# Is the Hydrogen Atom Ground State Wave Function Normalized?

• jg370
In summary, the conversation discusses the ground state wave function for an electron in a hydrogen atom and the normalization condition for any wave function. The wave function is given as \psi(r) = \frac{1}{\sqrt (\pi a_o^3)} e^\frac{-r}{a_o} and it is required to show that it is normalized. After attempting to integrate, it is found that a mistake was made and the correct integral should be \int_0^{\infty} e^{-\frac{2r}{a_0^2}} dr = \frac{a_0^2}{2}. It is suggested to use spherical coordinates for a better approach.
jg370

## Homework Statement

The ground state wave function for the electron in a hydrogen atom is:

$$\psi(r) = \frac{1}{\sqrt (\pi a_o^3)} e^\frac{-r}{a_o}$$

where r is the radial coordinate of the electron and a_o is the Bohr radius.

Show that the wave function as given is normalized.

## Homework Equations

Any wave function satisfying the following equation is said to be normalized:

$$\int_{-\infty}^{+\infty} |\psi|^2\dx = 1$$

## The Attempt at a Solution

Because the sum of all probabilities over all values of r must be 1,

$$\int_{-\infty}^{+\infty} (\frac{1}{\sqrt (\pi a_o^3)} e^\frac{-r}{a_o})^2 dr = \frac{1}{\pi a_o^3} \int_{-\infty}^{+\infty} e^\frac{-2r}{a_o^2} dr = 1$$

Since the integral can be expressed as the sum of two integrals, we have,

$$\frac{1}{\pi a_o^3} \int_{-\infty}^{+\infty} e^\frac{-2r}{a_o^2} dr = \frac{2}{\pi a_o^3} \int_0^{+\infty} e^\frac{-2r}{a_o^2} dr = 1$$

After integrating, I obtain,

$$\frac{1}{\pi a_o^3} = 1$$

which is definitely incorrect. However, I do not see any other way to proceed. could someone give some assitance.

Thanks for your kind assistance

jg370

you integrated wrong -
$\int_0^{\infty} e^{-\frac{2r}{a_0^2}} dr = [-\frac{a_0^2}{2} e^{-\frac{2r}{a_0^2}]_0^{\infty}=\frac{a_0^2}{2}$

which gives $\frac{1}{2 \pi a_0}=1$

the a_0 in your exponent shouldn't be squared, also the normalization condition says you should integrate over the bounds of your function. if r is the radial coordinate then how can it have a value of negative infinity? similarly your no longer normalizing in 1-d but 3D I recommend using spherical coordinates in which case your integral inherits a r^2 sin theta

Tks. I wil try again using spherical coordinates

## 1. What is a wave function?

A wave function is a mathematical description of a quantum system that contains all the information about its possible states. It is represented by a complex-valued function that describes the probability amplitude of finding a particle in a particular state.

## 2. Why is it important for the wave function to be normalized?

A normalized wave function ensures that the total probability of finding a particle in all possible states is equal to 1. This is a fundamental requirement in quantum mechanics as it guarantees that the system is in a valid physical state.

## 3. How do you normalize a wave function?

To normalize a wave function, you must divide it by the square root of the integral of its absolute square over all possible states. This process is known as the normalization constant and ensures that the total probability of finding a particle is equal to 1.

## 4. What happens if the wave function is not normalized?

If the wave function is not normalized, it means that the total probability of finding a particle in all possible states is not equal to 1. This can lead to incorrect predictions and violates the fundamental principles of quantum mechanics.

## 5. Is the wave function always normalized?

No, the wave function is not always normalized. In certain cases, such as when dealing with infinite systems or when using approximate methods, the wave function may not be normalized. However, it is important to normalize the wave function whenever possible to ensure accurate results.

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