Consider a particle in a 1-dimensional box defined by V(x)=0, for 0

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).

b).

c).

d).

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).

  and  

Therefore it is not an acceptable function.

b). The function in this case goes to 0 at but is non zero at . So this function is also not acceptable.

c). The function here goes to 0 at both and . But this wavelength can never satisfy the time independent Schrodinger equation.

Therefore, here can not be associated with the probability.

d). Here, the function goes to the infinite at  and . Thus this is also not valid choice for the wave function.

the fist one is not equalvalent but i think th second one isand the third isnt definetly or the last one is i think. hope this !

step-by-step explanation:

step-by-step explanation:

to find if any function is an inverse of the other, put the first equation in as x in the next one. then simplify and solve. if at the end you get the statement y = x, then it is an inverse. if you get anything else, it is not.