Análisis factorial conjunto

Solutions for Constant Potential Energy First let’s examine the solutions to the Schr¨odinger equation for the special case of a constant potential energy, equal to U₀. Then Eq. 5.5 becomes

−⁻h² 2m

d²ψ

dx² + U0ψ(x) = Eψ(x) (5.14) or (assuming for now that E> U₀)

d²ψ

dx² = −k²ψ(x) with k =

2m(E − U₀)

−h² (5.15)

The parameter k in this equation is equal to the wave number 2π/λ.

The solution to Eq. 5.15 is a function of x that, when differentiated twice, gives back the original function multiplied by the negative constant−k². The function that has this property is sin kx or cos kx. The most general solution to the equation is

ψ(x) = A sin kx + B cos kx (5.16) The constants A and B must be determined by applying the continuity and normalization requirements. We can demonstrate that Eq. 5.16 satisfies Eq. 5.15 by taking two derivatives:

dψ

dx = kA cos kx − kB sin kx d²ψ

dx² = −k²A sin kx− k²B cos kx= −k²(A sin kx + B cos kx) = −k²ψ(x) so the original equation is indeed satisfied.

To analyze the penetration of a particle into a forbidden region, we must consider the case in which the energy E of the particle is smaller than the potential energy U₀. For this case we can rewrite Eq. 5.14 as

d²ψ

dx² = k^2ψ(x) with k^=

2m(U0− E)

−h² (5.17)

In this case the general solution in the forbidden regions is

ψ(x) = Ae^kx+ Be^−kx (5.18) Once again, we can demonstrate that Eq. 5.18 is a solution of Eq. 5.17 by taking two derivatives:

dψ

dx = k^Ae^kx− k^Be^−kx d²ψ

dx² = k^2Ae^kx+ k^2Be^−kx= k^2(Ae^kx+ Be^−kx) = k^2ψ(x)

We will use Eqs. 5.16 and 5.18 as our solutions to the Schr¨odinger equation for constant potential energy in the allowed (E> U₀) and forbidden (E< U₀) regions.

The Free Particle

For a free particle, the force is zero and so the potential energy is constant. We may choose any value for that constant, so for convenience we’ll choose U₀= 0.

The solution is given by Eq. 5.16,ψ(x) = A sin kx + B cos kx. The energy of the particle is

E= ⁻h²k²

2m (5.19)

Our solution has placed no restrictions on k, so the energy is permitted to have any value (in the language of quantum physics, we say that the energy is not quan-tized). We note that Eq. 5.19 is the kinetic energy of a particle with momentum p=⁻hk or, equivalently, p= h/λ. This is as we would have expected, because the free particle can be represented by a de Broglie wave with any wavelength.

Solving for A and B presents some difficulties because the normalization integral, Eq. 5.9, cannot be evaluated from−∞to+∞for this wave function.

We therefore cannot determine probabilities for the free particle from the wave function of Eq. 5.16.

It is also instructive to write the wave function in terms of complex exponentials, using sin kx= (e^ikx− e^−ikx)/2i and cos kx = (e^ikx+ e^−ikx)/2:

ψ(x) = A

e^ikx− e^−ikx 2i

 + B

e^ikx+ e^−ikx 2



= A^e^ikx+ B^e^−ikx (5.20)

where A^= A/2i + B/2 and B^= −A/2i + B/2. To interpret this solution in terms of waves we form the complete time-dependent wave function using Eq. 5.6:

(x,t) = (A^e^ikx+ B^e^−ikx)e^−iωt= A^e^i(kx−ωt)+ B^e^−i(kx+ωt) (5.21) The dependence of the first term on kx− ωt identifies this term as representing a wave moving to the right (in the positive x direction) with amplitude A^, and the second term involving kx+ ωt represents a wave moving to the left (in the negative x direction) with amplitude B^.

If we want the wave to represent a beam of particles moving in the+x direction, then we must set B^= 0. The probability density associated with this wave is then, according to Eq. 5.7,

P(x) = |ψ(x)|² = |A^|²e^ikxe^−ikx= |A^|² (5.22) The probability density is constant, meaning the particles are equally likely to be found anywhere along the x axis. This is consistent with our discussion of the free-particle de Broglie wave in Chapter 4—a wave of precisely defined wavelength extends from x= −^∞ to x= +^∞ and thus gives a completely unlocalized particle.

Infinite Potential Energy Well

Now we’ll consider the formal solution to the problem we discussed in Section 5.2:

a particle is trapped in the region between x= 0 and x = L by infinitely high potential energy barriers. Imagine an apparatus like that of Figure 5.6, in which the particle moves freely in this region and makes elastic collisions with the perfectly rigid barriers that confine it. This problem is sometimes called “a particle in a box.” For now we’ll assume that the particle moves in only one dimension; later we’ll expand to two and three dimensions.

The potential energy may be expressed as:

U(x) = 0 0≤ x ≤ L

=∞ x< 0, x > L (5.23)

The potential energy is shown in Figure 5.9. We are free to choose any constant value for U in the region 0≤ x ≤ L; we choose it to be zero for convenience.

U = ∞

To ∞ To ∞

U = ∞ U = 0

x = 0 x = L

FIGURE 5.9 The potential energy of a particle that moves freely (U = 0) in the region 0≤ x ≤ L but is completely excluded (U =∞) from the regions x< 0 and x > L.

Because the potential energy is different in the regions inside and outside the well, we must find separate solutions in each region. We can analyze the outside region in either of two ways. If we examine Eq. 5.5 for the region outside the well, we find that the only way to keep the equation from becoming meaningless when U→∞is to requireψ = 0, so that Uψ will not become infinite. Alternatively, we can go back to the original statement of the problem. If the walls at the boundaries of the well are perfectly rigid, the particle must always be in the well, and the probability for finding it outside must be zero. To make the probability zero everywhere outside the well, we must makeψ = 0 everywhere outside. Thus we have

ψ(x) = 0 x< 0, x > L (5.24) The Schr¨odinger equation for 0≤ x ≤ L, when U(x) = 0, is identical with Eq. 5.14 with U₀ = 0 and has the same solution:

ψ(x) = A sin kx + B cos kx 0≤ x ≤ L (5.25) with

k=

2mE

h−² (5.26)

Our solution is not yet complete, for we have not evaluated A or B, nor have we found the allowed values of the energy E. To do this, we must apply the requirement thatψ(x) is continuous across any boundary. In this case, we require that our solutions for x< 0 and x > 0 match up at x = 0; similarly, the solutions for x> L and x < L must match at x = L.

Let us begin at x= 0. At x < 0, we have found that ψ = 0, and so we must set ψ(x) of Eq. 5.25 to zero at x = 0.

ψ(0) = A sin 0 + B cos 0 = 0 (5.27) which gives B= 0. Because ψ = 0 for x > L, the second boundary condition is ψ(L) = 0, so

ψ(L) = A sin kL + B cos kL = 0 (5.28) We have already found B= 0, so we must now have A sin kL = 0. Either A= 0, in which case ψ = 0 everywhere, ψ² = 0 everywhere, and there is no particle (a meaningless solution) or else sin kL= 0, which is true only when kL= π, 2π, 3π, . . ., or

kL= nπ n= 1, 2, 3, . . . (5.29)

With k= 2π/λ, we have λ = 2L/n; this is identical with the result obtained in introductory mechanics for the wavelengths of the standing waves in a string of length L fixed at both ends, which we already obtained in Section 5.2 (Eq. 5.1).

Thus the solution to the Schr¨odinger equation for a particle trapped in a linear region of length L is a series of standing de Broglie waves! Not all wavelengths are permitted; only certain values, determined from Eq. 5.29, may occur.

n = 1 n = 2 n = 3

n = 4 E₄ = 16E₀

E₃ = 9E₀

E₂ = 4E₀

E₁ = E₀

FIGURE 5.10 The first four energy levels in a one-dimensional infinite potential energy well.

From Eq. 5.26 we find that, because only certain values of k are permitted by Eq. 5.29, only certain values of E may occur— the energy is quantized! Solving Eq. 5.29 for k and substituting into Eq. 5.26, we obtain

E_n= ⁻h²k²

2m = ⁻h²π²n²

2mL² = h²n²

8mL² n= 1, 2, 3, . . . (5.30) For convenience, let E₀=⁻h²π²/2mL²= h²/8mL²; this unit of energy is determined by the mass of the particle and the width of the well. Then E_n= n²E₀, and the only allowed energies for the particle are E₀, 4E₀, 9E₀, 16E₀, etc. All intermediate values, such as 3E₀or 6.2E₀, are forbidden. Figure 5.10 shows the allowed energy levels. The lowest energy state, for which n= 1, is known as the ground state, and the states with higher energies(n > 1) are known as excited states.

Because the energy is purely kinetic in this case, our result means that only certain speeds are permitted for the particle. This is very different from the case of the classical trapped particle, in which the particle can be given any initial velocity and will move forever, back and forth, at the same speed. In the quantum case, this is not possible; only certain initial speeds can result in sustained states of motion; these special conditions represent the “stationary states.” Average values calculated according to Eq. 5.13 likewise do not change with time.

From one energy state, the particle can make jumps or transitions to another energy state by absorbing or releasing an amount of energy equal to the energy difference between the two states. By absorbing energy the particle will move to a higher energy state, and by releasing energy it moves to a lower energy state.

A similar effect occurs for electrons in atoms, in which the absorbed or released energy is usually in the form of a photon of visible light or other electromagnetic radiation. For example, from the state with n= 3 (E3 = 9E0), the particle might absorb an energy ofE = 7E0and jump upward to the n= 4 state (E4= 16E0) or might release energy of E = 5E0 and jump downward to the n= 2 state (E₂= 4E0).

Example 5.2

An electron is trapped in a one-dimensional region of length 1.00 × 10⁻¹⁰ m (a typical atomic diameter). (a) Find the energies of the ground state and first two excited states.

(b) How much energy must be supplied to excite the elec-tron from the ground state to the second excited state?

(c) From the second excited state, the electron drops down to the first excited state. How much energy is released in this process?

Solution

(a) The basic quantity of energy needed for this calculation is

E₀ = h²

8mL² = (hc)² 8mc²L²

= (1240 eV · nm)²

8(511, 000 eV)(0.100 nm)² = 37.6 eV

With E_n= n²E₀, we can find the energy of the states:

n= 1 : E1= E0 = 37.6 eV n= 2 : E₂= 4E₀= 150.4 eV n= 3 : E3= 9E0= 338.4 eV

(b) The energy difference between the ground state and the second excited state is

E = E₃− E₁= 338.4 eV − 37.6 eV = 300.8 eV

This is the energy that must be absorbed for the electron to make this jump.

(c) The energy difference between the second and first excited states is

E = E3− E2= 338.4 eV − 150.4 eV = 188.0 eV

This is the energy that is released when the electron makes this jump.

To complete the solution forψ(x), we must determine the constant A by using the normalization condition given in Eq. 5.9,_+∞

−∞ |ψ(x)|²dx= 1. The integrand is zero in the regions−∞< x ≤ 0 and L ≤ x < +∞, so all that remains is

 _L

0

A²sin² nπx

L dx= 1 (5.31)

from which we find A=

2/L. The complete wave function for 0 ≤ x ≤ L is then

ψn(x) =

2

Lsinnπx

L n= 1, 2, 3, . . . (5.32) In Figure 5.11, the wave functions and probability densitiesψ²are illustrated for the lowest several states.

In the ground state, the particle has the greatest probability to be found near the middle of the well (x= L/2), and the probability falls off to zero between the center and the sides of the well. This is very different from the behavior of a classical particle—a classical particle moving at constant speed would be found with equal probability at every location inside the well. The quantum particle also has constant speed but yet is still found with differing probability at various locations in the well. It is the wave nature of the quantum particle that is responsible for this very nonclassical behavior.

x = 0

n = 1 n = 3

n = 2 n = 4

x = L

x = 0 x = L x = 0 x = L

x = 0 x = L

FIGURE 5.11 The wave functions (solid lines) and probability densities (shaded regions) of the first four states in the one-dimensional infinite potential energy well.

Another example of nonclassical behavior occurs for the first excited state.

The probability density has a maximum at x= L/4 and another maximum at x= 3L/4. Between the two maxima, there is zero probability to find the particle in the center of the well at x= L/2. How can the particle travel from x = L/4 to x= 3L/4 without ever being at x = L/2? Of course, no classical particle could behave in such a way, but it is a common behavior for waves. For example, the first overtone of a vibrating string has a node at its midpoint and antinodes (vibrational maxima) at the¹/₄and³/₄locations.

The calculation of probabilities and average values is illustrated by the following examples.

Example 5.3

Consider again an electron trapped in a one-dimensional region of length 1.00 × 10⁻¹⁰m= 0.100 nm. (a) In the ground state, what is the probability of finding the electron in the region from x= 0.0090 nm to 0.0110 nm? (b) In the first excited state, what is the probability of finding the electron between x= 0 and x = 0.025 nm?

Solution

(a) When the interval is small, it is often simpler to use Eq. 5.7 to find the probability, instead of using the integration method. The width of the small inter-val is dx= 0.0110 nm − 0.0090 nm = 0.0020 nm. Evalu-ating the wave function at the midpoint of the interval (x= 0.0100 nm), we can use the n = 1 wave function with

(b) For this wide interval, we must use the integration method to find the probability:

P(x1: x₂) =

Evaluating this expression using the limits x₁ = 0 and x₂= 0.025 nm gives a probability of 0.25 or 25%. This result is of course what we would expect by inspection of the graph ofψ²for n= 2 in Figure 5.11. The interval from x= 0 to x = L/4 contains 25% of the total area under the ψ²curve.

Example 5.4

Show that the average value of x is L/2, independent of the quantum state.

Solution

We use Eq. 5.12; becauseψ = 0 except for 0 ≤ x ≤ L, the limits of integration are 0 and L:

x_av=

This can be integrated by parts or found in integral tables;

the result is

x_av= L 2

Note that, as required, this result is independent of n. Thus a measurement of the average position of the particle yields no information about its quantum state.

Let’s now look at how the uncertainty principle applies to the motion of this trapped particle. By solving Problems 34 and 35, you will find that the uncertainties in position and momentum for a particle in an infinite potential well arex = L

1/12 − 1/2π²n² andp = hn/2L. The product of the uncertainties is

xp = hn 2

 1 12− 1

2π²n² = h 2

 n² 12− 1

2π²

Clearly the product of the uncertainties grows as n grows. The minimum value occurs for n= 1, in which case xp = 0.090h = 0.57h. The ground state⁻ represents a fairly “compact” wave packet, but it is somewhat less compact than the minimum possible limit of 0.50h (Eq. 4.10). You can see from Figure 5.11⁻ how the wave becomes less compact (spreads out more) as n increases. Even for n= 2, the product of the uncertainties grows quickly to 1.67⁻h.

Finite Potential Energy Well

Because the infinite potential energy well is an idealization of a technique for confining a particle, we should examine the solution when the barriers at the sides of the well are finite rather than infinite. The potential energy well can be described by

U(x) = 0 0≤ x ≤ L

= U0 x< 0, x > L (5.33) and is sketched in Figure 5.12. We look for solutions in which the particle is confined to this well, and thus the energies that we deduce for the particle must be less than U₀.

x = 0 x = L

U = U₀ U = U₀

U = 0

FIGURE 5.12 The potential energy of a particle that is confined to the region 0≤ x ≤ L by finite barriers U0at x= 0 and x= L.

The solution in the center region (between x= 0 and x = L) is exactly the same as it was for the infinite well (Eq. 5.25):

ψ(x) = A sin kx + B cos kx (0 ≤ x ≤ L) (5.34) although the values that we deduced previously for the coefficients A and B are not valid in this calculation. The region x< 0 is an example of a situation in which the energy E of the particle is less than the potential energy U₀, and so we must use the solution in the form of Eq. 5.18,ψ(x) = Ce^kx+ De^−kxwith k^given in Eq. 5.17. This region includes x= −^∞, for which the term with the coefficient D becomes infinite. Because we cannot allow the probability to become infinite, we must discard this term by setting D= 0. The solution for x < 0 is then

ψ(x) = Ce^kx (x < 0) (5.35) In the region x> L, the energy E is once again smaller than U₀, and so the solution is also in the form of Eq. 5.18,ψ(x) = Fe^kx+ Ge^−kx. Here the region now includes x= +∞, for which the term with the coefficient F would become infinite. We must prevent that possibility by setting F= 0, so the solution in this region is

ψ(x) = Ge^−kx (x > L) (5.36) We now have 4 coefficients to determine (A, B, C, G) along with the energy E.

For this determination, we have 4 equations from the boundary conditions (the continuity of both ψ and dψ/dx at both x = 0 and x = L) and one equation

from the normalization condition. As you might imagine, solving 5 equations in 5 unknowns presents a straightforward but very tedious algebraic challenge.

Moreover, the resulting solution for the energy values cannot be obtained in terms of a direct equation such as Eq. 5.30, but instead must be found numerically by solving a transcendental equation. The result is a series of increasing energy values, but the number of energy values is finite rather than infinite, because the energy cannot be allowed to exceed the value of U₀.

n = 1 n = 2 n = 3

n = 4 E₄ = 375 eV

U₀ = 400 eV

E₃ = 227 eV

E₂ = 104 eV E₁ = 26 eV

FIGURE 5.13 The energy levels in a potential energy well of depth 400 eV.

There are only four energy states in this well.

As we did for the infinite potential energy well in Example 5.2, let’s consider a well of width L= 0.100 nm. We’ll choose the depth of the well to be U0= 400 eV.

Applying the boundary conditions at x= 0 and x = L, we can eliminate all of the coefficients and find an equation that involves only k and k^(both of which depend on the energy E). Solving that equation numerically, we find four possible values of the energy: E₁= 26 eV, E2= 104 eV, E3= 227 eV, E4 = 375 eV. Here the subscript just numbers the energy values, starting at the ground state; there is no simple functional dependence of the energies on the quantum number n as there was for the infinite well. The allowed energy levels are shown in Figure 5.13.

The probability densities (square of the wave functions) for these four states are shown in Figure 5.14. In some ways they are similar to the probability densities in the infinite well —note that each state has n maxima in its probability density, just like the infinite well (see Figure 5.11). Unlike the infinite well, these probability densities show the property of penetration into the classically forbidden region.

Look carefully at the continuity of the wave function and its slope at x= 0 and x= L; see how smoothly the sine and cosine function inside the well joins the exponentials in the forbidden regions.

The energy levels of the finite well are smaller than those of the infinite well of the same width (38 eV, 150 eV, 338 eV, 602 eV), and the differences increase as we go to higher states. This is consistent with the uncertainty principle—because of the penetration into the forbidden region,x is larger for the finite well and thusp_x must be smaller. As a result, the kinetic energies are smaller for the finite well. From Figure 5.14 we see that the penetration distance increases as we go up in energy, so the difference betweenx for the finite well and the infinite well increases and the energy discrepancy also increases.

−0.1 0 0.1 0.2

n = 1

x (nm) x (nm)

x (nm) x (nm)

−0.1 0 0.1 0.2

n = 2

−0.1 0 0.1 0.2

n = 3

−0.1 0 0.1 0.2

n = 4

FIGURE 5.14 The probability densities of the four states in the one-dimensional potential energy well of width 0.100 nm and depth 400 eV.

For an energy close to the top of the well such as E₄, a smaller uncertainty

E is necessary to reach the top of the well, giving a larger t ∼⁻h/E and thus a larger penetration distance. At the bottom of the well, the state E₁requires much more energy to reach the top of the well and thus needs a much largerE;

the smaller resultingt gives a smaller penetration distance into the forbidden region.

Two-Dimensional Infinite Potential Energy Well

When we extend the previous calculation to two and three dimensions, the principal features of the solution remain the same, but an important new feature is introduced. In this section we show how this occurs; this new feature, known as degeneracy, will turn out to be very important in our study of atomic physics.

To begin with, we need a Schr¨odinger equation that is valid in more than one dimension; our previous version, Eq. 5.5, included only one spatial dimension. If the potential energy is a function of x and y, we expect thatψ also depends on

To begin with, we need a Schr¨odinger equation that is valid in more than one dimension; our previous version, Eq. 5.5, included only one spatial dimension. If the potential energy is a function of x and y, we expect thatψ also depends on

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