Show linear combination is not Hermitian

• Shiz
In summary: Homework Statement In summary, the equation is given that it is not hermitian already, so we have to take the complex conjugate of the right hand side. There appears to be a mathematical reason for why the operator would become ∫ψj * \hat{A} ψi - i(∫ψj * \hat{B} ψi). If that's what you mean, then yes, Ahat and Bhat are hermitian operators. However, (Ahat -/+ iBhat) is not hermitian.
Shiz

Homework Statement

Linear combination is $\hat{A}$ + i$\hat{B}$. It's given that it is not Hermitian already.

Homework Equations

∫ψi * $\hat{Ω}$ ψj = (∫ψj * $\hat{Ω}$ ψi)*

The Attempt at a Solution

∫ψi * ($\hat{A}$ + i$\hat{B}$) ψj = (∫ψj * ($\hat{A}$ + i$\hat{B}$) ψi)*

I chose to work with the right hand side of the equation first.
∫ψi * ($\hat{A}$ + i$\hat{B}$) ψj = {∫ψj * $\hat{A}$ ψi + i(∫ψj * $\hat{B}$ ψi)}*

So I have to take the complex conjugate of the right hand side (not sure if that's the proper way to say it). What I don't understand is why the operator would become ∫ψj * $\hat{A}$ ψi - i(∫ψj * $\hat{B}$ ψi) and then ∫ψi * ($\hat{A}$ - i$\hat{B}$) ψj.

What are the mathematical reasons? The complex conjugate of + i$\hat{B}$ is -i$\hat{B}$. I would just be replacing the operator with its complex conjugate? That would give me the answer, but it doesn't seem that simple. Clarification at this would help! Thank you!

The reason is the wave functions $\Psi_{i,j}$ also get affected by conjugating. And when you want the hermitian conjugate of a producat of operators, you have to reverse the order of factors, since the hermitian conjugate (for example) of a matrix is the complex conjugate of the transpose of the original matrix: $(A^*)_{ij}=\bar{A_{ji}}$

If that's what you mean

Are ##\hat{A}## and ##\hat{B}## hermitian operators? If they are, it's not generally true that ##\hat{A} + i\hat{B}## is not hermitian (consider the case where ##\hat{B}## is the zero operator).

Ahat and Bhat are hermitian operators, yes. But (Ahat -/+ iBhat) is not hermitian. I think I understand but not sure. We have to take the complex conjugate to show that it is not hermitian. So (iBhat)* is (-iBhat). I was wondering if there was more to that step than what I am thinking.

Apologies for no code. I am on the iPhone app and am not sure how to do it.

The reason why the operator becomes ∫ψj * \hat{A} ψi - i(∫ψj * \hat{B} ψi) is because when taking the complex conjugate of a sum, each term in the sum must also be complex conjugated. In this case, the complex conjugate of i\hat{B} is -i\hat{B}, and when added to the complex conjugate of \hat{A}, it becomes ∫ψj * \hat{A} ψi - i(∫ψj * \hat{B} ψi). This is necessary to maintain the equality in the equation, as the complex conjugate is being taken on both sides.

In other words, when taking the complex conjugate of a sum, the order of the terms does not change, but each individual term is complex conjugated. This is why the operator becomes ∫ψj * \hat{A} ψi - i(∫ψj * \hat{B} ψi) and not ∫ψj * \hat{A} ψi + i(∫ψj * \hat{B} ψi).

Overall, this shows that the linear combination \hat{A} + i\hat{B} is not Hermitian because its complex conjugate is not equal to itself. In other words, the operator does not satisfy the Hermitian property, which states that the complex conjugate of an operator is equal to the operator itself. This is an important concept in quantum mechanics, as Hermitian operators represent physical observables in the system. Therefore, it is crucial to understand the mathematical reasons behind why certain operators are not Hermitian, as it affects the physical interpretations of the system.

1. What is a linear combination?

A linear combination is a mathematical operation that combines two or more vectors by multiplying each vector by a scalar and then adding the results. This results in a new vector that lies in the same vector space as the original vectors.

2. What does it mean for a linear combination to be Hermitian?

A linear combination is considered Hermitian if it satisfies the properties of a Hermitian matrix. This means that the linear combination must be both self-adjoint (equal to its own conjugate transpose) and real symmetric (equal to its own transpose).

3. How can I show that a linear combination is not Hermitian?

To show that a linear combination is not Hermitian, you can use the definition of a Hermitian matrix and check if the linear combination satisfies both the properties of self-adjointness and real symmetry. If the linear combination does not satisfy both properties, then it is not Hermitian.

4. What are the implications of a linear combination not being Hermitian?

If a linear combination is not Hermitian, it means that it does not have certain important properties that are required for many mathematical and physical applications. This can limit its usefulness and may require alternative methods or approaches to be used instead.

5. Can a linear combination be both Hermitian and non-Hermitian?

No, a linear combination cannot be both Hermitian and non-Hermitian at the same time. It either satisfies the properties of a Hermitian matrix or it does not. However, it is possible for a linear combination to be Hermitian with respect to one set of vectors and non-Hermitian with respect to another set of vectors.

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