Question

Let \(|u\rangle \in \mathcal{H}\) and let \(\mathcal{M}\) be a subspace of \(\mathcal{H}\). Show that the subset \(E=\) \(|u\rangle-\mathcal{M}\) is convex. Show that \(E\) is not necessarily a subspace of \(\mathcal{H}\).

Short answer

The subset \(E = |u\rangle - \mathcal{M}\) is convex as any line segment connecting two points in it also lies within \(E\). However, \(E\) is not necessarily a subspace of \(\mathcal{H}\), because if the vector \(|u\rangle\) is not in the subspace \(\mathcal{M}\), then \(E\) does not contain the zero vector, a requirement for it to be a subspace.

Step-by-step solution

Show That the Subset E is Convex

Let \(\mathcal{H}\) be a complex Hilbert space and \(\mathcal{M}\) be a subspace of \(\mathcal{H}\), let \(|u\rangle \in \mathcal{H}\). Also, let \(E = |u\rangle - \mathcal{M}\). To show that \(E\) is convex, consider any two elements, \(|e_{1}\rangle, |e_{2}\rangle \in E\). These elements can be written as \(|e_{1}\rangle = |u\rangle - |m_{1}\rangle\) and \(|e_{2}\rangle = |u\rangle - |m_{2}\rangle\) for some \(|m_{1}\rangle, |m_{2}\rangle \in \mathcal{M}\). Now, for any \(0 \leq t \leq 1\), we examine \(|e\rangle = t|e_{1}\rangle + (1 -t)|e_{2}\rangle\) and show that \(|e\rangle \in E\). This comes to \(|e\rangle = t|u\rangle + (1 - t)|u\rangle - t|m_{1}\rangle - (1 - t)|m_{2}\rangle = |u\rangle - (t|m_{1}\rangle + (1 - t)|m_{2}\rangle)\). Since \(\mathcal{M}\) is a subspace, by definition, \(t|m_{1}\rangle + (1 - t)|m_{2}\rangle \in \mathcal{M}\). Hence, \(|e\rangle \in E\), and the subset \(E\) is convex.

Show That E is Not Necessarily a Subspace.

A subset \(E\) of a vector space \(\mathcal{H}\) is a subspace if it satisfies three conditions: contents the zero vector, closed under our scalar multiplication, and closed under vector addition. We show that our set \(E = |u\rangle - \mathcal{M}\) does not have to satisfy these conditions, specifically the zero vector condition, hence it is not necessarily a subspace. Note that for the zero vector (|0\rangle) to be in \(E\), we would have \(|0\rangle = |u\rangle - |m\rangle\) for some \(|m\rangle \in \mathcal{M}\), this would imply that \(|u\rangle = |m\rangle \in \mathcal{M}\). Therefore, if \(|u\rangle \notin \mathcal{M}\), \(E\) does not contain the zero vector and hence is not a subspace.

Key concepts

Hilbert Space

A Hilbert space, named after the influential German mathematician David Hilbert, is a fascinating and crucial concept in mathematics and physics. Envision it as a vast, multidimensional canvas, where every point represents a state or configuration. Specifically, in the context of quantum mechanics, the space consists of 'quantum states'. But what makes a Hilbert space special is that it's complete, meaning it has all limit points of sequences within it, and is equipped with an inner product. This inner product provides a way to measure angles and lengths, giving us a robust geometrical framework to work with.

The completeness ensures that certain sequences, called Cauchy sequences, have a limit that is also within the Hilbert space. Due to this property, Hilbert spaces are the perfect playground for analysts and physicists, as they enable the application of rigorous mathematical techniques. Understanding Hilbert spaces unlocks a deeper comprehension of phenomena stretching from quantum physics to signal processing.

Vector Subspace

Delving into vector subspaces is like exploring a restricted zone within a more extensive territory. In the context of a Hilbert space, a vector subspace is a subset that has all the properties of the space itself, but is confined to a smaller dimension. This means that it's closed under addition and scalar multiplication.

In formal terms, for any two vectors within this subset, if you add them or multiply them by any scalar, they remain in the subset. Imagine a sheet of paper in three-dimensional space: no matter how you rotate or move the sheet, it always stays two-dimensional. Subspaces offer a simplified, more manageable setting that still retains the structural aspects of the larger space. They are incredibly useful in breaking down complex problems into smaller, more soluble chunks.

Convexity in Vector Spaces

Convexity in vector spaces is an intriguing property that provides a way to understand the geometry of these spaces. A subset of a vector space is convex if, for any two points within the set, the line segment connecting them lies entirely within the set. This concept is powerful because it implies predictability and a certain smoothness in the behavior of the set.

In our daily life, convex objects are those that have no indentations or holes—like a filled balloon. Convexity has immense implications in optimization theory, as it ensures that any local minimum is also a global minimum, reducing the complexities in finding 'the best solution'. The step-by-step solution we have about the convexity of a set formed by subtracting a subspace from an element of a Hilbert space illustrates an instance of this geometrical concept. Grasping convexity can give rise to better intuition when navigating the vast expanse of higher mathematics and physics.