Can Any Linear Operator Be Decomposed into Hermitian Components?

[dingo_d]

Homework Statement

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

Homework Equations

Operator is Hermitian if:

$$T=T^{\dagger}$$

The Attempt at a Solution

I don't know where to start :\ Should I try to see for some arbitrary vector $$|\psi\rangle$$, that I can write it in some basis, and see what it would do to write eigenvalue equation with those operators?

 

[arkajad]

Play with $$\hat{O}^\dagger+\hat{O}$$ and $$i(\hat{O}^\dagger-\hat{O})$$.

 

[dingo_d]

Hmmm.

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

$$\hat{O}^{\dagger}+\hat{O}=\hat{O}'-i\hat{O}''+\hat{O}'+i\hat{O}''=2\hat{O}'$$

Am I on the right track?

 

[arkajad]

Not too bad ...

 

[dingo_d]

XD Too bad? XD

For the $$i(\hat{O}^\dagger-\hat{O})$$ part I got $$2\hat{O}''$$

So I should somehow prove linear superposition or?

 

[arkajad]

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.