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

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

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 = 𝑉

Since inside wall 𝑉 = 0,

(5.23) 𝑑

(5.24) 𝑘 ≡ √(2𝑚/ℏ

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) = 𝑉

(5.26) 𝑑

(5.27) 𝜅 ≡ √[2𝑚/ℏ

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) = 𝐶𝑒

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, 𝐶

So 𝜓(𝑥) = 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): 𝐶𝑒

∂𝜓(𝑥)/∂𝑥: 𝜅𝐶𝑒

Considering the

(5.29) 𝐶𝑒

(5.30) 𝜅𝐶𝑒

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) 𝜅𝐶𝑒

(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

Now think about the wavenumber 𝑘 in the oscillatory region II. Since 𝑘 is proportional to the square root of the energy 𝐸 and 𝑉

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

Because Eq. 5.33 is a

(5.34) 𝑥/4 = cos (𝑥).

The solution th this equation can be read off the graph shown in

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

If we take particle to have the mass of an electron (𝑚 = 9.11 ⨯ 10

(5.34, 35) √(2𝑚/ℏ

(5.36) 𝐸 = [ℏ

So for the range of 𝑘𝑎/2 of approximately 0 to 3π shown in

𝐸 = 2 [ℏ

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 𝑉

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) 𝐸

Inserting 𝑚 = 9.11 ⨯ 10

𝐸

𝐸

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 𝐸

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

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

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/ℏ

Now define a reference wavenumber 𝑘

(5.38) 𝑘

(5.39) 𝑉

(5.27) 𝜅 ≡ √[2𝑚/ℏ

(5.40) 𝜅 = √[2𝑚/ℏ

This equation has the form of a circle of radius 𝑅:

𝑥

So plotting (𝑎/2)𝜅 on the 𝑦-axis and 𝑘𝑎/2 on the 𝑥-axis results in circles of radius 𝑘

Considering the

(5.41) 𝐶𝑒

and find the slope of the wavefunction by equating the first spatial derivatives

(5.42) 𝜅𝐶𝑒

As in the even solution case, dividing the Eq. 5.42 by Eq. 5.41, which gives

(5.43) 𝜅𝐶𝑒

(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 𝑉

When we consider the same condition of the even case with potential energy 𝑉

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, 𝐸

The wavefunctions for all six allowed energy levels for a finite potential well with 𝑉

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/ℏ^{2})]𝑎/2, or

(5.47) 𝐸 = (2/𝑎)^{2}(ℏ^{2}/2𝑚)𝑧^{2}. and

(5.48) 𝜅/(𝑧 2/𝑎/) = tan (𝑧). also

(5.49) 𝑧_{0} ≡ 𝑘_{0}𝑎/2, because 𝑘_{0} = √(𝑉_{0} 2𝑚/ℏ^{2}) in Eq. 5.38,

(5.50) 𝑧_{0} = √(𝑉_{0} 2𝑚/ℏ^{2}) a/2, or

(5.51) 𝑉_{0} = (2/𝑎)^{2}(ℏ^{2}/2𝑚)𝑧_{0}^{2}. Therefore

𝜅 ≡ √[2𝑚/ℏ^{2} (𝑉_{0} - 𝐸 )] = √[2𝑚/ℏ^{2}{(2/𝑎)^{2}(ℏ^{2}/2𝑚)𝑧_{0}^{2} - (2/𝑎)^{2}(ℏ^{2}/2𝑚)𝑧^{2}} = √[4/𝑎^{2} (𝑧_{0}^{2} - 𝑧^{2})] = 2/𝑎√(𝑧_{0}^{2} - 𝑧^{2}). So that substitution gives

(5.52) 𝜅/(𝑧 2/𝑎/) = 2/𝑎√(𝑧_{0}^{2} - 𝑧^{2})/(𝑧 2/𝑎/) = tan (𝑧) or √(𝑧_{0}^{2}/𝑧^{2} -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√(𝑧_{0}^{2}/𝑧^{2} -1) = 𝑧 tan (𝑧) or √(𝑧_{0}^{2} - 𝑧^{2}) = 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) √(𝑧_{0}^{2}/𝑧^{2} -1) = -cot (𝑧),

Multiplying both side by 𝑧 gives the alternative equation:

(5.54) √(𝑧_{0}^{2} - 𝑧^{2}) = 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.