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

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 \( z < -\frac{L_z}{2} \)
    • \( \psi{(z)} = G \exp{(\kappa z)} \)
  • for \( -\frac{L_z}{2} < z < \frac{L_z}{2} \)
    • \( \psi{(z)} = A \sin{(kz)} + B \cos{(kz)} \)
  • for \( z > \frac{L_z}{2} \)
    • \( \psi{(z)} = F \exp{(-\kappa z)} \)

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:

    • @ \( z = \frac{L_z}{2} \): \[ \begin{align} \psi{\left( \frac{L_z}{2} \right)} & = F \exp{ \left( \frac{-\kappa L_z}{2} \right) } \\
      & = A \sin{\left( \frac{k L_z}{2} \right)} + B \cos{\left( \frac{k L_z}{2} \right)} \\
      \end{align} \]

    • picking substitutions to consolidate the math: \[ \begin{align} X_L & = \exp{\left( \frac{-\kappa L_z}{2} \right)} \rightarrow \text{ the exponential part } \\
      S_L & = \sin{\left( \frac{-\kappa L_z}{2} \right)} \rightarrow \text{ the sinusoidal part } \\
      C_L & = \cos{\left( \frac{-\kappa L_z}{2} \right)} \rightarrow \text{ the co-sinusoidal part } \\
      \end{align} \]
    • plugging it back into the b.c. equation gives \[ \begin{align} FX_L & = AS_L + BC_L \\
      \end{align} \]


    • @ \( z = -\frac{L_z}{2} \):
      • using the above substitutions in the wave-equations \[ \begin{align} GX_L & = -AS_L + BC_L \\
        \end{align} \]

  • using continuity of wave slope at boundaries:

    • @ \( z = \frac{L_z}{2} \): \[ \begin{align} -\frac{\kappa}{k}GX_L = AC_L - BS_L \end{align} \]

    • @ \( z = -\frac{L_z}{2} \): \[ \begin{align} \frac{\kappa}{k}GX_L = AC_L + BS_L \end{align} \]

consolidating results of applying boundary conditions

\[ \begin{align} GX_L & = -AS_L + BC_L \tag{1} \\
FX_L & = AS_L + BC_L \tag{2}\\
-\frac{\kappa}{k}GX_L & = AC_L - BS_L \tag{3}\\
\frac{\kappa}{k}GX_L & = AC_L + BS_L \tag{4}\\
\end{align} \]

compatible solutions search

  • adding equations (1) and (2) \[ \begin{align} GX_L & = -AS_L + BC_L \tag{1} \\
    FX_L & = AS_L + BC_L \tag{2}\\
    \\
    \text{ gives: } & \\
    2BC_L & = (F+G)X_L \tag{5}\\
    \end{align} \]

  • subtracting equations (3) from (4) \[ \begin{align} -\frac{\kappa}{k}FX_L & = AC_L - BS_L \tag{3} \\
    \frac{\kappa}{k}GX_L & = AC_L + BS_L \tag{4}\\
    \\
    \text{ gives: } & \\
    2BS_L & = \frac{\kappa}{k}(F+G)X_L \tag{6} \\
    \end{align} \]


  • case 1: consider \( F \neq G \)
    • divide (6) by (5) \[ \begin{align} 2BS_L & = \frac{\kappa}{k}(F+G)X_L \tag{6} \\
      2BC_L & = (F+G)X_L \tag{5}\\
      \\
      & \text{ (6) divided by (5) gives: } \\
      \tan\left( \frac{kL_z}{2} \right) & = \frac{\kappa}{k} \tag{7} \\
      \end{align} \]
      • (7) is the condition for eigenvalues

    • subtracting (1) from (2) \[ \begin{align} FX_L & = AS_L + BC_L \tag{2}\\
      GX_L & = -AS_L + BC_L \tag{1} \\
      \\
      \text{ gives: } & \\
      2AS_L & = (F-G)X_L \tag{8}\\
      \end{align} \]

    • adding (3) and (4) \[ \begin{align} -\frac{\kappa}{k}FX_L & = AC_L - BS_L \tag{3} \\
      \frac{\kappa}{k}GX_L & = AC_L + BS_L \tag{4}\\
      \\
      \text{ gives: } & \\
      2AC_L & = -\frac{\kappa}{k}(F-G)X_L \tag{9} \\
      \end{align} \]
  • still keeping mind \( F \neq G \)
    • divide (9) by (8) \[ \begin{align} 2AC_L & = -\frac{\kappa}{k}(F-G)X_L \tag{9} \\
      2AS_L & = (F-G)X_L \tag{8} \\
      \\
      \text{ gives: } & \\
      -cot{\left( \frac{kL_z}{2} \right)} & = \frac{\kappa}{k} \tag{10} \\
      \end{align} \]
      • (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. \( F = G \) \[ \tan{\left( \frac{kL_z}{2} \right)} = \frac{\kappa}{k} \tag{7} \]
    2. \( F = -G \) \[ -cot{\left( \frac{kL_z}{2} \right)} = \frac{\kappa}{k} \tag{10} \]

exploring condition 1. \( F=G \)

  • the corresponding eigenvalue condition is: \[ \begin{align} \tan{\left( \frac{kL_z}{2} \right)} & = \frac{\kappa}{k} \tag{7} \\
    \text{ is derived from dividing } & (8) \text{ by } (9)\\
    2AS_L & = (F-G)X_L \tag{8} \\
    2AC_L & = -\frac{\kappa}{k}(F-G)X_L \tag{9} \\\
    \end{align} \]
    • 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: \( \psi(z) \propto \cos{(kz)} \)
      • since: \( \psi{(z)} = A \sin{(kz)} + B \cos{(kz)} \) and \(A = 0\)
      • even functions set
    • psi-cosine

exploring condition 2. \( F = -G \)

  • the corresponding eigenvalue condition is: \[ \begin{align} -\cot{\left( \frac{kL_z}{2} \right)} & = \frac{\kappa}{k} \tag{10} \\
    \text{ is derived from dividing } & (5) \text{ by } (6)\\
    2BC_L & = \frac{\kappa}{k}(F+G)X_L \tag{5} \\\
    2BS_L & = (F+G)X_L \tag{6} \\\
    \end{align} \]
    • 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: \( \psi(z) \propto \sin{(kz)} \)
      • since: \( \psi{(z)} = A \sin{(kz)} + B \cos{(kz)} \) and \(B = 0\)
      • odd functions sets
    • psi-cosine

nature of solutions

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

eigenwaves

  • to find eigenenergies, \(k\) and \(\kappa\) values have to be found, using \[ \begin{align} \tan{\left( \frac{kL_z}{2} \right)} & = \frac{\kappa}{k} \tag{7} \\
    -\cot{\left( \frac{kL_z}{2} \right)} & = \frac{\kappa}{k} \tag{10} \\
    \end{align} \]

  • change to ‘dimensionless’ units

borrowing the eigen energy expression

  • of the first level in the ‘infinite’ potential well width \( L_z \)** \[ \begin{align} E_{1}^{\infty} = \frac{h^2}{2m} \left( \frac{\pi}{L_z} \right)^2\\
    \end{align} \]
    • as a reference energy unit
    • so with this reference energy unit \[ \begin{align} \epsilon = \frac{E}{E_1^\infty} \end{align} \]
      • \( \epsilon \) : leading to dimensionless eigenenergy
    • dimensionless barrier height with respect to the same reference energy unit \[ \begin{align} \nu_0 = \frac{V_0}{E_1^\infty} \end{align} \]
  • also, expressing \(k\) and \(\kappa\) in terms of these energy units
    • k: \[ \begin{align} k & = \sqrt{ \frac{2mE}{h^2} } \\
      k & = \left( \frac{\pi}{L_z} \right) \sqrt{ \frac{E}{E_1^\infty} } \\
      k & = \left( \frac{\pi}{L_z} \right) \sqrt{ \epsilon } \\
      \end{align} \]
    • \( \kappa\): \[ \begin{align} \kappa & = \sqrt{ \frac{2m(V_0-E)}{h^2} } \\
      \kappa & = \left( \frac{\pi}{L_z} \right) \sqrt{ \frac{V_0 - E}{E_1^\infty} } \\
      \kappa & = \left( \frac{\pi}{L_z} \right) \sqrt{ \nu_0 - \epsilon } \\
      \end{align} \]
  • so, the ratio chunk \( \frac{\kappa}{k} \): \[ \frac{\kappa}{k} = \sqrt{\frac{V_0 - E}{E}} = \sqrt{ \frac{\nu_0 - \epsilon}{\epsilon} } \]


  • the expression chunk \( \frac{kL_z}{2} \): \[ \frac{kL_z}{2} = \frac{\pi}{2} \sqrt{ \frac{E}{E_1^\infty} } = \frac{ \pi }{ 2 } \sqrt{ \epsilon } \]
    • so \[ \tan{ \left( \frac{kL_z}{2} \right) } = \frac{\kappa}{k} \rightarrow \text{ (F = G eigenvalue condition) } \]
    • becomes \[ \begin{align} \tan{ \left[ \frac{\pi}{2} \sqrt{\epsilon} \right] } & = \sqrt{ \frac{\nu_0 - \epsilon}{\epsilon} } \\
      \sqrt{ \epsilon } \tan{ \left[ \frac{\pi}{2} \sqrt{\epsilon} \right] } & = \sqrt{ \left( \nu_0 - \epsilon \right) } \tag{11} \\
      \end{align} \]
  • the expression chunk \( \frac{\kappa L_z}{2} \): \[ \frac{\kappa L_z}{2} = \frac{\pi}{2} \sqrt{ \frac{V_0 - E}{E_1^\infty} } = \frac{ \pi }{ 2 } \sqrt{ \nu_0 - \epsilon } \]
    • similarly \[ -\cot{ \left( \frac{kL_z}{2} \right) } = \frac{\kappa}{k} \rightarrow \text{ (F = - G eigenvalue condition) } \]
    • becomes \[ \begin{align} -\cot{ \left[ \frac{\pi}{2} \sqrt{\epsilon} \right] } &= \sqrt{ \frac{\nu_0 - \epsilon}{\epsilon} } \\
      -\sqrt{ \epsilon } \cot{ \left[ \frac{\pi}{2} \sqrt{\epsilon} \right] } & = \sqrt{ \left( \nu_0 - \epsilon \right) } \tag{12} \\
      \end{align} \]

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:

\[ \begin{align} \sqrt{ \epsilon } \tan{ \left[ \frac{\pi}{2} \sqrt{\epsilon} \right] } & = \sqrt{ \left( \nu_0 - \epsilon \right) } \tag{11} \\
-\sqrt{ \epsilon } \cot{ \left[ \frac{\pi}{2} \sqrt{\epsilon} \right] } & = \sqrt{ \left( \nu_0 - \epsilon \right) }\tag{12} \\
\end{align} \]

  • for a given specific well depth \( \nu_0 \) plot of \(RHS\) \( \sqrt{ \left( \nu_0 - \epsilon \right) } \) is
    • \( \nu_0 = 2 \)
      • nu_0=2
    • \( \nu_0 = 5 \)
      • nu_0=5)
    • \( \nu_0 = 8 \)
      • nu_0=8
  • plotting LHS of equations (11) and (12) \[ \begin{align} \sqrt{ \epsilon } \tan{ \left( \frac{\pi}{\epsilon} \sqrt{\epsilon} \right) }: \textbf{ red } \\
    -\sqrt{ \epsilon } \cot{ \left( \frac{\pi}{\epsilon} \sqrt{\epsilon} \right) }: \textbf{ blue } \\
    \end{align} \]
    • nu_0=8
    • the solutions for the values of \( \epsilon \)
      • is found by the intersection of the specified \(\nu_0\) curve and the \(RHS\) equations
    • explicitly, for \( \nu_0 = 8 \)
      • nu_0=8
      • the values of \( \epsilon \) are
        • \( 0.663, 2.603, 5.609 \)
  • solutions:
    • finite-well-energies

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

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