Solutions for the Infinite Square Well

Section 4.2 Solutions for the Infinite Square Well

To demonstrate how we use Schrödinger's equation to derive wavefunction solutions, let's consider the simplest example of a “particle in a box,” i.e., a particle trapped in one dimension (and unable to move in the other two dimensions) between two impenetrable walls located at

x = 0

and

x = L .

Since the particle feels no force in the region between the walls, the potential energy in this region is constant, and therefore we will take it to be zero. Since the walls are impenetrable, the potential energy must become infinite at

x = 0

and

x = L .

Therefore the graph of the potential energy

U (x)

as a function of position is shown in Figure 4.1. This configuration is usually referred to as the infinite square well potential. For this example we will consider solutions for which the energy of the particle

E

is positive. (Solutions for

E < 0

for this potential actually do not exist due to the fact that the potential energy goes to infinity at the boundaries of the well.)

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Figure 4.1. Graph of the infinite square well potential as a function of position.

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Since there is no chance that the particle can penetrate either of the barriers at

x = 0

x = L,

the wavefunction

ψ (x)

must be exactly zero in the regions

x \leq 0

and

x > L .

Inside the well, where

U (x) = 0,

Schrödinger's equation can be written as

\begin{matrix} (4.1) & \frac{d^{2} ψ (x)}{d x^{2}} = - (\frac{2 m E}{ℏ^{2}}) ψ (x) . \end{matrix}

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Our job is to determine what mathematical function is consistent with (4.1), that is, can we find a function

ψ (x)

whose second derivative is equal to minus a positive constant

(2 m E / ℏ^{2})

times the original function? We already testing a solution to this equation in Section 3.5.2. In that section, we found that the wavefunction

ψ_{1} (x) = A \sin (k x)

works if

\begin{matrix} (4.2) & E = \frac{ℏ^{2} k^{2}}{2 m} . \end{matrix}

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from which it follows that

k = \pm \sqrt{2 m E} / ℏ .

This isn't the only possible solution — the wavefunction

ψ_{1} (x) = B \cos (k x)

also satisfies (4.1). Since either of these functions satisfies Schrödinger's equation, we can write a general solution as a linear combination of these two functions:

\begin{matrix} (4.3) & ψ (x) = A \sin (k x) + B \cos (k x) . \end{matrix}

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In order for this solution to work, we must adjust the values of the constants

A,

B,

and

k

such that the wavefunction inside the well matches up with the value of the wavefunction at positions

x = 0

and

x = L,

where we previously concluded that the value of the wavefunction should be zero. This process is called matching the boundary conditions. At

x = 0,

the wavefunction inside the well given by (4.3) becomes

ψ (0) = B,

so in order for the wavefunction to equal zero at the boundary

x = 0,

we must choose the value

B = 0 .

Thus the interior wavefunction must be given simply as

ψ (x) = A \sin (k x) .

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Continuing, we now look at the boundary condition at

x = L .

Again, the wavefunction must be zero here, so this requires that

\begin{matrix} (4.4) & A \sin (k L) = 0 . \end{matrix}

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From our knowledge of the sine function, we know that this function is zero for values such that

k L = 0, π, 2 π, 3 π, \dots = n π

for any integer

n .

Therefore this second boundary condition tells us that the possible values the constant

k

can assume are

\begin{matrix} (4.5) & k_{n} = \frac{n π}{L} . (n = 1, 2, 3, \dots) \end{matrix}

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We have thrown out the

k = 0

solution because that would correspond to a wavefunction

ψ (x) = 0

which corresponds to a particle that has zero probability of being found anywhere, (i.e., the particle doesn't exist). Combining these values for

k

with (4.2), we find that the possible energies

E_{n}

that the particle can have inside the well are given by

\begin{matrix} (4.6) & E_{n} = n^{2} \frac{π^{2} ℏ^{2}}{2 m L^{2}} = n^{2} \frac{h^{2}}{8 m L^{2}} (n = 1, 2, 3, \dots), \end{matrix}

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and the wavefunctions associated with each of these energies are

\begin{matrix} (4.7) & ψ_{n} (x) = A \sin (\frac{n π}{L} x) . \end{matrix}

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Figure 4.2 shows a graph of the allowed energies

E_{n}

for a particle in an infinite square well potential superimposed on the graph of potential energy

U (x)

and Figure 4.3 shows the wavefunctions

ψ_{n} (x)

for the four lowest allowed energies.

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A few general remarks are in order about these solutions.

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Figure 4.2. The allowed energies for a particle bound in an infinite square well potential. The energies are expressed in terms of the lowest (ground state) energy $E_{1} .$

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Figure 4.3. The wavefunction solutions for the infinite square well potential for the four lowest energy states.

Notice that the lowest allowed energy for a particle is not zero! This coincides with our discussion in chapter 3 in which Heisenberg's uncertainty principle requires a minimum non-zero energy for a particle that is confined to a finite region of space.
There are only certain values of the energy $E$ for which there are well-behaved solutions to Schrödinger's equation. When a particle is trapped in a finite region of space, its energy cannot be any continuous value but rather can only be one of many discrete values of energy. In the language of quantum mechanics, the energy is quantized.
The integer $n$ is referred to as the quantum number of the particle. For this problem, specifying a value for $n$ uniquely determines the state of the particle. As we shall see in more complicated systems, more than one quantum number will often be needed to completely specify a state.
The wavefunctions as given in (4.7) and Figure 4.3 should look familiar to you. Aren't these the same functions we used when describing the standing wave patterns on a vibrating string fixed at both ends? Indeed they are! In fact, standing waves on a string are determined by a differential equation similar to (4.1). Standing waves on a string exhibit their own form of quantization — in this case, the properties that are quantized are the wavelengths and frequencies.

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Example 4.4. Electron trapped in a 1-D nanotube.

Carbon atoms can be bonded into a cylindrical arrangement called a carbon nanotube. Carbon nanotubes can be fabricated with length-to-diameter ratios that are extremely large. These nanotubes have a broad range of applications including uses as conducting nano-wires and mechanical “scaffolding” for growing new bone cells.

A carbon nanotube of length $L$ can be approximated as an infinite square well of width $L .$ Consider a particular carbon nanotube of length $L = 10.0 nm$ and negligible diameter. Calculate the ground state energy for an electron trapped inside of this carbon nanotube.

Solution.

Approximating the carbon nanotube as an infinite square well, the possible allowed energies for the electron are given by (4.6). For the ground state energy, where $n = 1,$ the energy is given as

\begin{matrix} (4.8) & E_{1} = (1)^{2} \frac{h^{2}}{8 m_{e} L^{2}} . \end{matrix}

Because the length is given in nm, and eV's are convenient units for energy at the microscopic scale, it makes calculations easier if we multiply both the numerator and denominator by $c^{2},$ giving

\begin{matrix} (4.9) & E_{1} = (1)^{2} \frac{(h c)^{2}}{8 (m_{e} c^{2}) L^{2}} . \end{matrix}

Then we can insert the values $h c = 1240 eV \cdot nm,$ $L = 10 nm,$ and $m_{e} c^{2} = 0.511 MeV = 511 \times 10^{3} eV,$ giving

\begin{aligned} E_{1} = & (1)^{2} \frac{(1240 eV \cdot nm)^{2}}{8 \times 511 \times 10^{3} eV \times (10 nm)^{2}} . \\ = & 3.8 meV \end{aligned}

It is, of course, valid to use SI units for $h,$ $m_{e},$ and $L,$ but this entails calculations with large exponents and more unit conversions.

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We can use the idea of normalization (see Sec. Subsection 3.2.2) to assign a value to the constant

A

in the wavefunction solutions of the infinite square well potential given in (4.7). Given that the wavefunction is zero outside of the well, the total probability of finding the particle within the well should be one. Therefore,

\begin{matrix} (4.10) & \int_{0}^{L} | ψ_{2} (x) |^{2} d x = \int_{0}^{L} A^{2} \sin^{2} (\frac{n π}{L} x) d x = 1 . \end{matrix}

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Using the table of integrals in Appendix A of your Wolfson text, we can evaluate the integral using

\begin{matrix} (4.11) & \int_{0}^{L} \sin^{2} (\frac{n π}{L} x) = \frac{L}{2} . \end{matrix}

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Therefore

\begin{matrix} (4.12) & A^{2} (\frac{L}{2}) = 1 and A = \sqrt{\frac{2}{L}} . \end{matrix}

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Notice that this value of

A

is the same for all of the solutions, independent of

n .

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Example 4.5. Probability of finding a particle in an infinite square well potential.

A particle of mass $m$ is placed in an infinite square well of width $L$ in a quantum state for which $n = 2 .$ (a) In the vicinity of what position(s) is the particle most likely to be found within the well? (b) What is the probability of finding the particle between positions $x = L / 4$ and $x = L / 2 ?$

Solution.

(a)} The wavefunction for this particle is shown as $ψ_{2} (x)$ in Figure 4.3. Using (4.7), the probability density for this wavefunction is

\begin{matrix} (4.13) & | ψ_{2} (x) |^{2} = {(\sqrt{\frac{2}{L}})}^{2} \sin^{2} (\frac{2 π}{L} x), \end{matrix}

Figure 4.6. Probability density for a particle in the $n = 2$ state of an infinite square well.

as shown in Figure 4.6. The positions near which the particle is most likely to be found are where the probability density is the greatest. By examining the graph of probability density, these positions are located at $x = L / 4$ and $x = 3 L / 4 .$ (b) To calculate the probability of finding the particle between $x = L / 4$ and $x = L / 2,$ we use (3.8):

\begin{aligned} Prob [\frac{L}{4}, \frac{L}{2}] = & \int_{L / 4}^{L / 2} | ψ_{2} (x) |^{2} d x \\ = & \int_{L / 4}^{L / 2} (\frac{2}{L}) \sin^{2} (\frac{2 π}{L} x) d x \\ = & \frac{2}{L} {[\frac{x}{2} - \frac{L}{8 π} \sin (\frac{4 π}{L} x)]}_{L / 4}^{L / 2} \\ = & \frac{1}{4} \end{aligned}

where the integral can be evaluated using techniques that you learned in your calculus class or the table of integrals in Appendix A of your Wolfson text.

Because of the symmetry of the probability density, the area under the curve of ${| ψ |}^{2}$ between $x = L / 4$ and $x = L / 2$ is one-quarter of the total area, therefore it makes sense that the probability is $\frac{1}{4} .$