Wavefunctions in Momentum Space
- Quantum Mechanics 1.11
from 「Modern Quantum Mechanics」: Sakurai, J. J.
The position and momentum operators.
Momentum space
We’ve seen that the wavefunction in position space is given by as follows, where $\ket{\alpha}$ is an arbitrary state vector.
\[ \hat{\b{x}}\ket{\b{x}\rq} = \b{x}\rq\ket{\b{x}\rq} \] \[ \psi_\alpha(\b{x}) = \brkt{\b{x}}{\alpha} \]
Similarly, we can describe the wavefunction in momentum space, too.
\[ \hat{\b{p}}\ket{\b{p}\rq} = \b{p}\rq\ket{\b{p}\rq} \] \[ \phi_\alpha(\b{p}) = \brkt{\b{p}}{\alpha} \]
Let’s discuss eigenstate and eigenvalue of the momentum operator at 1-dimensional position space. We cet get such differential equation:
\[ p\brkt{x}{p} = -i\hbar\pdv{ }{x} \brkt{x}{p} \]
The solution is,
\[ \brkt{x}{p} = A \exp\left( \frac{ipx}{\hbar} \right) \]
To get the normalization constant $A$, let’s consider following inner product:
\[ \brkt{x_1}{x_2} = \int \d{p} \brkt{x_1}{p}\brkt{p}{x_2} \]
LHS is actually $\delta(x_1-x_2)$, and RHS can be evaluated using the solution above.
\[ \begin{align*} \delta(x_1-x_2) &= \abs{A}^2 \int \d{p} \exp\left[ \frac{ip(x_1-x_2)}{\hbar} \right] &= 2\pi\hbar\abs{A}^2 \delta(x_1-x_2) \end{align*} \]
Choosing $A$ to be real and positive by convention, we have
\[ \brkt{x}{p} = \frac{1}{\sqrt{2\pi\hbar}} \exp\left( \frac{ipx}{\hbar} \right) \]
We can now demonstrate how the position and momentum wavefunctions are related to each other. It’d be done just by substituting the momentum eigenstate into the following equations.
\[ \begin{gather*} \brkt{x}{\alpha} = \int \d{p} \brkt{x}{p}\brkt{p}{\alpha} \nl \brkt{p}{\alpha} = \int \d{x} \brkt{p}{x}\brkt{x}{\alpha} \end{gather*} \]
\[ \begin{gather*} \psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int \d{p} \exp\left( \frac{ipx}{\hbar} \right) \phi(p) \nl \phi(p) = \frac{1}{\sqrt{2\pi\hbar}} \int \d{x} \exp\left( -\frac{ipx}{\hbar} \right) \psi(x) \end{gather*} \]
The pair of equations are exactly the Fourier transform and the inverse Fourier transform.
Generalization to Three dimensions
Similarly, we get the 3-dimensional analogue of the equations above.
\[ \begin{gather*} \psi(\b{x}) = \frac{1}{\sqrt{2\pi\hbar}^3} \int \d^3{\b{p}} \exp\left( \frac{i\b{p}\cdot\b{x}}{\hbar} \right) \phi(\b{p}) \nl \phi(\b{p}) = \frac{1}{\sqrt{2\pi\hbar}^3} \int \d^3{\b{x}} \exp\left( -\frac{i\b{p}\cdot\b{x}}{\hbar} \right) \psi(\b{x}) \end{gather*} \]
