[QMSE] W03 - Finite Potential Well

potential well problems

1. particles in a finite well
   • a common practical problem when designing various modern optoelectronic devices
2. harmonic oscillator problem
   • particle in a parabolic shaped well
   • occurs in real world oscillating systems
   • will explore only non oscillating parts

• both problems will be solved mathematically
  • math more involved than particle in a box problem
  • the analytic solutions offer insight that is useful in practical QM systems

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finite potential well

• consider an electron in a finite depth well
• only a finite number of the energy levels exist within the limits of the well
  • because of the finite height of the well
• these energy levels have sinusoidal solutions
  • due to finite height of walls
  • the solutions penetrate the walls
  • non-zero at walls, unlike infinite height wells
• higher energy solutions penetrate further into the walls
• each higher energy has one zero crossing that the lower one
  • similar to the infinite height walls

note

• this problem is att the limit of complexity for analytical solutions
  • more complicated problems are solved numerically

setup

• height of the potential barrier is $V_0$
  • potential energy at bottom is $0$
• thickness of the well is $L_z$
  • position origin $0$ is halfway along the thickness
  • such that the well walls are at $-\frac{L_z}{2}$ and $\frac{L_z}{2}$

assumptions

• for an eigenenergy $E$ for which there is a solution
  • the solution has to take the form of a sinusoid in the middle of the well
  • and has to be exponentially decaying on both sides of the walls
  • the exponentially growing sides are discarded as it doesn’t agree with current understand of physics
    • the particle would not be in the middle if these are not dropped according to current assumptions

energy functions

• for some eigenenergy $E$
  • with $k=\sqrt{\frac{2mE}{h^2}}$
    • $k$: magnitude in the middle of the well
  • with $\kappa =\sqrt{\frac{2m(V_0-E)}{h^2}}$
    • $\kappa$: exponential decay constant for wave in the wall barrier
• for $z<-\frac{L_z}{2}$
  • $\psi \left(z\right)=G\exp \left(\kappa z\right)$
• for $-\frac{L_z}{2}<z<\frac{L_z}{2}$
  • $\psi \left(z\right)=A\sin \left(kz\right)+B\cos \left(kz\right)$
• for $z>\frac{L_z}{2}$
  • $\psi \left(z\right)=F\exp (-\kappa z)$

boundary conditions

• needed to solve for unknown coefficients
  • constants: $A,B,F$ and $G$
• atleast three has to obtained by solving boundary conditions
  • the fourth is found by normalization

• using continuity of wave at boundaries:

  • @ $z=\frac{L_z}{2}$: $$\begin{array}{cc}\psi \left(\frac{L_z}{2}\right)& =F\exp \left(\frac{-\kappa L_z}{2}\right)\\ & =A\sin \left(\frac{k{L}_z}{2}\right)+B\cos \left(\frac{k{L}_z}{2}\right)\end{array}$$

  • picking substitutions to consolidate the math: $$\begin{array}{cc}{X}_L& =\exp \left(\frac{-\kappa L_z}{2}\right)\to \text{the exponential part}\\ S_L& =\sin \left(\frac{-\kappa L_z}{2}\right)\to \text{the sinusoidal part}\\ C_L& =\cos \left(\frac{-\kappa L_z}{2}\right)\to \text{the co-sinusoidal part}\end{array}$$
  • plugging it back into the b.c. equation gives $$\begin{array}{cc}F{X}_L& =A{S}_L+B{C}_L\end{array}$$

  
  

  • @ $z=-\frac{L_z}{2}$:
    • using the above substitutions in the wave-equations $$\begin{array}{cc}G{X}_L& =-A{S}_L+B{C}_L\end{array}$$

• using continuity of wave slope at boundaries:

  • @ $z=\frac{L_z}{2}$: $$\begin{array}{c}-\frac{\kappa }{k}G{X}_L=A{C}_L-B{S}_L\end{array}$$

  • @ $z=-\frac{L_z}{2}$: $$\begin{array}{c}\frac{\kappa }{k}G{X}_L=A{C}_L+B{S}_L\end{array}$$

consolidating results of applying boundary conditions

$$\begin{array}{}G{X}_L& =-A{S}_L+B{C}_L& F{X}_L& =A{S}_L+B{C}_L& -\frac{\kappa }{k}G{X}_L& =A{C}_L-B{S}_L& \frac{\kappa }{k}G{X}_L& =A{C}_L+B{S}_L& \text{(1)}& \text{(2)}& \text{(3)}& \text{(4)}\end{array}$$

compatible solutions search

• adding equations (1) and (2) $$\begin{array}{cc}G{X}_L& =-A{S}_L+B{C}_L& F{X}_L& =A{S}_L+B{C}_L& & \\ \text{gives:}& \\ 2B{C}_L& =(F+G)X_L& \text{(1)}& \text{(2)}& \text{(5)}\end{array}$$

• subtracting equations (3) from (4) $$\begin{array}{cc}-\frac{\kappa }{k}F{X}_L& =A{C}_L-B{S}_L& \frac{\kappa }{k}G{X}_L& =A{C}_L+B{S}_L& & \\ \text{gives:}& \\ 2B{S}_L& =\frac{\kappa }{k}(F+G)X_L& \text{(3)}& \text{(4)}& \text{(6)}\end{array}$$

• case 1: consider $F\ne G$
  • divide (6) by (5) $$\begin{array}{cc}2B{S}_L& =\frac{\kappa }{k}(F+G)X_L& 2B{C}_L& =(F+G)X_L& & \\ & \text{(6) divided by (5) gives:}\\ \tan \left(\frac{k{L}_z}{2}\right)& =\frac{\kappa }{k}& \text{(6)}& \text{(5)}& \text{(7)}\end{array}$$
    • (7) is the condition for eigenvalues

  • subtracting (1) from (2) $$\begin{array}{cc}F{X}_L& =A{S}_L+B{C}_L& G{X}_L& =-A{S}_L+B{C}_L& & \\ \text{gives:}& \\ 2A{S}_L& =(F-G)X_L& \text{(2)}& \text{(1)}& \text{(8)}\end{array}$$

  • adding (3) and (4) $$\begin{array}{cc}-\frac{\kappa }{k}F{X}_L& =A{C}_L-B{S}_L& \frac{\kappa }{k}G{X}_L& =A{C}_L+B{S}_L& & \\ \text{gives:}& \\ 2A{C}_L& =-\frac{\kappa }{k}(F-G)X_L& \text{(3)}& \text{(4)}& \text{(9)}\end{array}$$
• still keeping mind $F\ne G$
  • divide (9) by (8) $$\begin{array}{cc}2A{C}_L& =-\frac{\kappa }{k}(F-G)X_L& 2A{S}_L& =(F-G)X_L& & \\ \text{gives:}& \\ -cot\left(\frac{k{L}_z}{2}\right)& =\frac{\kappa }{k}& \text{(9)}& \text{(8)}& \text{(10)}\end{array}$$
    • (10) is the condition for eigenvalues

• when $F=G$
  • leaves (9)
  • but not (10)
• when $F=-G$
  • leaves (10)
  • but not (9)
• other than these two cases
  • (10) and (9) eigenvalue conditions are contradictory

• so the two eigenvalue conditions are
  1. $F=G$ $$\begin{array}{}\tan \left(\frac{k{L}_z}{2}\right)=\frac{\kappa }{k}& \text{(7)}\end{array}$$
  2. $F=-G$ $$\begin{array}{}-cot\left(\frac{k{L}_z}{2}\right)=\frac{\kappa }{k}& \text{(10)}\end{array}$$

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exploring condition 1. $F=G$

• the corresponding eigenvalue condition is: $$\begin{array}{cc}\tan \left(\frac{k{L}_z}{2}\right)& =\frac{\kappa }{k}& \text{is derived from dividing}& \left(8\right)\text{by}\left(9\right)\\ 2A{S}_L& =(F-G)X_L& 2A{C}_L& =-\frac{\kappa }{k}(F-G)X_L& & \\ \text{(7)}& \text{(8)}& \text{(9)}\end{array}$$
  • here $F=G$ sets the RHS of both (8) and (9) to $0$
  • $S_L$ (sine) and $C_L$ (cosine) cannot both be $0$ at the same time
    • so $A=0$
  • inside the well: $\psi \left(z\right)\propto \cos \left(kz\right)$
    • since: $\psi \left(z\right)=A\sin \left(kz\right)+B\cos \left(kz\right)$ and $A=0$
    • even functions set
  • 

exploring condition 2. $F=-G$

• the corresponding eigenvalue condition is: $$\begin{array}{cc}-\cot \left(\frac{k{L}_z}{2}\right)& =\frac{\kappa }{k}& \text{is derived from dividing}& \left(5\right)\text{by}\left(6\right)\\ 2B{C}_L& =\frac{\kappa }{k}(F+G)X_L& 2B{S}_L& =(F+G)X_L& & \\ \text{(10)}& \text{(5)}& \text{(6)}\end{array}$$
  • here $F=-G$ sets the RHS of both (5) and (6) to $0$
  • $S_L$ (sine) and $C_L$ (cosine) cannot both be $0$ at the same time
    • so $B=0$
  • inside the well: $\psi \left(z\right)\propto \sin \left(kz\right)$
    • since: $\psi \left(z\right)=A\sin \left(kz\right)+B\cos \left(kz\right)$ and $B=0$
    • odd functions sets
  • 

nature of solutions

• without solving for the actual eigenenergies, the nature of the solution waves in the finite well are as follows
  • 

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eigenwaves

• to find eigenenergies, $k$ and $\kappa$ values have to be found, using $$\begin{array}{}\tan \left(\frac{k{L}_z}{2}\right)& =\frac{\kappa }{k}& -\cot \left(\frac{k{L}_z}{2}\right)& =\frac{\kappa }{k}& \text{(7)}& \text{(10)}\end{array}$$

• change to ‘dimensionless’ units

borrowing the eigen energy expression

• of the first level in the ‘infinite’ potential well width $L_z$** $$\begin{array}{c}{E}_1^{\infty }=\frac{h^2}{2m}{\left(\frac{\pi }{L_z}\right)}^2\end{array}$$
  • as a reference energy unit
  • so with this reference energy unit $$\begin{array}{c}ϵ=\frac{E}{E_1^{\infty }}\end{array}$$
    • $ϵ$ : leading to dimensionless eigenenergy
  • dimensionless barrier height with respect to the same reference energy unit $$\begin{array}{c}{\nu }_0=\frac{V_0}{E_1^{\infty }}\end{array}$$
• also, expressing $k$ and $\kappa$ in terms of these energy units
  • k: $$\begin{array}{cc}k& =\sqrt{\frac{2mE}{h^2}}\\ k& =\left(\frac{\pi }{L_z}\right)\sqrt{\frac{E}{E_1^{\infty }}}\\ k& =\left(\frac{\pi }{L_z}\right)\sqrt{ϵ}\end{array}$$
  • $\kappa$: $$\begin{array}{cc}\kappa & =\sqrt{\frac{2m(V_0-E)}{h^2}}\\ \kappa & =\left(\frac{\pi }{L_z}\right)\sqrt{\frac{V_0-E}{E_1^{\infty }}}\\ \kappa & =\left(\frac{\pi }{L_z}\right)\sqrt{{\nu }_0-ϵ}\end{array}$$
• so, the ratio chunk $\frac{\kappa }{k}$: $$\frac{\kappa }{k}=\sqrt{\frac{V_0-E}{E}}=\sqrt{\frac{{\nu }_0-ϵ}{ϵ}}$$

• the expression chunk $\frac{k{L}_z}{2}$: $$\frac{k{L}_z}{2}=\frac{\pi }{2}\sqrt{\frac{E}{E_1^{\infty }}}=\frac{\pi }{2}\sqrt{ϵ}$$
  • so $$\tan \left(\frac{k{L}_z}{2}\right)=\frac{\kappa }{k}\to \text{(F = G eigenvalue condition)}$$
  • becomes $$\begin{array}{cc}\tan \left[\frac{\pi }{2}\sqrt{ϵ}\right]& =\sqrt{\frac{{\nu }_0-ϵ}{ϵ}}\\ \sqrt{ϵ}\tan \left[\frac{\pi }{2}\sqrt{ϵ}\right]& =\sqrt{({\nu }_0-ϵ)}& \text{(11)}\end{array}$$
• the expression chunk $\frac{\kappa L_z}{2}$: $$\frac{\kappa L_z}{2}=\frac{\pi }{2}\sqrt{\frac{V_0-E}{E_1^{\infty }}}=\frac{\pi }{2}\sqrt{{\nu }_0-ϵ}$$
  • similarly $$-\cot \left(\frac{k{L}_z}{2}\right)=\frac{\kappa }{k}\to \text{(F = - G eigenvalue condition)}$$
  • becomes $$\begin{array}{cc}-\cot \left[\frac{\pi }{2}\sqrt{ϵ}\right]& =\sqrt{\frac{{\nu }_0-ϵ}{ϵ}}\\ -\sqrt{ϵ}\cot \left[\frac{\pi }{2}\sqrt{ϵ}\right]& =\sqrt{({\nu }_0-ϵ)}& \text{(12)}\end{array}$$

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graphical solution

• needs a computational graphing tool with ability to view and record points of intersection between curves
• consider the reduced equations (11) and (12) from above:

$$\begin{array}{}\sqrt{ϵ}\tan \left[\frac{\pi }{2}\sqrt{ϵ}\right]& =\sqrt{({\nu }_0-ϵ)}& -\sqrt{ϵ}\cot \left[\frac{\pi }{2}\sqrt{ϵ}\right]& =\sqrt{({\nu }_0-ϵ)}& \text{(11)}& \text{(12)}\end{array}$$

• for a given specific well depth ${\nu }_0$ plot of $RHS$ $\sqrt{({\nu }_0-ϵ)}$ is
  • ${\nu }_0=2$
    • 
  • ${\nu }_0=5$
    • 
  • ${\nu }_0=8$
    • 
• plotting LHS of equations (11) and (12) $$\begin{array}{c}\sqrt{ϵ}\tan \left(\frac{\pi }{ϵ}\sqrt{ϵ}\right):\text{red}\\ -\sqrt{ϵ}\cot \left(\frac{\pi }{ϵ}\sqrt{ϵ}\right):\text{blue}\end{array}$$
  • 
  • the solutions for the values of $ϵ$
    • is found by the intersection of the specified ${\nu }_0$ curve and the $RHS$ equations
  • explicitly, for ${\nu }_0=8$
    • 
    • the values of $ϵ$ are
      • $0.663,2.603,5.609$
• solutions:
  • 

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comparison to infinite well solutions

• compared to solutions for the infinitely deep well of the same width
  • finite well solutions are at lower energy
    • the heights lesser since they have lesser kinetic energy
    • due to tunneling into the potential barriers