Welch bounds

In

Mathematical statement

If ${\displaystyle \{x_{1},\ldots ,x_{m}\}}$ are unit vectors in ${\displaystyle \mathbb {C} ^{n}}$, define ${\displaystyle c_{\max }=\max _{i\neq j}|\langle x_{i},x_{j}\rangle |}$, where ${\displaystyle \langle \cdot ,\cdot \rangle }$ is the usual

inner product

on ${\displaystyle \mathbb {C} ^{n}}$. Then the following inequalities hold for ${\displaystyle k=1,2,\dots }$:

$${\displaystyle (c_{\max })^{2k}\geq {\frac {1}{m-1}}\left[{\frac {m}{\binom {n+k-1}{k}}}-1\right].}$$

Welch bounds are also sometimes stated in terms of the averaged squared overlap between the set of vectors. In this case, one has the inequality[2]^[3][4]

Applicability

If ${\displaystyle m\leq n}$, then the vectors ${\displaystyle \{x_{i}\}}$ can form an

in ${\displaystyle \mathbb {C} ^{n}}$. In this case, ${\displaystyle c_{\max }=0}$ and the bounds are vacuous. Consequently, interpretation of the bounds is only meaningful if ${\displaystyle m>n}$. This will be assumed throughout the remainder of this article.

Proof for k = 1

The "first Welch bound," corresponding to ${\displaystyle k=1}$, is by far the most commonly used in applications. Its proof proceeds in two steps, each of which depends on a more basic mathematical inequality. The first step invokes the Cauchy–Schwarz inequality and begins by considering the ${\displaystyle m\times m}$ Gram matrix ${\displaystyle G}$ of the vectors ${\displaystyle \{x_{i}\}}$; i.e.,

${\displaystyle G=\left[{\begin{array}{ccc}\langle x_{1},x_{1}\rangle &\cdots &\langle x_{1},x_{m}\rangle \\\vdots &\ddots &\vdots \\\langle x_{m},x_{1}\rangle &\cdots &\langle x_{m},x_{m}\rangle \end{array}}\right]}$

The trace of ${\displaystyle G}$ is equal to the sum of its eigenvalues. Because the rank of ${\displaystyle G}$ is at most ${\displaystyle n}$, and it is a

matrix, ${\displaystyle G}$ has at most ${\displaystyle n}$ positive with its remaining eigenvalues all equal to zero. Writing the non-zero eigenvalues of ${\displaystyle G}$ as ${\displaystyle \lambda _{1},\ldots ,\lambda _{r}}$ with ${\displaystyle r\leq n}$ and applying the Cauchy-Schwarz inequality to the inner product of an ${\displaystyle r}$-vector of ones with a vector whose components are these eigenvalues yields

${\displaystyle (\mathrm {Tr} \;G)^{2}=\left(\sum _{i=1}^{r}\lambda _{i}\right)^{2}\leq r\sum _{i=1}^{r}\lambda _{i}^{2}\leq n\sum _{i=1}^{m}\lambda _{i}^{2}}$

The square of the

(Hilbert–Schmidt norm) of ${\displaystyle G}$ satisfies

${\displaystyle ||G||^{2}=\sum _{i=1}^{m}\sum _{j=1}^{m}|\langle x_{i},x_{j}\rangle |^{2}=\sum _{i=1}^{m}\lambda _{i}^{2}}$

Taking this together with the preceding inequality gives

${\displaystyle \sum _{i=1}^{m}\sum _{j=1}^{m}|\langle x_{i},x_{j}\rangle |^{2}\geq {\frac {(\mathrm {Tr} \;G)^{2}}{n}}}$

Because each ${\displaystyle x_{i}}$ has unit length, the elements on the main diagonal of ${\displaystyle G}$ are ones, and hence its trace is ${\displaystyle \mathrm {Tr} \;G=m}$. So,

${\displaystyle \sum _{i=1}^{m}\sum _{j=1}^{m}|\langle x_{i},x_{j}\rangle |^{2}=m+\sum _{i\neq j}|\langle x_{i},x_{j}\rangle |^{2}\geq {\frac {m^{2}}{n}}}$

or

${\displaystyle \sum _{i\neq j}|\langle x_{i},x_{j}\rangle |^{2}\geq {\frac {m(m-n)}{n}}}$

The second part of the proof uses an inequality encompassing the simple observation that the average of a set of non-negative numbers can be no greater than the largest number in the set. In mathematical notation, if ${\displaystyle a_{\ell }\geq 0}$ for ${\displaystyle \ell =1,\ldots ,L}$, then

${\displaystyle {\frac {1}{L}}\sum _{\ell =1}^{L}a_{\ell }\leq \max a_{\ell }}$

The previous expression has ${\displaystyle m(m-1)}$ non-negative terms in the sum, the largest of which is ${\displaystyle c_{\max }^{2}}$. So,

${\displaystyle (c_{\max })^{2}\geq {\frac {1}{m(m-1)}}\sum _{i\neq j}|\langle x_{i},x_{j}\rangle |^{2}\geq {\frac {m-n}{n(m-1)}}}$

or

${\displaystyle (c_{\max })^{2}\geq {\frac {m-n}{n(m-1)}}}$

which is precisely the inequality given by Welch in the case that ${\displaystyle k=1}$.

Achieving the Welch bounds

In certain telecommunications applications, it is desirable to construct sets of vectors that meet the Welch bounds with equality. Several techniques have been introduced to obtain so-called Welch Bound Equality (WBE) sets of vectors for the ${\displaystyle k=1}$ bound.

The proof given above shows that two separate mathematical inequalities are incorporated into the Welch bound when ${\displaystyle k=1}$. The Cauchy–Schwarz inequality is met with equality when the two vectors involved are collinear. In the way it is used in the above proof, this occurs when all the non-zero eigenvalues of the Gram matrix ${\displaystyle G}$ are equal, which happens precisely when the vectors ${\displaystyle \{x_{1},\ldots ,x_{m}\}}$ constitute a

for ${\displaystyle \mathbb {C} ^{n}}$.

The other inequality in the proof is satisfied with equality if and only if ${\displaystyle |\langle x_{i},x_{j}\rangle |}$ is the same for every choice of ${\displaystyle i\neq j}$. In this case, the vectors are equiangular. So this Welch bound is met with equality if and only if the set of vectors ${\displaystyle \{x_{i}\}}$ is an equiangular tight frame in ${\displaystyle \mathbb {C} ^{n}}$.

Similarly, the Welch bounds stated in terms of average squared overlap, are saturated for all ${\displaystyle k\leq t}$ if and only if the set of vectors is a ${\displaystyle t}$-design in the complex projective space ${\displaystyle \mathbb {CP} ^{n-1}}$.^[4]

See also

• Spherical design
• Quantum t-design
• Equiangular lines

References