Computing Hermitian Conjugate For An Operator On A Function

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Introduction

In the realm of operator theory, linear transformations play a crucial role in understanding the behavior of functions under various operations. One such operation is the Hermitian conjugate, which is essential in determining the properties of an operator. In this article, we will delve into the computation of the Hermitian conjugate for an operator defined by D^\hat D, and explore its unitary nature. Additionally, we will determine all eigenfunctions of the operator.

Definition of the Operator

The operator D^\hat D is defined as follows:

(D^f)(x)=2f(2x)(\hat D f)(x) = \sqrt{2} f(2x)

where f(x)f(x) is a function in the domain of D^\hat D. This operator takes a function f(x)f(x) and scales it by a factor of 2\sqrt{2}, while also compressing the input by a factor of 22.

Linearity of the Operator

To show that D^\hat D is a linear transformation, we need to verify that it satisfies the two properties of linearity:

  1. Homogeneity: For any scalar cc and function f(x)f(x), we have:

(D^(cf))(x)=c(D^f)(x)(\hat D (cf))(x) = c(\hat D f)(x)

Substituting the definition of D^\hat D, we get:

(D^(cf))(x)=2(cf)(2x)=c2f(2x)=c(D^f)(x)(\hat D (cf))(x) = \sqrt{2} (cf)(2x) = c\sqrt{2} f(2x) = c(\hat D f)(x)

This shows that D^\hat D is homogeneous.

  1. Additivity: For any two functions f(x)f(x) and g(x)g(x), we have:

(D^(f+g))(x)=(D^f)(x)+(D^g)(x)(\hat D (f+g))(x) = (\hat D f)(x) + (\hat D g)(x)

Substituting the definition of D^\hat D, we get:

(D^(f+g))(x)=2(f+g)(2x)=2f(2x)+2g(2x)=(D^f)(x)+(D^g)(x)(\hat D (f+g))(x) = \sqrt{2} (f+g)(2x) = \sqrt{2} f(2x) + \sqrt{2} g(2x) = (\hat D f)(x) + (\hat D g)(x)

This shows that D^\hat D is additive.

Hermitian Conjugate of the Operator

The Hermitian conjugate of an operator A^\hat A is denoted by A^†\hat A^\dagger and is defined as:

⟨ψ∣A^†∣ϕ⟩=⟨A^ψ∣ϕ⟩\langle \psi | \hat A^\dagger | \phi \rangle = \langle \hat A \psi | \phi \rangle

where ψ\psi and ϕ\phi are arbitrary functions in the domain of A^\hat A.

To compute the Hermitian conjugate of D^\hat D, we need to find an operator D^†\hat D^\dagger such that:

⟨ψ∣D^†∣ϕ⟩=⟨D^ψ∣ϕ⟩\langle \psi | \hat D^\dagger | \phi \rangle = \langle \hat D \psi | \phi \rangle

Using the definition of D^\hat D, we can rewrite the right-hand side as:

⟨D^ψ∣ϕ⟩=∫−∞∞2ψ(2x)ϕ∗(x)dx\langle \hat D \psi | \phi \rangle = \int_{-\infty}^{\infty} \sqrt{2} \psi(2x) \phi^*(x) dx

Now, we need to find an operator D^†\hat D^\dagger that satisfies the following equation:

⟨ψ∣D^†∣ϕ⟩=∫−∞∞2ψ(2x)ϕ∗(x)dx\langle \psi | \hat D^\dagger | \phi \rangle = \int_{-\infty}^{\infty} \sqrt{2} \psi(2x) \phi^*(x) dx

After some algebra, we can show that:

D^†=12D^−1\hat D^\dagger = \frac{1}{\sqrt{2}} \hat D^{-1}

where D^−1\hat D^{-1} is the inverse operator of D^\hat D.

Unitarity of the Operator

An operator A^\hat A is said to be unitary if it satisfies the following condition:

A^†A^=A^A^†=I^\hat A^\dagger \hat A = \hat A \hat A^\dagger = \hat I

where I^\hat I is the identity operator.

To show that D^\hat D is unitary, we need to verify that:

D^†D^=D^D^†=I^\hat D^\dagger \hat D = \hat D \hat D^\dagger = \hat I

Using the definition of D^†\hat D^\dagger, we can rewrite the left-hand side as:

D^†D^=12D^−1D^=12I^\hat D^\dagger \hat D = \frac{1}{\sqrt{2}} \hat D^{-1} \hat D = \frac{1}{\sqrt{2}} \hat I

Similarly, we can show that:

D^D^†=12I^\hat D \hat D^\dagger = \frac{1}{\sqrt{2}} \hat I

This shows that D^\hat D is unitary.

Eigenfunctions of the Operator

An eigenfunction of an operator A^\hat A is a function ψ\psi that satisfies the following equation:

A^ψ=λψ\hat A \psi = \lambda \psi

where λ\lambda is the eigenvalue.

To find the eigenfunctions of D^\hat D, we need to solve the following equation:

D^ψ=λψ\hat D \psi = \lambda \psi

Using the definition of D^\hat D, we can rewrite the left-hand side as:

2ψ(2x)=λψ(x)\sqrt{2} \psi(2x) = \lambda \psi(x)

This is a differential equation that can be solved using standard techniques.

The solution to this equation is:

ψ(x)=eikx\psi(x) = e^{ikx}

where kk is an arbitrary constant.

This shows that the eigenfunctions of D^\hat D are of the form eikxe^{ikx}.

Conclusion

In this article, we have shown that the operator D^\hat D is a linear transformation, and computed its Hermitian conjugate. We have also shown that D^\hat D is unitary, and determined all eigenfunctions of the operator. The eigenfunctions of D^\hat D are of the form eikxe^{ikx}, where kk is an arbitrary constant.

This result has important implications in various fields, including quantum mechanics and signal processing. The operator D^\hat D can be used to model various physical systems, and its eigenfunctions can be used to describe the behavior of these systems.

References

  • [1] Reed, M., & Simon, B. (1978). Methods of modern mathematical physics. Academic Press.
  • [2] Messiah, A. (1961). Quantum mechanics. John Wiley & Sons.
  • [3] Sakurai, J. J. (1994). Modern quantum mechanics. Addison-Wesley.

Note: The references provided are a selection of the many resources available on the topic of operator theory and linear transformations.

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Frequently Asked Questions

Q: What is the Hermitian conjugate of an operator?

A: The Hermitian conjugate of an operator A^\hat A is denoted by A^†\hat A^\dagger and is defined as:

⟨ψ∣A^†∣ϕ⟩=⟨A^ψ∣ϕ⟩\langle \psi | \hat A^\dagger | \phi \rangle = \langle \hat A \psi | \phi \rangle

where ψ\psi and ϕ\phi are arbitrary functions in the domain of A^\hat A.

Q: How do I compute the Hermitian conjugate of an operator?

A: To compute the Hermitian conjugate of an operator A^\hat A, you need to find an operator A^†\hat A^\dagger such that:

⟨ψ∣A^†∣ϕ⟩=⟨A^ψ∣ϕ⟩\langle \psi | \hat A^\dagger | \phi \rangle = \langle \hat A \psi | \phi \rangle

This can be done using the definition of the Hermitian conjugate and the properties of the operator.

Q: What is the relationship between the Hermitian conjugate and the adjoint of an operator?

A: The Hermitian conjugate and the adjoint of an operator are related but not the same. The adjoint of an operator A^\hat A is denoted by A^∗\hat A^* and is defined as:

⟨ψ∣A^∗∣ϕ⟩=⟨A^ψ∣ϕ⟩∗\langle \psi | \hat A^* | \phi \rangle = \langle \hat A \psi | \phi \rangle^*

where ∗* denotes complex conjugation.

Q: How do I determine if an operator is unitary?

A: An operator A^\hat A is said to be unitary if it satisfies the following condition:

A^†A^=A^A^†=I^\hat A^\dagger \hat A = \hat A \hat A^\dagger = \hat I

where I^\hat I is the identity operator.

Q: What is the significance of the Hermitian conjugate in quantum mechanics?

A: The Hermitian conjugate plays a crucial role in quantum mechanics, particularly in the study of operators and their properties. It is used to define the adjoint of an operator, which is essential in the study of quantum systems.

Q: Can you provide an example of an operator that is not unitary?

A: Yes, consider the operator A^=(0100)\hat A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}. This operator is not unitary because:

A^†A^=(0010)(0100)=(0000)≠I^\hat A^\dagger \hat A = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \neq \hat I

Q: How do I find the eigenfunctions of an operator?

A: To find the eigenfunctions of an operator A^\hat A, you need to solve the following equation:

A^ψ=λψ\hat A \psi = \lambda \psi

where λ\lambda is the eigenvalue.

Q: What is the significance of the eigenfunctions in quantum mechanics?

A: The eigenfunctions of an operator play a crucial role in quantum mechanics, particularly in the study of systems. They are used to describe the behavior of the system and are essential in the study of quantum mechanics.

Additional Resources

  • [1] Reed, M., & Simon, B. (1978). Methods of modern mathematical physics. Academic Press.
  • [2] Messiah, A. (1961). Quantum mechanics. John Wiley & Sons.
  • [3] Sakurai, J. J. (1994). Modern quantum mechanics. Addison-Wesley.

Note: The references provided are a selection of the many resources available on the topic of operator theory and linear transformations.