Qubits as hypermatrices and entanglement

In this paper, we represent n-qubits as hypermatrices and consider various applications to quantum entanglement. In particular, we use the higher-order singular value decomposition of hypermatrices to prove that the π-transpose is an LU invariant. Additionally, through our construction we show that the matrix representation of the combinatorial hyperdeterminant of 2n-qubits can be expressed as a product of the second Pauli matrix, allowing us to derive a formula for the combinatorial hyperdeterminant of 2n-qubits in terms of the n-tangle.


INTRODUCTION
For the last few decades, classifying entangled states has been a major endeavor for researchers in theoretical quantum information [1,2,3,4].For bipartite quantum systems, the theory of entanglement is well understood and established [5], however for multipartite systems, the very notion of entanglement is still being worked out [6,7,8,9].As such, much of the focus has been on better understanding and expanding the theory of entanglement in multipartite systems [6].
Two pure states are considered equivalently entangled if they are locally unitarily (LU) equivalent; if |ψ and |ϕ are n-qubit states, then this means that there exist U 1 , ..., U n ∈ SU (2) such that Thus, it is of great importance to find operations on states that are invariant under local unitary equivalence in the classification of entangled states.In this paper, we represent pure n-qubits as hypermatrices and apply the theory of multilinear algebra to these states to study LU invariants.Specifically, we consider the higher-order singular value decomposition of hypermatrices [10,11] and show from our representation that the π-transpose is an LU-invariant.
Next, we prove a formula relating the matrix of the hyperdeterminant of an arbitrary 2n-qubit to the tensor product of the second Pauli matrix, which then allows us to express the n-tangle [12] in terms of the hyperdeterminant.This shows that in some sense, the hyperdeterminant provides a measurement of entanglement.

PRELIMINARIES
Let C n be the complex n-dimensional vector space.Let v i ∈ C ni be N vectors, where (v i ) j are the jth coordinates of v i .The outer product of v 1 ,v 2 , ..., v N is defined to be the hypermatrix v 1 • v 2 • ... • v N ∈ C n1×n2×...×nN of order N whose (i 1 i 2 ...i N )-coordinate is given by (v 1 ) i1 (v 2 ) i2 ...(v N ) iN .Now, let H ∈ C n1×n2×...×nN be a hypermatrix and A i ∈ C mi×ni be N rectangular matrices.The multilinear multiplication of (A 1 , A 2 , ..., A N ) with H is defined to be the hypermatrix Multilinear multiplication is linear in terms of the matrices in both parts; that is, if Multilinear multiplication interacts with the outer product in the following way: (2.4) Date: (where For more details on these and other operations on hypermatrices, the reader is referred to [13].
We also need the notion of higher-order singular value decomposition.The higher-order singular value decomposition, first discovered in [10], states that any hypermatrix H ∈ C n1×n2•••×nN can be written as where each V k is an n k × n k unitary matrix for 1 ≤ k ≤ N and Σ is an n 1 × n 2 × ... × n N hypermatrix such that for each Σ i k =α , obtained by fixing the k th index to α, satisfies: (1) the all-orthogonality that Σ i k =α , Σ i k =β = 0 for all 1 ≤ k ≤ N and α = β, where , is the Frobenius inner product, and (2) the ordering that We call Σ a core tensor of H, and Σ i k =j subtensors of Σ.We also call the k-mode singular values of H.It is known that the k-mode singular values are unique [10]; that is, (2.6) H → {k-mode singular values of H} is a well-defined function.Note that When N = 2, the higher-order singular value decomposition reduces to the typical matrix singular value decomposition.Indeed, we may express the higher-order singular value entirely in terms of matrices by considering the k-mode unfolding.Recall that the k-mode unfolding [14] of a hypermatrix H ∈ C n1×n2×...×nN is the n k ×(n k+1 ...n N n 1 ...n k−1 ) matrix, denoted H (k) , whose (i k , j) entry is given by (i 1 , ..., i N )-entry of H, with (2.7) or in the case where the index starts at 0, It was shown in [10] that if H ∈ C n1×n2×...×nN has the higher-order singular value decomposition (2.10) then H (n) has the matrix singular value decomposition (2.11) where . We also review the notion of the π-transpose [13] of a hypermatrix H = [H i1...iN ] ∈ C n1×n2×...×nN , which is defined as the hypermatrix (2.12) where π ∈ S N .We also note that if n 1 = n 2 = ... = n N =: n, then we say that H is a cuboid hypermatrix of order N with length n.Lastly, we review the Cayley's first hyperdeterminant, also known as the combinatorial hyperdeterminant.Suppose H is a cuboid hypermatrix of order N with side length n, i.e.H ∈ C N times n × ... × n .For a permutation σ ∈ S n , let l(σ) = l denote the smallest number of transpositions needed to form σ: σ = s i1 . . .s i l .Then the combinatorial hyperdeterminant [13], of H is defined to be A σ1(j)σ2(j)...σN (j) .
Note that hdet is identically 0 for all hypermatrices of odd order, and for hypermatrices of even order it is equal to ( [13,15]).The next result is well-known and referenced in this paper.

CORRESPONDANCE BETWEEN QUBITS AND HYPERMATRICES
Suppose we have two strings a = a 1 a 2 ...a n and b = b 1 b 2 ...b n .Recall that a < b in the lexicographic order if a i < b i , where i is the first position where the two strings differ.For example, in the lexicographic order, 000 < 001 < 010 < 011 < 100 < 101 < 110 < 111.
Let ψ be any pure n-qubit state ψ = i1,...,in∈{0,1} |i n , and the amplitudes satisfy We can order the amplitudes of ψ in the lexicographic order and define the 2 n -dimensional vector In the following, we will identify the pure state ψ with the vector |ψ .We also consider the following outer product a cuboid hypermatrix of length 2 and order n, whose Frobenius norm is 1.In other words, with the ((i 1 + 1), ..., (i n + 1))-entry of ψ entry being ψ i1...in .Consequently, we have an isomorphism between the Hilbert spaces of pure n-qubits ψ and their corresponding hypermatrices ψ.
Let ψ and ϕ be two pure n-qubit states with corresponding hypermatrices ψ and ϕ.We say that two hypermatrices ψ and ϕ are LU equivalent if there exists U 1 , ..., U n ∈ SU (2) such that Lemma 3.1.The pure states ψ and ϕ are LU equivalent if and only if ψ and ϕ are LU equivalent.
Proof.Suppose ψ and ϕ are LU equivalent, then there exists U 1 , ..., U n ∈ SU (2) such that where the last equality follows from the linearity of Kronecker products.The isomorphism constructed above maps this vector to the hypermatrix and by the linearity of multilinear matrix multiplication, this is equal to Thus, ψ and ϕ are LU equivalent if and only if ψ and ϕ are LU equivalent.
We note that our construction and this result straightforwardly extends to n-qudits, however, for this paper we only focus on n-qubits.
Consequently, if ψ and ϕ are LU equivalent, then , and so if ψ has the higher-order singular value decomposition is the higher-order singular value decomposition for ϕ, showing that they share the same core tensor.Hence, they have the same k-mode singular values (by uniqueness).On the other hand, in [16,17], Liu et.al. proved that if two states ψ and ϕ have the same core tensor, then they are LU equivalent.We thus have the following theorem.
Proof.Suppose ψ has the higher-order singular value decomposition In particular, this implies that the ((i 1 + 1), ..., (i n + 1))-entry of ψ is given by the sum 2 j1,...,jn=1 and so for any π ∈ S n , the ((i 1 + 1), ..., (i n + 1))-entry of ψ π is given by , the above sum is well-defined; moreover, since we are just permuting the rows in the sum, it follows that (3.12) where P 1 , ..., P n are some 2 × 2 permutation matrices (which recall are orthogonal, hence unitary).Thus, ψ and ψ π have the same core tensor in their higher-order singular value decomposition, proving that they are LU equivalent.
which in matrix form can be represented as For π 1 = (13) ∈ S 3 , the corresponding 2 × 2 × 2 hypermatrix is given by which in matrix form can be represented as Similarly, for π 2 = (132), the corresponding 2 × 2 × 2 hypermatrix is given by which in matrix form can be represented as The core tensor for each of these states is the same, which in matrix form is Consequently, ψ, ψ π1 , and ψ π2 are LU equivalent.Switching back quantum states, this is the same as saying that the following 3-qubits are LU equivalent On the other hand, consider the quantum state |ϕ That is, ϕ is obtained from ψ by switching the (211)-coordinate of ψ with its (121)-coordinate and vice versa, and leaving everything else fixed.Hence, there is no permuation relating ψ with ϕ.Indeed, the matrix form of ϕ can be represented as and so it follows that the matrix form of its core tensor is From this we see that ψ and ϕ have different 1-mode singular values, proving that ψ and ϕ (and hence ψ and ϕ) are not LU equivalent.
It just so happened in our example that the two states ψ and ϕ, which were not related by a permutation, were not LU equivalent.We ask the following question: for any two quantum states that are not related by a permutation, are they necessarily not LU equivalent?If this is indeed true, then this would allow us to fully characterize entangled states in terms of the π-transpose, and additionally, it would make it very easy and quick to determine whether or not two states are LU equivalent.

HYPERDETERMINANTS AND n-TANGLES
Recall that the n-tangle, a proposed measure of entanglement for pure 2n-qubit states proposed in [12], is defined as Let ψ be a 2n-qubit and ψ be its corresponding hypermatrix as described in Section 3. We now introduce an important matrix Ent n as follows.Recall that the hyperdeterminant of ψ is given by (−1) m ψ 0σ2(0)...σ2n(0) ψ 1σ2(1)...σ2n (1) where m denotes the number of permutations σ i ∈ S 2 which are transpositions, and we will simply refer to the hyperdeterminant by Ent(ψ).Note in particular that since each σ i is in S 2 , they are either the identity permutation (which takes 0 to 0 and 1 to 1) or they are the transposition which takes 0 to 1 and 1 to 0. Note also that hdet( ψ) gives a quadratic form in the coefficients of ψ.Also, recall that for an arbitrary quadratic form q ij x i x j , the matrix of the quadratic form q is the matrix Q = [q ij ] ∈ C n×n .Denoting the vector x 1 ... x n t as x, we have that (4.3) x t Qx = q(x 1 , ..., x n ).
We will denote the matrix of the quadratic form given by Ent(ψ) as Ent n .
since σ 2 is either the identity or the only transposition in S 2 .The matrix of this quadratic form is where e i is the i th vector in the standard ordered basis for C 16 .
From the above examples, we notice a few patterns.In general, each term in hdet( ψ) is of the form ±ψ i...i2n ψ i1...i2n where i j = 1 − i j .Equivalently, each term is of the form ±|ψ j |ψ 4 n −j+1 for 1 ≤ j ≤ 4 n .So after factoring out 1 2 (which for the rest of this section we will assume we have already done), it follows that in general Ent n is an anti-diagonal matrix with 1's and −1's on its main anti-diagonal.
Going from left to right, we represent each entry of the main anti-diagonal of Ent n as a + or −, with 1 being identified as a + and −1 being identified as a −.We then have that the main anti-diagonal of Ent 1 is given by the string In particular, the first entry gives the sign of the term ψ 00 ψ 11 , the 2 nd entry gives the sign of the term ψ 01 ψ 10 , the 3 rd entry gives the sign of the term ψ 01 ψ 10 , and the 4 th entry gives the sign of the term ψ 00 ψ 11 .Similarly, the main anti-diagonal of Ent 2 is given by the string The first entry gives the sign of the term ψ 0000 ψ 1111 , the 2 nd entry gives the sign of the term ψ 0001 ψ 1110 ,..., the 8 th entry gives the sign of the term ψ 0111 ψ 1000 , the 9 th entry gives the sign of the term ψ 0111 ψ 1000 ,..., and the 16 th entry gives the sign of the term ψ 0000 ψ 1111 .By the hyperdeterminant formula, the sign of is positive if there are an even number of 0's and 1's in either factor; likewise, the sign of is negative if there is an odd number of 0's and 1's in either factor.Thus, + corresponds to a coefficient of |ψ with an even number of 0's and 1 ′ s, and − corresponds to a coefficient of |ψ with an odd number of 0's and 1's.
Identify the coefficient ψ i1...i2n with the binary string i 1 ...i 2n , and let B = b 1 b 2 ...b 4 n denote the sequence consisting of all binary strings of length 2n ordered via the lexicographic order.We call a binary string b i "even" if it has an even number of 0's and 1's, and we call it "odd" if it has an odd number of 0's and 1's.Let χ be a function given by (4.4) χ Lastly, set So in particular, 3 occurs after a sequence of P , 4 occurs after a sequence of P N , 5 occurs after a sequence of P N N P , 6 occurs after a sequence of P N N P N P P N , and so on.Indeed, in general, we have the following result.
Lemma 4.1.For k ≥ 3, the binary string k occurs after a sequence of P 's and N 's, which we denote as S.Moreover, the k + 1 string occurs after the sequence SS, where S is obtained after switching all P 's in S to N , and likewise flipping all N 's in S to P .Consequently, if it takes a sequence of S (consisting of some ordering of +'s and −'s, which we assume nothing about) to get from b 1 up to but not including k = b 2 k−1 +1 , then it takes a sequence of S to get from k = b 2 k−1 +1 up to but not including k + 1 = b 2 k +1+i .That is, k + 1 occurs after a string of SS.From this and the fact that to get to 3 it takes a sequence of P , it follows that S is a sequence of P 's and N 's.
To recap, Ent n is the matrix of the hyperdeterminant of the 2n-qubit |ψ , whose coefficients ψ i1...i2n we have identified with the binary string i 1 ...i 2n , and each such string we have assigned a + or − to based on its parity.Ent n is an anti-diagonal matrix whose main anti-diagonal can be represented as a sequence of +'s and −'s.
Recall that the main anti-diagonal of Ent 1 is given by P .After a sequence of P , we end up at the string 3 = 0...000100.Therefore, by the lemma, after a sequence of P P = P N , we end up at the string 4 = 0...001000, and consequently after a sequence of P N P N = P N N P , we end up at the string 5 = 0...010000.Hence, the main anti-diagonal of Ent 2 is given by (4.8) (P P )(P P ) = P N N P Applying the same reasoning, it follows that the main anti-diagonal of Ent 3 is given by (4.9) (P N N P P N N P )(P N N P P N N P ) = P N N P N P P N N P P N P N N P.
Indeed, continuing with this reasoning, in general, we have the following result.
Proposition 4.2.The main anti-diagonal of Ent n is given by a sequence of P 's and N 's.Moreover, denoting its main anti-diagonal as S, we have that the main anti-diagonal of Ent n+1 is given by Since the second quarter of the main anti-diagonal of Ent is the negation of the first quarter, and since the second half of the main anti-diagonal of the negation of the first half, we have the following consequence.2 ) Ent n is a symmetric anti-diagonal matrix whose main anti-diagonal consists of 1's and −1's, and this holds for all positive integers n.Now we would like to study the relationship between the matrix of the hyperdeterminant of an arbitrary 2n-qubit state with the spin-flip transformation.
To start with, we consider the structure of σ ⊗2n y .First note that which (like Ent) is a symmetric anti-diagonal matrix consisting of 1's and −1's.Going from left to right and representing each entry of the main anti-diagonal of σ ⊗2 y as a + or −, with 1 being identified as + and −1 being identified as −, we have that the main anti-diagonal of σ ⊗2 y is given by − + +−, which in our previous notation is just N .Proof.This is a straightforward calculation: A similar formula for the n-tangle involving determinants of the coefficients of ψ was proven in [18], however by linking the n-tangle to the hyperdeterminant we can apply the theory of multilinear algebra to the n-tangle and more broadly the study of entanglement.For instance, it is known that the n-tangle is an LU-invariant, in fact, more generally a SLOCC invariant [19], and indeed this fact immediately follows from the above corollary since the hyperdeterminant is invariant under multilinear multiplication of matrices in the special linear group (Proposition 2.1).

2 where | ψ = σ ⊗2n y ,
with σ y being the second Pauli matrix 0 −i i 0 .The product σ ⊗2n y is sometimes known as the spin-flip transformation on 2n-qubits.We now consider the relation between the n-tangle and the combinatorial hyperdeterminant via the hypermatrix of pure 2n-qubit states.

Fact 1 .For 1
The binary string with a 1 in only its k th position occurs in the (2 k−1 + 1) th position of B. ≤ k ≤ 2n, call a binary string with only a 1 in the k th position k.From Fact 1, in our notation, we have that k = b 2 k−1 +1 .