Schrödinger Equation (6)

Daniel Fleish  A Student's Guide to the Schrödinger Equation  (Cambridge University Press  2020)

          5.2  Finite Rectangular Potential Well      

The finite rectangular well is an example of a configuration with piecewise constant potential energy of zero value inside the wall and a finite value outside the well shown in Fig. 5.10.
   The solutions to the Schrödinger equation within and outside the finite rectangular well will have some similarities and differences. Similarities includs the oscillatory nature of wavefunction 𝜓(𝑥) within the well. And the slope of the wavefunction ∂𝜓(𝑥)/∂𝑥 to be continuous across the walls. For the potential energy defined as in Fig. 5.10, the particle will be bound if 𝐸 < 𝑉0 and free if 𝐸 > 𝑉0. In this section, the energy will be taken as 0 < 𝐸 < 𝑉0. The wavefunction solutions are oscillatory inside the well, but they do not go to zero at the edges of the well. instead, they decay exponentially in that region, often called the "evanescent" region.
   Remembering in  Section 4.3 the basics of wavefunction behavior in a region of piecewise constant potential, in classically allowed regions (𝐸 > 𝑉) wavefunctions exhibit oscillatory behavior, but in classically forbidden regions (𝐸 < 𝑉) those exhibit exponentially decaying behavior. Applying these concepts to a quantum particle in a finite rectangular well with potential energy 𝑉 = 0  inside the well and 𝑉 = 𝑉0 outside the well tells us the wavefunctions of a particles with energy 𝐸 > 0 within the well will oscillate sinusoidally.
   Since inside wall 𝑉 = 0, TISE-the time independent Schrödinsr equation is
    (5.23)     𝑑2[𝜓(𝑥)]/𝑑𝑥2 = -2𝑚/ℏ2 𝐸𝜓(𝑥) = -𝑘2𝜓(𝑥),
    (5.24)     𝑘 ≡ √(2𝑚/ℏ2 𝐸),
exactly as in the case of the infinite rectangular well, constant 𝑘 represents the wavenumber and 𝑘 = 2π/𝜆. The solutions to Eq. 5.23 may be written using either exponentials or sinusoidal functions. So the general solution may be written as
    (5.25)     𝜓(𝑥) = 𝐴 cos (𝑘𝑥) + 𝐵 sin (𝑘𝑥),
where constants 𝐴 and 𝐵 are determined by boundary conditions.
   In the regions to the left and right potential well the potential energy 𝑉(𝑥) = 𝑉0 exceeds the total energy 𝐸, so the 𝐸 < 𝑉0, and these are classically forbidden regions. In those regions, the TISE can be written as
    (5.26)     𝑑2[𝜓(𝑥)]/𝑑𝑥2 = -2𝑚/ℏ2 (𝐸 - 𝑉0) 𝜓(𝑥) = +𝜅2𝜓(𝑥),
    (5.27)     𝜅 ≡ √[2𝑚/ℏ2 (𝑉0 - 𝐸 )],
where the constant 𝜅 is a "decay constant" that determines the rate at which the wavenumber tends towards zero as section 4.3 explains. The larger the decay constant 𝜅, and the faster the wavefunction decays over 𝑥. The general solution in exponential functions is
    (5.28)     𝜓(𝑥) = 𝐶𝑒𝜅𝑥 + 𝐷𝑒-𝜅𝑥,
where constants 𝐶 and 𝐷 are determined by boundary conditions.
   Before applying the boundary condition, we can determine something about 𝐶 and 𝐷. Calling those constants, 𝐶left and 𝐷left to the left of the well (𝑥 < -𝑎/2) which we will call Region I and the well Region II, 𝐷left𝑒-𝜅𝑥 will become infinitely large unless 𝐷left is zero there. Likewise, calling the constants 𝐶right and 𝐷right to the right of the well (𝑥 < 𝑎/2) which we will call Region III, 𝐶right𝑒𝜅𝑥 will become infinitely large unless 𝐶right is zero there.
   So 𝜓(𝑥) = 𝐶left𝑒𝜅𝑥 in Region I where 𝑥 is negative, and 𝜓(𝑥) = 𝐷right𝑒-𝜅𝑥 in Region III where 𝑥 is positive. And since the symmetry of the potential 𝑉(𝑥) about 𝑥 = 0 means that the wavefunction 𝜓(𝑥) must have either even or odd parity across all values of 𝑥 (referring to Fig. 5.2), we also know that 𝐶left must equal 𝐷right for even solutions and that 𝐶left must equal -𝐷right for odd solutions. So for even solutions we can write 𝐶left = 𝐷right = 𝐶, and for odd solutions 𝐶left = 𝐶 and 𝐷right = -𝐶.
   these conclusions about wavefunction 𝜓(𝑥) are summarized in the following table, which also shows the first spatial derivative ∂𝜓(𝑥)/∂𝑥 in each region.
              Behavior/Region:     Evanescent (I)               Oscillatory (II)                       Evanescent (III) 
                 𝜓(𝑥):                        𝐶𝑒𝜅𝑥                  𝐴 cos (𝑘𝑥) or 𝐵 sin (𝑘𝑥)               𝐶𝑒-𝜅𝑥 or -𝐶𝑒-𝜅𝑥
                ∂𝜓(𝑥)/∂𝑥:                 𝜅𝐶𝑒𝜅𝑥               -𝑘𝐴 sin (𝑘𝑥) or 𝑘𝐵 cos (𝑘𝑥)         -𝜅𝐶𝑒-𝜅𝑥 or 𝜅𝐶𝑒-𝜅𝑥

   Considering the even solutions first, matching the amplitude of 𝜓(𝑥) across the wall at the left edge of the wall, 𝑥 = 𝑎/2, gives
    (5.29)     𝐶𝑒𝜅(-𝑎/2) = 𝐴 cos [𝑘(-𝑎/2)],
    (5.30)     𝜅𝐶𝑒𝜅(-𝑎/2) = -𝑘𝐴 sin [𝑘(-𝑎/2)].
   If we now divide Eq. 5.30 by Eq. 5.29. (forming a quantity called the logarithmic derivative, which is 1/𝜓 ∂𝜓/∂𝑥), the result is
    (5.31)     𝜅𝐶𝑒𝜅(-𝑎/2)/𝐶𝑒𝜅(-𝑎/2) =  -𝑘𝐴 sin [𝑘(-𝑎/2)]/𝐴 cos [𝑘(-𝑎/2)], or
    (5.32)     𝜅 = -𝑘 tan (-𝑘𝑎/2) = 𝑘 tan (𝑘𝑎/2),
    (5.33)     𝜅/𝑘 = tan (𝑘𝑎/2).
This is the mathematical expression of the requirement that the amplitude and slope of the wavefunction 𝜓(𝑥) must be continuous accross the boundary between regions.
   As we can see in Fig. 5.10, this equation tell us that the constant 𝜅 determines the curvature (and thus the decay rate) of 𝜓(𝑥) in the region I and III, and the 𝜅 is proportional to the square root of 𝑉0 - 𝐸. So in the region I and III, the value of 𝜓(𝑥) and its slope are determined by the "depth" of the particles's energy level in the well.
   Now think about the wavenumber 𝑘 in the oscillatory region II. Since 𝑘 is proportional to the square root of the energy 𝐸 and 𝑉0 = 0 there, the value of 𝜓(𝑥) and the slope on the inside of the potential well boundaries re determined by the "height" of the particle's energy level in the well.
   And here's the payoff of this logic: only certain value of energy will cause both 𝜓(𝑥) and its first derivative ∂𝜓(𝑥)/∂𝑥 to be continuous across the boundaries of the finite potential well. For even solutions, those ratios are given by Eq. 5.33. Sketches of wavefunctions with matching or mismatching slopes at the edges of the well are shown in Fig. 5.10.
   Because Eq. 5.33 is a transcendental equation which cannot be solved analytically, we can imagine solving this equation graphically. To do this it may help to start by considering a simple transcendendtal equation, such as
    (5.34)     𝑥/4 = cos (𝑥).
The solution th this equation can be read off the graph shown in Fig. 5.11. This graph makes the solutions to the equation clear: just look for the values of 𝑥 at which  the lines cross. In this example, those values are near 𝑥 = -3.595, 𝑥 = -2.133, and 𝑥 = +1.252.
   For Eq. 5.33 things are a bit more complex. So we better take a look at the graphical solutions for several finite potential wells with specified width and depth. In Fig. 5.12 we cn see the graphical solution process at work for three finite rectangular wells, all with width 𝑎 = 2.5 ⨯ 10-10m and potential energy 𝑉0 = 0 of 2, 60, and 160 eV. The three solid curves represent the ratio 𝜅/𝑘 for the three values of 𝑉0 and the three dot lines do the values of tan (𝑘𝑎/2). The horizontal axis of this graph represents a range of values of 𝑘𝑎/2 and so they increases from just about zero to approximately 3π, the ratio 𝜅/𝑘 decreases because 𝑘 is getting bigger.
   If we take particle to have the mass of an electron (𝑚 = 9.11 ⨯ 10-31 kg). Solving Eq. 5.24 for 𝐸 gives
    (5.34, 35)     √(2𝑚/ℏ2 𝐸) = 𝑘   or   𝐸 = ℏ2𝑘2/2𝑚.
    (5.36)     𝐸 = [ℏ2(𝑘𝑎/2)2(2/𝑎)2]/2𝑚 = 2 [ℏ2(𝑘𝑎/2)2]/𝑚𝑎2.
   So for the range of 𝑘𝑎/2 of approximately 0 to 3π shown in Fig. 5.12, the energy range on the plot extends from 𝐸 = 0 to
                𝐸 = 2 [ℏ2(3π)2]/𝑚𝑎2-34 Js)2(3π)2]/[(9.11 ⨯ 10-31 kg)(2.5 ⨯ 10-10 m2] = 3.47 ⨯ 10-17 J = 216.6 eV.
   Knowing how to convert of 𝑘𝑎/2 to values of energy allows us to determine the allowed energy levels of finite potential well. So in case of 𝑉0 = 160 eV, the intersections occur at 𝑘𝑎/2 values of 0.445π, 1.33π, and 2.18π. Plugging those values into Eq. 5.36 gives the allowed energy values of 4.76 eV, 42.4 eV, and 114.3 eV less than 𝑉0 = 160 eV.
   It's also instructive to compare these energies to the allowed energies of the infinite rectangular well, given in the previous section as
    (5.7)     𝐸𝑛 = 𝑘n22/2𝑚 = 𝑛2π22/2𝑚𝑎2.
Inserting 𝑚 = 9.11 ⨯ 10-31 kg and 𝑎 = 2.5 ⨯ 10-10m into this equation gives the lowest six energy levels (𝑛 = 1 to 6):
               𝐸1 = 6.02 eV         𝐸3 = 54.2 eV         𝐸5 = 150.4 eV
               𝐸2 = 24.1 eV         𝐸4 = 96.3 eV         𝐸6 = 216.6 eV
where the ∞ superscript is a reminder of the infinite rectangular well.
   Thus the ground-state energy for this potential well (𝐸 ≈ 4.8 eV) compares to 𝐸1 = 6.02 eV; the ratio is 4.8/6.02 = 0.8. Comparing the next two even solution energy levels for the finite well to 𝐸3 and 𝐸5 gives ratio of 42.4/54.2 = 0.78 and 114.3/150.4 = 0.76. 
   Also with a given depth on the number of allowed energies varying the width of a finite potential well causes some change of solutions, for example, for the depth of the well 2 eV we can compare the widths of the wells which are 𝑎 = 2.5 ⨯ 10-10 m, 𝑎 = 10 ⨯ 10-10m, and 𝑎 = 25 ⨯ 10-10 m respectively. We can see that for finite rectangular wells of the same depth, wider wells may support a large number of allowed energy levels and interestingly, the result looks similar to Fig. 5.12, if we imagine 60 eV line for 𝑎 = 10 ⨯ 10-10 m case and 160 eV line for 𝑎 = 25 ⨯ 10-10 m case in terms of shape. 
   Now we may consider an alternative form of this equation by multiplying both sides of Eq. 5.33 by the factor 𝑘𝑎/2: 
       (5.37)     (𝑘𝑎/2)𝜅/𝑘 = (𝑘𝑎/2)tan (𝑘𝑎/2)     or     (𝑎/2)𝜅 = (𝑘𝑎/2)tan (𝑘𝑎/2).
We can see the effect in Fig. 5.15.
   To understand why the (𝑎/2)𝜅 function produces circles when plotted with 𝑘𝑎/2 on horizontal axis we should recall that the wavenumber and the total energy are related by the equations  
    (5.24, 36)     𝑘 ≡ √[2𝑚/ℏ2)𝐸]     and      𝐸  = 2ℏ2(𝑘𝑎/2)2/𝑚𝑎2.
Now define a reference wavenumber 𝑘0 as the wavenumber that the particle under consideration would have if its total energy were 𝑉0:
    (5.38)     𝑘0 ≡ √[𝑉0 2𝑚/ℏ2)],  which means
    (5.39)     𝑉0 = ℏ2𝑘02/2𝑚 = 2ℏ2(𝑘0𝑎/2)2/𝑚𝑎2. Also recalling
    (5.27)     𝜅 ≡ √[2𝑚/ℏ2 (𝑉0 - 𝐸 )], so
    (5.40)     𝜅 = √[2𝑚/ℏ2 {2ℏ2(𝑘0𝑎/2)2/𝑚𝑎2 - 2ℏ2(𝑘𝑎/2)2/𝑚𝑎2}] = √[4/𝑎2 {(𝑘0𝑎/2)2 -(𝑘𝑎/2)2}]     or     (𝑎/2)𝜅 = √[(𝑘0𝑎/2)2 -(𝑘𝑎/2)2].
This equation has the form of a circle of radius 𝑅:
                𝑥2 + 𝑦2 = 𝑅2     or     𝑦 = √(𝑅2 - 𝑥2).
So plotting (𝑎/2)𝜅 on the 𝑦-axis and 𝑘𝑎/2 on the 𝑥-axis results in circles of radius 𝑘0𝑎/2, as seen in Fig. 5.15.

   Considering the odd solutions now, matching the amplitude of 𝜓(𝑥) across the wall at the left edge of the wall, 𝑥 = 𝑎/2, gives
    (5.41)     𝐶𝑒𝜅(-𝑎/2) = 𝐵 sin [𝑘(-𝑎/2)],
and find the slope of the wavefunction by equating the first spatial derivatives
    (5.42)     𝜅𝐶𝑒𝜅(-𝑎/2) = 𝑘𝐵 cos [𝑘(-𝑎/2)].
   As in the even solution case, dividing the Eq. 5.42 by Eq. 5.41, which gives
    (5.43)     𝜅𝐶𝑒𝜅(-𝑎/2)/𝐶𝑒𝜅(-𝑎/2) = {𝑘𝐵 cos [𝑘(-𝑎/2)]}/{𝐵 sin [𝑘(-𝑎/2)]}, or
    (5.44)     𝜅 = 𝑘 cot (-𝑘𝑎/2) = -𝑘 cot (𝑘𝑎/2).
Dividing both sides of this equation by 𝑘 gives
    (5.45)     𝜅/𝑘 = -cot (𝑘𝑎/2).
which is the odd-solution version of the transcendental equation Eq. 5.33. We should note that its left side is identical to the even-solution case, but the right side involves the negative cotangent rather than the positive tangent of 𝑘𝑎/2.
   The graphical approach to solving Eq. 5.45 is possible and an alternative form of the transcendental equation can also be found for the odd-party solutions. One striking difference between the graphics for-odd parity solution and the graphics for even-parity solution is the lack of odd-parity solutions for the 𝑉0 = 2 eV potential well. That's one sequence of the shifting of the negative-cotangent curves by π/2 to the right relative to the tangent curves of the even- parity case. However, if we increase the well width, odd solutions become possible even for a shallow well. So the conclusion is that "small" or "weak" potential well may not support any odd-parity solutions, but they always support at least one even-parity solution.
   When we consider the same condition of the even case with potential energy 𝑉0 = 2 eV, 60 eV, and 160 eV, the intersections ocuur at 𝑘𝑎/2 values of 0.888π, 1.76π, and 2.55π. Those values results the allowed energy values of 19.0 eV, 74.6 eV, and 156.3 eV.
   Also the corresponding energy levels of the same particle in an inifinite rectangluar well with the same width are, remembering that the lowest energy level comes from the first even solution, 𝐸2 = 24.1 eV, 𝐸4 = 96.3 eV, and 𝐸6 = 216.6 eV. So as for the even solutions, the energy levels of the 160 eV finite well are 70% to 80% of the energies of the corresponding infinite well.
   The wavefunctions for all six allowed energy levels for a finite potential well with 𝑉0 = 160 eV and width 𝑎 = 2.5 ⨯ 10-10 m are shown in Fig. 5.20. As we can see, the ground-stae wavefunction is an even function with low curvature (due to small value of 𝐸, which means small wavenumber 𝑘) and fast decay in the evanescent region (due to large value of 𝑉0 - 𝐸, which means large decay constant 𝜅). As energy 𝐸 and wavenumber 𝑘 increase, the curvature becomes larger, meaning the more cycles fit within the well. But larger 𝐸 means smaller values of 𝑉0 - 𝐸, so 𝜅 decreases, and smaller decay rates means larger penetration into the classically forbidden region.

   The last bit of business for the finite potential well is the substitution of variables. That is to make a new variable 𝑧, defined as the wavenumber 𝑘 and the half-width of the potential well 𝑎/2, so 𝑧 ≡ 𝑘𝑎/2. Since 𝑘 = 2π/𝜆, where 𝜆 is wavelength, 𝑧 represents the number of wavelengths in the half width 𝑎/2, converted to radian by the factor of 2π.
   Then we can repeat the previous process with 𝑧 as follows
    (5.46)     𝑧 ≡ 𝑘𝑎/2 = [√(𝐸 2𝑚/ℏ2)]𝑎/2, or
    (5.47)     𝐸 = (2/𝑎)2(ℏ2/2𝑚)𝑧2. and
    (5.48)     𝜅/(𝑧 2/𝑎/) = tan (𝑧). also
    (5.49)     𝑧0 ≡ 𝑘0𝑎/2, because 𝑘0 = √(𝑉0 2𝑚/ℏ2) in Eq. 5.38,
    (5.50)     𝑧0 = √(𝑉0 2𝑚/ℏ2) a/2, or
    (5.51)     𝑉0 = (2/𝑎)2(ℏ2/2𝑚)𝑧02. Therefore
                𝜅 ≡ √[2𝑚/ℏ2 (𝑉0 - 𝐸 )] = √[2𝑚/ℏ2{(2/𝑎)2(ℏ2/2𝑚)𝑧02 - (2/𝑎)2(ℏ2/2𝑚)𝑧2} = √[4/𝑎2 (𝑧02 - 𝑧2)] = 2/𝑎√(𝑧02 - 𝑧2). So that substitution gives
    (5.52)     𝜅/(𝑧 2/𝑎/) = 2/𝑎√(𝑧02 - 𝑧2)/(𝑧 2/𝑎/) = tan (𝑧)     or     √(𝑧02/𝑧2 -1) = tan (𝑧).
This is form of the even-solution transcendetal equation is entirely equivalent to Eq. 5.33. And an alternative form of this equation, equivalent to Eq. 5.37 (𝑘𝑎/2)𝜅/𝑘 = (𝑘𝑎/2)tan (𝑘𝑎/2), becomes
    (5.53)     𝑧√(𝑧02/𝑧2 -1) = 𝑧 tan (𝑧)     or     √(𝑧02 - 𝑧2) = 𝑧 tan (𝑧). This "𝑧-version" of Eq. 5.37 looks simple and makes clear that the circular nature of the curve produced by plotting on the vertical axis of 𝑧 tan (𝑧) with the horizontal axis of 𝑧.
   And we can continue to apply the same substitution of variables to the odd-parity solutions for the finite potential well.  Then Eq. 5.45 gives
    (5.53)     √(𝑧02/𝑧2 -1) = -cot (𝑧),  
   Multiplying both side by 𝑧 gives the alternative equation:
    (5.54)     √(𝑧02 - 𝑧2) = 𝑧 cot (𝑧).

   The payback for the extra effort required for the finite well is more realistic representation of physically realizable condition than the infinite well. But the use of piecewise-constant potentials limits the applicability of the finite-well model.   

go to top