KAHIBARO
Table of Contents
Why normalization matters
In quantum mechanics, the wave function $\psi$ is used to calculate probabilities. The chapter on probability interpretation explains that $|\psi|^2$ gives probability density. Normalization is the rule that makes the total probability equal to 1.
If a particle must be somewhere in the allowed region of space, then the sum of all probabilities of finding it anywhere must be 100 percent. In mathematics, that means the integral of the probability density over all space must be 1.
For a normalized wave function,
$$
\int_{\text{all space}} |\psi(\mathbf{r},t)|^2 \, d\tau = 1
$$
In one dimension, this becomes
$$
\int_{-\infty}^{\infty} |\psi(x,t)|^2 \, dx = 1
$$
Here, $d\tau$ means a tiny volume element. In one dimension, it is just $dx$.
Probability density and total probability
The quantity $|\psi(x,t)|^2$ tells us how probability is distributed in space. It does not by itself have to equal 1 at each point. Instead, its integral over a region gives the probability of finding the particle in that region.
If we want the probability of finding the particle between $x=a$ and $x=b$, we calculate
$$
P(a \le x \le b) = \int_a^b |\psi(x,t)|^2 \, dx
$$
If the wave function is normalized, then taking the whole allowed space gives probability 1.
A wave function is physically acceptable only if its total probability is finite. To represent a single particle, it is usually required that
$$
\int |\psi|^2 \, d\tau = 1
$$
after normalization.
How to normalize a wave function
Often a wave function is given with an unknown constant, such as $A$. We determine $A$ by imposing the normalization condition.
Suppose
$$
\psi(x) = A f(x)
$$
Then
$$
|\psi(x)|^2 = |A|^2 |f(x)|^2
$$
To normalize, we require
$$
\int |\psi(x)|^2 dx = |A|^2 \int |f(x)|^2 dx = 1
$$
So the constant must satisfy
$$
|A|^2 = \frac{1}{\int |f(x)|^2 dx}
$$
If $A$ is chosen real and positive, then
$$
A = \frac{1}{\sqrt{\int |f(x)|^2 dx}}
$$
Example in one dimension
Consider a wave function inside the interval $0 \le x \le L$:
$$
\psi(x) = A \sin\left(\frac{\pi x}{L}\right)
$$
and $\psi(x)=0$ outside this interval.
To normalize it, we require
$$
\int_0^L |\psi(x)|^2 dx = 1
$$
Substitute the function:
$$
A^2 \int_0^L \sin^2\left(\frac{\pi x}{L}\right) dx = 1
$$
Using
$$
\int_0^L \sin^2\left(\frac{\pi x}{L}\right) dx = \frac{L}{2}
$$
we get
$$
A^2 \frac{L}{2} = 1
$$
so
$$
A = \sqrt{\frac{2}{L}}
$$
Therefore the normalized wave function is
$$
\psi(x) = \sqrt{\frac{2}{L}} \sin\left(\frac{\pi x}{L}\right)
$$
A simple exponential example
Now consider
$$
\psi(x) = A e^{-\alpha x}, \quad x \ge 0
$$
with $\alpha > 0$, and $\psi(x)=0$ for $x<0$.
Normalization gives
$$
\int_0^\infty |A|^2 e^{-2\alpha x} dx = 1
$$
So
$$
|A|^2 \left(\frac{1}{2\alpha}\right) = 1
$$
which means
$$
|A|^2 = 2\alpha
$$
If $A$ is real and positive,
$$
A = \sqrt{2\alpha}
$$
Thus the normalized wave function is
$$
\psi(x) = \sqrt{2\alpha} e^{-\alpha x}
$$
Normalizable and non-normalizable functions
Not every mathematical function can represent a physical state of a particle. Some functions cannot be normalized because the integral of $|\psi|^2$ diverges.
For example, the constant function
$$
\psi(x)=A
$$
for all $x$ from $-\infty$ to $\infty$ gives
$$
\int_{-\infty}^{\infty} |A|^2 dx = \infty
$$
unless $A=0$. So it cannot be normalized as an ordinary wave function over all space.
This is an important physical test. If the total probability is infinite, the function does not describe a localized single-particle state in the usual sense.
A wave function is normalizable if
$$
\int |\psi|^2 d\tau
$$
is finite.
If this integral is infinite, the function cannot be normalized in the usual way.
Normalization in different dimensions
The idea is the same in one, two, or three dimensions. Only the integration element changes.
| Dimension | Normalization condition | ||
|---|---|---|---|
| 1D | $\int_{-\infty}^{\infty} | \psi(x,t) | ^2 dx = 1$ |
| 2D | $\int\!\!\int | \psi(x,y,t) | ^2 dx\,dy = 1$ |
| 3D | $\int\!\!\int\!\!\int | \psi(x,y,z,t) | ^2 dx\,dy\,dz = 1$ |
In compact notation for three dimensions,
$$
\int |\psi(\mathbf{r},t)|^2 \, d\tau = 1
$$
Normalization and time
For a properly behaving isolated quantum system, normalization is preserved in time. This means that if the wave function is normalized at one instant, it remains normalized as it evolves.
So although $\psi(x,t)$ may change shape with time, the total probability remains 1.
Normalization expresses conservation of total probability.
A normalized wave function stays normalized during correct quantum time evolution.
Geometry of probability density
A normalized wave function can be pictured as a curve whose squared magnitude has total area 1 in one dimension.
The height of the curve at one point is not itself a probability. Probability comes from the area under the curve over an interval.
Common mistakes
A common mistake is to require $\psi$ itself to integrate to 1. That is wrong. It is $|\psi|^2$ that must integrate to 1.
Another mistake is forgetting the complex conjugate. Since wave functions can be complex, the probability density is
$$
|\psi|^2 = \psi^* \psi
$$
not simply $\psi^2$.
For complex wave functions,
$$
|\psi|^2 = \psi^*(x,t)\psi(x,t)
$$
This quantity is always real and nonnegative.
Summary
Normalization is the condition that the total probability of finding a quantum particle somewhere in space is 1. It is written as the integral of $|\psi|^2$ over all space equal to 1. When a wave function contains an unknown constant, normalization determines that constant. A physically acceptable wave function must be normalizable, meaning the integral of $|\psi|^2$ is finite. Normalization is one of the most basic and necessary requirements for a quantum state.