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A separable state in $\mathcal{H}_{a}\otimes\mathcal{H}_{b}$ is given by

$$\rho_{s}=\sum_{\alpha,\beta}p(\alpha,\beta)|\alpha\rangle\!\langle\alpha|\otimes|\beta\rangle\!\langle\beta|.$$

Now, my question is, is there a suitable choice of $\{|\alpha \rangle\}$ and $\{|\beta \rangle\}$ such that all of them are elements from a complete basis (possibly non-unique) in individual subsystem?

A reason I think the bases $\{|\alpha \rangle\}$ and $\{|\beta \rangle\}$ will form a complete basis is because separable state space is the convex hull of tesnor products of symmetric rank-$1$ projectors $|\alpha\rangle\!\langle \alpha|\otimes|\beta\rangle\!\langle \beta|$. The extreme points are orthonormal sets $\{|\alpha\rangle\!\langle \alpha|\}$ and $\{|\beta \rangle\!\langle \beta|\}$. Is it true? Any help is appreciated.

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  • $\begingroup$ What does 'taking the pairs $(i,j)=\alpha$ and $(i,k)=\beta$' mean? $\endgroup$
    – narip
    Jul 13, 2021 at 8:37
  • $\begingroup$ Its a type of multi-indexing. $\endgroup$
    – WInterfell
    Jul 13, 2021 at 8:40
  • $\begingroup$ @WInterfell you cannot delete it because there is an upvoted answer. I can close it and redirect to the version on physics.SE, if you so wish (I wouldn't delete it because there are comments here that might be relevant) $\endgroup$
    – glS ♦
    Jul 13, 2021 at 15:43
  • $\begingroup$ @gIS alright, as you wish. Thanks for the patience and understanding. $\endgroup$
    – WInterfell
    Jul 13, 2021 at 15:49
  • $\begingroup$ Closing this question by OP's request. See the post on physics.SE. $\endgroup$
    – glS ♦
    Jul 13, 2021 at 15:54

2 Answers 2

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In general, no. It can be that the sets of states $\{|\alpha\rangle\}$ and $\{|\beta\rangle\}$ span their respective spaces. However, the size of each set can be larger than the dimension of the space, so the states are not all linearly independent and therefore not a basis.

For example, $\rho^{a/b}_1=p_1|0\rangle\langle 0|+(1-p_1)|1\rangle\langle 1|$ and $\rho^{a/b}_2=p_2|+\rangle\langle +|+(1-p_2)|-\rangle\langle -|$. So, $\{|\alpha\rangle\}=\{|0\rangle,|1\rangle,|+\rangle,|-\rangle\}$ and is not a basis.

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  • $\begingroup$ I understood, then the extreme points on the compact, convex separable state space neither form basis for individual subsystem nor they form a basis for the entire system? But it seems odd, shouldn't the extreme points be linearly independent i.e. orthogonal? What I meant to say is that, the combined state $\{|\alpha>|\beta>\}$ are not the extreme points for the quantum states in question as they are linearly dependent. Once, we write is in terms of extreme points do they form a basis for individual subsystem? $\endgroup$
    – WInterfell
    Jul 13, 2021 at 9:30
  • $\begingroup$ To be more precise, consider $\rho_{1}^{a}\otimes\rho_{1}^{b}=p_{1}^{2}|00><00|+(1-p_{1})^{2}|11><11|+p_{1}(1-p_{1})(|01><01|+|10><10|)$ ans $\rho_{2}^{a}\otimes\rho_{2}^{b}=p_{2}^{2}|++><++|+(1-p_{2})^{2}|--><--|+p_{2}(1-p_{2})(|+-><+-|+|-+><-+|)$. Then, we can make a basis transformation $(|++>,|-->,|+->,|-+>)$ to $(|00>,|11>,|01>,|10>)$ in $\rho_{2}^{a}\otimes\rho_{2}^{b}$ and will have the necessary decomposition. $\endgroup$
    – WInterfell
    Jul 13, 2021 at 11:08
  • $\begingroup$ You can look at the spectral decomposition of $\rho_s$. However, even if $\rho_s$ is separable, its eigenstates need not be. For example $p|B\rangle\langle B|+(1-p)I/4$ where $|b\rangle$ is the Bell state. This is diagonal in the Bell basis, and for $p\leq 1/3$, is separable. $\endgroup$
    – DaftWullie
    Jul 13, 2021 at 11:47
  • $\begingroup$ I'm not sure if part of your problem is that you're considering the individual components (the sum over $i$) which really tell you very little about the composite. The only basis that's really worth considering comes from the reduced density matrix $\rho_A=\sum_ip_i\rho^a_i$. You can certainly find the eigenvectors of that and they form an orthonormal basis but, by the above comment, they still tell you very little about the overall system. $\endgroup$
    – DaftWullie
    Jul 13, 2021 at 11:50
  • $\begingroup$ Thanks, I understood, in your example, the reduced states are $I/2$ thus, we won't have much information about the composite. However, for the separable part of the above state (which is Werner state, I suppose) shouldn't we find a separable decomposition of the form $\sum_{x,y}p(x,y)|x><x|\otimes|y><y|$ and then the distinct eigenvectors of the form $|\alpha>|\beta>$ would be the suitable representation? $\endgroup$
    – WInterfell
    Jul 13, 2021 at 12:39
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As a trivial counterexample, let $\mathcal H$ be a Hilbert space with $\dim(\mathcal H)=299792458$, and let $\{|i\rangle\}_{i=1}^{299792458}$ be a basis for it. Then $\rho_s\equiv|1\rangle\!\langle1|\otimes|1\rangle\!\langle1|$ is separable and is already decomposed like you mention, but clearly $|1\rangle$ does not span the full space.

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  • $\begingroup$ That's a product state, which is an extreme point, thus it can be thought of as a part of suitable basis system. In particular, we have $\rho_{s}=\sum_{\alpha,\beta}p(\alpha,\beta)|\alpha><\alpha|\otimes|\beta><\beta|$ where $P(\alpha,\beta)=\delta_{\alpha,1}\delta_{\beta,1}$. I agree with your comment that, it won't always span the full space but my question dealt with whether we can always make suitable changes on the state such that the $\{|\alpha>\}$ and $\{|\beta>\}$ can be part of complete basis in each susbsystem. $\endgroup$
    – WInterfell
    Jul 13, 2021 at 11:14
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    $\begingroup$ well, if you allow for elements of the sum attached with vanishing coefficients, then isn't it obvious that you can always do it? Just decompose an arbitrary $\rho_s$ like you do in the question, complete the bases on each side, and "add them to the sum" attaching them with vanishing coefficients. Or are you asking whether any separable state can be decomposed in such a way for some choice of bases on both sides? In that case, this question might be of interest $\endgroup$
    – glS ♦
    Jul 13, 2021 at 11:17
  • $\begingroup$ My question was essentially the second one. The link is useful, thanks, perhaps I can follow the procedure mentioned there and construct the suitable decomposition. $\endgroup$
    – WInterfell
    Jul 13, 2021 at 11:30
  • $\begingroup$ @WInterfell in that case, this question here is also related: quantumcomputing.stackexchange.com/q/13031/55. In particular, the section in Watrous' book linked there might be of interest. But I should remark that these results rely on Caratheodory's thm, which doesn't control the orthogonality of bases on each side, so it's not quite the same as what you seem to be asking here $\endgroup$
    – glS ♦
    Jul 13, 2021 at 11:46
  • $\begingroup$ The Watrous book indeed helpful. Specifically, Eq.(6.15) gives a representation of the pure decomposition in terms of a observable basis, do you think that the orthogonality is of the basis is ensured in such a decomposition? $\endgroup$
    – WInterfell
    Jul 13, 2021 at 12:43

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