The dual of the completely bounded norm is called the diamond norm, which plays an important role in quantum information theory, as it can be used to measure the distance between quantum channels. The diamond norm of is typically denoted . For properties of the completely bounded and diamond norms, see [1,2,3].

A method for efficiently computing the completely bounded and diamond norms via semidefinite programming was recently presented in [4]. The purpose of this post is to provide MATLAB scripts that implement this algorithm and demonstrate its usage.

Download and Install

In order to make use of these scripts to compute the completely bounded or diamond norm, you must download and install two things: the SeDuMi semidefinite program solver and the MATLAB scripts themselves.

1. SeDuMi – Please follow the instructions on the SeDuMi website to download and install it. If possible, you should install SeDuMi 1.1R3, not SeDuMi 1.21 or SeDuMi 1.3, since there is a bug with the newer versions when dealing with complex matrices.
2. CB Norm MATLAB Package – Once SeDuMi is installed, download the CB norm MATLAB scripts, unzip them, and place them in your MATLAB scripts directory. The zip file contains 10 MATLAB scripts.

Usage Examples

The representation of the linear map that the CBNorm and DiamondNorm functions take as input is a pair of arrays of its left- and right- generalized Choi-Kraus operators. That is, an array of operators and such that for all .

Basic Examples

If we wanted to compute the completely bounded and diamond norms of the map

the MATLAB input and output would be as follows:

>> PhiA(:,:,1) = [1,1;1,0];
>> PhiA(:,:,2) = [1,0;1,2];
>> PhiB(:,:,1) = [1,0;0,1];
>> PhiB(:,:,2) = [1,2;1,1];
>> CBNorm(PhiA,PhiB)

ans =

    7.2684

>> DiamondNorm(PhiA,PhiB)

ans =

    7.4124

So we see that its completely bounded norm is 7.2684 and its diamond norm is 7.4124.

If we instead want to compute the completely bounded or diamond norm of a completely positive map, we only need to provide its Kraus operators – i.e., operators  such that for all . Furthermore, in this case semidefinite programming isn’t used at all, since [1, Proposition 3.6] tells us that and , and computing is trivial. The following example demonstrates the usage of these scripts in this case, via a completely positive map with four (essentially random) Kraus operators:

>> PhiA(:,:,1) = [1 0 0;0 1 1];
>> PhiA(:,:,2) = [-3 0 1;5 1 1];
>> PhiA(:,:,3) = [0 2 0;0 0 0];
>> PhiA(:,:,4) = [1 1 3;0 2 0];
>> CBNorm(PhiA)

ans =

   42.0000

>> DiamondNorm(PhiA)

ans =

   38.7303

Transpose Map

Suppose we want to compute the completely bounded or diamond norm of the transpose map on . A generalized Choi-Kraus representation is given by defining , where is the standard basis of (i.e., and are the operators with matrix representation in the standard basis with a one in the -entry and zeroes elsewhere). It is known that the completely bounded and diamond norms of the n-dimensional transpose map are both equal to n, which can be verified in small dimensions as follows:

>> % 2-dimensional transpose
>> PhiA(:,:,1) = [1 0;0 0];
>> PhiA(:,:,2) = [0 1;0 0];
>> PhiA(:,:,3) = [0 0;1 0];
>> PhiA(:,:,4) = [0 0;0 1];
>> PhiB = PhiA;
>> CBNorm(PhiA,PhiB)

ans =

    2.0000

>> DiamondNorm(PhiA,PhiB)

ans =

    2.0000

>> % 3-dimensional transpose
>> I = eye(3);
>> for i=1:3
for j=1:3
PhiA(:,:,3*(i-1)+j) = I(:,i)*I(j,:);
end
end
>> PhiB = PhiA;
>> CBNorm(PhiA,PhiB)

ans =

    3.0000

>> DiamondNorm(PhiA,PhiB)

ans =

    3.0000

>> PhiA(:,:,1) = [1 0;0 1];
>> PhiA(:,:,2) = [1 1;-1 1]/sqrt(2);
>> PhiB(:,:,1) = [1 0;0 1];
>> PhiB(:,:,2) = -[1 -1;1 1]/sqrt(2);
>> CBNorm(PhiA,PhiB)

ans =

    1.4142

>> DiamondNorm(PhiA,PhiB)

ans =

    1.4142

References

1. V. I. Paulsen. Completely bounded maps and operator algebras. Cambridge University Press, 2003.
2. N. Johnston, D. W. Kribs, and V. I. Paulsen. Computing stabilized norms for quantum operations via the theory of completely bounded maps. Quantum Inf. Comput., 9:16-35, 2009.
3. J. Watrous. Theory of quantum information lecture notes.
4. J. Watrous. Semidefinite programs for completely bounded norms. Theory Comput., 5:217–238, 2009.

Separable Pure State Preservers and Entangling Gates

In the design of quantum algorithms, entangling gates play a very important role. Entangling gates are unitary operators that are able to generate entanglement. A bit more specifically, a unitary operator U ∈ M_n ⊗ M_n (where M_n is the space of n × n complex matrices) is called an entangling gate if there exists a separable pure state v = a ⊗ b ∈ Cⁿ ⊗ Cⁿ such that Uv is entangled. Conversely, we will say that a unitary operator U preserves separability if Uv is separable whenever v is separable.

In order to answer the question of what unitaries preserve separability, it is instructive to consider some simple examples (this is often a useful way to formulate conjectures regarding preserver problems). For example, it is clear that if U = A ⊗ B for some unitary operators A, B ∈ M_n, then U preserves separability (because U(a ⊗ b) = Aa ⊗ Bb is separable). Another example of a unitary operator that preserves separability is the swap (or flip) operator S defined on separable states by S(a ⊗ b) = b ⊗ a (the action of S on the rest of Cⁿ ⊗ Cⁿ is determined by extending linearly). It turns out that these are essentially the only operators that preserve separability [1,2,3]:

Theorem 1. Let U ∈ M_n ⊗ M_n be a unitary operator. Then U preserves separability (i.e., U is not an entangling gate) if and only if there exist unitary operators A, B ∈ M_n such that either U = A ⊗ B or U = S(A ⊗ B).

As we already saw, the “if” direction of the above result is trivial – the meat and potatoes of the theorem comes from the “only if” direction (as is typically the case with results about linear preservers). Theorem 1 was first proved in [1] essentially by case analysis and checking the action of a separability-preserving unitary on a basis of Cⁿ ⊗ Cⁿ, and was subsequently re-proved using similar techniques (but with different motivations and connections) in [2]. The result was proved in [3] by using the vector-operator isomorphism and the fact that a linear map Φ : M_n → M_n preserves the set of rank-1 operators if and only if there exist A, B ∈ M_n such that either Φ(X) ≡ AXB or Φ(X) ≡ AX^tB [4].

Theorem 1 also follows as a simple corollary of several related results that have recently been proved in [5,6]. A version of Theorem 1 for multipartite systems (i.e., systems that are the tensor product of more than two copies of Cⁿ) can be found in [3] and [7].

Universal Entangling Gates

A universal entangling gate is, as its name suggests, a stronger form of an entangling gate – it is a unitary operator U such that U(a ⊗ b) is entangled for all a, b ∈ Cⁿ (contrast this with entangling gates, which require only that U(a ⊗ b) is entangled for some a, b ∈ Cⁿ). The structure of universal entangling gates is much less well-understood than that of entangling gates, though we can still at least say when they exist.

It is not difficult to convince yourself that universal entangling gates can’t exist in small dimensions. Let’s begin by supposing n = 2. The set of pure states in C² ⊗ C² can be regarded as a 7-dimensional real manifold (7 = 2 × (n × n) – 1, where we subtract one because pure states all have unit length), while the set of separable pure states in C² ⊗ C² can be regarded as a 5-dimensional real manifold (5 = (2 × n – 1) + (2 × n – 1) – 1, where the final one is subtracted because the overall phase of the first system relative to the second system is irrelevant). Thus, if U ∈ M₂ ⊗ M₂ were a universal entangler, it would have to send a 5-dimensional manifold into the 7 – 5 = 2 remaining dimensions of the space, which seems unlikely. Similarly, if n = 3 and U ∈ M₃ ⊗ M₃ were a universal entangler, it would have to send a 9-dimensional manifold into the 17 – 9 = 8 remaining dimensions of the space, which also seems unlikely.

Indeed, this type of argument was made rigorous via methods of algebraic geometry in [8], where the following result was proved:

Theorem 2. There exists a universal entangling gate in M_n ⊗ M_n if and only if n ≥ 4.

Despite knowing when universal entangling gates exist, we still don’t have a characterization of such operators, nor do we even have many explicit examples (does anyone have an explicit example for 3 ⊗ 4 or 4 ⊗ 4 systems?). Similar techniques to those used in the proof of Theorem 2 should also shed light on when universal entangling gates exist in multipartite systems M_n1 ⊗ M_n2 ⊗ … ⊗ M_nk, but to my knowledge this calculation has not been explicitly carried out.

References:

1. M. Marcus and B. N. Moyls, Transformations on tensor product spaces. Pacific Journal of Mathematics 9, 1215–1221 (1959).
2. F. Hulpke, U. V. Poulsen, A. Sanpera, A. Sen De, U. Sen, and M. Lewenstein, Unitarity as preservation of entropy and entanglement in quantum systems. Foundations of Physics 36, 477–499 (2006). E-print: arXiv:quant-ph/0407118
3. N. Johnston, Characterizing Operations Preserving Separability Measures via Linear Preserver Problems. To appear in Linear and Multilinear Algebra (2011). E-print: arXiv:1008.3633 [quant-ph]
4. L. Beasley, Linear operators on matrices: the invariance of rank k matrices. Linear Algebra and its Applications 107, 161–167 (1988).
5. E. Alfsen and F. Shultz, Unique decompositions, faces, and automorphisms of separable states. Journal of Mathematical Physics 51, 052201 (2010). E-print: arXiv:0906.1761 [math.OA]
6. S. Friedland, C.-K. Li, Y.-T. Poon, and N.-S. Sze, The automorphism group of separable states in quantum information theory. Journal of Mathematical Physics 52, 042203 (2011). E-print: arXiv:1012.4221 [quant-ph]
7. R. Westwick, Transformations on tensor spaces. Pacific Journal of Mathematics 23, 613–620 (1967).
8. J. Chen, R. Duan, Z. Ji, M. Ying, J. Yu, Existence of Universal Entangler. Journal of Mathematical Physics 49, 012103 (2008). E-print: arXiv:0704.1473 [quant-ph]

1. On an infinite square grid, draw a quarter circle from one corner of a square to the opposite corner of that square:
2. Call an endpoint of a quarter circle (or a “Q-toothpick”) exposed if it does not touch the endpoint of any other quarter circle.
3. From each exposed endpoint, draw two more quarter circles, each of the same size as the first quarter circle you drew. Furthermore, the two quarter circles that you draw are the ones that can be drawn “smoothly” (without creating a 90° or 180° corner). Thus the next two generations of the automaton are (already-placed quarter circles are green, newly-added quarter circles are red):
   

The name “Q-toothpick” comes from its analogy to the more well-studied toothpick automaton (see Sloane’s A139250 and this paper), in which toothpicks (rather than quarter circles) are repeatedly placed on a grid where exposed ends of other toothpicks lie. In this post, we will examine how this automaton evolves over time, and in particular we will investigate the types of shapes that it produces.

Counting Q-Toothpicks

While the Q-toothpick automaton appears quite random and unpredictable for the first few generations, evolving past generation 6 or so reveals several patterns. The following image depicts the evolution of the automaton for its first 19 generations.

The first 19 generations of the Q-toothpick cellular automaton (red segments are pieces that are newly added in the current generation)

Perhaps the most notable pattern is that the grid is more or less filled up in an expanding square starting from the initial Q-toothpick. In fact, by inspecting generations 4, 6, 10, 18, we see that at generation 2ⁿ + 2 (n = 1, 2, 3, …) the automaton has roughly filled in a square of side length 2ⁿ⁺¹ + 1, and then evolution continues from there on out of the corners of that square. Also, the number of cells added (A187211) at these generations can now easily be computed:

A187211(2ⁿ + 2) = 16 + 8(2^n-1 – 1) for n ≥ 3.

Furthermore, the growth in the following generations repeats itself. In particular, we have:

A187211(2ⁿ + 3) = 22 for n ≥ 1,
A187211(2ⁿ + 4) = 40 for n ≥ 2,
A187211(2ⁿ + 5) = 54 for n ≥ 2.

Similarly, for n ≥ 3, the four values of A187211(2ⁿ + 6) through A187211(2ⁿ + 9) are similarly constant (their values are 56, 70, 120, and 134). In general, for n ≥ k the 2^k-1 values of A187211(2ⁿ + 2^k-1 + 2) through A187211(2ⁿ + 2^k + 1) are constant in n, though I am not aware of a general formula for what these constants are. If we ignore the first four generations and arrange the number of Q-toothpicks added in each generation in rows of length 2ⁿ, we obtain a table that begins as follows:

22, 20
22, 40, 54, 40
22, 40, 54, 56, 70, 120, 134, 72
22, 40, 54, 56, 70, 120, 134, 88, 70, 120, 150, 168, 246, 360, 326, 136

C scripts are provided at the end of this post for computing the values of A187210 and A187211 (and hence the values in the above table).

Shapes Traced Out by Q-Toothpicks

By far the most common of these shapes are circles, diamonds and hearts. The fourth shape appears only on the diagonal and it’s not difficult to see that it forever will make up the entirety of the diagonal (with the exception of the circle in the center). The fifth and sixth objects are the first two members of an infinite family of objects that appear as the automaton evolves. The fifth object first appears in generation 9, and sixth object (which is basically two copies of the fifth object) first appears in generation 17. The following object, which is basically made up of two copies of the sixth object (i.e., four copies of the fifth object) first appears in generation 33:

In general, a new object of this type (made of 2ⁿ copies of the fifth object above) first appears in generation 2ⁿ⁺³ + 1. In fact, these objects are the only ones that are traced out by this automaton. [Edit: this final claim is not true! See ebcube's great post that shows a double-heart shape in generation 31.]

Update [March 28, 2011]: I have added a script that counts the number of circles, diamonds, and hearts in the nth generation of the Q-toothpick automaton, and another script that computes Sloane’s A187212.

Download:

• A187210.c – computes the total number of Q-toothpicks present in the nth generation
• A187211.c – computes the number of Q-toothpicks added in the nth generation
• A187212.c – computes the number of Q-toothpicks if we restrict them to the positive quadrant
• count_shapes.c – computes the number of circles, diamonds, and hearts in the nth generation

• You are given a hexagonal tile with six paths printed on it, with two path ends touching each side of the hexagon. One such tile is as follows:

• You may rotate, but not move the hexagon that has been provided to you.
• Once you have selected an orientation of the hexagon, a path is traced along that hexagon, and you are provided a new hexagon that you may rotate at the end of your current path.
• The goal of the game is to create the longest path possible without running into either the centre hexagon or the outer edge of the game board.

To make things a bit more interesting, the game was updated in November 2010 to include a new scoring system that gives you 1 + 2 + 3 + … + n (the nth triangular number) points on a turn if you extend the length of your path by n on that turn. This encourages clever moves that significantly extend the length of the path all at once. The question that I am going to answer today is what the maximum score in Entanglement is under this scoring system (inspired by this reddit thread).

On a Standard-Size Game Board

The standard Entanglement game board is made up of a hexagonal ring of 6 hexagons, surrounded by a hexagonal ring of 12 hexagons, surrounded by a hexagonal ring of 18 hexagons, for a total of 36 hexagons. In order to maximize our score, we want to maximize how much we increase the length of our path on our final move. Thus, we want to just extend our path by a length of one on each of our first 35 moves, and then score big on the 36th move.

Well, each hexagon that we lay has six paths on it, for a total of 6*36 = 216 paths on the board. 35 of those paths will be used up by our first 35 moves. It is not possible to use all of the remaining 181 paths, however, because many of them lead into the edge of the game board or the central hexagon, and connecting to such a path immediately ends the game. Because there are 12 path ends that touch the central hexagon and 84 path ends that touch the outer border, there must be at least (12+84)/2 – 1 = 47 unused paths on the game board (we divided by 2 because each unused path takes up two path ends and we subtracted 1 because one of the paths will be used by us).

Thus we can add a length of at most 181 – 47 = 134 to our path on the 36th and final move of the game, giving a total score of at most 35 (from the first 35 moves of the game) + 1 + 2 + 3 + … + 134 = 35 + 9045 = 9080. Not only is this an upper bound of the possible scores, but it is actually attainable, as demonstrated by the following optimal game board:

Paths in red are unused, the green line depicts the portion of the path laid by the first 35 moves of the game, and the blue line depicts the portion of the path (of length 134) gained on the 36th move. One fun property of the above game board is that it is actually completely “unentangled” – no paths cross over any other paths.

On a Larger or Smaller Game Board

Other than being a good size for playability purposes, there is no reason why we couldn’t play Entanglement on a game board of larger or smaller radius (by radius I mean the number of rings of hexagons around the central hexagon – the standard game board has a radius of 3). We will compute the maximum score simply by mimicking our previous analysis for the standard game board. If the board has radius n, then there are 6 + 12 + 18 + … + 6n = 3n(n+1) hexagons, each of which contains 6 paths. Thus there are 18n(n+1) lengths of path, 3n(n+1)-1 of which are used in the first 3n(n+1)-1 moves of the game, and we want to add as many as possible of the remaining 15n(n+1)+1 lengths of path in the final move of the game. There are 12 path ends that touch the central hexagon and 12 + 24n path ends that touch the outer edge of the game board. Thus there are at least (12 + 12 + 24n)/2 – 1 = 11 + 12n unused paths on the game board.

Tallying the numbers up, we see that on the final move, we can add at most 15n(n+1)+1 – (11 + 12n) = 15n² + 3n – 10 length of path. If T(n) = n(n+1)/2 is the nth triangular number, then we see that it’s not possible to obtain more than 3n(n+1)-1 + T(15n² + 3n – 10) = (225/2)n⁴ + 45n³ – 135n² – (51/2)n + 44 points. In fact, this score is obtainable via the exact same construction as the optimal board in the n = 3 case – just extend the (counter)clockwise rotation of the path in the obvious way. Thus, the maximum score for a game of Entanglement on a board of radius n for n = 1, 2, 3, … is given by the sequence 41, 1613, 9080, 29462, 72479, …

As with the regular look-and-say sequence, the way we will attack these sequences is by constructing a “periodic table” of elementary non-interacting subsequences that all terms in the sequence are made up of. Then standard recurrence relation techniques will allow us to determine the rate of growth of the length of the terms in the sequences as well as the limiting distribution of the different digits in the sequence.

The Ternary Look-and-Say Sequence

Since we have already looked at the regular (i.e., decimal) look-and-say sequence, which is equivalent to the base-4 version of the sequence since it never contains a digit of 4 or larger, and we have also looked at the binary version of the sequence, it makes sense to ask what happens in the intermediate case of the ternary (base-3) version of the sequence: 1, 11, 21, 1211, 111221, 1012211, … (see A001388).

As always, we begin by listing the noninteracting subsequences that make this version of the sequence tick. Not surprisingly, it is more complicated than the corresponding table (of 10 subsequences) in the binary case, but not as complicated as the corresponding table (of 92 subsequences) in the decimal case.

The (27×27) transition matrix for this evolution rule is included in the text file at the end of this post. Its characteristic polynomial is

The maximal eigenvalue of the transition matrix is thus the largest root of x³ – x – 1, which is approximately 1.324718. It follows that the number of digits in the terms of this sequence grows on average by about 32.5% from one term to the next.

The Look-and-Say Sequence with Digits 1 and 2

Closely related to the ternary version of the sequence is the sequence obtained by reading the previous term in the sequence, but with the restriction that you can never use a number larger than 2 (see A110393). This sequence begins 1, 11, 21, 1211, 111221, 21112211, …, and the sixth term is obtained by reading the fifth term as “two ones, one one, two twos, one one”. Because only two different digits appear in this sequence, it is perhaps not surprising that its table of noninteracting subsequences is quite simple:

The transition matrix associated with this evolution rule is

As before, the average rate of growth of the number of digits in the terms of this sequence is determined by the magnitude of the largest eigenvalue of this matrix. A simple calculation reveals that this eigenvalue is √φ = 1.272…, where φ = (1 + √5)/2 is the golden ratio. Furthermore, we can answer the question of how many 1s there are in the terms of this sequence compared to 2s by looking at the eigenvector corresponding to the maximal eigenvalue:

What this means is, for example, that the second elementary subsequence (11) occurs φ times as frequently as the fourth elementary subsequence (1211). By weighting the subsequences by the entries in this vector appropriately, we can calculate the limiting ratio of the number of ones to the number of twos as

Download: Transition matrices [plaintext file]

In this post, I am going to investigate the related binary version of the sequence, which starts off 1, 11 much like the regular sequence. But then when reading 11, we read it as “two ones”. Since two in binary is 10, the next term in the sequence is 101. When reading that term, we read it as “one one, one zero, one one”, so the next term is 111011. That term is read as “three ones, one zero, two ones”, and since three is 11 in binary and two is 10 in binary, the next term is 11110101, and so on. In this post we will answer two questions in particular about this sequence:

1) On average, how much longer is the (n+1)^th term in the sequence than the n^th term in the sequence?

2) On average, what is the ratio of the number of ones to the number of zeroes in the sequence?

Non-Interacting Subsequences

Much like the regular look-and-say sequence, we are able to study this sequence by constructing a “basis” of non-interacting subsequences that every term in the binary look-and-say sequence is made up of. Fortunately, constructing such a family of subsequences for the binary version of the look-and-say sequence is much simpler than it is for the decimal version of the sequence – here we only need ten different basic subsequences (whereas we needed 92 different subsequences for the regular look-and-say sequence!). These ten subsequences, and the subsequences they evolve into, are summarized in the following table.

So for example, the first term in the sequence, 1, evolves into the subsequence (2), which is 11. That term then evolves into subsequence (3) followed by subsequence (1), or 101. That term then evolves into the subsequence (5) followed by the subsequence (2), or 111011, and so on. The reason that this representation of the sequence is useful is we can use it to describe the evolution of the binary look-and-say sequence entirely within a matrix T. In particular, we let T be the matrix with 1 in its (i,j) entry if the subsequence (i) appears in the evolution rule for subsequence (j), and 0 in its (i,j) entry otherwise:

Now if v is a 10-dimensional vector whose ith entry indicates how many times the subsequence (i) appears in a particular term of the binary look-and-say sequence, it follows that the entries of Tv tell us how many times each subsequence appears in the next term of the binary look-and-say sequence. So it follows from standard theory of linear homogeneous recurrence relations that we can now read off all of the long-term behaviour of the binary look-and-say sequence from the eigenvalues and eigenvectors of T.

Rate of Growth of the Sequence

The asymptotic rate of growth of the number of digits in the terms of the binary look-and-say sequence is simply the magnitude of the largest eigenvalue of the transition matrix T above. Using Maple it is simple to derive this value. If L_n is the number of digits in the n^th term of the binary look-and-say sequence, then

This limit is approximately 1.465571, which means that the binary version of this sequence grows much faster than the decimal version of the sequence (recall that the growth rate of the number of digits of the regular look and say sequence is approximately 1.303577). This limit is also the unique real root of the cubic x³ – x² – 1, which follows from the fact that the characteristic polynomial of T is

Ratio of Number of Ones to Zeroes

If we let N_n denote the number of ones in the n^th term of the binary look-and-say sequence, and if we let Z_n denote the number of zeroes in the n^th term of the sequence, what is

In other words, what is the average ratio of ones to zeroes in this sequence? The following table shows the value of N_n/Z_n for n = 3, 4, …, 25, which might give some intuition to the problem:

Based on numerical estimates like those given in the table above, it has been conjectured that the limiting ratio is 5/3 (or some nearby value). We will now show that the limit does indeed exist, but its value is not 5/3 — it just happens to be really close to 5/3.

Much like the maximal eigenvalue of T tells us the overall growth rate of the sequence, the corresponding eigenvector tells us the distribution of the different subsequences that are present in the limit. Once we know the distribution of the individual subsequences, it is not difficult to find out the overall ratio of ones to zeroes by weighing the different subsequences appropriately. So our first step is to find the eigenvector corresponding to the maximal eigenvalue. To this end, it will be convenient to let

α is the same as in the previous section, and β is exactly the growth rate limit that we computed. Then the eigenvector corresponding to the maximal eigenvalue of T is:

What this means is that, in the limit, the fifth subsequence, 1110, is β times as frequently-occurring as the sixth subsequence, 11110 (for example). Now we just weigh each subsequence according to how many zeros and ones they contain, and we find the limiting ratio of ones to zeroes is

In particular, this ratio does not equal 5/3, but rather its decimal expansion begins 1.6657272222676… (which is less than 1/1000 away from 5/3).