Eigenfunctions of the Position Operator 📂Quantum Mechanics

Eigenfunctions of the Position Operator

Theorem¹

The eigenvalues of the position operator $\hat{x}$ are all real numbers, and the eigenfunction $\phi_{y}$ corresponding to the eigenvalue $y \in \mathbb{R}$ is the Dirac delta function.

$$ \phi_{y}(x) = \delta (x - y) $$

Explanation

Computing the eigenfunctions of the position operator yields the peculiar result of the Dirac delta function. This is because every continuous value over the entire real line is possible as an eigenvalue. The case where eigenvalues exist as continuous values in this way is called the continuous spectrum. In the case of the continuous spectrum, the eigenfunctions do not belong to the Hilbert space, so the orthonormality condition $\braket{\phi_{y^{\prime}} | \phi_{y}} = \delta_{y^{\prime} y}$ cannot be applied. The eigenfunctions of the continuous spectrum must satisfy the following Dirac normalization.

$$ \braket{\phi_{y^{\prime}} | \phi_{y}} = \delta (y - y^{\prime}) $$

The completeness of eigenfunctions also holds in the form of an integral rather than a series, so an arbitrary wave function $f$ is expanded as an integral of eigenfunctions as follows.

$$ f(x) = \int_{-\infty}^{\infty} c(y) \phi_{y}(x) \d y = \int_{-\infty}^{\infty} f(y) \delta (x - y) \d y $$

Here, the expansion coefficient is $c(y) = \braket{\phi_{y} | f} = f(y)$, that is, the value of the wave function itself. The Dirac delta function is a function whose value is $0$ everywhere except at a single point. Since a single particle cannot exist in several places at the same time, its position must be exactly one point, and therefore it can be seen as natural that it is represented by the Dirac delta function.

Proof

Let $\hat{x}$ be the position operator. Let $y$ be an eigenvalue and $\phi_{y}$ the eigenfunction corresponding to it.

$$ \hat{x} \phi_{y} = y \phi_{y} $$

Since the position operator $\hat{x}$ is the same as multiplying by $x$, we obtain the following.

$$ x \phi_{y} = y \phi_{y} \implies (x-y)\phi_{y}(x) = 0 $$

If $x \ne y$, then $(x-y) \ne 0$, so $\phi_{y}(x) = 0$ must hold. That is, $\phi_{y}$ must have the value $0$ everywhere with $x \ne y$. At $x = y$ any value is fine, so let us assume the following.

$$ \phi_{y}(x) = \begin{cases} a, & x = y \\ 0, & x \ne y \end{cases} $$

For this function to be an eigenfunction, that is, a wave function, its square integral must be greater than $0$. Only then is normalization possible and a probabilistic interpretation feasible. However, the integral of the above function is $0$. That is, from the standpoint of integration, $\phi_{y}$ is in effect the zero function.

$$ \int_{-\infty}^{\infty} |\phi_{y}(x)|^{2} \d x = 0 $$

Then we need to introduce a function whose value is $0$ everywhere except at a single point yet whose integral is not $0$, and that is precisely the Dirac delta function. If we regard $\phi_{y}$ as the Dirac delta function below, all of the above conditions are satisfied.

$$ \phi_{y}(x) = \delta(x - y) $$

In this case, as explained above, the normalization condition is as follows.