Finite potential well
Finite potential well
Particle confined to an finite potential well an interval [-a, +a].
The particle has energy < 0.
The potential is :
-U0 (U0 > 0) for x ∈ [-a, +a],
0 Outside that range, that is x < -a or x > a
The particle has energy < 0.
The potential is :
-U0 (U0 > 0) for x ∈ [-a, +a],
0 Outside that range, that is x < -a or x > a
In[]:=
ClearAll["Global`*"](*Thedimensionsoftheimagesproducedareset:400isthewidthand200istheheight.*)MyImageSize={400,200};$Assumptions=x∈Reals&&(*VariablePositiononthexaxis*)p∈Reals&&(*VariableMomentum*)ℏ∈Reals&&ℏ>0&&(*Planckconstant*)a∈Reals&&a>0&&(*Boxwidth*)U0∈Reals&&U0>0&&(*Potentialenergyinthewellis-U0*)m∈Reals&&m>0&&(*Massoftheparticle*)n∈Integers&&n>0(*Integerfornumberofsolutions*);
SolutionsoftheSchröedingerequation: φn(x)withn=0,1,2,3,4
SolutionsoftheSchröedingerequation: (x)withn=0,1,2,3,4
φ
n
In this notebook, for simplicity, we do not solve the Schröedinger equation but we immediately write the solutions and show that they satisfy the Schröedinger equation.
With Energy < 0, it is not possible to solve the Schröedinger equation exactly, so we show a numerical solution for the following parameters:
ℏ = 1; m = 1; a = 5; = 1;
The Schröedinger equation is solvable only for the discrete values of the Energy En[n], they are all possible values for energy. Each solution φ[n,x] represents a physical state and it is associated with its respective energy En[n].
The values in the energy Energy[n] were found numerically with n = 0, 1, 2, 3, 4 (we have omitted the procedure).
Each wave function φ[n, x] is associated with its respective energy En[n] .
As already mentioned above, below we immediately write the solutions φ[n,x] omitting the details of the resolution of the Schröedinger equation.
Now let’s set the parameters that will determine the specific numerical solution:
With Energy < 0, it is not possible to solve the Schröedinger equation exactly, so we show a numerical solution for the following parameters:
ℏ = 1; m = 1; a = 5;
U0
The Schröedinger equation is solvable only for the discrete values of the Energy En[n], they are all possible values for energy. Each solution φ[n,x] represents a physical state and it is associated with its respective energy En[n].
The values in the energy Energy[n] were found numerically with n = 0, 1, 2, 3, 4 (we have omitted the procedure).
Each wave function φ[n, x] is associated with its respective energy En[n] .
As already mentioned above, below we immediately write the solutions φ[n,x] omitting the details of the resolution of the Schröedinger equation.
Now let’s set the parameters that will determine the specific numerical solution:
In[]:=
ℏ=1;m=1;a=5;U0=1;
We define the potential function that takes on different values in two different areas:
- take zero value for x<-a and x>a, ie outside the hole;
- takes value -Uo for all other cases, that is, for -a ≤ x ≤ +a, ie inside the hole;
- take zero value for x<-a and x>a, ie outside the hole;
- takes value -Uo for all other cases, that is, for -a ≤ x ≤ +a, ie inside the hole;
In[]:=
V[x_]:=Piecewise[{{0,x<-a||x>a}},-U0];
Here we define the following functions λ e q that we will use in the solutions φ[n,x] of the Schröedinger equation:
Here we define the following functions λ e q that we will use in the solutions φ[n,x] of the Schröedinger equation:
In[]:=
λ:=;q:=;
2mAbs[Energy]
ℏ
2m(U0-Abs[Energy])
ℏ
We do not prove it, but t he Schröedinger equation has two types of different solutions : the EVEN functions and the ODD functions
We do not prove it, but t he Schröedinger equation has two types of different solutions : the EVEN functions and the ODD functions
As already mentioned above, below we write the solutions φ[n,x] omitting the details of the resolution of the Schröedinger equation.
The equation Schröedinger has been solved in each single zone x<-a, -a≤x≤+a, x>a and then the function is required to be continuous. The g1[x], g2[x], g3[x] functions have been found in the three different areas. Each of the functions is defined exclusively in one of the three zones.
The functions g1[x] and g2[x] assume the same value at the boundary point of the two zones x=-a.
The functions g2[x] and g3[x] assume the same value at the boundary point of the two zones x=+a.
The equation Schröedinger has been solved in each single zone x<-a, -a≤x≤+a, x>a and then the function is required to be continuous. The g1[x], g2[x], g3[x] functions have been found in the three different areas. Each of the functions is defined exclusively in one of the three zones.
The functions g1[x] and g2[x] assume the same value at the boundary point of the two zones x=-a.
The functions g2[x] and g3[x] assume the same value at the boundary point of the two zones x=+a.
(*--------------------------------------------------------------------------*)(*-------------------------EVENsolutions,,-----------------------*)(*--------------------------------------------------------------------------*)
φ
0
φ
2
φ
4
In[]:=
g1[x_]:=;(*Thefunctionisdefinedforx<-a*)g2[x_]:=Cos[qx];(*Thefunctionisdefinedfor-a≤x≤+a*)g3[x_]:=;(*Thefunctionisdefinedforx>a*)
λx
-λa
Cos[qa]
-λx
We define the one function φ(x) by putting together the three previous functions.Now the following function φ[x] is defined for each x from -∞ to +∞.
φ(x):=Piecewise
g1(x) | x<-a |
g2(x) | -a≤x∧x≤a |
g3(x) | x>a |
In[]:=
φ[x_]:=Piecewise[{{g1[x],x<-a},{g2[x],-a≤x&&x≤a},{g3[x],x>a}}]
We find the normalization constant. Thus the new function CostNorm*φ(x) is normalized, that is, it results x=1
+∞
∫
-∞
2
CostNorm*φ(x)
In[]:=
CostNorm:=
1
NIntegrate[,{x,-2a,2a}]
2
Abs[φ[x]]
We write the EVEN solutions normalized. n = 0, 2 , 4:
We write the EVEN solutions normalized. n = 0, 2 , 4:
In[]:=
Energy=-0.9622;En[0]=Energy;φ[0,x_]=CostNormφ[x];Energy=-0.6647;En[2]=Energy;φ[2,x_]=CostNormφ[x];Energy=-0.1193;En[4]=Energy;φ[4,x_]=CostNormφ[x];
(*------------------------------------------------------------------------*)(*--------------------------ODDsolutions,----------------------*)(*------------------------------------------------------------------------*)
φ
1
φ
3
g1[x_]:=;(*Thefunctionisdefinedforx<-a*)g2[x_]:=-Sin[qx];(*Thefunctionisdefinedfor-a≤x≤+a*)g3[x_]:=-;(*Thefunctionisdefinedforx>a*)
λx
-λa
Sin[qa]
-λx
We define the one function φ(x) by putting together the three previous functions.Now the following function φ[x] is defined for each x from -∞ to +∞.
φ(x):=Piecewise
g1(x) | x<-a |
g2(x) | -a≤x∧x≤a |
g3(x) | x>a |
In[]:=
φ[x_]:=Piecewise[{{g1[x],x<-a},{g2[x],-a≤x&&x≤a},{g3[x],x>a}}]
We find thenormalization constant . Thus the new function CostNorm*φ (x) is normalized, that is, it results x=1
+∞
∫
-∞
2
CostNorm*φ(x)
In[]:=
CostNorm:=
1
NIntegrate[,{x,-2a,2a}]
2
Abs[φ[x]]
Now we write the ODD solutions normalized. n = 1, 3
Now we write the ODD solutions normalized. n = 1, 3
In[]:=
Energy=-0.8494;En[1]=Energy;φ[1,x_]=CostNormφ[x];Energy=-0.4142;En[3]=Energy;φ[3,x_]=CostNormφ[x];
From now on, φ[n,x] gives us the solutions with n = 0, 1, 2, 3, 4.
From now on, φ[n,x] gives us the solutions with n = 0, 1, 2, 3, 4.
The φ(n,x) satisfy the Schröedinger equation
The φ(n,x) satisfy the Schröedinger equation
In this section we prove that the φ[n, x] satisfy the Schröedinger equation .
Recall that, since the schro equation has been solved numerically (approximated), for which the solutions Phi are approximate and do not solve the eq of schr analytically (exactly).
The Schröedinger equation is
Recall that, since the schro equation has been solved numerically (approximated), for which the solutions Phi are approximate and do not solve the eq of schr analytically (exactly).
The Schröedinger equation is
-φ(n,x)+V(x)φ(n,x)-Energy(n)φ(n,x) =0withn=0,1,2,3,4
2
ℏ
2m
2
∂
∂
2
x
In what follows n is the number of the solution φ[n,x] we are testing.
For now we have set n = 0, but you can try changing the value of n (0, 1, 2, 3, 4) and do the following calculations again.
For now we have set n = 0, but you can try changing the value of n (0, 1, 2, 3, 4) and do the following calculations again.
In[]:=
n=0;
I define the following EQ[x] function using the first member of the previous Schröedinger equation.
If φ[n,x] satisfies the Schröedinger equation, then it must turn out that the following function EQ[x] is zero or close to zero for every value of x
If φ[n,x] satisfies the Schröedinger equation, then it must turn out that the following function EQ[x] is zero or close to zero for every value of x
In[]:=
EQ[x_]=-φ[n,x]+V[x]φ[n,x]-En[n]φ[n,x];
2
ℏ
2m
∂
x
∂
x
To check the values of EQ[x] for each x, let us graph the function EQ[x]
In[]:=
Plot[EQ[x],{x,-2a,2a},PlotRange{-,}]
-16
10
-16
10
Out[]=
For any n mean value of x is zero or close to zero that is the center of the well, as in the case of the classical particle.
Below we have set n = 2, but you can try changing the value of n (0, 1, 2, 3, 4) and do the following calculations again.
Below we have set n = 2, but you can try changing the value of n (0, 1, 2, 3, 4) and do the following calculations again.
The ProbP[n,p] function is the probability density of the momentum p, for any given n