7.5 Bound Particles

7.5.1 A Packet in a Box

We considered our first bound system, the infinite square well, in a discussion of quantum mechanical standing waves in Section 7.2.4. In fact, all bound states are standing waves.

We noted in Section 7.2.4 that it is a mistake to think of the stationary states of the well as describing particles bouncing back and forth between the ``walls'' of the well.

We return here briefly to the infinite square well to address the question of whether and how one would describe such a particle. We will trap the free Gaussian wave packet we developed in Section 7.4.4 in an infinite square well and observe its behavior.

The continuous wavenumber spectrum of Eq. 7.72 cannot exist in the trap. Instead, we have a Gaussian distribution of discrete states of the well,^7.7

$\begin{displaymath} b_n = e^{-\frac{\sigma^2 \pi^2}{2L^2} (n - \bar{n})^2 } \end{displaymath}$

(7.88)

The distribution peaks at

$\begin{displaymath} \bar{n} = \sqrt{\frac{2 m L^2 \bar{E}}{\hbar^2 \pi^2}} \end{displaymath}$

(rounded to an integer) which for a 1 eV electron in a 10 nm well gives $\bar{n} = 16$ .

The time dependent wavefunction is a sum rather than an integral,

$\begin{displaymath} \Psi(x,t) = C \sum_{n=0}^{\infty} e^{-\frac{\sigma^2 \pi^2}... ...})^2 } \sin \left( \frac{n \pi x}{L} \right) e^{-i \omega_n t} \end{displaymath}$

(7.89)

The infinite sum of Eq. 7.89 may look like trouble, but with $\sigma = 1.0$ nm, the wavenumber distribution includes significant contributions from only $6 \leq n \leq 26$ , and we can have a computer calculate and plot the probability density.

Animation : QMboxedPacket.mw [7.5 MB] (see Appendix A on running animations)

The probability density does not initially resemble a Gaussian wave packet, because we have (rudely) placed it at , where it interacts with the ``wall'' of the well. However, it does look familiar as it traverses the center of the well.
Observing the evolution of the probability distribution of the packet, we find that it exhibits interference effects that resemble ``beats'' but that once the initial packet has spread, the distribution is relatively flat.
If we wait long enough, several components of the packet come back into phase and show brief packet-like behavior.

You will consider for homework the question of whether this packet behaves periodically.

7.5.2 Finite Square Well Review

Our work with scattering from barriers and wells gives us a head start on finding the bound states of the finite square well,

$\begin{displaymath} V(x) = \left\{ \begin{array}{lc} 0 & (\vert x\vert < \frac{L}{2}) \\ V_o & (\vert x\vert > \frac{L}{2}) \end{array}\right. \end{displaymath}$

(7.90)

where

is a positive constant.

However, we will not recycle the well scattering calculation of Sec. 7.4.2, because for bound states, we can simplify the solutions somewhat.

In the classical region, $\vert x\vert < \frac{L}{2}$ , the wavefunctions are standing waves with either even or odd symmetry,

$\begin{displaymath} \Psi_\mathrm{odd} \propto \cos(k x) e^{-i\omega t} \end{displaymath}$ (7.91)

or

$\begin{displaymath} \Psi_\mathrm{odd} \propto \sin(k x) e^{-i\omega t} \end{displaymath}$ (7.92)

where

$\begin{displaymath} k = \frac{\sqrt{2mE}}{\hbar} \end{displaymath}$ (7.93)
In the classically forbidden region, $\vert x\vert > \frac{L}{2}$ , we have , giving imaginary , and the wavefunctions become declining exponentials,

$\begin{displaymath} \Psi \propto \left\{ \begin{array}{lc} e^{\alpha x} & (x < -... ...L}{2})\\ e^{-\alpha x} & (x > \frac{L}{2}) \end{array}\right. \end{displaymath}$ (7.94)

where

$\begin{displaymath} \alpha = \frac{\sqrt{2m(V_o - E)}}{\hbar} \end{displaymath}$ (7.95)

**Figure 7.3:** A graphical solution of Eq.'s 7.96 and 7.97 giving the energies of the six bound states of an electron in a square well 10 eV deep and 1 nm wide. The dashed curves correspond to odd symmetry states.
$\scalebox{0.7}{ \includegraphics{FSWenergies.eps} }$

Matching $\Psi$ and $\Psi'$ at either boundary yields a pair of transcendental equations, one for the even

$\begin{displaymath} \alpha = \sqrt{\alpha_o^2 - k^2} = k \tan\left(\frac{kL}{2}\right) \end{displaymath}$

(7.96)

and one for the odd

$\begin{displaymath} \alpha = \sqrt{\alpha_o^2 - k^2} = -k \cot\left(\frac{kL}{2}\right) \end{displaymath}$

(7.97)

symmetry solutions, where

$\begin{displaymath} \alpha_o = \frac{\sqrt{2mV_o}}{\hbar} \end{displaymath}$

(7.98)

A graphical solution of Eq.'s 7.96 and 7.97 for an electron in a square well 10 eV deep and 1 nm wide is shown in Fig. 7.3.

The expression on the left hand sides of Eq.'s 7.96 and 7.97 describes a circle of radius $\alpha_o$ . This radius determines the number of solutions and hence the number of bound states. A deeper well has more bound states.

For our example, there are six bound states, three with even symmetry and three with odd symmetry at

eV. Plots of the wavefunctions are shown in Fig. 7.4.

**Figure 7.4:** Even (upper panel) and odd (lower panel) symmetry bound state wavefunctions of an electron confined to a finite square well of depth 10 eV and width 1 nm.
$\scalebox{0.65}{ \includegraphics{FSWeven.eps} }$ $\scalebox{0.65}{ \includegraphics{FSWodd.eps} }$

7.5.3 The Quantum Harmonic Oscillator

The Hamiltonian of the harmonic oscillator is^7.8

$\begin{displaymath} \hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2} m \omega^2 \hat{x}^2 \end{displaymath}$

(7.100)

It appears that we can factor the Hamiltonian with the raising operator or creation operator

$\begin{displaymath} \hat{a}_+ = \frac{1}{\sqrt{2m}} (\hat{p} + i m \omega \hat{x}) \end{displaymath}$

(7.101)

and the lowering operator or annihilation operator

$\begin{displaymath} \hat{a}_- = \frac{1}{\sqrt{2m}} (\hat{p} - i m \omega \hat{x}) \end{displaymath}$

(7.102)

(the names will become clear)

However, it is unclear whether

$\begin{displaymath} \hat{H} = \hat{a}_+ \hat{a}_- \hspace{0.5cm}? \end{displaymath}$

$\begin{displaymath} \hat{H} = \hat{a}_- \hat{a}_+ \hspace{0.5cm}? \end{displaymath}$

and, since $\hat{a}_-$ and $\hat{a}_+$ contain differentiation operators, order matters.

We define the commutator of two operators as

$\begin{displaymath}[\hat{A},\hat{B}]= \hat{A}\hat{B} - \hat{B}\hat{A} \end{displaymath}$

(7.103)

which is nonzero if order matters. Where $[\hat{A},\hat{B}] = 0$ , operators $\hat{A}$ and $\hat{B}$ are said to commute.

When evaluating commutators, it is helpful to include a wavefunction as a reminder that the operators operate on functions to the right. For example,

$\begin{eqnarray*}[\hat{p},\hat{x}]\psi(x) &=& \hat{p}\hat{x} \psi(x) - \hat{x} ... ...x)\\ &=& -i\hbar \left(\psi(x) + x \psi'(x) - x \psi'(x)\right) \end{eqnarray*}$

the fact that we must apply the product rule in the first term above was made clearer by the explicit inclusion of $\psi(x)$ . We conclude that $\hat{p}$ and $\hat{x}$ do not commute. Instead,

$\begin{displaymath}[\hat{p},\hat{x}]= -i \hbar \end{displaymath}$

(7.104)

Noting that

$\begin{displaymath}[\hat{A},\hat{B}]= -[\hat{B},\hat{A}] \end{displaymath}$

we also have

$\begin{displaymath}[\hat{x},\hat{p}]= i \hbar \end{displaymath}$

(7.105)

You will show for homework that order does matter here and that in particular,

$\begin{displaymath}[\hat{a}_+, \hat{a}_-]= - \hbar \omega \end{displaymath}$

(7.106)

Choosing an order at random, and evaluating the product,

$\begin{eqnarray*} \hat{a}_+ \hat{a}_- &=& \frac{1}{2m} (\hat{p} + i m \omega \h... ...}^2 +i m \omega i \hbar)\\ &=& \hat{H} - \frac{\hbar \omega}{2} \end{eqnarray*}$

we find that

$\begin{displaymath} \hat{H} = \hat{a}_+ \hat{a}_- + \frac{\hbar \omega}{2} \end{displaymath}$

(7.107)

Eq.'s 7.106 and 7.107 give an alternate form,

$\begin{displaymath} \hat{H} = \hat{a}_- \hat{a}_+ - \frac{\hbar \omega}{2} \end{displaymath}$

(7.108)

Note that since we evaluated the derivatives in the commutator of $\hat{p}$ and $\hat{x}$ , we have been able to work completely with operator notation. We will proceed to use operator algebra to find the energy spectrum of the system.

We will assume that bound state wavefunctions $\psi_o, \psi_1, \psi_2, ...$ exist, and that they correspond to a set of discrete energies

The Schrödinger equation takes the form

$\begin{displaymath} \hat{H} \psi_n = E_n \psi_n \end{displaymath}$

(7.109)

Consider the effect $\hat{H}$ has on the function $(\hat{a}_+ \psi(x))$

$\begin{eqnarray*} \hat{H}(\hat{a}_+ \psi_n) &=& \left( \hat{a}_+ \hat{a}_- + \f... ...ega ) \psi_n(x)\\ &=& ( E_n + \hbar \omega ) (\hat{a}_+ \psi_n) \end{eqnarray*}$

What we have found is that the function $(\hat{a}_+ \psi_n)$ also satisfies the Schrödinger equation. It is another bound state wavefunction, and $E_n + \hbar \omega$ is its energy.

If we investigate the state $(\hat{a}_- \psi_n)$ , we find

$\begin{displaymath} \hat{H}(\hat{a}_- \psi_n) = ( E_n - \hbar \omega ) (\hat{a}_- \psi_n) \end{displaymath}$

Apparently, the raising and lowering operators turn one of the $\psi_n$ into another one with higher and lower energy, respectively.

If there is a lowest energy state described by $\psi_o$ and

, we should expect nothing to result if we attempt to ``lower'' it,

$\begin{displaymath} \hat{a}_- \psi_o = 0 \end{displaymath}$

(7.110)

If we operate on this ground state with the Hamiltonian, we have

$\begin{eqnarray*} \hat{H} \psi_o &=& E_o \psi_o\\ &=& \left(\hat{a}_+ \hat{a}_-... ... &=& \hat{a}_+ \hat{a}_- \psi_o + \frac{\hbar \omega}{2} \psi_o \end{eqnarray*}$

but with Eq. 7.110, the first term above vanishes, and we find

$\begin{displaymath} E_o = \frac{\hbar \omega}{2} \end{displaymath}$

(7.111)

Further, Eq. 7.110 is a differential equation that we can solve for the ground state wavefunction,

$\begin{displaymath} \left(-i\hbar \frac{d}{d x} - i m \omega x \right) \psi_o(x) = 0 \end{displaymath}$

$\begin{displaymath} \frac{d}{dx} \psi_o(x) = -\frac{m \omega}{\hbar} x \psi_o(x) \end{displaymath}$

which can be integrated to find

$\begin{displaymath} \psi_o(x) = C e^{-\frac{m\omega}{2 \hbar} x^2} \end{displaymath}$

(7.112)

The ground state probability density of the harmonic oscillator is a Gaussian of width

$\begin{displaymath} \sigma = \sqrt{\frac{\hbar}{2 m \omega}} = \sqrt{\frac{\hbar}{2 \sqrt{\kappa m}} } \end{displaymath}$

(7.113)

where $\kappa$ is the ``spring constant'' of the potential.

**Figure 7.5:** The wavefunctions of the first four states of an electron in a harmonic oscillator potential with $\hbar \omega = 1$ eV.
$\scalebox{0.7}{ \includegraphics{sho1-4.eps} }$

If we proceed with the hypothesis that the raising and lowering operators connect nearest neighbors, we have

$\begin{displaymath} \hat{a}_\pm \psi_n = A_{n\pm} \psi_{n\pm1} \end{displaymath}$

(7.114)

where the $A_{n\pm}$ are constants we need to evaluate to preserve normalization.

Starting with the ground state wavefunction, we can construct all of the others using the raising operator. The wavefunctions of the first four states of an electron in a harmonic oscillator potential with $\hbar \omega = 1$ eV are shown in Fig. 7.5.

Adjacent energy levels of a quantum mechanical harmonic oscillator are separated by $\hbar \omega$ . Hence, the energy level diagram is ladder-like.

7.5.4 WKB Thinking : Qualitative Sketches

We can apply the ideas suggested by the WKB approximation (See Sections 3.2.1 and 7.4.3) to bound states. In classically allowed regions (

Greater $\Leftrightarrow$ greater momentum $\Leftrightarrow$ smaller amplitude $\Psi$ and smaller peak $\varrho$
Greater $\Leftrightarrow$ greater momentum $\Leftrightarrow$ larger local , shorter local $\lambda$