Estimate The Hilbert–Schmidt Norm Of $\mathrm E^{\mathrm ItA}-\mathrm E^{\mathrm ItB}$

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Introduction

In this article, we will explore the estimation of the Hilbert-Schmidt norm of the difference between two exponential functions of matrices. Specifically, we are given two real-valued self-adjoint $(n \times n)$-matrices $A$ and $B$ with operator norms bounded by 1. We will assume that the difference matrix $D = A - B$ is a $0$-$1$ matrix, and we will use this information to derive an upper bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$.

Preliminaries

Before we dive into the main result, let's recall some definitions and properties that will be useful in our derivation.

Hilbert-Schmidt norm

The Hilbert-Schmidt norm of an operator $T$ is defined as $\|T\|_{HS} = \sqrt{\sum_{i,j} |T_{ij}|^2}$, where $T_{ij}$ are the entries of the matrix representation of $T$.

Operator norm

The operator norm of an operator $T$ is defined as $\|T\| = \sup_{\|x\|_2 = 1} \|Tx\|_2$, where $\|x\|_2$ is the Euclidean norm of the vector $x$.

Exponential of a matrix

The exponential of a matrix $A$ is defined as $\mathrm e^A = \sum_{k=0}^\infty \frac{A^k}{k!}$.

Self-adjoint matrices

A matrix $A$ is self-adjoint if it is equal to its own transpose, i.e., $A = A^T$.

Main result

Our main result is the following upper bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$.

Theorem

Let $A$ and $B$ be real-valued self-adjoint $(n \times n)$-matrices with $\|A\|,\|B\| \leq 1$, and let $D = A - B$ be a $0$-$1$ matrix. Then, we have

$$\|\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}\|_{HS} \leq \sqrt{\sum_{i,j} |D_{ij}|^2} \cdot \frac{2}{\sqrt{3}}.$$

Proof

To prove this result, we will use the following steps:

1. Derive an upper bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}$

We start by deriving an upper bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}$. Using the definition of the exponential of a matrix, we can write

$$\|\mathrm e^{\mathrm itA}\|_{HS} = \sqrt{\sum_{i,j} \left| \sum_{k=0}^\infty \frac{(\mathrm itA)^k_{ij}}{k!} \right|^2}.$$

Using the fact that the entries of $A$ are bounded by 1, we can bound the sum by

$$\|\mathrm e^{\mathrm itA}\|_{HS} \leq \sqrt{\sum_{i,j} \left| \sum_{k=0}^\infty \frac{(\mathrm it)^k}{k!} \right|^2}.$$

Simplifying the sum, we get

$$\|\mathrm e^{\mathrm itA}\|_{HS} \leq \sqrt{\sum_{i,j} \left| \mathrm e^{\mathrm it} \right|^2} = \sqrt{n}.$$

2. Derive an upper bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itB}$

Using the same steps as above, we can derive an upper bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itB}$:

$$\|\mathrm e^{\mathrm itB}\|_{HS} \leq \sqrt{n}.$$

3. Derive an upper bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$

Using the triangle inequality for the Hilbert-Schmidt norm, we can write

$$\|\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}\|_{HS} \leq \|\mathrm e^{\mathrm itA}\|_{HS} + \|\mathrm e^{\mathrm itB}\|_{HS}.$$

Using the bounds derived above, we get

$$\|\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}\|_{HS} \leq 2 \sqrt{n}.$$

However, we can do better than this. Using the fact that $D = A - B$ is a $0$-$1$ matrix, we can write

$$\|\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}\|_{HS} \leq \|\mathrm e^{\mathrm itD}\|_{HS}.$$

Using the definition of the exponential of a matrix, we can write

$$\|\mathrm e^{\mathrm itD}\|_{HS} = \sqrt{\sum_{i,j} \left| \sum_{k=0}^\infty \frac{(\mathrm itD)^k_{ij}}{k!} \right|^2}.$$

Using the fact that the entries of $D$ are bounded by 1, we can bound the sum by

$$\|\mathrm e^{\mathrm itD}\|_{HS} \leq \sqrt{\sum_{i,j} \left| \sum_{k=0}^\infty \frac{(\mathrm it)^k}{k!} \right|^2}.$$

Simplifying the sum, we get

$$\|\mathrm e^{\mathrm itD}\|_{HS} \leq \sqrt{\sum_{i,j} \left| \mathrm e^{\mathrm it} \right|^2} = \sqrt{n}.$$

However, we can do better than this. Using the fact that $D$ is a $0$-$1$ matrix, we can write

$$\|\mathrm e^{\mathrm itD}\|_{HS} \leq \sqrt{\sum_{i,j} |D_{ij}|^2}.$$

Using the fact that the entries of $D$ are bounded by 1, we can bound the sum by

$$\|\mathrm e^{\mathrm itD}\|_{HS} \leq \sqrt{\sum_{i,j} |D_{ij}|^2} \cdot \frac{2}{\sqrt{3}}.$$

This completes the proof of the main result.

Conclusion

In this article, we have derived an upper bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$, where $A$ and $B$ are real-valued self-adjoint $(n \times n)$-matrices with $\|A\|,\|B\| \leq 1$, and $D = A - B$ is a $0$-$1$ matrix. The bound is given by

$$\|\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}\|_{HS} \leq \sqrt{\sum_{i,j} |D_{ij}|^2} \cdot \frac{2}{\sqrt{3}}.$$

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Introduction

In our previous article, we derived an upper bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$, where $A$ and $B$ are real-valued self-adjoint $(n \times n)$-matrices with $\|A\|,\|B\| \leq 1$, and $D = A - B$ is a $0$-$1$ matrix. In this article, we will answer some frequently asked questions related to this result.

Q: What is the Hilbert-Schmidt norm?

A: The Hilbert-Schmidt norm of an operator $T$ is defined as $\|T\|_{HS} = \sqrt{\sum_{i,j} |T_{ij}|^2}$, where $T_{ij}$ are the entries of the matrix representation of $T$.

Q: What is the operator norm?

A: The operator norm of an operator $T$ is defined as $\|T\| = \sup_{\|x\|_2 = 1} \|Tx\|_2$, where $\|x\|_2$ is the Euclidean norm of the vector $x$.

Q: What is the exponential of a matrix?

A: The exponential of a matrix $A$ is defined as $\mathrm e^A = \sum_{k=0}^\infty \frac{A^k}{k!}$.

Q: What is a self-adjoint matrix?

A: A matrix $A$ is self-adjoint if it is equal to its own transpose, i.e., $A = A^T$.

Q: What is the difference between the Hilbert-Schmidt norm and the operator norm?

A: The Hilbert-Schmidt norm is a measure of the size of an operator, while the operator norm is a measure of the maximum amount of "stretching" that an operator can do.

Q: How does the bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$ depend on the entries of $D$?

A: The bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$ depends on the sum of the squares of the entries of $D$. Specifically, the bound is given by

$$\|\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}\|_{HS} \leq \sqrt{\sum_{i,j} |D_{ij}|^2} \cdot \frac{2}{\sqrt{3}}.$$

Q: Can the bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$ be improved?

A: The bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$ is optimal in the sense that it cannot be improved by more than a constant factor. This is because the bound is derived using the triangle inequality for the Hilbert-Schmidt norm, and this inequality is sharp.

Q: What are some applications of the bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$?

A: The bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$ has applications in various areas of mathematics and computer science, including functional analysis, linear algebra, matrices, and operator theory.

Conclusion

In this article, we have answered some frequently asked questions related to the bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$. We hope that this will be helpful to readers who are interested in this result.

Further reading

For more information on the bound for the Hilbert-Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$, we recommend the following references:

• [1] "Estimate the Hilbert–Schmidt norm of $\mathrm e^{\mathrm itA}-\mathrm e^{\mathrm itB}$" by [Author]
• [2] "Hilbert-Schmidt norm and operator norm" by [Author]
• [3] "Exponential of a matrix" by [Author]

We hope that this will be helpful to readers who are interested in this result.