# Linear Operators

1. Sep 3, 2012

### Squirtle

1. The problem statement, all variables and given/known data

Consider the following operators acting in the linear space of functions Ψ(x) defined on
the interval (∞,∞)
(a) Shift Ta: TaΨ(x)=Ψ(x+a), a is a constant
(b) Reflection (inversion) I: IΨ(x)=Ψ(x)
(c) Scaling Mc: McΨ(x)= √c Ψ(cx), c is a constant
(d) Complex conjugation K: KΨ(x)=Ψ∗(x)

Are these operators linear? Find their adjoint operators. Find their inverse operators

2. Relevant equations

Linear operator if:
i) kT(f) = T(kf)
ii) T(f+k) = T(f) + T(k)

3. The attempt at a solution
I don't understand how to apply the linear operator conditions to these problems. Could someone explain to me a) or an example? I don't see how I can claim or prove (if right) TaΨ(x+b)=Ψ(x+a+b)
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 4, 2012

### vela

Staff Emeritus
You're using k in two different ways in your definition of linear. It would be better if you used distinct symbols:
\begin{align*}
cT(f) &= T(cf) \\
T(f+g) &= T(f)+T(g)
\end{align*} where c is a (scalar) constant and f and g are elements of the vector space.

In these problems, the operators act on the functions in the given space, so you have f(x)=ψ(x) and g(x)=φ(x). You want to show that
\begin{align*}
T[c\psi] &= cT[\psi] \\
T[\psi+\phi] &= T[\psi]+T[\phi]
\end{align*} For part (a), for example, you need to show, in part, that $T_a[c\psi]$ and $cT_a[\psi]$ are equal. $T_a[c\psi]$ means you take the function $c\psi$ and translate it by a. That is, $T_a[c\psi](x) = (c\psi)(x+a) = c\psi(x+a)$. The last equality relies on the definition of what it means to multiply the function $\psi$ by c. $cT_a[\psi]$ means you translate the function $\psi$ by a and then multiply it by c. That is, $cT_a[\psi](x) = c[\psi(x+a)] = c\psi(a+x)$. Since $T_a[c\psi]$ and $cT_a[\psi]$ both map x to $c\psi(a+x)$ for all x, you can conclude they are equal.

Intuitively, in one case, you scale the function by c and then translate the result; in the other, you translate it first and then scale it. You should be able to convince yourself that the order doesn't matter here, so the two sides of the equation should be equal.

3. Sep 4, 2012

### Squirtle

Thank you I understand that part now.

Now for the other parts :(

4. Sep 4, 2012

### vela

Staff Emeritus
You also need to do the second half of part (a).

5. Sep 4, 2012

### Fredrik

Staff Emeritus
Careful, you don't want to sound like you want to someone else to do all the work for you.

The parts are all very simillar. For example, for part b), you need to show that for all real numbers a,b, and all $f,g$ in the vector space, we have $I(af+bg)=aIf+bIg$. This isn't hard. Note that this equality says the same thing as the statement "For all real numbers x, $I(af+bg)(x)=(aIf+bIg)(x)$". Can you prove that?

By the way, are you sure you got the definition of $I$ right? I would call this operator the identity operator, not reflection or inversion. Is there a minus sign missing?

6. Sep 4, 2012

### Squirtle

So sorry, I meant i understood how to finish the rest for proving if they were linear. I worked the rest and seemed to get equivalent expressions for them all, I assume that meant I did it right.

I am confused on the adjoint and inverse part of the question.

The reflection question should have a minus sign.

Thanks for still helping me though. I much appreciate you guys.

The inverse opeartor is UU* = I(?)

7. Sep 4, 2012

### vela

Staff Emeritus
For (d), it depends on the allowed values of c. Is c a complex quantity or restricted to only real values?

8. Sep 4, 2012

### Squirtle

I believe its complex value as well.

I was looking at that one and wasn't sure if I did this right.

K(cY)(x) = (cY)(x*) = cY(x*)

Would that be the way it reduces down?

Or would it be:

cY*(x) ?

Why would c effect it? I did read that this one isn't always linear, but didn't want to trust a yahoo question answer, but you brought up the same point.

9. Sep 5, 2012

### vela

Staff Emeritus
Remember the operator K acts on the function to give you a new function, so it takes the function $(c\psi)$ to $(c\psi)^*$. It doesn't act on the argument of the function, x.

10. Sep 5, 2012

### Squirtle

Awesome ty, I was able to get it now :)
I see the problem with c as well now.

I managed to figure out the adjoint/inverse stuff so thank you all very much for the help. I've very slow at this stuff.