Random subspaces of a tensor product (I)

This is the first post in a series about a problem inside RMT \cap QIT that I have been working on for some time now [cn2,bcn]. Since I find it to be very simple and interesting, I will present it in a series of blog notes that should be accessible to a large audience. I will also use this material to prepare the talks I will be giving this summer on this topic ;).

In what follows, all vector spaces shall be assumed to be complex and k \leq n are fixed constants. For a vector y \in \mathbb R^k, the symbol y^\downarrow denotes its ordered version, i.e. y and y^\downarrow are the same up to permutation of coordinates and y^\downarrow_1 \geq \cdots \geq y^\downarrow_k.

1. Singular values of vectors in a tensor product

Using the non-canonical isomorphism \mathbb C^k \otimes \mathbb C^n \simeq \mathbb C^k \otimes (\mathbb C^n)^*, one can see any vector

\mathbb C^k \otimes \mathbb C^n \ni x = \sum_{i=1}^k \sum_{j=1}^n x_{ij} e_i\otimes f_j


as a matrix

\mathcal M_{k \times n} \ni X = \sum_{i=1}^k \sum_{j=1}^n x_{ij} e_if_j^*.


In this way, by using the singular value decomposition of the matrix X (keep in mind that we assume k \leq n), one can write

x = \sum_{i=1}^k \sqrt{\lambda_i} e'_i \otimes f'_i,


where (f'_i), resp. (g'_i) are orthonormal families in \mathbb C^k, resp. \mathbb C^n. The vector \lambda = \lambda(x) \in \mathbb R_+^{k} is the singular value vector of x and we shall always assume that it is ordered \lambda(x) = \lambda(x)^\downarrow. It satisfies the normalization condition

\sum_{i=1}^k \lambda_i(x)= |x|^2.


In particular, if x is a unit vector, then \lambda(x) \in \Delta^\downarrow_k, where \Delta_k is the probability simplex

 \Delta_k = \left\{ y \in \mathbb R_+^k \, : \, \sum_{i=1}^k y_i = 1\right\}


and \Delta^\downarrow_k is its ordered version.

In QIT, the decomposition of x above is called the Schmidt decomposition and the numbers \lambda_i(x) are called the Schmidt coefficients of the pure state |x \rangle.

2. The singular value set of a vector subspace

Consider now a subspace V \subset \mathbb C^k \otimes \mathbb C^n of dimension \dim V = d and define the set

 K_V = \{\lambda(x) \, : \, x \in V \text{ and } |x| = 1 \} \subseteq \Delta^\downarrow_k,


called the singular value subset of the subspace V.

Below are some examples of sets K_{V}, in the case k=3, where the simplex \Delta_{3} is two-dimensional. In all the four cases, k=n=3 and d=2. In the last two pictures, one of the vectors spanning the subspace V has singular values (1/3,1/3,1/3).

3. Basic properties

Below is a list of very simple properties of the sets K_{V}.

Proposition 1. The set K_V is a compact subset of the ordered probability simplex \Delta_k^\downarrow having the following properties:

  1. Local invariance: K_{(U_1 \otimes U_2)V} = K_V, for unitary matrices U_1 \in \mathcal U(k) and U_2 \in \mathcal U(n).
  2. Monotonicity: if V_1 \subset V_2, then K_{V_1} \subset K_{V_2}.
  3. If d=1, V=\mathbb C x, then K_V={\lambda(x)}.
  4. If d > (k-1)(n-1), then (1,0,\ldots,0) \in K_V.

Proof: The first three statements are trivial. The last one is contained in [cmw], Proposition 6 and follows from a standard result in algebraic geometry about the dimension of the intersection of projective varieties.

4. So, what is the problem ?

The question one would like to answer is the following:

How does a typical K_V look like ?

In order to address this, I will introduce random subspaces in the next post future. In the next post, I look at the special case of anti-symmetric tensors.

References

[bcn] S. Belinschi, B. Collins and I. Nechita, Laws of large numbers for eigenvectors and eigenvalues associated to random subspaces in a tensor product, to appear in Invent. Math.

[cn2] B. Collins and I. Nechita, Random quantum channels II: Entanglement of random subspaces, RĂ©nyi entropy estimates and additivity problems, Adv. in Math. 226 (2011), 1181--1201.

[cmw] T. Cubitt, A. Montanaro and A. Winter, On the dimension of subspaces with bounded Schmidt rank, J. Math. Phys. 49, 022107 (2008).

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2 Responses to Random subspaces of a tensor product (I)

  1. teobanica says:

    That was nice, waiting forward for part two! Btw a typo - which caused me lots of troubles when first reading - when definining $latex y^downarrow$ you forgot a key "1" index in the definition. Actually, this ordering seems to play a key role in everything, but, is it really a truly unavoidable trick, or can you do it as well without ordering? (conceptual question coming from a former pure mathematician, of course 🙂

    • ionnechita says:

      Thanks Teo, typo corrected. Ordering is convenient, you know the spectrum is a "set with multiplicities", so it makes sense to order everything to keep track of things. Or, you might just add all the permutations of vectors, but think about all the black ink wasted when printing the papers!

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