# Good wavefunction (Griffiths)

Two different wave functions are graphed at the right.To the left there is a parabola $Psi_I=A_I(L x-x^2)$, to the right the wave function is made up two segments. For the two wave functions $L=2nm$

a. Explain why $psi_2$ cannot be a realistic wave function.What has to be changed to make it a good wave function?

b. Normalize each wave function.

c. What is the expectation value $<x>$ of each wave function?

The wave function need to be “smooth”, that is to be continuous and also its derivative continuous. Second triangular wave function is not smooth (does not have a continuous derivative at its peak). The necessity of the wave function being smooth comes from the Schrodinger equation

$-ℏ^2/2m*∇^2 ψ+Vψ=Eψ$

so that $∇^2 ψ$ EXIST.

In order for the second graph to be a wave function you need to impose the continuity of the first derivative in all points in space.

The mathematical shapes of the given wave functions are

$ψ_1=A_1*(L x-x^2 )$

$ψ_2=$

$=(A_2/0.67L)*x$ for$ 0<x<0.67*L/A$

$=A_2-A_2/0.33L*(x-0.67L)$ for $0.67L≤x<L$

Normalization condition is

$1=∫_0^L |ψ|^2*d x$

$1=A_1^2*∫_0^L (L x-x^2 )^2*d x= A_1^2*(x^5/5-(L x^4)/2+(l^2 x^3)/3)(from 0 to L)=$

$=A_1^2*l^5/30 or A_1=√(30/L^5)$

$1=A_2^2 ∫_0^0.67L x^2/(0.67L)^2 d x+A_2^2*∫_0.67L^L (1-1/0.33L*(x-0.67L) )^2$

$d x=⋯=(A_2^2)/(0.67L)^2 *(0.67L)^3/3+A_2^2*11L/100=(0.223L+0.110L) A_2^2=0.333L*A_2^2$

$A_2=√(1/(0.333*L))=√(3/L)$

c)

For the first wave function

$<x>=∫_0^L ψ_1^* xψ_1 d x=A_1^2 ∫_0^L x(L x-x^2)d x=$

$=A_1^2*0.0167L^6=30*0.01666*L=0.500*L=1 nm$

For second wave function

$<x>=∫_0^L ψ_2^* xψ_2 d x=$

$= A_2^2 ∫_0^0.67L (x*x^2)/(0.67L)^2 d x +A_2^2 ∫_0.67L^L x*(1-1/0.33L*(x-0.67L) )^2 d x=$

$=(A_2^2)/(0.67L)^2 *(0.67L)^4/4+A_2^2*0.08277L^2=$

$=(0.1122L^2+0.08277L^2 )*3/L=0.5850L=1.17 nm$