2. The bound states of a particle in a finite rectangular well

active investigation, mainly due to the quantum size effects (QSEs), like the oscillatory behavior of the film stability [1], of the lattice deformation [2], of the work function [3], etc., in dependence of the number of atomic monolayers. The QSEs, predicted in the pioneering papers of Sandomirskii [4] and Schulte [5], are important for both practical and theoretical reasons. The ultrathin metallic films have a special relevance for ferromagnetic materials, as they are responsible for the giant magnetoresistivity of the Fe/Cr antiferromagnetic lattice [6]. Also, the possibility of obtaining ultrathin metallic films, having a specific number of monolayers, allows the experimentalist to tune the work function, controlling the chemistry of the metallic surface [3]. All these effects can be satisfactorily explained with a quite simple physics, whose basic ingredient is the different quantization imposed to electrons moving on longitudinal and transversal directions. Namely, the electrons moving parallel to the surface of the metallic film are quantized by cyclic conditions; the result is that the wave vectors are quasi-continuous. The electrons moving perpendicular to the film are considered as confined in a rectangular well, so they are quantized according to the theory of quantum wells; the

How simple can the model of the well be, in order to provide a quantitative understanding of the physics of ultrathin metallic films? In spite of its simplicity, even the model of the infinite rectangular well gives sometimes good results, for instance, for the calculation of lattice deformation [2] or of Fermi energy [7]. These successes can be explained by the fact that, if the number of monolayers is not very small, ð Þ n≳25 , the deep levels play a dominant role, and the difference between the corresponding levels (i.e., having the same quantum number) of the finite and of the infinite well is negligible. However, for a small number of monolayers ð Þ n≲5 , this approximation does not work anymore. This is why it is important to obtain the exact

In this chapter, we shall present exact or approximate analytic results for the energy levels of a finite square well and show how they can improve the simple theoretical models which give a quantitative understanding of the behavior of ultrathin metallic films, especially the QSEs. Its structure is the following: in the second section, we shall discuss the quantum problem of the finite square well, mainly in order to put the eigenvalue equations in an appropriate form. The next one is a short review of the various attempts of solving these transcendental eigenvalue equations. The fourth section describes a simple algebraic approximation of the solution of the eigenvalue equations—the parabolic approximation—mentioning also similar but more precise approaches. In the next one, we put the eigenvalue equations in differential form and obtain the exact solution as a series expansion. The sixth section is devoted to the applications in the quantum statistical physics of the ultrathin metallic films of the analytic results obtained for the bound-state energy in a finite square well. By analyzing the predictions of the three models frequently used in the physics of ultrathin metallic films (infinite, semi-infinite, and finite square well) for the Fermi wave vector, we show the key role played by the finitude of the well, in the evaluation of QSEs. In the last section, we describe how our results can improve the current theory of this class of

value of the energy levels in the finite well or at least a precise approximation.

result is that the spectrum is discrete.

84 Heterojunctions and Nanostructures

metallic films.

Until the mid-1980 of the previous century, the finite rectangular well was just an elementary problem of quantum mechanics, with applications in finding the energy levels of the quasi-free electrons on long molecules [8] or of the Ramsauer-Townsend effect [9]. The progress of solidstate physics, which finally led to the fabrication of quantum wells [10], quantum dots, or ultrathin metallic films [11, 12] and to the observation of QSEs associated with them, transformed these simple systems from problems of elementary quantum mechanics into theoretical models of devices of great practical interest.

We shall study now the movement of a particle in a finite rectangular well. There are, in principle, two ways of defining the potential of the well, choosing the origin of the energy ð Þ E ¼ 0 at the top or at the bottom of the well. In the first case, the advantage is that the energy of the bound states ("inside the well") is negative, as usual in quantum mechanics; in the second one, that is, in the limit of a very deep well, the energy level tends to the energy of the corresponding level of the infinite well. Even elementary, this distinction might be useful, in order to avoid confusions. We shall examine in detail the first case, so we shall consider a potential having the form (Figure 1):

$$V(\mathbf{x}) = -\mathcal{U} \cdot \mathcal{O}\left(\frac{\mathfrak{a}}{2} - |\mathbf{x}|\right) \tag{1}$$

where θ is the Heaviside function. The second case is shortly mentioned later on (Eqs. (24) and (25)). The Schroedinger equation for a particle of mass m moving in the potential (1) is

$$\left[-\frac{\hbar^2}{2m}\frac{d^2}{d\mathbf{x}^2} + V(\mathbf{x}) - E\right]\psi(\mathbf{x}) = 0\tag{2}$$

Figure 1. The square well potential (Eq. (1)).

As the potential is invariant at spatial inversion, V xð Þ¼ Vð Þ �x , the solutions have welldefined parity. Let us put

$$E = -\frac{\hbar^2 \varkappa^2}{2m}, \quad \mathcal{U} = \frac{\hbar^2 k\_0^2}{2m}; \quad k^2 = k\_0^2 - \varkappa^2 \tag{3}$$

sometimes called potential strength, which actually characterizes both the particle ð Þ m and the

q

q

2P

The sign must be chosen in agreement with Eqs. (11) and (12), so to satisfy the conditions

In other words, to solve the eigenvalue, Eq. (14) means to find the functions ζð Þp , ξð Þp ,

cos ξð Þp ξð Þp

<sup>2</sup> > 0 for even states and < 0 for odd states, as we shall indicate explicitly in the forthcom-

sin ka 2 ka 2

¼ � 1

¼ �p, p <sup>¼</sup> <sup>1</sup>

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>P</sup><sup>2</sup> � <sup>k</sup> 2 ð Þ <sup>a</sup>=<sup>2</sup> <sup>2</sup>

> ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi <sup>P</sup><sup>2</sup> � <sup>k</sup> 2 ð Þ <sup>a</sup>=<sup>2</sup> <sup>2</sup>

ka=<sup>2</sup> , even states (11)

Quantum Wells and Ultrathin Metallic Films http://dx.doi.org/10.5772/intechopen.74150 87

ka=<sup>2</sup> , odd states (12)

<sup>P</sup> ð Þ odd states (14)

<sup>P</sup> (15)

<sup>x</sup> <sup>¼</sup> y xð Þ (16)

� �<sup>2</sup> " # ==<sup>13</sup> (13)

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi

Using well-known trigonometric identities, Eqs. (11) and (12) take the form:

<sup>P</sup> ð Þ even states ;

¼ �p,

This is, of course, a difficult task. If we write Eq. (15) in a slightly different form

<sup>x</sup> <sup>¼</sup> y xð Þ, cos <sup>x</sup>

to solve Eq. (15) means to invert the function y xð Þ defined by Eq. (16), i.e., to obtain the function x yð Þ: Clearly, x in Eq. (16)—and in the rest of the chapter—has nothing to do with the space coordi-

The functions ζð Þp , ξð Þp correspond to the intersections of the plots of the functions sin x=x, cos x=x with the line y ¼ �p, which satisfy the sign rule mentioned, after Eq. (14). The number of solutions depends on the value of p: If there is at least one solution ξð Þp for any p, the solution ζð Þp exists only for p < 1: In Figure 2, the functions sin x=x, cos x=x and the line y ¼ �p, for p ¼ 0:1, are plotted. The x�coordinate of the intersections corresponds to the

sin x

ka ¼

ka ¼ �

<sup>E</sup> ¼ �<sup>U</sup> <sup>1</sup> � ka

k 2 <sup>0</sup>a<sup>2</sup> � <sup>k</sup><sup>2</sup> a2

k 2 <sup>0</sup>a<sup>2</sup> � k 2 a2

q

q

well ð Þ a; U , Eqs. (8) and (9) become

tan ka 2 ¼

cot ka 2 ¼ �

> cos ka 2 ka 2

¼ � 1

sin ζð Þp ζð Þp

Also, the energy is

tan ka

ing paragraphs.

satisfying the equations:

nate x, as initially used in Eqs. (1)–(7).

functions ζð Þp , ξð Þp , as we shall explain later on.

where the quantities k, k0, ϰ have the dimension of wave vectors. With these notations, the Schroedinger equation for the particle inside the well takes the form:

$$
\left(\frac{d^2}{dx^2} + k^2\right)\psi(\mathbf{x}) = 0, \ |\mathbf{x}| < \frac{a}{2} \tag{4}
$$

For the particle outside the well, it is

$$
\left(\frac{d^2}{dx^2} - \varkappa^2\right)\psi(x) = 0, \ |x| > \frac{a}{2} \tag{5}
$$

The even solutions are

$$u\_{+}(\mathbf{x}) = A\_{+} \cos k\mathbf{x}, \qquad \qquad 0 \leqslant \mathbf{x} \leqslant a/2$$

$$u\_{+}(\mathbf{x}) = A\_{+} \cos k\mathbf{a} \quad e^{\varkappa(a-\mathbf{x})}, \qquad \qquad \mathbf{x} > a/2 \tag{6}$$

$$u\_{+}(-\mathbf{x}) = u\_{+}(\mathbf{x})$$

and the odd ones are

$$u\_{-}(\mathbf{x}) = A\_{-} \sin k\mathbf{x}, \qquad \qquad 0 \leqslant \mathbf{x} \leqslant a/2$$

$$u\_{-}(\mathbf{x}) = A\_{-} \sin k\mathbf{a} \; e^{\varkappa(a-\mathbf{x})}, \qquad \qquad \mathbf{x} > a/2 \tag{7}$$

u�ð Þ¼ �x u�ð Þx

The continuity of the derivative in x ¼ a=2 gives, for even states

$$\tan\frac{ka}{2} = \frac{\varkappa}{k} \tag{8}$$

and for odd states

$$
\cot \frac{ka}{2} = -\frac{\varkappa}{k} \tag{9}
$$

Defining the dimensionless parameter

$$P = k\_0 a / 2 = \sqrt{2mL} \frac{a}{2\hbar} = \frac{1}{p} \tag{10}$$

sometimes called potential strength, which actually characterizes both the particle ð Þ m and the well ð Þ a; U , Eqs. (8) and (9) become

$$\tan\frac{ka}{2} = \frac{\sqrt{k\_0^2 a^2 - k^2 a^2}}{ka} = \frac{\sqrt{P^2 - k^2 \left(a/2\right)^2}}{ka/2}, \text{ even states} \tag{11}$$

$$\cot \frac{ka}{2} = -\frac{\sqrt{k\_0^2 a^2 - k^2 a^2}}{ka} = -\frac{\sqrt{P^2 - k^2 \left(a/2\right)^2}}{ka/2}, \text{ odd states} \tag{12}$$

Also, the energy is

As the potential is invariant at spatial inversion, V xð Þ¼ Vð Þ �x , the solutions have well-

where the quantities k, k0, ϰ have the dimension of wave vectors. With these notations, the

k2 0 2m

ψð Þ¼ x 0, xj j <

ψð Þ¼ x 0, xj j >

uþð Þ¼ x A<sup>þ</sup> cos kx, 0⩽ x⩽a=2

uþð Þ¼ �x uþð Þx

u�ð Þ¼ x A� sin kx, 0⩽ x⩽ a=2

u�ð Þ¼ �x u�ð Þx

tan ka <sup>2</sup> <sup>¼</sup> <sup>ϰ</sup>

cot ka <sup>2</sup> ¼ � <sup>ϰ</sup>

<sup>P</sup> <sup>¼</sup> <sup>k</sup>0a=<sup>2</sup> <sup>¼</sup> ffiffiffiffiffiffiffiffiffiffi

<sup>2</sup>mU <sup>p</sup> <sup>a</sup>

<sup>2</sup><sup>ℏ</sup> <sup>¼</sup> <sup>1</sup>

; k<sup>2</sup> <sup>¼</sup> <sup>k</sup>

2

a

a

<sup>u</sup>þð Þ¼ <sup>x</sup> <sup>A</sup><sup>þ</sup> cos ka e<sup>ϰ</sup>ð Þ <sup>a</sup>�<sup>x</sup> , x <sup>&</sup>gt; <sup>a</sup>=<sup>2</sup> (6)

<sup>u</sup>�ð Þ¼ <sup>x</sup> <sup>A</sup>� sin ka e<sup>ϰ</sup>ð Þ <sup>a</sup>�<sup>x</sup> , x <sup>&</sup>gt; <sup>a</sup>=<sup>2</sup> (7)

<sup>0</sup> � <sup>ϰ</sup><sup>2</sup> (3)

<sup>2</sup> (4)

<sup>2</sup> (5)

<sup>k</sup> (8)

<sup>k</sup> (9)

<sup>p</sup> (10)

, U <sup>¼</sup> <sup>ℏ</sup><sup>2</sup>

<sup>E</sup> ¼ � <sup>ℏ</sup><sup>2</sup>

ϰ2 2m

Schroedinger equation for the particle inside the well takes the form:

d2 dx<sup>2</sup> <sup>þ</sup> <sup>k</sup> 2 � �

d2 dx<sup>2</sup> � <sup>ϰ</sup><sup>2</sup> � �

The continuity of the derivative in x ¼ a=2 gives, for even states

defined parity. Let us put

86 Heterojunctions and Nanostructures

For the particle outside the well, it is

The even solutions are

and the odd ones are

and for odd states

Defining the dimensionless parameter

$$E = -U\left[1 - \left(\frac{ka}{2P}\right)^2\right] \;//13\tag{13}$$

Using well-known trigonometric identities, Eqs. (11) and (12) take the form:

$$\frac{\cos\frac{ka}{2}}{\frac{ka}{2}} = \pm \frac{1}{P} \text{ (even states)}; \quad \frac{\sin\frac{ka}{2}}{\frac{ka}{2}} = \pm \frac{1}{P} \text{ (odd states)}\tag{14}$$

The sign must be chosen in agreement with Eqs. (11) and (12), so to satisfy the conditions tan ka <sup>2</sup> > 0 for even states and < 0 for odd states, as we shall indicate explicitly in the forthcoming paragraphs.

In other words, to solve the eigenvalue, Eq. (14) means to find the functions ζð Þp , ξð Þp , satisfying the equations:

$$\frac{\sin \zeta(p)}{\zeta(p)} = \pm p, \quad \frac{\cos \xi(p)}{\xi(p)} = \pm p, \quad p = \frac{1}{P} \tag{15}$$

This is, of course, a difficult task. If we write Eq. (15) in a slightly different form

$$\frac{\sin x}{x} = y(x), \quad \frac{\cos x}{x} = y(x) \tag{16}$$

to solve Eq. (15) means to invert the function y xð Þ defined by Eq. (16), i.e., to obtain the function x yð Þ: Clearly, x in Eq. (16)—and in the rest of the chapter—has nothing to do with the space coordinate x, as initially used in Eqs. (1)–(7).

The functions ζð Þp , ξð Þp correspond to the intersections of the plots of the functions sin x=x, cos x=x with the line y ¼ �p, which satisfy the sign rule mentioned, after Eq. (14). The number of solutions depends on the value of p: If there is at least one solution ξð Þp for any p, the solution ζð Þp exists only for p < 1: In Figure 2, the functions sin x=x, cos x=x and the line y ¼ �p, for p ¼ 0:1, are plotted. The x�coordinate of the intersections corresponds to the functions ζð Þp , ξð Þp , as we shall explain later on.

Figure 2. The x� coordinate of the intersection points between the functions sin x=x (solid) and cos x=x (dashed) with the lines y xð Þ¼ p (dots) and y xð Þ¼�p (dash-dots), marked with a point, corresponds to the functions ζ1ð Þp , ζ2ð Þp , respectively, ξ1ð Þp , ξ2ð Þp , for p ¼ 0:1:

We shall write in a more explicit form Eq. (15), taking into account both the sign of the tan function (or of the cot function, which is, evidently, the same thing), as already mentioned, and the intervals of monotony of the functions sin x=x, cos x=x [13]. The extremum points of the function cos x=x are given by the roots rcn of the equation:

$$
\tan x = -\frac{1}{x} \tag{17}
$$

<sup>x</sup><sup>∈</sup> ð Þ rs, <sup>2</sup>; <sup>2</sup><sup>π</sup> : sin <sup>x</sup>

<sup>x</sup><sup>∈</sup> ð Þ rs, <sup>3</sup>; <sup>3</sup><sup>π</sup> : sin <sup>x</sup>

<sup>x</sup>∈ð Þ <sup>0</sup>; <sup>π</sup> : sin <sup>x</sup>

If the particle moves not in potential V xð Þ given by (1), but in a potential

Eð Þ<sup>1</sup>

<sup>n</sup> <sup>¼</sup> <sup>U</sup> kna 2P <sup>2</sup>

According to the parity of n, kna=2 corresponds to the functions ξ and ζ, for instance, k1a=2 ¼

As already mentioned, the advantage of using the potential (1) is that the energy of a particle "inside the well," so in a bound state, is negative, corresponding to the most usual convention of quantum mechanics. However, the form (26) of the potential has the advantage that its levels approach, in the limit of a very deep well, the levels of the infinite well. Indeed, for n ! ∞, so for very deep wells, the quantization condition for the wave vector becomes kna≃ nπ, so

kn <sup>≃</sup> <sup>n</sup><sup>π</sup>

and Eq. (27) gives the expression of the wave vector corresponding to the n�th state in an

Eð Þ <sup>∞</sup> <sup>n</sup> <sup>¼</sup> <sup>π</sup><sup>2</sup>ℏ<sup>2</sup>

ζ2ð Þp , ζ3ð Þp : The function ζ1ð Þp satisfies the equation:

According to Eq. (13), the energy eigenvalues are

then the energy levels will be given by

ξ1ð Þp , k2a=2 ¼ ζ1ð Þp , etc.

infinite well:

and so on. Each of Eqs. (18)–(20) and (22)–(23) has a unique solution, ξ1ð Þp , ξ2ð Þp , ξ3ð Þp , respectively, ζ2ð Þp , ζ3ð Þp : On the aforementioned intervals, the functions cos x=x, sin x=x are monotonic and have an inverse. The inverse functions are ξ1ð Þp , ξ2ð Þp , ξ3ð Þp , respectively,

En ¼ �<sup>U</sup> <sup>þ</sup> <sup>U</sup> kna

2P <sup>2</sup>

<sup>x</sup> ¼ �p; x � <sup>ζ</sup>2ð Þ<sup>p</sup> (22)

Quantum Wells and Ultrathin Metallic Films http://dx.doi.org/10.5772/intechopen.74150

<sup>x</sup> <sup>¼</sup> p; x � <sup>ζ</sup>3ð Þ<sup>p</sup> (23)

<sup>x</sup> <sup>¼</sup> p; x � <sup>ζ</sup>1ð Þ<sup>p</sup> (24)

<sup>V</sup>ð Þ<sup>1</sup> ð Þ¼ <sup>x</sup> V xð Þþ <sup>U</sup> (26)

<sup>a</sup> (28)

<sup>2</sup>ma<sup>2</sup> <sup>n</sup><sup>2</sup> (29)

(25)

89

(27)

where rcn is the root closest to ð Þ n � 1 π: The eigenvalue equations for the even states are

$$\mathbf{x} \in \left(0, \frac{\pi}{2}\right) : \quad \frac{\cos \mathbf{x}}{\mathbf{x}} = p ; \ \mathbf{x} \equiv \xi\_1(p) \tag{18}$$

$$\mathbf{x} \in \left( r\_{\varepsilon 2}, \frac{3\pi}{2} \right) : \quad \frac{\cos \mathbf{x}}{\mathbf{x}} = -p; \ \mathbf{x} \equiv \xi\_2(p) \tag{19}$$

$$\mathbf{x} \in \left( r\_{\text{c3}}, \frac{5\pi}{2} \right) : \quad \frac{\cos \mathbf{x}}{\mathbf{x}} = p; \ \mathbf{x} \equiv \xi\_3(p) \tag{20}$$

and so on.

Similarly, the extremum points of the function sin x=x are the roots rsn of the equation:

$$
\tan \mathfrak{x} = \mathfrak{x} \tag{21}
$$

where rsn is the root closest to <sup>n</sup> � <sup>1</sup> 2 π: The eigenvalue equations for the odd states are

Quantum Wells and Ultrathin Metallic Films http://dx.doi.org/10.5772/intechopen.74150 89

$$\mathbf{x} \in (r\_{s,2}, 2\pi): \quad \frac{\sin \mathbf{x}}{\mathbf{x}} = -p; \ \mathbf{x} \equiv \mathbb{Q}\_2(p) \tag{22}$$

$$\mathbf{x} \in (r\_{s,3}, 3\pi): \quad \frac{\sin \mathbf{x}}{\mathbf{x}} = p; \ \mathbf{x} \equiv \zeta\_3(p) \tag{23}$$

and so on. Each of Eqs. (18)–(20) and (22)–(23) has a unique solution, ξ1ð Þp , ξ2ð Þp , ξ3ð Þp , respectively, ζ2ð Þp , ζ3ð Þp : On the aforementioned intervals, the functions cos x=x, sin x=x are monotonic and have an inverse. The inverse functions are ξ1ð Þp , ξ2ð Þp , ξ3ð Þp , respectively, ζ2ð Þp , ζ3ð Þp : The function ζ1ð Þp satisfies the equation:

$$\mathbf{x} \in (0, \pi): \quad \frac{\sin \mathbf{x}}{\mathbf{x}} = p; \ \mathbf{x} \equiv \zeta\_1(p) \tag{24}$$

According to Eq. (13), the energy eigenvalues are

$$E\_n = -\mathcal{U} + \mathcal{U} \left(\frac{k\_n a}{2P}\right)^2\tag{25}$$

If the particle moves not in potential V xð Þ given by (1), but in a potential

$$V^{(1)}(\mathbf{x}) = V(\mathbf{x}) + \mathcal{U} \tag{26}$$

then the energy levels will be given by

We shall write in a more explicit form Eq. (15), taking into account both the sign of the tan function (or of the cot function, which is, evidently, the same thing), as already mentioned, and the intervals of monotony of the functions sin x=x, cos x=x [13]. The extremum points of the

Figure 2. The x� coordinate of the intersection points between the functions sin x=x (solid) and cos x=x (dashed) with the lines y xð Þ¼ p (dots) and y xð Þ¼�p (dash-dots), marked with a point, corresponds to the functions ζ1ð Þp , ζ2ð Þp , respec-

tan <sup>x</sup> ¼ � <sup>1</sup>

where rcn is the root closest to ð Þ n � 1 π: The eigenvalue equations for the even states are

: cos <sup>x</sup>

: cos <sup>x</sup>

: cos <sup>x</sup>

Similarly, the extremum points of the function sin x=x are the roots rsn of the equation:

<sup>x</sup> (17)

x

<sup>x</sup> <sup>¼</sup> p; x � <sup>ξ</sup>1ð Þ<sup>p</sup> (18)

<sup>x</sup> ¼ �p; x � <sup>ξ</sup>2ð Þ<sup>p</sup> (19)

<sup>x</sup> <sup>¼</sup> p; x � <sup>ξ</sup>3ð Þ<sup>p</sup> (20)

tan x ¼ x (21)

π: The eigenvalue equations for the odd states are

function cos x=x are given by the roots rcn of the equation:

y

88 Heterojunctions and Nanostructures

tively, ξ1ð Þp , ξ2ð Þp , for p ¼ 0:1:

x ∈ 0; π 2 

> 3π 2

5π 2 

2

x∈ rc2;

x∈ rc3;

and so on.

where rsn is the root closest to <sup>n</sup> � <sup>1</sup>

$$E\_n^{(1)} = \mathcal{U}\left(\frac{k\_n a}{2P}\right)^2\tag{27}$$

According to the parity of n, kna=2 corresponds to the functions ξ and ζ, for instance, k1a=2 ¼ ξ1ð Þp , k2a=2 ¼ ζ1ð Þp , etc.

As already mentioned, the advantage of using the potential (1) is that the energy of a particle "inside the well," so in a bound state, is negative, corresponding to the most usual convention of quantum mechanics. However, the form (26) of the potential has the advantage that its levels approach, in the limit of a very deep well, the levels of the infinite well. Indeed, for n ! ∞, so for very deep wells, the quantization condition for the wave vector becomes kna≃ nπ, so

$$k\_n \simeq \frac{n\pi}{a} \tag{28}$$

and Eq. (27) gives the expression of the wave vector corresponding to the n�th state in an infinite well:

$$E\_n^{(\ast \ast)} = \frac{\pi^2 \hbar^2}{2ma^2} n^2 \tag{29}$$
