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...
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.
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