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

 

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