Probabilistic interpretation and normalization of the wave function in quantum mechanics
Wave Function
The Wave Function is a function in quantum mechanics that represents the state of motion of a particle with respect to time and position. In a sushi restaurant, the wave function with respect to position and time is denoted as $\psi (x,t)$, and the wave function with respect to position and independent of time is denoted as $u(x)$.
Probabilistic Interpretation
The method of understanding the state of a particle through the wave function is based on the statistical (probabilistic) interpretation of Max Born. Here, the integral of the square of the magnitude of the wave function over a certain interval is given the meaning of the probability of finding the particle in that interval.
$$ \int _{a} ^b |\psi (x,t)|^2dx \\[1em] = \text{The probability that a particle exists in the interval } [a,b] \text{ at time } t $$
In quantum mechanics, $\left| \psi (x,\ t) \right|^2$ is treated as the probability density function of finding a particle at a certain point $x$ when the time is $t$. Therefore, the above equation represents the probability of the particle existing in the interval $[a, b]$ when the time is $t$. Since the particle must exist somewhere, the integral over the entire interval must be $1$.
$$ \int_{-\infty}^{\infty} |\psi (x,\ t)|^2 dx=1 $$
The above condition comes from the perspective of interpreting the wave function probabilistically.
Normalization
However, if we look at the equation below, we see that while $\psi$ satisfies the Schrödinger Equation, its constant multiple $a\psi$ also satisfies the Schrödinger equation.
$$ H\psi = E\psi \implies aH\psi = aE\psi \implies H(a\psi) = E(a\psi) $$
If we apply the above interpretation to $a\psi$, it becomes ${\displaystyle \int _{-\infty}^{\infty}} |a\psi|^2 dx=a^2 \ne 1$, and this value cannot be interpreted as a probability. Therefore, we need to adjust the magnitude of the wave function so that the integral over the entire interval of the wave function becomes $1$, giving it a probabilistic interpretation. This is called normalization.
When dealing with wave functions in quantum mechanics, normalization is mandatory. For example, suppose the integral concerning a wave function $\psi$ is as follows.
$$ \begin{equation} \int_{-\infty}^{\infty} |\psi|^2dx=9 \end{equation} $$
Then, it is not treated as it is. By dividing both sides by $9$, it becomes ${\displaystyle \int_{-\infty}^{\infty} }|\frac{1}{3}\psi|^2dx=1$, making a probabilistic interpretation possible. Here, when $\psi$ is normalized, it becomes $\frac{1}{3}\psi$, which we call the normalized wave function. The functions dealt with in quantum mechanics are the normalized $\frac{1}{3}\psi$.
Inner Product
Expressing this in terms of Inner Product, it is as follows. If $\psi$ is a normalized wave function,
$$ \braket{\psi | \psi} = \int\limits_{-\infty}^{\infty} \psi^{\ast}(x) \psi(x) dx = \int\limits_{-\infty}^{\infty} |\psi(x)|^{2} dx = 1 $$
Square-Integrable
On the other hand, if the integral value of the wave function’s probability density is not $1$, as in $(1)$, it’s not an issue. This is because the magnitude can be adjusted through normalization. The problem occurs when the integral value diverges. Therefore, the wave function satisfying the Schrödinger equation must satisfy the equation below.
$$ \int_{-\infty}^{\infty} |\psi (x,\ t)|^2dx <\infty $$
A wave function that satisfies the above condition is called a square-integrable function. A square-integrable wave function should converge to $0$ as the function value approaches $x \rightarrow \pm \infty$. If not, it means the area under the graph of the wave function does not converge, and hence it is not square-integrable.