Decomposition of linear operator

In summary, any linear operator \hat{O} can be decomposed into a sum of two Hermitian operators, \hat{O}' and \hat{O}'', where \hat{O}=\hat{O}'+i\hat{O}''. The process of finding these operators involves playing and guessing, and then carefully verifying that the guessed operators have all the required properties.
  • #1
dingo_d
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0

Homework Statement



Show that any linear operator [tex]\hat{O}[/tex] can be decomposed as [tex]\hat{O}=\hat{O}'+i\hat{O}''[/tex], where [tex]\hat{O}'[/tex] and [tex]\hat{O}''[/tex] are Hermitian operators.

Homework Equations



Operator is Hermitian if:

[tex]T=T^{\dagger}[/tex]

The Attempt at a Solution



I don't know where to start :\ Should I try to see for some arbitrary vector [tex]|\psi\rangle[/tex], that I can write it in some basis, and see what it would do to write eigenvalue equation with those operators?
 
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  • #2
Play with [tex]\hat{O}^\dagger+\hat{O}[/tex] and [tex]i(\hat{O}^\dagger-\hat{O})[/tex].
 
  • #3
Hmmm.

So [tex]\hat{O}^{\dagger}+\hat{O}=\hat{O}'^{\dagger}-i\hat{O}''^{\dagger}+\hat{O}'+i\hat{O}''[/tex] and since those with ' and '' are hermitian it follows:

[tex]\hat{O}^{\dagger}+\hat{O}=\hat{O}'-i\hat{O}''+\hat{O}'+i\hat{O}''=2\hat{O}'[/tex]

Am I on the right track?
 
  • #4
Not too bad ...
 
  • #5
XD Too bad? XD

For the [tex]i(\hat{O}^\dagger-\hat{O})[/tex] part I got [tex]2\hat{O}''[/tex]

So I should somehow prove linear superposition or?
 
  • #6
Well, so what is your guess for O' and O''? Guess what they should be, then check carefully if your guess has all the required properties.

This is how we often solve problems: we play, we guess, and the we check carefully, in all detail, if our guess really constitutes a solution of the original problem. The way towards a solution does not have to be logical. What needs to be logical is the verification.
 
Last edited:

Related to Decomposition of linear operator

What is the decomposition of a linear operator?

The decomposition of a linear operator is the process of breaking down a linear transformation into simpler components. This is done in order to better understand and analyze the behavior of the operator.

What are the different types of decomposition for linear operators?

There are several types of decomposition for linear operators, including eigenvalue decomposition, singular value decomposition, and spectral decomposition. Each type has its own unique properties and applications.

What is the purpose of decomposing a linear operator?

The main purpose of decomposing a linear operator is to simplify its representation and make it easier to analyze. This can help in solving complex problems and understanding the behavior of the operator in different situations.

How is the decomposition of a linear operator related to matrix diagonalization?

Matrix diagonalization is a form of decomposition for square matrices that involves transforming the matrix into a diagonal form. This is often used in the decomposition of linear operators, as it helps in simplifying the operator and understanding its properties.

What are some real-world applications of decomposition of linear operators?

Decomposition of linear operators has numerous applications in various fields, including signal processing, image processing, quantum mechanics, and data analysis. It is used to solve complex problems and understand the behavior of systems in these fields.

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