choose mixed density operators but how I am reading it, I would think the pure state density operators are "more general" for perhaps the very naive reason that there are more indices giving more possible choices?

Hopefully this clarifies my overall question: can either pure or mixed states be seen as "more general" than one or the other in some sense, specifically with how many coefficients there are for each state represented?

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    $\begingroup$ I don't understand the notation. But note that the Schmidt decomposition only works if you bipartite your system (again, I don't know if you did that since I really don't get the notation). Anyway, what is your last line actually showing? Of course a convex combination of (different) pure states is a mixed state, but why'd you need the Schmidt decomposition? Could you elaborate on what the question actually is? I don't understand the need for the Schmidt decomposition here at all. What do you mean with 'more general'? $\endgroup$ Apr 25 at 16:03
  • $\begingroup$ @JasonFunderberker My understanding is that Schmidt decomposition can be generalised to a multipartite system, which is what I'm considering above, a K-partite system. I apologize for the notation, I've been working on trying to keep it concise. Also realised I made a mistake in the last line, should be fixed now. I'm using Schmidt decomposition as a way of considering states in large, factorisable Hilbert spaces. The goal is to compute various entropies for various configurations of entangled states with generalised expression for mixed and pure states! $\endgroup$ Apr 25 at 16:12
  • $\begingroup$ But what is the question? Sorry, but I really don't understand.You haven't put any question mark and I still don't see the need for the Schmidt decomposition when writing a mixed state in terms of pure states or in terms of its eigenstates. Please put some sentences where you elaborate what you're trying to ask and specify what you mean with "more general". $\endgroup$ Apr 25 at 16:17
  • $\begingroup$ @JasonFunderberker I've tried to restate it now, hopefully it makes more sense? $\endgroup$ Apr 25 at 16:40

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The Schmidt decomposition is just a confusing red herring here.

Given a basis $\lvert \psi_i\rangle$ of our Hilbert space, we may expand any vector in that basis, and hence a normalized state is $\lvert \psi\rangle = \sum c_i\lvert \psi_i\rangle$ with $\sum_i \lvert c_i\rvert^2 = 1$. So its associated density matrix is

$$ \rho_\psi = \lvert \psi\rangle\langle \psi\rvert = \sum_{i,j} c_ic^\ast_j \lvert \psi_i\rangle\langle \psi_i\rvert = \sum_{i,j} \rho_{ij}\lvert \psi_i\rangle\langle \psi_j\rvert. \tag{1}$$

You seem to be confused because this "looks" as if there is more freedom here in the $\rho_{ij}$ than in the $\lambda_i$ when we write a generic mixed state as $$ \rho = \sum_i \lambda_i \lvert \phi_i\rangle \langle \phi_i\rvert,\tag{2}$$ in terms of its eigenstates $\lvert \phi_i\rangle$ - after all there are $n^2$ of the $\rho_{ij}$ and only $n$ of the $\lambda_i$.

There are a few different ways to note that eq. (2) defines a more general class of operators than eq. (1):

  • Eq. (1) is an expansion of an arbitrary $\rho_\psi$ in a fixed basis $\lvert \psi_i\rangle$, something that works for all pure state matrices, while eq. (2) is specifically in the eigenbasis of $\rho$ - some other $\rho'$ would not have the same form in that same basis in general.

  • There isn't actually more independent parameters in eq. (1) because there are only as much of the $c_i$ as there are of the $\lambda_i$ in eq. (2), and while the $\lambda_i$ are constrained by $\sum_i \lambda_i = 1$, the $c_i$ are constrained by $\sum_i \lvert c_i\rvert^2 = 1$, so in both cases you essentially have $n-1$ independent parameters, but in eq. (2) you can additionally choose the basis differently and this gives you different $\rho$s (see the first point again), so there is more freedom in eq. (2) than in eq. (1).

  • A very "manifest" difference is simply that the pure state matrix fulfills the purity condition $\rho_\psi^2 = \rho_\psi$ by construction, while the mixed state matrix does not - it does not follow from eq. (2) and $\sum_i \lambda_i = 1$, while it does follow from eq. (1) and $\sum_i \lvert c_i\rvert^2 = 1$, so that alone shows you the second equation is more general.

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  • $\begingroup$ Thank you! The fixed basis you refer to, would that be a tensor product state in a multipartite system? While the eigenbasis is a linear combination of tensor product states? This is why I am using Schidmt decomposition, constructing a basis as a linear combination of tensor product states in a multipartite system! $\endgroup$ Apr 25 at 16:58
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    $\begingroup$ @CouchAnalysis You can fix any basis you want (that's the nature of bases)... $\endgroup$
    – ACuriousMind
    Apr 25 at 16:59
  • $\begingroup$ Alright, so in my case, the constraints on $a_{IJ}$ mean that there are more degrees of freedom to $b_I$? $\endgroup$ Apr 25 at 17:08
  • $\begingroup$ Another question that comes to mind. In eq (1), you mean to write $\rho_\psi = \sum_{i,j} \rho_{ij} |\psi_i\rangle\langle\psi_j|$ I presume? $\endgroup$ Apr 25 at 18:56

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