Particle confined in one-dimensional box
Particle confined in one-dimensional box
Particle confined to an interval (one - dimensional box) of the x axis, in the range [0, L]
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(*Thedimensionsoftheimagesproducedareset:400isthewidthand200istheheight.*)MyImageSize={400,200};$Assumptions=x∈Reals&&(*VariablePositiononthexaxis*)p∈Reals&&(*VariableMomentum*)ℏ∈Reals&&ℏ>0&&(*Planckconstant*)L∈Reals&&L>0&&(*Boxwidth*)m∈Reals&&m>0&&(*Massoftheparticle*)n∈Integers&&n>0(*Integerfornumberofsolutions*);
Solutions of the Schröedinger equation: φ1(x),φ2(x),φ3(x)
Solutions of the Schröedinger equation: (x),(x),(x)
φ
1
φ
2
φ
3
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.
Since the function must be continuous and nothing outside the range [0,L], then to solve the Schröedinger equation it was imposed that wave function φ(x) must vanish for x=0 and x=L. It turns out that 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].
Since the function must be continuous and nothing outside the range [0,L], then to solve the Schröedinger equation it was imposed that wave function φ(x) must vanish for x=0 and x=L. It turns out that 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 energy levels are numbered with n=1,2,3...
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En[n_]:=;
1
2m
2
nπℏ
L
Thefollowingsolutionφ(n,x)=
isRealanditisnothingoutsidetherange[0,L]
2 L πnx L | 0≤x≤L |
0 | x<0∨x>L |
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φ[n_,x_]:=PiecewiseSin,0≤x≤L,{0,x<0||x>L};
2
L
nπx
L
Solution for n=1
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φ[1,x]
Out[]=
|
Solution for n = 2
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φ[2,x]
Out[]=
|
Solution for n = 3
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φ[3,x]
Out[]=
|
Theφ(n,x)satisfytheSchroedingerequation-φ(n,x)En(n)φ(n,x)
2
ℏ
2m
2
∂
∂
2
x
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-φ[n,x]En[n]φ[n,x]//Simplify
2
ℏ
2m
∂
x
∂
x
Out[]=
x<0||0<x<1||x>1
Theφ[n,x]arenormalizedforanyn,iethefollowingrelationisvalid
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∞
∫
-∞
2
φ[n,x]
Out[]=
True
Foranyn,theaveragevalueofxisobtainedwiththeformulaxxanditturnsoutthatisthatisthecenterofthebox,asinthecaseoftheclassicalparticle:
+∞
∫
-∞
2
φ(n,x)
L
2
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x
+∞
∫
-∞
2
Abs[φ[n,x]]
Out[]=
1
2
In order to create the graphs we set the value of L
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L=1;
The following V(x) function is used exclusively to display the potential in the graph
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V[x_]=Piecewise[{{10,x<0},{10,x>L}},-10];
Wedefinethefollowingfunction"GeneratePlot1"todrawthegraphsofrealfunction(x)
φ
n
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MainColorPlot=ColorData[97,"ColorList"][[1]];GeneratePlot1[f_,fLabel_]:=Plot{f,V[x]},x,-,L,PlotRange{-2,2},PlotStyle->{Default,Orange},Filling{2->Bottom},Ticks{Automatic,None},AxesStyleDirective[Thickness[0.003],Arrowheads[{0.04}]],AxesLabel{"x",Style["U(x)",Orange]},PlotPoints100,Epilog{Text[Style[fLabel,MainColorPlot],{0.5,1.8}]},BaseStyle12,ImageSizeMyImageSize,AspectRatioFull;
L
8
9
8
Weplotofrealwavefunction(x)relativetothevaluesn=1,2,3.
φ
n
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GeneratePlot1[φ[1,x],"(x)"]GeneratePlot1[φ[2,x],"(x)"]GeneratePlot1[φ[3,x],"(x)"]
φ
1
φ
2
φ
3
Out[]=
Out[]=
Out[]=
Let's fix the value of the following three constants
The following V(x) function is used exclusively to display the potential in the graph