Question

Prove that

${ \bf \displaystyle \int_{- \infty}^{\infty} | \bf \psi ( x ) |² dx = 1 }$

Topic :- Wave Functions ^^​

Answer 1

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Answer 2

To prove that:

$\rm \int_{-\infty}^{\infty}|\psi(x)|^2dx = 1$

we need to use the properties of the wave function (x).

The wave function describes the probability density of finding a particle at a certain position in space. The probability density is defined as the square of the absolute value of the wave function, i.e., |(x)|². The total probability of finding the particle in the entire space is equal to 1, which means that the integral of the probability density over all space must be equal to 1.

Mathematically, this can be expressed as:

$\rm \int_{-\infty}^{\infty}|\psi(x)|^2dx = 1$

This integral is known as the normalization integral, and it ensures that the probability density is properly normalized to one.

To prove this, we can use the fact that the wave function must be square integrable, which means that:

$\rm \int_{-\infty}^{\infty}|\psi(x)|^2dx < \infty$

This condition ensures that the wave function is well-behaved and can be integrated over all space.

Next, we can use the identity:

$\rm \int_{-\infty}^{\infty}f(x)dx = \int_{-\infty}^{\infty}f^*(x)dx$

where f*(x) is the complex conjugate of f(x).

Using this identity, we can write:

$\rm\int_{-\infty}^{\infty}|\psi(x)|^2dx = \int_{-\infty}^{\infty}\psi^*(x)\psi(x)dx$

Since the product of the complex conjugate of

$\rm\psi(x) \: and \: \psi(x) \: is \: a \: real \\ \rm number, we \: can \: write:$

$\rm \int_{-\infty}^{\infty}\psi^*(x)\psi(x)dx = \int_{-\infty}^{\infty}\psi(x)\psi^*(x)dx$

Now, we can use the fact that the product of

$\rm\psi(x) \: and \: \psi^*(x) \: is \: equal \: to \: |\psi(x)|^2, \\ \rm s o \: we \: have :$

$\rm \int_{-\infty}^{\infty}\psi(x)\psi^*(x)dx = \int_{-\infty}^{\infty}|\psi(x)|^2dx$

Therefore, we have:

$\rm \int_{-\infty}^{\infty}|\psi(x)|^2dx = \int_{-\infty}^{\infty}\psi(x)\psi^*(x)dx = 1$

This proves that the integral of the probability density over all space is equal to 1, which means that the wave function is properly normalized.