In this problem you will show that the number of nodes of the stationary states of a one dimensional potential always increases with energy Consider two (real, normalized) solutions ( n and m ) to the time independent Schrdinger equation (for a given potential V (x) ), with energies E n E m (a) Show that (b) Let x 1 and x 2 be two adjacent nodes of the function m (x) Show that (c) If n (x) has no nodes between and , then it must have the same sign everywhere in the interval Show that (b) then leads to a contradiction Therefore, between every pair of nodes of m (x), n (x) must have at least one node, and in particular the number of nodes increases with energy d (dve v v dite ) ( Vm dx dx dx 2m 24 (En Em) VmVn

Question: In this problem you will show that the number of nodes of the stationary states of a one-dimensional potential always increases with energy. Consider two

In this problem you will show that the number of nodes of the stationary states of a one-dimensional potential always increases with energy. Consider two (real, normalized) solutions (Ψ_n and Ψ_m) to the time-independent Schrödinger equation (for a given potential V (x) ), with energies E_n m.

(a) Show that

d & (dve v. - v. dite ) = ( Vm dx

(b) Let x₁ and x₂ be two adjacent nodes of the function Ψ_m (x). Show that

dx dx 2m 24 (En Em) VmVn.

(c) If Ψ_n (x) has no nodes between and , then it must have the same sign everywhere in the interval. Show that (b) then leads to a contradiction. Therefore, between every pair of nodes of Ψ_m (x), Ψ_n (x) must have at least one node, and in particular the number of nodes increases with energy.

d & (dve v. - v. dite ) = ( Vm dx dx dx 2m 24 (En Em) VmVn.

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