Hermitian Matrix and Commutation relations

In summary, the conversation discusses the property of Hermitian matrices and how it applies to the matrix C (AB - BA). It is shown that C must be multiplied by a complex number in order for it to be Hermitian. The conversation ends with a suggestion to take the Hermitian conjugate of both sides to prove that C is Hermitian.
  • #1
dykuma
56
7
Homework Statement
Consider two hermitian matrices who do not commute. Show that the result of the commutation equals a hermitian matrix times the complex number I.
Relevant Equations
[A,B] = AB- BA = iC
I think I roughly see what's happening here.

> First, I will assume that AB - BA = C, without the complex number.

>Matrix AB equals the transpose of BA. (AB = (BA)t)

>Because AB = (BA)t, or because of the cyclic property of matrix multiplication, the diagonals of AB equals the diagonals of BA. Thus, when you subtract the two matrices, the diagonals are zero.

>After subtracting AB- BA, because AB = (BA)t, the off diagonal elements (AB)i,j will equal the off diagonal elements (AB-BA)j,i in magnitude, but will have a sign difference. (so element (AB-BA)1,2 = -(AB-BA)2,1 )

>Further, (AB-BA) equals the negative of it's transpose. So (AB-BA)= -(AB-BA)t

>Thus, for matrix C (AB - BA) to be hermitian, all that's required is it would need to be multiplied by the complex number -I (so that way the complex conjugate transpose equals the original matrix). As such, for C to be Hermitian, C = -I(AB-BA)

>However, there is a rule stating that "the product of two hermitian matrices A and B is also hermitian if and only if AB = BA". But I originally assumed that AB- BA = C, meaning that C can not be Hermitian. As such, AB- BA must equal C times a complex number if matrix C is to be hermitian. Doing so ensures the final result of AB - BA is not hermitian, and that C alone is hermitian.What I'm having trouble with is showing this. I'm not great at doing proofs in linear algebra, and I don't how to begin to show this. I feel like all I've shown up to this point is that AB-BA = I(-I(AB-BA)).
 
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  • #2
You want to show that the matrix ##C=\frac{1}{i}(AB-BA)## is Hermitian, meaning that ##C^T=\overline{C}.## Can you take the transpose of ##\frac{1}{i}(AB-BA),## using the fact that ##A,B## are Hermitian?
 
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  • #3
Define ##iC = AB-BA## and take the Hermitian conjugate of both sides. Then reinsert the definition to get rid of the A and B. Watch magic happen.
 
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Related to Hermitian Matrix and Commutation relations

1. What is a Hermitian matrix?

A Hermitian matrix is a square matrix that is equal to its own conjugate transpose. This means that the elements on the main diagonal are real numbers, and the elements above and below the diagonal are complex conjugates of each other.

2. What is the significance of a Hermitian matrix?

Hermitian matrices have several important properties, including being diagonalizable and having real eigenvalues. They are also closely related to the concept of self-adjoint operators in linear algebra and have various applications in quantum mechanics and signal processing.

3. How do you determine if a matrix is Hermitian?

To determine if a matrix is Hermitian, you can check if it is equal to its own conjugate transpose. This can be done by taking the complex conjugate of each element and then transposing the matrix. If the resulting matrix is equal to the original matrix, then it is Hermitian.

4. What are commutation relations?

Commutation relations refer to the order in which two operators are applied to a given system. In the context of Hermitian matrices, it refers to the order in which two matrices are multiplied. The commutator of two matrices is defined as the difference between the product of the two matrices in different orders.

5. How are commutation relations related to Hermitian matrices?

In quantum mechanics, the commutation relation between two observables is related to the uncertainty principle. In the context of Hermitian matrices, commutation relations are used to determine if two matrices commute (i.e. if their commutator is equal to zero). This is an important property when dealing with quantum systems and can provide insights into the behavior of the system.

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