📂Quantum Mechanics

Eigenfunctions of the Momentum Operator

Theorem¹

The eigenvalues of the momentum operator $\hat{p}$ are all real numbers, and the eigenfunction $f_{p}$ corresponding to an eigenvalue $p \in \mathbb{R}$ is as follows.

$$ f_{p}(x) = \frac{1}{\sqrt{2\pi\hbar}} e^{\i p x / \hbar} $$

Explanation

When we compute the eigenfunctions of the momentum operator, they turn out to have the form of complex exponential functions, that is, plane waves. As with the eigenfunctions of the position operator, this is because every continuous value across the entire real line is possible as an eigenvalue. The case where eigenvalues exist as continuous values in this way is called a continuous spectrum. In the case of a continuous spectrum, the eigenfunctions do not belong to the Hilbert space, so the orthonormality condition $\braket{f_{p^{\prime}} | f_{p}} = \delta_{p^{\prime} p}$ cannot be applied. The eigenfunctions of a continuous spectrum must satisfy the Dirac normalization as follows.

$$ \braket{f_{p^{\prime}} | f_{p}} = \delta (p - p^{\prime}) $$

The completeness of the 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.

$$ f(x) = \int_{-\infty}^{\infty} c(p) f_{p}(x) \d p $$

Here, the expansion coefficient $c(p) = \braket{f_{p} | f}$ is precisely the Fourier transform of $f$, and it is called the wave function in momentum space. The reason only real eigenvalues are physically allowed is that if the eigenvalue $p$ were complex, the eigenfunction $e^{\i p x / \hbar}$ would diverge as $x \to \pm\infty$, making normalization impossible.

Proof

Let $\hat{p}$ be the momentum operator. Let $p$ be an eigenvalue of the operator and $f_{p}$ the eigenfunction corresponding to it.

$$ \hat{p} f_{p} = p f_{p} $$

Since the momentum operator is $\hat{p} = -\i\hbar \dfrac{\d}{\d x}$, the eigenvalue equation becomes the following differential equation.

$$ -\i\hbar \frac{\d}{\d x} f_{p}(x) = p f_{p}(x) \implies f_{p}^{\prime}(x) = \frac{\i p}{\hbar} f_{p}(x) $$

Since this is a first-order linear differential equation, its solution, with respect to a normalization constant $A$, is as follows.

$$ f_{p}(x) = A e^{\i p x / \hbar} $$

If the eigenvalue $p$ were not real but a complex number $p = a + b\i$ ($b \ne 0$), the eigenfunction would be as follows.

$$ f_{p}(x) = A e^{\i a x / \hbar} e^{- b x / \hbar} $$

In this case, due to the factor $e^{-bx/\hbar}$ multiplied at the back, the function diverges as $x \to +\infty$ or $x \to -\infty$, so normalization is impossible. Therefore, the only physically allowed eigenvalues are the real ones.

Now let us determine the normalization constant $A$. As with the eigenfunctions of the position operator, $f_{p}$ is not square-integrable, so ordinary normalization does not work; hence we apply Dirac normalization.

$$ \braket{f_{p^{\prime}} | f_{p}} = \int_{-\infty}^{\infty} f_{p^{\prime}}^{\ast}(x) f_{p}(x) \d x = |A|^{2} \int_{-\infty}^{\infty} e^{\i (p - p^{\prime}) x / \hbar} \d x $$

Here, this integral does not converge in the ordinary sense, but using somewhat difficult mathematics, we can show that the integral representation of the Dirac delta function is $\int_{-\infty}^{\infty} e^{\i (p - p^{\prime}) x / \hbar} \d x = 2\pi\hbar \delta(p - p^{\prime})$. Using this, we obtain the following.

$$ \braket{f_{p^{\prime}} | f_{p}} = 2\pi\hbar |A|^{2} \delta(p - p^{\prime}) $$

For the Dirac normalization condition $\braket{f_{p^{\prime}} | f_{p}} = \delta(p - p^{\prime})$ to hold, we must have $2\pi\hbar |A|^{2} = 1$, that is, $A = 1 / \sqrt{2\pi\hbar}$. Therefore, the normalized eigenfunction of the momentum operator is as follows.

$$ f_{p}(x) = \frac{1}{\sqrt{2\pi\hbar}} e^{\i p x / \hbar} $$

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