Infinite Potential Wall

The energy eigenfunctions and eigenvalues of the infinite potetial wall:

ψ[x_,n_] = (Sqrt[(2/L)])Sin[((n π)/L)x];En[n_] = (((ℏ^2)(π^2))/(2 m (L^2)))(n^2);

$Assumptions = n∈Integers && l∈Integers && n>0 && m > 0 && L>0 && ℏ>0 && l>0

Normalization:

(∫_0^L)ψ[x,n] ψ[x,n]\[DifferentialD]x

Calculate  <x>, <(x^2)>, Δx:

xav[n_] = (∫_0^L)ψ[x,n] x ψ[x,n] \[DifferentialD]x

x2av[n_] = (∫_0^L)ψ[x,n] (x^2) ψ[x,n] \[DifferentialD]x

Δx[n_] = (Sqrt[x2av[n] - (((xav[n]))^2)])//Simplify

Calculate  <p>, <(p^2)>, Δp:

pav[n_] = (∫_0^L)ψ[x,n] (-I ℏ) (∂_x)ψ[x,n] \[DifferentialD]x

p2av[n_] = (∫_0^L)ψ[x,n] (- (ℏ^2)) (∂_(x,x))ψ[x,n] \[DifferentialD]x

Δp[n_] = (Sqrt[p2av[n] - (((pav[n]))^2)])//Simplify

Heisenberg Uncertainty principle:

Δx[n] Δp[n]//Simplify

The product is always larger than ℏ/2:

Table[Δx[n] Δp[n]//Simplify//N,{n,1,10}]

Problem 1:

Consider the particle in the initial state ϕ(x,0) = A x (L-x). Compute the state ϕ(x,t) and the corresponding average values of position , momentum etc.

Solution:

Consider the expression of the wave function interms of the energy eigenfunctions:
ϕ(x,0) ≡ϕ(x) =  (∑_(n=1)^∞)(Subscript[c, n])(Subscript[ψ, n])(x), (Subscript[c, n])= (∫_0^L)(Subscript[ψ, n])*(x)ϕ(x)dx
ϕ(x,t) = (∑_(n=1)^∞)(Subscript[c, n])(Subscript[ψ, n])(x) (e^(-i (Subscript[E, n])t/ℏ))
< x > (t) = (∑_(n,m=1)^∞)(Subscript[c, m])*(Subscript[c, n])(e^(-i ((Subscript[E, n])-(Subscript[E, m]))t/ℏ))(∫_0^L)(Subscript[ψ, m])*(x)x (Subscript[ψ, n])(x)dx
< (x^2) > (t) = (∑_(n,m=1)^∞)(Subscript[c, m])*(Subscript[c, n])(e^(-i ((Subscript[E, n])-(Subscript[E, m]))t/ℏ))(∫_0^L)(Subscript[ψ, m])*(x)(x^2) (Subscript[ψ, n])(x)dx

We compute the normalization factor A and define the wave function at t=0

A = (1/(Sqrt[(∫_0^L)(((x(L-x)))^2)\[DifferentialD]x]))

ϕ[x_,0] = A x (L-x);(∫_0^L)((ϕ[x,0])^2)\[DifferentialD]x

We compute the coeeficients (Subscript[c, n]):

cn[n_] = (∫_0^L)ψ[x,n] ϕ[x,0]\[DifferentialD]x /. {Cos[n π] -> (((-1))^n),Sin[n π] -> 0}

Even order n  coefficients are zero. Odd eigenfunctions are absent since ϕ(x,0) is even.

cn[2 n]//Simplify

cm[n_] = cn[2 n+1]//Simplify

Of course M->∞ but we will use it as a cutoff:

ϕ[x_,t_,M_] := (∑_(l=0)^M)cm[l] ψ[x,2l+1] (E^(-I En[2l+1]t/ℏ))

These are the periods (Subscript[T, n])= ((2π)/(Subscript[ω, n]))=((2π)/((Subscript[E, n])/ℏ))= ((4 m (L^2))/(π ℏ (n^2)))

((4 m (L^2))/(π ℏ (n^2)))/.{L-> 1,m-> 1, ℏ-> 1}//N

Plot[Re[{ϕ[x,0,20],ϕ[x,0.3,20],ϕ[x,0.5,20]}]/.{L-> 1,m-> 1, ℏ-> 1},{x,0,1}]

We compute the coefficients (Subscript[x, nm])= (∫_0^L)(Subscript[ψ, n])*(x)x (Subscript[ψ, m])(x)dx and (Subscript[(x^2), nm])= (∫_0^L)(Subscript[ψ, n])*(x)(x^2) (Subscript[ψ, m])(x)dx

xnm[n_,l_] = (∫_0^L)ψ[x,n]x ψ[x,l]\[DifferentialD]x

The n=m case must be treated differently:

xnm[n_,n_] = (∫_0^L)ψ[x,n]x ψ[x,n]\[DifferentialD]x

Table[xnm[i,j],{i,1,10},{j,1,10}]//MatrixForm

x2nm[n_,l_] = (∫_0^L)ψ[x,n](x^2) ψ[x,l]\[DifferentialD]x

x2nm[n_,n_] = (∫_0^L)ψ[x,n](x^2) ψ[x,n]\[DifferentialD]x

Table[x2nm[i,j],{i,1,10},{j,1,10}]//MatrixForm

xavϕ[t_, NN_, MM_] := (∑_(n=0)^NN)(∑_(l=0)^MM) cm[n] cm[l] xnm[2n+1,2l+1] (E^(-I(En[2l+1]-En[2n+1])t/ℏ))

Plot[xavϕ[t,6,6]/.{L-> 1,m-> 1, ℏ-> 1},{t,0,6}]

x2avϕ[t_, NN_, MM_] := (∑_(n=0)^NN)(∑_(l=0)^MM) cm[n] cm[l] x2nm[2n+1,2l+1] (E^(-I(En[2l+1]-En[2n+1])t/ℏ))

One should choose some values of the cutoff in order to compute. 6->10 makes no visible difference

Plot[{x2avϕ[t,6,6]}/.{L-> 1,m-> 1, ℏ-> 1},{t,0,1}]

Momenta now:

pnm[n_,l_] = (∫_0^L)ψ[x,n](-I ℏ) (∂_x)ψ[x,l]\[DifferentialD]x /. Cos[l π] -> (((-1))^l)

pnm[n_,n_] = (∫_0^L)ψ[x,n](-I ℏ) (∂_x)ψ[x,n]\[DifferentialD]x

Table[pnm[i,j],{i,1,10},{j,1,10}]//MatrixForm

p2nm[n_,l_] = (∫_0^L)ψ[x,n](- (ℏ^2)) (∂_(x,x))ψ[x,l]\[DifferentialD]x

p2nm[n_,n_] = (∫_0^L)ψ[x,n](- (ℏ^2)) (∂_(x,x))ψ[x,n]\[DifferentialD]x

Table[p2nm[i,j],{i,1,10},{j,1,10}]//MatrixForm

pavϕ[t_, NN_, MM_] := (∑_(n=0)^NN)(∑_(l=0)^MM) cm[n] cm[l] pnm[2n+1,2l+1] (E^(-I(En[2l+1]-En[2n+1])t/ℏ))

As can be seen by inspection all coefficients are 0:

Plot[pavϕ[t,6,6]/.{L-> 1,m-> 1, ℏ-> 1},{t,0,6}]

Since (Subscript[p, n m]) are non zero only at the diagonal, <p>(t) does not change with time:

p2avϕ[t_, NN_, MM_] := (∑_(n=0)^NN)(∑_(l=0)^MM) cm[n] cm[l] p2nm[2n+1,2l+1] (E^(-I(En[2l+1]-En[2n+1])t/ℏ))

Plot[p2avϕ[t,6,6]/.{L-> 1,m-> 1, ℏ-> 1},{t,0,6}]

Compare with the exact formula for <(p^2)>(0):

(∫_0^L)ϕ[x,0] (-(ℏ^2))(∂_(x,x))ϕ[x,0]\[DifferentialD]x