9
$\begingroup$

I've rewritten the question in math notation, and I've left the old version in physics bra-ket notation here.

Background

A simple consequence of the singular value decomposition is that any vector $v$ in a vector space $V$ formed by the tensor product of two smaller spaces ("subsystems") $U$ and $W$ of dimension $d_U$ and $d_W$,

$v \in V = U \otimes W$,

has a special decomposition in terms of rank-one tensors (aka product states)

$v = \sum_{i=1}^d \lambda_i u_i \otimes w_i $, $\qquad u_i \in U$, $\qquad w_i \in W$, $\qquad d = \mathrm{min}(d_U,d_W)$

built from the fixed orthonormal bases $\{ u_i \}$ and $\{ w_i \}$.

This decomposition has many nice properties, and it's simple to see that there's no way to generalize it to the case of 3 or more subsystems while keeping all of them. For instance, a generic state in $V = \bigotimes_{n=1}^N V_n$ cannot be expressed as a sum of fewer than $\tilde{d}$ simple tensors, with $\tilde{d} \gg d_n = \mathrm{dim}V_n$ for all $n=1,\ldots,N$.

Vague Question

Is there a useful/natural/canonical decomposition of a vector into a "small" number of orthogonal simple tensors for the case of 3 or more subsystems? At the very least, the GHZ state should take it's canonical form in such a decompostion:

$v_{GHZ} = a_1 \otimes \cdots \otimes a_N + b_1 \otimes \cdots \otimes b_N$, $\qquad a_n, b_n \in V_n$, $\qquad \langle a_n ; b_n \rangle = 0$ for all $n$.

Specific Question

If we guess that the entropy function

$H[\{ p_i \} ] = -\sum_i p_i \ln p_i$

is appropriate, we can define the "minimum-entropy product-state decomposition" to be the decomposition

$v = \sum_{i=1}^{\tilde{d}} \lambda_i v_i$, $\qquad v_i = \bigotimes_{n=1}^N \psi_i^n$, $\qquad \psi_i^n \in V_n$

with the minimum value for the entropy $H[\{ p_i = \lambda_i^2 \} ]$, under the condition that the $\{ v_i \}$ are orthonormal. Note that we allow $\langle \psi_i^n ; \psi_j^n \rangle \neq 0$ for $i \neq j$.

The natural questions to ask are: Is this decomposition generically unique (i.e. unique except for some set of measure zero in the global vector space)? Is it continuous? (What other properties should this decomposition have to satisfy to be useful?)

I am 99% sure that in the case of $N=2$, this reduces to the Schmidt decomposition and that the answers to both questions is "yes".

Is any of this sensitive to our choice of the entropy function $H$, as opposed to some other permutation-invariant and majoritization-preserving function of the spectrum?

$\endgroup$
5
  • 1
    $\begingroup$ the question sounds interesting, but the notation totally throws me off! $\endgroup$
    – Suvrit
    Apr 26, 2011 at 7:44
  • $\begingroup$ Fixed as best I could. Let me know if there's anything else non-standard. $\endgroup$ Apr 26, 2011 at 17:14
  • 1
    $\begingroup$ Since you asked: the mathematical terminology for product state is rank-one tensor, simple tensor or pure tensor. You may want to avoid the last one since it overlaps with the notion of pure states and generates some confusion. $\endgroup$ Apr 27, 2011 at 7:00
  • $\begingroup$ Suggested reading is Charles van Loan's "The Ubiquitous Kronecker Product" (2000). For a snapshot of recent developments, see the NSF Workshop "Directions in Tensor-Based Computation and Modeling" (2009). $\endgroup$ May 9, 2011 at 17:30
  • $\begingroup$ @Federico, @John: Thank you both. I have adjusted the terminology, and I will look at those references. $\endgroup$ May 11, 2011 at 17:47

4 Answers 4

3
$\begingroup$

Vague question:

The hierarchical higher order SVD would lead you to the canonical form for the GHZ state. When you have $H_1 \otimes H_2 \otimes \ldots \otimes H_n$ you basically first Schmidt decompose your state $\psi$ relative to $H_1 \otimes (H_2 \otimes \ldots \otimes H_N)$ and then continue like this until you finish.

In more generality, the Tucker decomposition is the most general multilinear generalization of the SVD. It is not, however, unique. The hierarchical higher order SVD I mentioned above is a special case of the Tucker decomposition.

Specific question:

I don't know the answer to your question, but I did notice this: If you start with $\psi = \sum_{jk} \gamma_{jk} \alpha_j \otimes \beta_k$ then your entropy is essentially the entropy of the post measurement mixed state with measurement basis being $\alpha_j \otimes \beta_k$. So what you want to look at is which local measurement scheme gives you the minimum post-measurement entropy. So in the bipartite case all you need to do is take a partial trace and whatever the entropy of that matrix, that is the minimum post-measurement entropy. I.e. you get entanglement, which confirms your suspicion that the Schmidt decomposition is what you get in bipartite case. A similar strategy might work in the multipartite case.

$\endgroup$
2
  • 1
    $\begingroup$ I don't think your argument for the Schmidt decomposition minimizing entropy goes through because I am considering the set of all possible orthonomal bases of product states, not just those canonically constructed from local bases of the subsystems. Nonetheless, I have proven the claim is true by constructing explicitly a sequence of decomps, from an arbitrary product-state decomp to the Schmidt decomp, which are ordered by the majorization partial order. This proves the Schmidt decomposition minimizes all functions F[{p_i}] = sum_i f(p_i) of the spectrum, f(p) concave, not just entropy. $\endgroup$ May 17, 2011 at 2:15
  • $\begingroup$ This answer seems to be best currently available, so I'm accepting it. If I make any progress, or find more comprehensive info elsewhere, I'll post it as a separate answer. $\endgroup$ May 17, 2011 at 2:17
3
$\begingroup$

An answer to your "vague question" may be provided by http://en.wikipedia.org/wiki/Higher-order_singular_value_decomposition, but as far as I know "finding a good generalization of the SVD for 3-tensors" is kind of a research problem in linear algebra, too. Check out Kolda, Bader "Tensor decompositions and applications". Your idea of minimizing an entropy function is new, as far as I can tell, but I am not a tensor expert.

$\endgroup$
2
$\begingroup$

There is some recent work on tensors (and also the special cases of symmetric and alternating tensors) that admit orthogonal or unitary decompositions (resp., symmetric or alternating decompositions). I will point out, particularly, the following three papers by Elina Robeva and coauthors:

$\endgroup$
1
$\begingroup$

You may wish to check out the Matrix Product States ansatz http://en.wikipedia.org/wiki/Matrix_product_state that vaguely answers both of your questions. I especially recommend the review "The density-matrix renormalization group in the age of matrix product states" by Prof Ulrich Schollwöck, see doi: 10.1016/j.aop.2010.09.012

$\endgroup$
1
  • $\begingroup$ Thanks Dmitry! Matrix product states are something I've heard a bit about but have always meant to try to understand more thoroughly. $\endgroup$ Jun 15, 2014 at 23:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.