Find the value of A to normalize the wave function: {A, -L `<=` x `<=` L {0, for all other x ...there is no given L I'm so confused is this infinity or a function like 1/L or something? I...

Find the value of A to normalize the wave function:

{A, -L `<=` x `<=` L

{0, for all other x

...there is no given L I'm so confused is this infinity or a function like 1/L or something? I know it's an integral. But even for -L or -`oo` to L or `oo`  that equals zero. 

Expert Answers
Borys Shumyatskiy eNotes educator| Certified Educator

Hello!

There is no given `L` and no given `A.` But, this is not a problem, we can write expressions with variables.

To normalize a wave function means to find the expression for `A` (depending on `L`) such that the probability of finding a particle somewhere on the entire axis is exactly `1.` It is known that a probability density function `pd(x)` is the square of a wave function `Psi(x).`

Also, by the definition of a probability density function, the probability of a particle being (at least) somewhere is  `int_(-oo)^(+oo) pd(x) dx = int_(-oo)^(+oo) |Psi(x)|^2 dx.` 

 

In our case `Psi(x) = 0` for `x` outside `[-L,L],` therefore the integral is equal to

`int_(-L)^(+L) |Psi(x)|^2 dx = int_(-L)^(+L) |A|^2 dx = 2LA^2,`

and this must be `1,` so `A^2 = 1/(2L),`  `A = 1/sqrt(2L).` This is the answer.