Question 2

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Question 2.

Hey! Could you also please help me with an exercise (Notes 5, eq 21): show that a separable bipartite state can always be expressed as convex combination of pure product states: \[ \rho_{AB} = \sum\limits_{i=1} p_i |\psi_i\rangle\langle\psi_i|\otimes |\phi_i\rangle\langle\phi_i|. \] I get confused when I expand the density matrices and try to confine the sum to only one variable.

Hi! Good question. The definition of a separable state is that we can write it in the form \[ \rho_{AB} = \sum_{i=1}^n p_i \omega_i^A \otimes \sigma_i^B \] for some probability distribution \(\{p_i\}_{i=1}^n\) and density matrices \(\omega_i^A\) and \(\sigma_i^B\). As you mentioned, the trick in in the number of index variables. Each state \(\omega_i^A\) has a decomposition into at most \(d_A:= \dim \mathcal{H}_A\) pure states: \[ \omega_i^A = \sum_{j=1}^{d_A} q^{(i)}_j | \psi^{(i)}_j \rangle\langle \psi^{(i)}_j |. \](1) We know we can restrict to \(d_A\) pure states because the eigendecomposition gives a decomposition with that many pure states, and we don’t need to restrict the sum to fewer than \(d_A\) states because we can always have the associated probability \(q^{(i)}_j = 0\). So for each \(i\), we have a probability distribution \(\{q^{(i)}_j\}_{j=1}^{d_A}\) and set of pure states \(\{| \psi^{(i)}_j \rangle\}_{j=1}^{d_A}\) which satisfy (1). Likewise, we can decompose each \(\sigma_B^{(i)}\) as \[ \sigma_i^B = \sum_{k=1}^{d_B} r^{(i)}_k | \phi^{(i)}_k \rangle\langle \phi^{(i)}_k |. \] for some probability distribution \(\{r^{(i)}_k\}_{k=1}^{d_B}\) and pure states \(\{| \phi^{(i)}_k \rangle\}_{k=1}^{d_B}\). Putting this together, we have \[ \rho_{AB} = \sum_{i=1}^n \sum_{j=1}^{d_A} \sum_{k=1}^{d_B} p_i q^{(i)}_j r^{(i)}_k | \psi^{(i)}_j \rangle\langle \psi^{(i)}_j |\otimes | \phi^{(i)}_k \rangle\langle \phi^{(i)}_k |. \](2) This is probably where you got to in your expansions. The trick is to realize that the bound variables \(i,j,k\) don’t have inherent meaning; the sum is simply \[ \rho_{AB} = p_1 q^{(1)}_1 r^{(1)}_1 | \psi^{(1)}_1 \rangle\langle \psi^{(1)}_1 |\otimes | \phi^{(1)}_1 \rangle\langle \phi^{(1)}_1 | + p_1 q^{(1)}_2 r^{(1)}_k | \psi^{(1)}_2 \rangle\langle \psi^{(1)}_2 |\otimes | \phi^{(1)}_1 \rangle\langle \phi^{(1)}_1 | + \ldots \] where we have a term for each value of \(i,j,k\). So we can biject the set \(\{1,2,\dotsc,n\} \times \{1,2,\dotsc,d_A\} \times \{1,2,\dotsc,d_B\}\) with the set \(\{1,2,\dotsc, n*d_A*d_B\}\), by for example1.] \[ \alpha(i,j,k) = (i-1)*d_Ad_B + (j-1)*d_B + k \] which runs from \(\alpha(1,1,1) = 1\) to \(\alpha(n,d_A,d_B) = nd_Ad_B\), and in fact is a bijection as we could check. So we can define a probability distribution \(\{t_\beta\}_{\beta=1}^{nd_Ad_B}\) by the inverse mapping: if \(\alpha(i,j,k) = \beta\), then \(t_\beta = p_i q^{(i)}_j r^{(i)}_j\). Likewise, we can define pure states \[ | \chi_\beta \rangle_A = | \psi^{(i)}_j \rangle_A \] where \(i\) and \(j\) are such that \(\beta= \alpha(i,j,k)\) for some \(k\). Likewise, we define pure states \[ | \tilde \chi_\beta \rangle_B = | \phi^{(i)}_k \rangle_B \] where \(i\) and \(k\) are such that \(\beta= \alpha(i,j,k)\) for some \(j\). Putting it all together, \[ \rho_{AB} = \sum_{\beta=1}^{nd_Ad_B} t_\beta | \chi_\beta \rangle\langle \chi_\beta |\otimes | \tilde \chi_\beta \rangle\langle \tilde \chi_\beta |. \](3) Why is this equal? Since each term in the sum on the right-hand side of (3) appears exactly once as a term in the sums on the right-hand side of (2).

This trick often isn’t written out, so I hope it helps to have it here. In scientific programming languages, the bijection \(\alpha\) is often called sub2ind, since it converts “subscripts” like \(i,j,k\) to a single index, \(\beta\). For example, Julia is an open-source scientific programming language and documents the function here, if you’re interested.

  1. Thanks to this stackoverflow answer