Problem 18 - A Course in Modern Mathematical Physics: Groups, Hilbert Space and Differential Geometry

Question

Let \(A\) be a bounded openitor on a Hilbert space \(\mathcal{H}\) with a one- damensional rangc. (a) Show that there exist vectors \(u, v\) such that \(A x=\langle v \mid x\rangle u\) for all \(x \in \mathcal{H}\). (b) Show that \(A^{2}=\lambda A\) for some scalar \(\lambda\), and that \(\|A\|=\|u\|\|v\|\). (c) Prove that \(A\) is hermitian, \(A^{*}=A\), If and only if there exists a real number \(a\) such that \(v=a u\).

Step-by-Step Solution

Answer

The vectors \(u\), \(v\) are found by considering the scalar function on \(x\) that gives the proportionality between \(Ax\) and \(u\). The operator \(A\) squared equals \(\lambda A\) for a scalar \(\lambda\), and the norm of \(A\) equals the product of the norms of \(u\) and \(v\). The operator \(A\) is Hermitian if and only if \(v = a u\) for some real number \(a\).

Step 1: Find out \(u\) and \(v\)

Since the operator \(A\) has a one-dimensional range, all vectors \(Ax\) are proportional. Let \(u\) be a non-zero vector in the range of \(A\), then for any vector \(x\) there exists a scalar function \(\phi:(x)\) such that \(Ax=\phi(x)u\). Now let \(v\) be the vector in \(\mathcal{H}\) defined by \(v=\phi\). So, the operator \(A\) can be expressed as \(Ax = \langle v | x \rangle u \) for all \(x\) in \( \mathcal{H}\)

Step 2: Proof of \(A^{2} = \lambda A\) and \(\|A\|=\|u\|\|v\|\)

By substituting \(Ax\) in the equation for \(A^{2}x\), one obtains: \(A^{2}x = A \langle v | x \rangle u = \langle v | x \rangle A u\). Since \(A u\) is proportional to \(u\), let it be \(\lambda u\) where \(\lambda\) is a scalar. Thus, \(A^{2}x = \langle v | x \rangle \lambda u = \lambda A x\), which means \(A^{2} = \lambda A\). The norm of \(A\) is given by \(\|A\| = sup_{\|x\|=1} \|Ax\| = \|u\| \|v\|\)

Step 3: Conditions for \(A\) to be Hermitian

By definition, \(A\) is Hermitian if \(A\) equals its adjoint \(A^{*}\). This means \(A x = A^{*} x\), which translates to \(\langle v | x \rangle u = \langle x | v \rangle^{*} u\). If this is to hold for all \(x\)'s in the Hilbert space \(\mathcal{H}\), then it means that \(\langle v | x \rangle = \langle x | v \rangle^{*}\), which implies that \(v = a u\) where \(a\) is a real number

Key Concepts

Bounded Operator

In the realm of functional analysis, particularly when dealing with Hilbert spaces, understanding operators such as a "bounded operator" is crucial. These operators are functions that take vectors from one space and map them to another with specific properties.
A bounded operator ensures that when applied to a vector, the resulting output is always within a predictable range of length.
In other words, it doesn't stretch the vector infinitely but keeps it "bounded." This property is mathematically expressed by:

• There exists a constant \( M \) such that for every vector \( x \) in the Hilbert space \( \mathcal{H} \), \( \|Ax\| \leq M \|x\| \).
• This implies that the operator is "limited" or "controlled" in its behavior.

This concept plays a vital role in ensuring stability and predictability when operators are used in solving equations or transforming data within the space.

One-dimensional Range

The idea of a one-dimensional range in the context of operators on a Hilbert space revolves around the simplicity of the operator's effect.
When an operator has a one-dimensional range, it means that any vector you input will always produce an output that lies along a single line determined by a vector \( u \).
Mathematically, if \( A \) is such an operator, then:

• For every vector \( x \) in the space \( \mathcal{H} \), there exists a scalar \( \phi(x) \) such that \( Ax = \phi(x) \cdot u \).
• This indicates that all outputs of the operator are merely scaled versions of the vector \( u \).

This property simplifies the analysis significantly, as it reduces the complexity drastically, essentially "flattening" the n-dimensional problem into one-dimension.

Hermitian Operators

Hermitian operators, also known as self-adjoint operators, are a fundamental concept in linear algebra and quantum mechanics.
These operators possess a symmetry that is both algebraically and geometrically significant.
A Hermitian operator satisfies the condition:

• \( A = A^* \), where \( A^* \) is the "adjoint" or "transpose complex conjugate" of \( A \).

This implies that applying the operator to a vector and then taking an image from it is the same as taking the vector's image first and then applying the operator.
Hermitian operators have properties like:

• **Real eigenvalues:** Every eigenvalue of a Hermitian operator is real.
• **Orthogonal eigenvectors:** Eigenvectors associated with distinct eigenvalues are orthogonal.
• **Representations in quantum mechanics:** They represent observable quantities where measurements yield real numbers, reflecting their physical significance.

Understanding Hermitian operators provides insight into the real-world implications and mathematical elegance within the space.

Scalar Function

In mathematical analysis, particularly in the context of operators on Hilbert spaces, a scalar function comes into play as a simple yet fundamental concept.
A scalar function is essentially a function that maps vectors from the Hilbert space to scalar field values, generally real or complex numbers.
In the context of the original exercise, this function is used to indicate proportional scaling factors for vectors. When discussing an operator \( A \) with a one-dimensional range, if the output from an operator can be expressed as \( Ax = \phi(x) u \), then \( \phi(x) \) is a scalar function.
The beauty of scalar functions is in their simplicity, allowing complex transformations to be understood in terms of easily manageable numbers. This makes them a powerful tool in simplifying analysis and computation within Hilbert spaces.