Question

How to prove if this function is linear, \[\hat{K}[a\psi] = a*\psi*\] ?

Answer 1

Are you sure this is all the question? I think something is missing!

Answer 2

oops, the * are complex conjugate and the \(\hat{K}\) is the complex conjugation operator and \(\psi = \psi(x)\) is defined for all x in a Hilbert space.

Answer 3

Can you tell me what \(a\) is? a real constant?!

Answer 4

It's not specified in the question.

Answer 5

I am not so sure of my answer, but here what I have. If \(\hat K\) is the complex conjugation operator then \(\hat K(v)=v^*\), where \(v^*\) is the complex conjugate of \(v\). Now let's check the two properties of a linear transformation.
1) \(\hat K(cv)=c\hat K(v)\), where c is a real number.
2) \(\hat K(u+v)=\hat K(u)+\hat K(v)\), where u and v are two vector spaces.
I think the first property is very clear since the conjugate of a real number is itself. So \(\hat K(cv)=c^*v^*=cv*=c\hat K(v)\).

Answer 6

As for the second property. Let \(u=a+bi\) and \(v=c+di\), where \(a,b,c\) and \(d\) are all real numbers. Let's see if property (2) is applicable for the function \(\hat K\).
L.H.S=\(\hat K(u+v)=\hat K((a+c)+(b+d)i)=(a+b)-(b+d)i\)
R.H.S=\(\hat K(u)+\hat K(v)= \hat K(a+bi)+ \hat K(c+di)=a-bi+c-di=(a+b)-(b+d)i\).
You can see that they are equal, and hence the function is linear!

Answer 7

Thanks:) Thats really helpful.

Answer 8

I think I prematurely said that you were right, if \(u = a+ ib \) and one of the properties of a linear operator is that \(f(ux) = uf(x)\) then
\[\hat{K}[u\psi(x)] = u^*\psi^*(x)\]
\[\hat{K}[u\psi(x)] = (a-ib)\psi^*(x) \neq u\hat{K}[\psi(x)]\]