Translation operator

(personal thinking)

In a Hilbert space like $\mathbb C^5$ we have the position operator

$$ \hat{X} = \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & 3 & 0 & 0 \\ 0 & 0 & 0 & 4 & 0 \\ 0 & 0 & 0 & 0 & 5 \\ \end{bmatrix} $$

and we can consider the matrix which is something like the translation operator:

$$ T=\left(\begin{array}{lllll} 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right) $$

Given a vector $v=(a,b,c,d,e)$, the result of the operator is "translate" the coordinates:

$$ Tv=(e,a,b,c,d). $$

We can consider the matrix $P$ such that the matrix exponential $e^P=T$, son $P$ can be seen like the generator of translations. I think that $P$ is something like the momentum operator in this setup. This operator has a special basis of eigenvectors, representing the pure momentum states.

Given a vector $v$ in the initial basis, the coordinates in this new basis is a kind of Laplace transform.

In the continuous case everything is analogous: the vectors are functions $f$ and the Hilbert space is a space of functions. The translation operator is

$$ T: f(x) \mapsto f(x-1) $$

According to Taylor's theorem,

$$ T(f)(x)=e^{-\partial_x} (f) $$

son we have the momentum operator $P=e^{-\partial_x}$. The eigenvectos of this operator are

$$ e^{sx}, s\in \mathbb C $$

So Laplace transform is nothing but a change of basis. But what I don't understand yet why we choose only some eigenvectors in Quantum Mechanics, i.e., only

$$ e^{i\omega x}, \omega \in \mathbb R $$

i.e., why we take Fourier transform.

Why Only Certain Eigenvectors?:

In quantum mechanics, the states of a system are described by wavefunctions. The wavefunctions that are physically meaningful are those that are square-integrable (bound states), meaning their absolute square integrates to 1 over all space. This ensures that the total probability of finding a particle somewhere in space is 1.

The functions $e^{sx}$ are not square-integrable for all $s \in \mathbb{C}$. However, the functions $e^{i\omega x}$ are orthogonal over all space, and they form a basis for the space of square-integrable functions. This is why the Fourier transform, which uses these functions, is so useful in quantum mechanics.

________________________________________

Author of the notes: Antonio J. Pan-Collantes

antonio.pan@uca.es

INDEX: