# Translating a Wave Function

In algebra, or pre-calc, you learn that you can change the position of a function by modifying the argument. In quantum physics this idea is used to displace wave functions. If a function starts off at one position, and moves to another position, all that is needed is a change in argument. However, quantum physics likes to use linear operators to alter functions. What would an operator look like if it can change the argument of a function? In this post, I will construct a 1-D example of such an operator.

A general wave function can be written as: $\Psi (x)$, where the shape of $\Psi$ is dependent on the spatial variable $x$.
To translate a function by distance $a$, modify the argument of $\Psi$,
To move the function right by $a$, $\Psi(x) \rightarrow \Psi (x-a)$
To move the function left by $a$, $\Psi(x) \rightarrow \Psi(x+a)$
Let’s just take the $\Psi(x+a)$ example, and without loss of generality say that $a$ can be positive or negative.
Next we can take advantage of Taylor expansions.
A function $f(x)$ can be expanded around a point $a$: $f(x) = f(a) + (x-a)\frac{d}{dx}f(a) + \frac{(x-a)^2}{2!} \frac{d^2}{dx^2}f(a) + ...$
Here, in our example, we want to expand $\Psi(x+a)$ around $x$, to express the translated function $\Psi(x+a)$ in terms of the original function $\Psi(x)$:
$\Psi(x+a) = \Psi(x) + a\frac{d}{dx}\Psi(x) + \frac{a^2}{2!} \frac{d^2}{dx^2} \Psi(x) + ...$
Note: $\frac{d}{dx} = \frac{i}{\hbar} p$.
A more complete version of this expression would be:
$\Psi(x+a) = \sum\limits_{n=0}^{\infty} \frac{a^n}{n!} \frac{d^n}{dx^n} \Psi(x) = \Psi(x) + \sum\limits_{n=1}^{\infty} \frac{a^n}{n!} \frac{d^n}{dx^n} \Psi(x)$
This sum’s structure is similar to the Taylor expansion of the exponential function.
$e^x = 1 + x + x^2/2! + ... = \sum\limits_{n=1}^{\infty} \frac{x^n}{n!}$
Every operator in the $\Psi(x+a)$ expansion can be reduced into an simplified operator: $\Psi(x+a) = e^{a\frac{d}{dx}} \Psi(x)$
$= e^{\frac{ai}{\hbar} p} \Psi(x).$
This new operator $e^{a\frac{d}{dx}}$ can be expanded to return to what we had before: $e^{a\frac{d}{dx}} = \sum\limits_{n=0}^{\infty} \frac{a^n}{n!} \frac{d^n}{dx^n}$.
So the translated function is a version of the original function with a specific type of interference. Such that the structure is: $Translated = Original + Interference$.
$\Psi(x+a) = \Psi(x) + \sum\limits_{n=1}^{\infty} \frac{a^n}{n!} \frac{d^n}{dx^n} \Psi(x) = \Psi(x) + \Delta(a)\Psi(x)$
We can reduce the operators in the interference terms into an exponential in the usual way:
$\Delta(a) = \sum\limits_{n=1}^{\infty} \frac{a^n}{n!} \frac{d^n}{dx^n}$
$= e^{a \frac{d}{dx} } - 1$

The expectation value $\left< e^{a\frac{d}{dx}} \right>$ characterizes the average overlap between $\Psi(x)$ and $\Psi(x+a)$.
$\left< e^{a\frac{d}{dx}} \right> = \left< \Psi(x) \right| e^{a\frac{d}{dx}} \left| \Psi(x) \right>$
$= \left< \Psi(x) | \Psi(x+a) \right>$
$= \left< \Psi(x) \right| 1 + \Delta(a) \left| \Psi(x) \right>$
$= 1 + \left< \Delta(a) \right>$
The expectation value of the translation operator is then unity plus the expectation value of the interference operator.
$\left< \Delta (a) \right> = \sum\limits_{n=1}^{\infty} \frac{a^n}{n!} \left< \frac{d^n}{dx^n} \right>$
$= \sum\limits_{n=1}^{\infty} \frac{(ia)^n}{(\hbar)^n n!} \left< p^n \right>$
The interference expectation value is shown to be an expectation value of a function of the momentum operator.
In the case of localized waves, as $a$ gets much greater than the width of $\Psi(x)$, the interference term approaches -1, since the overlap between the original and translated wave function decreases. This is equivalent to the original and displaced wave function becoming more and more orthogonal.
limit $_{a\rightarrow \infty} \left< \Delta (a) \right> = -1$
limit $_{a\rightarrow \infty} \left< \Psi(x) | \Psi(x+a) \right> = 0$