A simple model of a chemical bond: A particle in a one-dimensional box

In this section, we will consider a very simple model that describes an electron in a chemical bond. This is the so-called particle in a box model. We imagine a particle strictly confined between two ``walls'' by a potential energy $V(x)$ that is shown in the figure below. $V(x)$ confines the particle to the region $x\in [0,L]$. Such a model, although simple, is physically reasonable. For example, imagine an electron in a CC single bond. Since carbon atoms are roughly 24000 times more massive than an electron, thinking classically for a moment, this would be like a ball-bearing between two wrecking balls, which would certainly seem like two infinitely high walls. The box length $L$ for a CC single bond would be roughly 1.5 Å.

Figure 1: Illustration of a particle in a box

Mathematically, the potential energy is expressed as

$$V(x) = \left\{\matrix {0\;\;\;{\rm for}\;\;\;0\leq x \leq L \cr \cr \infty\;\;\;{\rm for} \;\;\;x < 0, x > L}\right.$$

If the potential is infinite outside the box, then there is zero probability that the particle will be found there. Thus, we require $\psi(x)=0$ outside the box, and this can only happen if at the boundaries $x=0$ and $x=L$, $\psi(0)=\psi(L)=0$. These, then, are the boundary conditions on $\psi(x)$. We will also require that the wavefunction be normalized to 1. Since $\psi(x)=0$ outside the box, we only need to integrate between 0 and $L$, so that

$$\int_0^L \vert\psi(x)\vert^2 dx = 1$$

Since we already know that $\psi(x)=0$ outside the box, we only need to solve the Schrödinger inside the box where $V(x)=0$. In this case, the Schrödinger looks like the free-particle equation we already wrote down

$$-{\hbar^2 \over 2m}{d^2 \psi \over dx^2} = E\psi(x)$$

Rearranging gives

$${d^2 \psi \over dx^2} = -{2mE \over \hbar^2}\psi(x)$$

We have already seen that the solutions are sin or cos functions. In fact, we can easily show that the solution of this differential equation must be either

$$\psi(x) = A\sin\left(\sqrt{2mE \over \hbar^2} x\right) \;\;\... ...;\;\;\; \psi(x) = B\cos\left(\sqrt{2mE \over \hbar^2} x\right)$$

where $A$ and $B$ are arbitrary constants. But since we need $\psi(0)=0$, the only solution that is consistent with the boundary condition is the sin solution. Thus, we take

$$\psi(x) = A\sin\left(\sqrt{2mE \over \hbar^2} x\right)$$

It's a simple matter to verify that this function satisfies the differential equation:

$${d\psi \over dx} = A \sqrt{2mE \over \hbar^2}\cos\left(\sqrt{2mE \over \hbar^2} x\right)$$

$${d^2\psi \over dx^2} = -A\left({2mE \over \hbar^2}\right)\cos\left(\sqrt{2mE \over \hbar^2} x\right)$$

$$= -\left({2mE \over \hbar^2}\right)\psi(x)$$

At this point, all we have is a general solution to the equation, but we still do not know the allowed values of $E$. But, there is also still one boundary condition we have not yet enforced. We need $\psi(L)=0$. This requires

$$A\sin\left(\sqrt{2mE \over \hbar^2} L\right)=0$$

In order to satisfy this, we can take any point where the sin function vanishes. We know that $\sin(n\pi)=0$ for any value of $n$. Thus, our boundary condition is equivalent to

$$\sqrt{2mE \over \hbar^2}L = n\pi$$

Solving this equation for $E$ gives us the allowed energies:

$$E = {\hbar^2 \pi^2 \over 2mL^2}n^2 \equiv E_n$$

Note, first, that there are many allowable values of $E$, so we denote these by $E_n$ according to the value of $n$ that we choose. For each allowable energy, there is also a wavefunction $\psi_n(x)$ obtained by substituting $E_n$ into the expression for $\psi(x)$. This yields

$$\psi_n(x) = A\sin\left({n\pi x \over L}\right)$$

We still need to determine the constant $A$, but before we do this, a comment is in order. Specifically, we are going to place a restriction on $n$. We will restrict $n$ to be the natural numbers $n=1,2,3,...$. The value $n=0$ gives a wavefunction $\psi_0(x)$ that is everywhere 0, and since the particle must be somewhere, this solution is not allowed. In addition, although the negative numbers $n=-1,-2,...$ also satisfy the boundary condition, since $\psi_{-n}(x) = -\psi_n(x)$, because $\sin(-y)=-\sin(y)$, we do not need to keep the negative values of $n$ because an overall minus sign on $\psi(x)$ does not change the probability of finding the particle in a region of the box becuase $p(x) = \vert\psi(x)\vert^2$. Thus, those wavefunctions that give unique probability densities are those for which $n=1,2,3,....$.

Let us now determine the constant $A$. There is still one piece of information we have not used and that is the normalization of $\psi_n(x)$. For each $n$, this function must be normalized to 1. This requires

$$\int_0^L \vert\psi_n(x)\vert^2 dx = 1$$

$$A^2 \int_0^L \sin^2\left({n\pi x \over L}\right)dx = 1$$

$$A^2\int_0^L\left[{1 \over 2} - {1 \over 2} \cos\left({2n\pi x \over L}\right)\right] = 1$$

$$A^2 \left[\left.{x \over 2}\right\vert _0^L - \left.{L \over 4n\pi}\cos\left({2n\pi x \over L}\right)\right\vert _0^L\right] = 1$$

$$A^2{L \over 2} = 1$$

$$A = \sqrt{2 \over L}$$

Thus, each wavefunction takes the form

$$\psi_n(x) = \sqrt{2 \over L}\sin\left({n\pi x \over L}\right)$$

The figure below shows the energy level diagram together with some of the wavefunctions and their associated probability densities $p_n(x) = \psi_n^2(x)$:

Figure: Energy level diagram, lowest order wavefunctions and lowest order probability densities for the particle in a box.

That the $\psi_n(x)$ satisfy the Schrödinger equation can be expressed compactly as

$$\hat{H}\psi_n(x) = E_n\psi_n(x)$$

Let us now make a few important points that help us with a physical interpretation of the functions $\psi_n(x)$ and energies $E_n$.

1.

First what do we really mean when we refer to the $E_n$ as the ``allowed'' energies? ``Allowed'' here means that if we perform an experiment to measure the energy of the system, the only possible outcome of such a measurement is one of the $E_n$. That is, the $E_n$ are the only possible values we can obtain in a measurement of the energy. Thus, we often refer to the $E_n$ as the eigenvalues of $\hat{H}$, eigen being the German word for ``own'' or ``characteristic''.

2.

What do we mean by the ``allowed'' wave functions $\psi_n(x)$? In general, the wave function of a particle is not restricted to be one of the $\psi_n(x)$. We can prepare the particle in a state or with a wave function $\Psi(x)$ that is different from $\psi_n(x)$, however, $\Psi(x)$ must be a combination of the allowed $\psi_n(x)$, and it must satisfy $\Psi(0) = 0$, $\Psi(L) = 0$, and it must be normalized:

$$\int_0^L \vert\Psi(x)\vert^2 dx = 1$$

3.

If we do prepare the system such that $\Psi(x) = \psi_n(x)$ for some value of $n$, then a measurement of the energy will yield the value $E_n$ with probability 1. For example, if $\Psi(x) = \psi_2(x)$, we would then obtain $E_2$ with probability 1 if we were to measure the energy. If $\Psi(x)$ were a combination of more than one of the $\psi_n(x)$, then identical energy measurements will have different outcomes that cannot be predicted. For example, if $\Psi(x) = (1/\sqrt{2})(\psi_1(x) + \psi_2(x))$, then a measurement of energy would yield $E_1$ with probability 1/2 and $E_2$ with probability 1/2.

Several other things are worth noting before we proceed with this example. First, as $n$ increases, the number of nodes in the wave function increases. A node is a point at which the wavefunction cross the $x$-axis. For $n=1$, $\psi_1(x)$ has zero nodes, for $n=2$, $\psi_2(x)$ has one node, for $n=3$, $\psi_3(x)$ has two nodes, and in general, $\psi_n(x)$ has $n-1$ nodes. Also note that $\psi_n^2(x) \geq 0$ for all $x$, even though $\psi_n(x)$ can be either positive or negative. This distinction will become particularly important when we consider the quantum theory of chemical bonding.

Another important property of the allowed wave functions is known as the orthogonality property. It is straightforward to show that

$$\int_0^L \psi_n(x)\psi_m(x) dx = 0\;\;\;\;\;{\rm if}\;\; m\neq n$$

Obviously, the integral must be 1 if $m=n$. To prove the orthogonality property, one only needs to substitute the form of $\psi_n(x)$ and $\psi_m(x)$ and do the integral. Again, we will make use of this property when we consider chemical bonding.

Finally, note that the lowest energy a particle can have is not 0! The lowest energy corresponds to $n=1$

$$E_1 = {\hbar^2 \pi^2 \over 2mL^2} > 0$$

This is true for any bound system in quantum mechanics. This lowest energy is called the zero-point energy of the system. This zero-point energy is consistent with the uncertainty principle. Since there is always energy in the system, it will never be possible to obtain $\Delta x=0$ and $\Delta p=0$ under any circumstances.

Let us now proceed to work through an example that illustrates what the wavefunctions mean.

Example: A particle in a one-dimensional box of length $L$ has an energy

$$E = {2\hbar^2 \pi^2 \over mL^2}$$

What is the probability that a measurement of the location of the particle will find it between $x=0$ and $x=L/2$?

From the given energy, the particle must be in the $n=2$ state so that its wavefunction is

$$\psi_2(x) = \sqrt{2 \over L}\sin\left({2\pi x \over L}\right)$$

The probability that the particle will be found in the region $x\in[0,L/2]$ when its position is measured is then

$$P(x\in[0,L/2]) = \int_0^{L/2}\psi_2^2(x) dx$$

$$= {2 \over L}\int_0^{L/2}\sin^2\left({2\pi x \over L}\right) dx$$

$$= {2 \over L}\int_0^{L/2} \left[{1 \over 2} - {1 \over 2}\cos\left({4\pi x \over L}\right)\right]dx$$

$$= {2 \over L} \left[\left.{x \over 2}\right\vert _0^{L/2} - \left.{L \over 8\pi}\sin\left({4\pi x \over L}\right) \right\vert _0^{L/2}\right]$$

$$= {2 \over L}\cdot {L \over 4} = {1 \over 2}$$

Therefore, the probability is exactly $1/2$ that the particle will be found in the left half of the box.

It is important to note that if the particle is prepared with one of the allowable wavefunctions $\psi_n(x)$, then a measurement of its energy will always yield the same value $E_n$. Thus, since the energy has a well-defined value, this value will be obtained in all independent measurements in which the system is prepared the same way. This is to be contrasted with a measurement of position, which can yield any value between $x=0$ and $x=L$ with varying probability. Moreover, preparing the system in the same energy state and repeating the measurement of position will yield different values in different repetitions of the experiment. All we can determine from the theory is the probability that the particle will be found in a certain region.

Example: A particle of mass $m$ in a box of length $L$ is prepared in a quantum state such that its wave function is

$$\Psi(x) = {1 \over \sqrt{2}}\left[\psi_1(x) + \psi_2(x)\right]$$

a.

Show that this state is correctly normalized.

b.

What is the probability that a measurement of the particle's position will yield a value between $x=0$ and $x=L/4$?

c.

What is the probability that a measurement of the particle's energy yields a value $\hbar^2\pi^2/(2mL^2)$?

a.

The normalization condition requires that

$$\int_0^L \vert\Psi(x)\vert^2 dx = 1$$

Taking the square of the given wave function and integrating, we have

$$\int_0^L {1 \over 2}\left[\psi_1(x) + \psi_2(x)\right]^2 dx = {1 \over 2}\left[\int_0^L \psi_1^2(x) dx + 2\int_0^L \psi_1(x)\psi_2(x)dx + \int_0^L \psi_2^2(x)dx\right]dx$$

$$= {1 \over 2}[1 + 0 + 1]$$

$$= 1$$

b.

The probability is given by

$$P(x\in[0,L/4]) = \int_0^{L/4}\vert\Psi(x)\vert^2 dx$$

Substituting in gives

$$P(x\in[0,L/4]) = {1 \over 2}{2 \over L} \int_0^{L/4} \left[\... ...over L}\right) + \sin^2\left({2\pi x \over L}\right)\right]dx$$

In order to calculate this, the following trig identities are useful:

$$\sin^2(ax) = {1 \over 2} - {1 \over 2}\cos(2ax)$$

$$\sin(ax)\sin(bx) = {1 \over 2} \left[\cos(a-b)x - \cos(a+b)x\right]$$

Using these, the probability becomes

$$P(x\in[0,L/4]) = {1 \over L}\int_0^{L/4} \left[{1 \over 2} - {1 \over 2}\cos\left(... ...L}\right) + {1 \over 2} - {1 \over 2}\cos\left({4\pi x \over L}\right)\right]dx$$

$$= {1 \over L} \left.\left[{L \over 4} - {L \over 4\pi}\sin{\pi \ove... ...ver 3\pi}\sin{3\pi \over 4} - {L \over 8\pi}\sin\pi\right] \right\vert _0^{L/4}$$

$$= {1 \over 4} - {1 \over 4\pi} + {\sqrt{2} \over 3\pi} \approx 0.32 = 32\%$$

c.

Just as we cannot predict the outcome of a measurement of the particle's position, we cannot predict the outcome of a measurement of a particle's energy. We can, however, predict the probability for obtaining any of the allowed energy values. Since these are the only allowed values, such a measurement must yield one of these values can cannot yield any other value. This is what is actually meant by the term ``allowed'' for an arbitrary quantum state.

The formula that we need for predicting the probabilities of energy measurements is the following: If $\Psi(x)$ is the quantum state of a system, the probability that a measurement of energy yields the allowed value $E_n$ is calculated using the corresponding wave function $\psi_n(x)$ from

$$P(E = E_n) = \left\vert\int_0^L \psi_n^*(x)\Psi(x)dx\right\vert^2$$

This is a general formula. For any other problem, the only possible change is that the integration region $x\in [0,L]$ would be replaced by whatever the spatial domain of the problem is. Despite how it looks, this formula is actually much easier to use than the one for predicting the probabilities of position measurements, thanks to the orthogonality property.

In order to anser the specific question, note that the energy $\hbar^2\pi^2/(2mL^2)$ is actually the energy $E_1$, so the question is really asking what is the probability that a measurement of energy yields the value $E_1$. According to the formula, this requires that we use the wave function $\psi_1(x)$. Substituting this and $\Psi(x)$ into the formula yields

$$P(E = E_1) = \left\vert\int_0^L \psi_1(x) {1 \over \sqrt{2}}\left[\psi_1(x) + \psi_2(x)\right]dx\right\vert^2$$

Now, the integral itself can be solved without integrating anything. Note that because of orthogonality, the integral is just

$$\int_0^L \psi_1(x) {1 \over \sqrt{2}}\left[\psi_1(x) + \psi_2(x)\right]dx = {1 \over \sqrt{2}}\int_0^L\psi_1^2(x)dx + {1 \over \sqrt{2}}\int_0^L \psi_1(x)\psi_2(x)dx$$

$$= {1 \over \sqrt{2}}\cdot 1 + {1 \over \sqrt{2}}\cdot 0$$

$$= {1 \over \sqrt{2}}$$

Hence, squaring this result give the probability of 1/2 to find obtain the result $E=E_1$. Note that this is just the square of the coefficient of $\psi_1(x)$ in $\Psi(x)$. From this result, it can be seen that the probability of obtaining the result $E=E_2$ is also $1/2$.

As a question for thought, what would be the probability that a measurement of energy would yield the value $\hbar^2\pi^2/(\sqrt{2}mL^2)$?