Normalizing a One-Dimensional Wavefunction

Hello. In the previous lesson, the Born rule turned the wavefunction into a position probability density:

$$\rho (x)=|\psi (x){|}^2.$$

Probabilities for regions are areas under that density. This creates a necessary consistency condition: if we search everywhere the particle is allowed to be, we must find it with probability $1$. Normalization is the process of choosing the overall scale of a wavefunction so that this is true.

In this lesson, you will normalize simple one-dimensional wavefunctions defined over finite intervals. The calculations establish a habit needed throughout quantum chemistry: determine the allowed domain first, square the magnitude of the wavefunction, integrate, and set the result equal to one.

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Total probability fixes the wavefunction’s scale

For a particle whose wavefunction is defined over all space, normalization requires

$${\int }_{-\infty }^{\infty }|\psi (x){|}^{2}dx=1.$$

If the particle is confined to a stated domain, such as $0\le x\le L$, and the wavefunction is zero outside that domain, only integrate where it is nonzero:

$${\int }_0^L|\psi (x){|}^{2}dx=1.$$

This is not merely a mathematical convention. Since $|\psi (x){|}^{2}dx$ is the probability of finding the particle in a tiny interval of width $dx$, the complete integral is the probability of finding it anywhere in its permitted space.

Suppose a proposed wavefunction has a known shape $f(x)$, but an unknown overall constant $A$:

$$\psi (x)=Af(x).$$

For a domain from $a$ to $b$, normalization gives

$${\int }_a^b|Af(x){|}^{2}dx=1.$$

Because $A$ does not depend on $x$, its magnitude squared can be taken outside the integral:

$$|A{|}^2{\int }_a^b|f(x){|}^{2}dx=1.$$

Thus, if we define

$$I={\int }_a^b|f(x){|}^{2}dx,$$

then

$$|A|=\frac{1}{\sqrt{I}}.$$

The core rule is therefore:

$$\boxed{{\psi }_{\mathrm{normalized}}(x)=\frac{f(x)}{\sqrt{{\int }_a^b|f(x){|}^{2}dx}}}$$

when $f(x)$ is nonzero only on $a\le x\le b$.

For introductory real-valued wavefunctions, we usually choose $A$ positive. Strictly, normalization fixes $|A|$, not the sign or overall complex phase. Multiplying every value of a wavefunction by $e^{i\varphi }$, where $\varphi$ is a constant real phase, leaves $|\psi {|}^2$ unchanged.

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A reliable normalization procedure

For the kinds of functions used in early quantum chemistry, follow these five steps in order.

1. State the domain. Identify the interval where the wavefunction is nonzero. If it is zero outside a box, those exterior regions contribute zero to the integral.

2. Write the normalization condition over that domain.

   $${\int }_a^b{\psi }^{\ast }(x)\psi (x)dx=1.$$

3. Substitute the wavefunction and form the magnitude squared. For a real wavefunction, this is simply its square. For a complex wavefunction, use the complex conjugate.

4. Perform the integral and solve for the unknown constant. Remember that an amplitude constant becomes $|A{|}^2$ after taking the magnitude squared.

5. Check the result. Substitute the constant back into the normalization integral. A correct result gives exactly $1$.

Two common errors explain most incorrect answers:

• Integrating $\psi (x)$ rather than $|\psi (x){|}^2$. A wavefunction can have positive and negative lobes that cancel in $\int \psi (x)dx$, but probabilities cannot cancel.

• Forgetting that the constant is squared:

  $$|Af(x){|}^2=|A{|}^2|f(x){|}^2.$$

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Example 1: a constant wavefunction in a finite region

Consider a particle equally likely to be found anywhere in a one-dimensional region of length $L$:

$$\psi (x)=\{\begin{array}{ll}C,& 0\le x\le L,\\ 0,& \text{otherwise}.\end{array}$$

The constant $C$ is unknown. Since the particle is confined to the interval from $0$ to $L$, normalization is

$${\int }_0^L|C{|}^{2}dx=1.$$

Because $|C{|}^2$ is constant,

$$|C{|}^2{\int }_0^{L}dx=1.$$

The integral is just $L$:

$$|C{|}^{2}L=1.$$

Therefore,

$$|C|=\frac{1}{\sqrt{L}}.$$

Choosing the conventional positive real constant gives

$$\boxed{\psi (x)=\{\begin{array}{ll}\frac{1}{\sqrt{L}},& 0\le x\le L,\\ 0,& \text{otherwise}.\end{array}}$$

The corresponding density is constant:

$$|\psi (x){|}^2=\frac{1}{L}.$$

This makes physical sense. Every equal-width interval has the same probability. For example, the probability of locating the particle in the left half is

$$P\left(0\le x\le \frac{L}{2}\right)={\int }_0^{L/2}\frac{1}{L}dx=\frac{1}{2}.$$

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Example 2: a sinusoidal wavefunction

A form that will soon become central for particles confined in one-dimensional boxes is

$$\psi (x)=\{\begin{array}{ll}A\sin \left(\frac{n\pi x}{L}\right),& 0\le x\le L,\\ 0,& \text{otherwise},\end{array}$$

where $n$ is a positive integer and $A$ is unknown.

Begin with normalization:

$${\int }_0^L{\left|A\sin \left(\frac{n\pi x}{L}\right)\right|}^{2}dx=1.$$

The sine function and $A$ are real in this example, so

$$A^2{\int }_0^L{\sin }^2\left(\frac{n\pi x}{L}\right)dx=1.$$

For any positive integer $n$, the sine-squared function has average value $1/2$ across the box. Equivalently, using

$${\sin }^{2}u=\frac{1-\cos (2u)}{2},$$

shows that the oscillating cosine contribution integrates to zero over the full interval. Therefore,

$${\int }_0^L{\sin }^2\left(\frac{n\pi x}{L}\right)dx=\frac{L}{2}.$$

Substitute that result:

$$A^2\frac{L}{2}=1.$$

Solving for the positive choice of $A$,

$$A=\sqrt{\frac{2}{L}}.$$

The normalized wavefunction is

$$\boxed{{\psi }_n(x)=\{\begin{array}{ll}\sqrt{\frac{2}{L}}\sin \left(\frac{n\pi x}{L}\right),& 0\le x\le L,\\ 0,& \text{otherwise}.\end{array}}$$

Its probability density is

$$|{\psi }_n(x){|}^2=\frac{2}{L}{\sin }^2\left(\frac{n\pi x}{L}\right).$$

Notice what normalization has and has not changed. It sets the total area under this density equal to one, but it does not remove nodes or alter the spatial pattern of peaks. The quantum number $n$ determines that pattern; the normalization constant supplies the physically necessary scale.

The figure also reinforces a point from the prior lesson: a node remains a zero in the probability density. In a sine-based box state, the density may be divided into several separated lobes, but normalization concerns the sum of all their areas, not the area of one lobe.

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Example 3: a polynomial shape

Normalization is not restricted to sine waves. Suppose

$$\psi (x)=\{\begin{array}{ll}Ax(L-x),& 0\le x\le L,\\ 0,& \text{otherwise}.\end{array}$$

This wavefunction is zero at both walls and largest around the center. To find $A$, normalize:

$${\int }_0^L|Ax(L-x){|}^{2}dx=1.$$

Everything is real, so

$$A^2{\int }_0^L{x}^2(L-x{)}^{2}dx=1.$$

Expand the polynomial before integrating:

$$x^2(L-x{)}^2=L^2{x}^2-2L{x}^3+x^4.$$

Therefore,

$$A^2{\int }_0^L\left(L^2{x}^2-2L{x}^3+x^4\right)dx=1.$$

Evaluate term by term:

$$A^2{\left[\frac{L^2{x}^3}{3}-\frac{L{x}^4}{2}+\frac{x^5}{5}\right]}_0^L=1.$$

At $x=L$, the bracket becomes

$$\frac{L^5}{3}-\frac{L^5}{2}+\frac{L^5}{5}=\frac{L^5}{30}.$$

So,

$$A^2\frac{L^5}{30}=1,$$

and the positive normalization constant is

$$A=\frac{\sqrt{30}}{L^{5/2}}.$$

Hence,

$$\boxed{\psi (x)=\{\begin{array}{ll}\frac{\sqrt{30}}{L^{5/2}}x(L-x),& 0\le x\le L,\\ 0,& \text{otherwise}.\end{array}}$$

A useful dimensional check is available here. In one dimension, since $|\psi (x){|}^{2}dx$ is dimensionless probability, $\psi$ must have units of length to the power $-1/2$. The factor $x(L-x)$ has units of length squared, while $A$ has units of length to the power $-5/2$. Their product indeed has units of length to the power $-1/2$.

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What normalization does not mean

Normalization is a scaling condition, not an additional physical force or a claim that every location is equally likely.

It does not mean:

• the wavefunction itself must have values between $0$ and $1$;
• the probability density must be below $1$;
• the wavefunction must be positive;
• every part of the domain has equal probability;
• a normalized state has no nodes.

A probability density can exceed $1$ if it is concentrated in a sufficiently short region. Only an integrated probability must lie between $0$ and $1$.

Also, not every mathematical function can represent a normalizable single-particle state. The integral

$$\int |\psi (x){|}^{2}dx$$

must be finite. If it diverges, no finite multiplicative constant can make the total probability equal to one. The simple confined functions in this lesson are well behaved, so the normalization process works directly.

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Key takeaways

Normalization gives a wavefunction the correct probabilistic scale:

$${\int }_{\text{allowed domain}}|\psi (x){|}^{2}dx=1.$$

For a wavefunction written as $Af(x)$, determine the constant from

$$|A{|}^2{\int }_{\text{allowed domain}}|f(x){|}^{2}dx=1.$$

Always identify the domain first, use ${\psi }^{\ast }\psi$ rather than $\psi$ alone, and remember that the overall constant is squared in the probability density. Normalization fixes total probability, while the functional shape still determines where the particle is more or less likely to be found.

Next, you will use normalized wavefunctions to calculate expectation values of position and momentum: the statistical averages predicted for many identically prepared particles.