### Archive

Posts Tagged ‘Unextendible Product Bases’

## In Search of a 4-by-11 Matrix

October 1st, 2013

IMPORTANT UPDATE [January 30, 2014]: I have managed to solve the 4-by-11 case: there is no such matrix! Details of the computation that led to this result, as well as several other related results, are given in . See Table 3 in that paper for an updated list of which cases still remain open (the smallest open cases are now 5-by-11 and 6-by-10).

After spinning my wheels on a problem for far too long, I’ve decided that it’s time to enlist the help of the mathematical and programming geniuses of the world wide web. The problem I’m interested in asks for a 4-by-11 matrix whose columns satisfy certain relationships. While the conditions are relatively easy to state, the problem size seems to be just slightly too large for me to solve myself.

### The Problem

The question I’m interested in (for reasons that are explained later in this blog post) is, given positive integers p and s, whether or not there exists a p-by-s matrix M with the following three properties:

1. Every entry of M is a nonzero integer;
2. The sum of any two columns of M contains a 0 entry; and
3. There is no way to append a (s+1)th column to M so that M still has property 2.

In particular, I’m interested in whether or not such a matrix M exists when p = 4 and s = 11. But to help illustrate the above three properties, let’s consider the p = 3, s = 4 case first, where one such matrix M is: $M = \begin{bmatrix}1 & -1 & 2 & -2 \\ 1 & -2 & -1 & 2 \\ 1 & 2 & -2 & -1\end{bmatrix}.$

The fact that M satisfies condition 2 can be checked by hand easily enough. For example, the sum of the first two columns of M is [0, -1, 3]T which contains a 0 entry, and it is similarly straightforward to check that the other 5 sums of two columns of M each contain a 0 entry as well.

Checking property 3 is slightly more technical (NP-hard, even), but is still doable in small cases such as this one. For the above example, suppose that we could add a 5th column (which we will call z = [z1, z2, z3]T) to M such that its sum with any of the first 4 columns has a 0 entry. By looking at M’s first column, we see that one of z’s entries must be -1 (and by the cyclic symmetry of the entries of the last 3 columns of M, we can assume without loss of generality that z1 = -1). By looking at the last 3 columns of M, we then see that either z2 = 2 or z3 = -2, either z2 = 1 or z3 = 2, and either z2 = -2 or z3 = 1. Since there is no way to simultaneously satisfy all 3 of these requirements, no such column z exists.

### What’s Known (and What Isn’t)

As I mentioned earlier, the instance of this problem that I’m really interested in is when p = 4 and s = 11. Let’s first back up and briefly discuss what is known for different values of p and s:

• If s ≤ p then M does not exist. To see this, simply note that property 3 can never be satisfied since you can always append one more column. If we denote the (i,j)-entry of M by mij and the i-th entry of the new column z by zi, then you can choose zi = -mii for i = 1, 2, …, s.
• Given p, the smallest value of s for which M exists is: (a) s = p+1 if p is odd, (b) s = p+2 if p = 4 or p ≡ 2 (mod 4), (c) s = p+3 if p = 8, and (d) s = p+4 otherwise. This result was proved in  (the connection between that paper and this blog post will be explained in the “Motivation” section below).
• If s > 2p then M does not exist. In this case, there is no way to satisfy property 2. This fact is trivial when p = 1 and can be proved for all p by induction (an exercise left to the reader?).
• If s = 2p then M exists. To see this claim, let the columns of M be the 2p different columns consisting only of the entries 1 and -1. To see that property 2 is satisfied, simply notice that each column is different, so for any pair of columns, there is a row in which one column is 1 and the other column is -1. To see that property 3 is satisfied, observe that any new column must also consist entirely of 1’s and -1’s. However, every such column is already a column of M itself, and the sum of a column with itself will not have any 0 entries.
• If s = 2p – 4 (and p ≥ 3) then M exists. There is an inductive construction (with the p = 3, s = 4 example from the previous section as the base case) that works here. More specifically, if we let Mp denote a matrix M that works for a given value of p and s = 2p – 4, we let Bp be the matrix from the s = 2p case above, and 1k denotes the row vector with k ones, then $M_{p+1} = \begin{bmatrix}M_p & B_p \\ 1_{2^p-4} & -1_{2^p}\end{bmatrix}$

is a solution to the problem for p’ = p+1 and s’ = 2p+1 – 4.
• If 2p – 3 ≤ s ≤ 2p – 1 then M does not exist. This is a non-trivial result that follows from .

Given p, the above results essentially tell us the largest and smallest values of s for which a solution M to the problem exists. However, we still don’t really know much about when solutions exist for intermediate values of s – we just have scattered results that say a solution does or does not exist in certain specific cases, without really illuminating what is going on. The following table summarizes what we know about when solutions do and do not exist for small values of p and s (a check mark ✓ means that a solution exists, a dash – means no solution exists, and ? means we don’t know).

s \ p 1 2 3 4 5
1
2
3
4
5
6
7
8
9 ?
10 ?
11 ? ?
12
13
14
15
16
17 – 26
27 ?
28
29
30
31
32

The table above shows why I am interested in the p = 4, s = 11 case: it is the only case when p ≤ 4 whose solution still is not known. The other unknown cases (i.e., p = 5 and s ∈ {9,10,11,27}, and far too many to list when p ≥ 6) would be interesting to solve as well, but are a bit lower-priority.

### Some Simplifications

Some assumptions about the matrix M can be made without loss of generality, in order to reduce the search space a little bit. For example, since the values of the entries of M don’t really matter (other than the fact that they come in positive/negative pairs), the first column of M can always be chosen to consist entirely of ones (or any other value). Similarly, permuting the rows or columns of M does not affect whether or not it satisfies the three desired properties, so you can assume (for example) that the first row is in non-decreasing order.

Finally, since there is no advantage to having the integer k present in M unless -k is also present somewhere in M (i.e., if M does not contain any -k entries, you could always just replace every instance of k by 1 without affecting any of the three properties we want), we can assume that the entries of M are between -floor(s/2) and floor(s/2), inclusive.

### Motivation

The given problem arises from unextendible product bases (UPBs) in quantum information theory. A set of pure quantum states $|v_1\rangle, \ldots, |v_s\rangle \in \mathbb{C}^{d_1} \otimes \cdots \otimes \mathbb{C}^{d_p}$ forms a UPB if and only if the following three properties hold:

1. (product) Each state $|v_j\rangle$ is a product state (i.e., can be written in the form $|v_j\rangle = |v_j^{(1)}\rangle \otimes \cdots \otimes |v_j^{(p)}\rangle$, where $|v_j^{(i)}\rangle \in \mathbb{C}^{d_i}$ for all i);
2. (basis) The states are mutually orthogonal (i.e., $\langle v_i | v_j \rangle = 0$ for all i ≠ j); and
3. (unextendible) There does not exist a product state $|z\rangle$ with the property that $\langle z | v_j \rangle = 0$ for all j.

UPBs are useful because they can be used to construct quantum states with very strange entanglement properties , but their mathematical structure still isn’t very well-understood. While we can’t really expect an answer to the question of what sizes of UPBs are possible when the local dimensions $d_1, \ldots, d_p$ are arbitrary (even just the minimum size of a UPB is still not known in full generality!), we might be able to hope for an answer if we focus on multi-qubit systems (i.e., the case when $d_1 = \cdots = d_p = 2$).

In this case, the 3 properties above are isomorphic in a sense to the 3 properties listed at the start of this post. We associate each state $|v_j\rangle$ with the j-th column of the matrix M. To each state in the product state decomposition of $|v_j\rangle$, we associate a unique integer in such a way that orthogonal states are associated with negatives of each other. The fact that $\langle v_i | v_j \rangle = 0$ for all i ≠ j is then equivalent to the requirement that te sum of any two columns of M has a 0 entry, and unextendibility of the product basis corresponds to not being able to add a new column to M without destroying property 2.

Thus this blog post is really asking whether or not there exists an 11-state UPB on 4 qubits. In order to illustrate this connection more explicitly, we return to the p = 3, s = 4 example from earlier. If we associate the matrix entries 1 and -1 with the orthogonal standard basis states $|0\rangle, |1\rangle \in \mathbb{C}^2$ and the entries 2 and -2 with the orthogonal states $|\pm\rangle := (|0\rangle \pm |1\rangle)/\sqrt{2}$, then the matrix M corresponds to the following set of s = 4 product states in $\mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \mathbb{C}^2$: $|0\rangle|0\rangle|0\rangle, \quad |1\rangle|-\rangle|+\rangle, \quad |+\rangle|1\rangle|-\rangle, \quad|-\rangle|+\rangle|1\rangle.$

The fact that these states form a UPB is well-known – this is the “Shifts” UPB from , and was one of the first UPBs found.

References

1. N. Johnston. The minimum size of qubit unextendible product bases. In Proceedings of the 8th Conference on the Theory of Quantum Computation, Communication and Cryptography (TQC), 2013. E-print: arXiv:1302.1604 [quant-ph], 2013.
2. L. Chen and D. Ž. Ðjoković. Separability problem for multipartite states of rank at most four. J. Phys. A: Math. Theor., 46:275304, 2013. E-print: arXiv:1301.2372 [quant-ph]
3. C. H. Bennett, D. P. DiVincenzo, T. Mor, P. W. Shor, J. A. Smolin, and B. M. Terhal. Unextendible product bases and bound entanglement. Phys. Rev. Lett., 82:5385–5388, 1999. E-print: arXiv:quant-ph/9808030
4. N. Johnston. The structure of qubit unextendible product basesJournal of Physics A: Mathematical and Theoretical, 47:424034, 2014. E-print: arXiv:1401.7920 [quant-ph], 2014.

## How to Construct Minimal Unextendible Product Bases

March 14th, 2013

In quantum information theory, a product state $|v\rangle \in \mathbb{C}^{d_1}\otimes\mathbb{C}^{d_2}$ is a quantum state that can be written as an elementary tensor: $|v\rangle=|v_1\rangle\otimes|v_2\rangle\text{ with }|v_i\rangle\in\mathbb{C}^{d_i}\ \text{ for } i=1,2,$

while states that can not be written in this form are called entangled. In this post, we will be investigating unextendible product bases (UPBs), which are sets $S\subset\mathbb{C}^{d_1}\otimes\mathbb{C}^{d_2}$ of mutually orthogonal product states with the property that no other product state is orthogonal to every member of $S$.

In this post, we will be looking at how to construct small UPBs. Note that UPBs can more generally be defined on multipartite spaces (i.e., $\mathbb{C}^{d_1}\otimes\mathbb{C}^{d_2}\otimes\cdots\otimes\mathbb{C}^{d_p}$ for arbitrary $p\geq 2$), but for simplicity we stick with the bipartite (i.e., $p= 2$) case in this blog post.

### Simple Examples

The most trivial unextendible product basis is simply the computational basis: $S:=\big\{|0\rangle\otimes|0\rangle,\ldots,|0\rangle\otimes|d_2-1\rangle,\ldots,|d_1-1\rangle\otimes|0\rangle,\ldots,|d_1-1\rangle\otimes|d_2-1\rangle\big\}.$

However, the above UPB is rather trivial – the unextendibility condition holds vacuously because $S$ spans the entire Hilbert space, so of course there is no product state (or any state) orthogonal to every member of $S$.

It is known that when $\min\{d_1,d_2\}\leq 2$, the only UPBs that exist are trivial in this sense – they consist of a full set of $d_1d_2$ states. We are more interested in UPBs that contain fewer vectors than the dimension of the Hilbert space (since, for example, these UPBs can be used to construct bound entangled states ). One of the first such UPBs to be constructed was called “Pyramid” . To construct this UPB, define $h:=\tfrac{1}{2}\sqrt{1+\sqrt{5}}$ and $N:=\tfrac{1}{2}\sqrt{5+\sqrt{5}}$, and let $|\phi_j\rangle:=\tfrac{1}{N}[\cos(2\pi j/5),\sin(2\pi j/5),h]\text{ for }0\leq j\leq 4.$

Then the following set of 5 states in $\mathbb{C}^3\otimes\mathbb{C}^3$ is a UPB: $S_{\textup{pyr}}:=\big\{|v_0\rangle,|v_1\rangle,|v_2\rangle,|v_3\rangle,|v_4\rangle\big\},$

where $|v_i\rangle:=|\phi_i\rangle\otimes|\phi_{2i(\text{mod }5)}\rangle$.

It is a straightforward calculation to verify that the members of $S_{\textup{pyr}}$ are mutually orthogonal (and thus form a product basis). To verify that there is no product state orthogonal to every member of $S_{\textup{pyr}}$, we first observe that any 3 of the $|\phi_j\rangle$‘s form a linearly independent set (verification of this claim is slightly tedious, but nonetheless straightforward). Thus there is no state $|w\rangle\in\mathbb{C}^3$ that is orthogonal to more than 2 of the $|\phi_j\rangle$‘s. Thus no product state $|w_1\rangle\otimes|w_2\rangle\in\mathbb{C}^3\otimes\mathbb{C}^3$ is orthogonal to more than 2 + 2 = 4 members of $S_{\textup{pyr}}$, which verifies unextendibility.

### Minimum Size

One interesting question concerning unextendible product bases asks for their minimum cardinality. It was immediately noted that any UPB in $\mathbb{C}^{d_1}\otimes\mathbb{C}^{d_2}$ must have cardinality at least $d_1+d_2-1$. To see this, suppose for a contradiction that there existed a UPB $S$ containing $(d_1-1)+(d_2-1)$ or fewer product states. Then we could construct another product state that is orthogonal to $d_1-1$ members of $S$ on $\mathbb{C}^{d_1}$ and another $d_2-1$ members of $S$ on $\mathbb{C}^{d_2}$, for a total of $(d_1-1)+(d_2-1)$ members of $S$, which shows that $S$ is extendible.

Despite being such a simple lower bound, it is also attainable for many values of $d_1,d_2$  (and very close to attainable in the other cases [3,4]). The goal of this post is to focus on the case when there exists a UPB of cardinality $d_1+d_2-1$, which is characterized by the following result of Alon and Lovász:

Theorem . There exists a UPB in $\mathbb{C}^{d_1}\otimes\mathbb{C}^{d_2}$ of (necessarily minimal) size $d_1+d_2-1$ if and only if $d_1,d_2\geq 3$ and at least one of $d_1$ or $d_2$ is odd.

In spite of the above result that demonstrates the existence of a UPB of the given minimal size in many cases, how to actually construct such a UPB in these cases is not immediately obvious, and is buried throughout the proofs of  and its references. The goal of the rest of this post is to make the construction of a minimal UPB in these cases explicit.

### Orthogonality Graphs

The orthogonality graph of a set of $s$ product states in $\mathbb{C}^{d_1}\otimes\mathbb{C}^{d_2}$ is graph with coloured edges (there are 2 colours) on $s$ vertices (one for each product state), such that there is an edge connecting two vertices with the $i$th colour if and only if the two corresponding product states are orthogonal on the $i$th party.

For example, the orthogonality graph of the Pyramid UPB introduced earlier is illustrated below. Black edges represent states that are orthogonal on the first party, and red dotted edges represent states that are orthogonal on the second party. If the product states under consideration are mutually orthogonal, then their orthogonality graph is the complete graph $K_s$. Unextendibility is a bit more difficult to determine, but nonetheless a useful technique for constructing UPBs is to first choose a colouring of the edges of $K_s$, and then try to construct product states that lead to that colouring.

### A Minimal Construction

In the orthogonality graph of the Pyramid UPB, all of the edges that connect a vertex to a neighbouring vertex are coloured black, and all other edges are coloured red. We can construct minimal UPBs by generalizing this graph in a natural way. Suppose without loss of generality that $d_1$ is odd, and we wish to construct a UPB of size $s := d_1 + d_2 - 1$. We construct the orthogonality graph by arranging $s$ vertices in a circle and connecting any vertices that are a distance of $(d_1-1)/2$ or less from each other via a black edge. All other edges are coloured red. For example, in the $d_1 = d_2 = 3$ case, this gives the orthogonality graph above. In the $d_1 = 5, d_2 = 4$ case, this gives the orthogonality graph below. Our goal now is to construct product states that have the given orthogonality graph. This is straightforward to do, since every state must be orthogonal to $d_1-1$ of the other states on $\mathbb{C}^{d_1}$ and orthogonal to the $d_2-1$ other states on $\mathbb{C}^{d_2}$. Thus, we can just pick $|v_0\rangle$ arbitrarily, then pick $|v_1\rangle$ randomly subject to the constraint that it is orthogonal to $|v_0\rangle$ on the first subsystem, and so on, working our way clockwise around the orthogonality graph, generating each product state randomly subject to the orthogonality conditions.

Furthermore, it can be shown (but will not be shown here – the techniques are similar to those of  and are a bit technical) that this procedure leads to a product basis that is in fact unextendible with probability 1. In order to verify unextendibility explicitly, one approach is to check that any subset of $d_1$ of the product states are linearly independent on $\mathbb{C}^{d_1}$ and any subset of $d_2$ of the product states are linearly independent on $\mathbb{C}^{d_2}$.

References

1. C. H. Bennett, D. P. DiVincenzo, T. Mor, P. W. Shor, J. A. Smolin, and B. M. Terhal. Unextendible product bases and bound entanglement. Phys. Rev. Lett., 82:5385–5388, 1999. E-print: arXiv:quant-ph/9808030
2. N. Alon and L. Lovász. Unextendible product bases. J. Combinatorial Theory, Ser. A, 95:169–179, 2001.
3. K. Feng. Unextendible product bases and 1-factorization of complete graphs. Discrete Appl. Math., 154:942–949, 2006.
4. J. Chen and N. Johnston. The minimum size of unextendible product bases in the bipartite case (and some multipartite cases). Comm. Math. Phys., 333(1):351–365, 2015. E-print: arXiv:1301.1406 [quant-ph]