Respuesta :
The question is incomplete. The complete question is :
Consider a particle in a one-dimensional box defined by V(x)=0, a>x>0 and V(x)=infinity, x a, x 0. Explain why each of the following un-normalized functions is or is not an acceptable wave function based on criteria such as being consistent with the boundary conditions, and with the association of psi^+(x)psi(x)dx with the probability.
a). [tex]$A \cos \frac{n \pi x}{a}$[/tex]
b). [tex]$B(x+x^2)$[/tex]
c). [tex]$Cx^3(x-a)$[/tex]
d). [tex]$\frac{D}{\sin \frac{n \pi x}{a}}$[/tex]
Solution :
a). The function of cosine here do not go to the value 0 at the boundary of the box (that is to be expected as V is infinite there).
[tex]$ \Psi (0) = A$[/tex] and [tex]$ \Psi (a) = A \cos(n \pi) = \pm 1$[/tex]
Therefore it is not an acceptable function.
b). The function in this case goes to 0 at [tex]$x=0, $[/tex] but is non zero at [tex]$x=a$[/tex]. So this function is also not acceptable.
c). The function here goes to 0 at both [tex]$x=a$[/tex] and [tex]$x=b$[/tex] . But this wavelength can never satisfy the time independent Schrodinger equation.
[tex]-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\Psi(x)\neq E\Psi(x)\quad \text{for} \ \quad \Psi(x)=Cx^3(x-a)[/tex]
Therefore, [tex]$\Psi + \Psi $[/tex] here can not be associated with the probability.
d). Here, the function goes to the infinite at [tex]$x=0 $[/tex] and [tex]$x=a$[/tex]. Thus this is also not valid choice for the wave function.
