Eigenfunctions of translation operator and transposed operator property proof

[Muchacho]

Homework Statement

Find the eigenfunctions and eigenvalues of the translation operator $$\widehat{T_{a}}$$
Translation operator is defined as $$\widehat{T_{a}}\psi(x)=\psi(x+a)$$ (you all know that, probably you just call it differently)

Homework Equations

The eigenvalue/eigenfunction equation is given like
$$\widehat{T_{a}}\psi_{n}(x)=f_{n}\psi_{n}(x)$$

The Attempt at a Solution

I write the eigenvalues $$f_{n}$$ are in the form of $$c_{n}(a)$$ (c noting that it is a complex coefficient)
And I don't know how to proove it correctly and clearly, but I get that only possible eigenfunctions are exponentials $$\psi_{n}(x)=e\ ^{x+2n\pi\i}$$ and the eigenvalues for $$\widehat{T_{a}}$$ are $$c_{n}(a)=e\ ^{a+2n\pi\i}$$

I don't have any ideas how to proove it more clearly, because this "solution" involves more thinking and assuming than solving.

I also did similar solution for inversion operator $$\widehat{I_{x}}\psi(x)=\psi(-x)$$ by finding eigenvalues just looking at the properties of an equation which was like $$\psi(-x)=f\psi(x)$$ and using the property of odd and even functions and thus finding 2 eigenvalues of $$f_{1}=1$$ and $$f_{2}=-1$$ and thus getting infinite number of eigenfuntions. The real solution to this problem was using the property that $$\widehat{I_{x}}^2=\widehat{1}$$ and modify the equation $$\widehat{I_{x}}\psi(x)=f\psi(x)$$ by multiplying both sides from the left with $$\widehat{I_{x}}$$ and thus getting $$\psi(x)=f^{2}\psi(x)$$ and just needed to solve the quadratic equation of $$f^{2}=1$$ thus getting the same values of $$f_{1}=1$$ and $$f_{2}=-1$$ This was just an example of possible ways to solve eigenvalue equations, but in this case - the real solution shows that no other solutions are possible.


And also I have this second problem which is more like proof of a formula which is usually given as a property.

Homework Statement

Proove that $$(AB)^{T}=B^{T}A^{T}$$

Homework Equations

The transposed operator is given in bra-ket notation as $$\left\langle\varphi\left|\widehat{A}^{T}\right|\psi\right\rangle=\left\langle\psi^{*}\left|\widehat{A}\right|\varphi^{*}\right\rangle$$ or in integral form as $$\int\varphi(x)\widehat{A}^{T}\psi(x)dx=\int\psi(x)\widehat{A}\varphi(x)dx$$

The Attempt at a Solution

Well, I have completely no ideas on where to start with this one.
Because I started with it like $$\left\langle\varphi\left|\widehat{AB}\right|\psi\right\rangle=\left\langle\varphi\left|\widehat{A}\right|\widehat{B}\psi\right\rangle=\left\langle\psi^{*}\left|\widehat{A}\right|\widehat{B}^{T}\varphi^{*}\right\rangle$$ but I really doubt that the last operation is correct and I'm allowed to do so.

I hope I made my doubts and problems clear and thanks for help in advance!

 

[tiny-tim]

Welcome to PF!

Hi Muchacho! Welcome to PF!

(have a psi: ψ )

Where did you get n from?

Remember, this works for any a.

Hint: take logs of the equation.