Use polarization identity to prove a linear operator over a finite-dimensional Hilbert space is completely specified by its expectations on a sphere

Question

This is from the book Quantum Theory from First Principles: An Informational Approach, which I thought I'd give a read as I found the authors' two papers on the same subject to be rather impenetrable. The book begins with a review of quantum theory from a mathematical perspective of Hilbert spaces and quantum information theory. The very first exercise in the book is this: Now the book does say it assumes a certain familiarity with Hilbert spaces, but also contains a brief review of concepts the authors consider important. And normally, when a book asks me to prove something, I assume it means with information already provided, as if I am allowed to go online and look for any known properties about Hilbert spaces to use in a proof, then just about any proof not worthy of a Nobel prize becomes trivial. However, the sum-total of information the book has given us about Hilbert spaces prior to this point, is this: Obviously getting from that to even understanding the notation used for the polarization identity here requires some prior knowledge, and I'm doing my best to remember my Dirac notation from undergraduate quantum mechanics, but I still don't see how we get from the identity they provide to a linear operator being completely specified by its expectations of the set of all vectors of some fixed positive length. The solution provided for this exercise at the end of the chapter doesn't exactly help either:

Now I will admit that I'm jumping in at the deep end here on subjects that are definitely more than a little rusty for me, but I think this explanation could really benefit from some elaboration. I believe I understand the second sentence, it is saying basically that since we are in a linear space, if we know any property of a particular vector, then we also know that property of any scaled multiple of that vector.

Where they lost me was in the first sentence though. I almost wonder if the printers put the wrong equation in by mistake. How do we get from that elaborate sum in (2.6) do being able to express any matrix-element $\langle y|A|x \rangle$ of $A$ in terms of its diagonal elements? How does that help us to completely specify $A$ for any $|\lambda \rangle$? And what does that i.e. at the end mean? Is proving that $\langle \lambda|A|\lambda \rangle = 0$ for every $\lambda \in \mathbf{S}_s(\mathcal{H})$ iff $A = 0$ sufficient to completely specify $A$, or is that unrelated?

Am I missing something obvious here or is this just a very badly written exercise?

Answer 1

Okay, after spending far too much time studying polarization identities and trying to make sense of an exercise that I presume was adapted from an entry for the International Obfuscated Math Notation Competition, I believe I finally understand what this question is asking and how to get the answer they provided. First of all, it is sufficient to prove that $\langle \lambda | A | \lambda \rangle = 0 \ \forall | \lambda \rangle \in \mathsf S_s(\mathcal H)$ iff $A = 0$ to show that an operator $A \in \mathsf{Lin}(\mathcal{H})$ is completely defined by its expectations. To see this, refer to this similar proof for a dense subspace and apply linearity to get it from a non-zero spherical shell.

The authors of this book in their answer seem to be making a much stronger statement though, that we can express any matrix-element $\langle y | A | x \rangle$ in terms of its diagonal elements. This is true, and can be shown simply enough using polarization identities of inner products. However, for some reason that I cannot for the life of me comprehend, in this book, they decided to express the polarization identity in terms of outer products, and while this is mathematically valid, it does not seem especially useful. I spent quite a lot of time fiddling with products of two matrix-elements to see if I could exploit the outer product that appears in the middle there, and while I did eventually reach a solution using that method, I do not care to reproduce the several pages of algebra this required here. Here then is a much simpler proof directly applying the complex polarization identity to a matrix-element calculation. If anyone would like to attempt a proof using the specific equation this book provides, be my guest. This proof takes heavy inspiration from the discussion on this question and makes the derivation a bit more explicit.

Lemma 1. For any $x$ in any linear space, we have $$\sum_{k=0}^3 i^kx = 0$$ Proof. Comes from cancellation. $$\sum_{k=0}^3 i^kx = x + ix - x - ix = 0 \ \Box$$ Lemma 2. For any Hilbert space $\mathcal H$, any $| \mu \rangle \in \mathcal H \setminus \{0\}$, any operator $A \in \mathsf{Lin}(\mathcal H)$, and any $s > 0$, there exists a real number $r > 0$ and $| \lambda \rangle \in \mathsf S_s(\mathcal H)$ such that $$| \mu \rangle = r | \lambda \rangle$$ Proof. Let $r = \frac{||\mu||}{s}$ and $| \lambda \rangle = \frac{| \mu \rangle}{r}$ and this follows directly from linearity. $\Box$

Theorem. For any $| x \rangle, |y \rangle \in \mathcal H$, any $s > 0$, and any operator $A \in \mathsf{Lin}(\mathcal H)$, it is possible to express the matrix-element $\langle y | A | x \rangle$ entirely in terms of expectations $\langle \lambda | A | \lambda \rangle$ with $| \lambda \rangle \in \mathsf S_s(\mathcal H)$.

Proof. Let $| x \rangle, |y \rangle \in \mathcal H$, $s > 0$, and $A \in \mathsf{Lin}(\mathcal H)$. Define $| \mu_k \rangle = | x \rangle + i^k | y \rangle$. Then $\langle \mu_k | = | \mu_k \rangle^\dagger = \langle x | + (-i)^k \langle y |$. By Lemma 2, if $| \mu_k \rangle \ne 0$, then there exist $r_k > 0$ and $| \lambda_k \rangle \in \mathsf S_s(\mathcal H)$ such that $| \mu_k \rangle = r_k | \lambda_k \rangle$. If $| \mu_k \rangle = 0$, then we set $r_k = | \lambda_k \rangle = 0$. In either case, $\langle \mu_k | A | \mu_k \rangle = r_k^2 \langle \lambda_k | A | \lambda_k \rangle$. We can then take the sum $$\frac{1}{4} \sum_{k=0}^3 i^k r_k^2 \langle \lambda_k | A | \lambda_k \rangle = \frac{1}{4} \sum_{k=0}^3 i^k \langle \mu_k | A | \mu_k \rangle \\ = \frac{1}{4} \sum_{k=0}^3 i^k \left( \langle x | + (-i)^k \langle y | \right) A \left( | x \rangle + i^k | y \rangle \right) \\ = \frac{1}{4} \sum_{k=0}^3 \left[ i^k \langle x | A | x \rangle + (-1)^k \langle x | A | y \rangle + \langle y | A | x \rangle + i^k \langle y | A | y \rangle \right]$$ By Lemma 1, all of the terms of this sum are $0$ except for $\langle y | A | x \rangle$, leaving $$\langle y | A | x \rangle = \frac{1}{4} \sum_{k=0}^3 i^k r_k^2 \langle \lambda_k | A | \lambda_k \rangle \ \Box$$

Since a linear operator in a finite-dimensional space is completely defined by its matrix-elements, this suffices to show that it is also completely defined by its expectations $\langle \lambda | A | \lambda \rangle$ over $| \lambda \rangle \in \mathsf S_s(\mathcal H)$ for any $s > 0$.