Squirtle said:
Homework Statement
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
Homework Equations
Linear operator if:
i) kT(f) = T(kf)
ii) T(f+k) = T(f) + T(k)
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)
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.
