Wave function

The square of the wave function $\Psi(x,t)$ can be interpreted as the probability density of finding a particle at a certain position and time. We can say that the probability of finding the particle between $a$ and $b$ is $\int_a^b |\Psi(x,t)|^2 \d x$ .

The probability must sum to one, so $\int_{-\infty}^\infty |\Psi(x,t)|^2 \d x = 1$ .

Since the Schrödinger equation is linear, if $\Psi$ is a solution, $A\Psi$ is too. We can choose this normalization factor such that the square of $\Psi$ makes sense as a probability density.

Note

We can differentiate to find $\dd{t}{} \int_{-\infty}^\infty |\Psi(x,t)|^2 \d x = 0$ , so a wave function will stay normalized.

Momentum space

$\Psi$ is the wave function in position space, but we can just as well describe the state of the system in other spaces. Commonly we use the momentum-space wave function $\phi(p)$ .

We can find the frequency components of the time-independent wave function $\psi$ by taking the Fourier transform, to find $\tilde\phi(k)$ . Then by the de Broglie relation we can say $p=\hbar k$ , and so $\phi(p) = \hbar \tilde\phi(k)$ . This gives us

\begin{align*} \tilde\phi(k) &= \frac{1}{\sqrt{2\pi}} \int_{-\infty}^\infty \psi(x) e^{-ikx} \d x \\ \phi(p) &= \frac{1}{\sqrt{2\pi \hbar}} \int_{-\infty}^\infty \psi(x) e^{ipx/\hbar} \d x. \end{align*}

In terms of $\phi(p)$ , we can write $\psi$ as the inverse Fourier transform

\psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^\infty \phi(p) e^{-ipx/\hbar} \d p.

In terms of $\tilde\phi(k)$ it’s just the usual inverse Fourier transform.