Particle in Box

As an example of a three-dimensional problem in wave mechanics, consider a particle trapped in a square potential well of infinite depth, such that

$$U(x,y,z) = \left\{ \begin{array}{lll} 0&\mbox{\hspace{0.5cm}}... ...z\leq a\\ [0.5ex] \infty &&\mbox{otherwise} \end{array}\right..$$ (11.163)

Within the well, the stationary wavefunction, $\psi(x,y,z)$, satisfies

$$- \frac{\hbar^{\,2}}{2\,m}\,\nabla^{\,2}\psi = E\,\psi,$$ (11.164)

subject to the boundary conditions

$$\psi(0,y,z) = \psi(x,0,z)=\psi(x,y,0) =0,$$ (11.165)

and

$$\psi(a,y,z) = \psi(x,a,z)=\psi(x,y,a) =0,$$ (11.166)

because $\psi=0$ outside the well. Let us try a separable wavefunction of the form

$$\psi(x,y,z) =\psi_0\,\sin(k_x\,x)\,\sin(k_y\,y)\,\sin(k_z\,z).$$ (11.167)

This expression automatically satisfies the boundary conditions (11.165). The remaining boundary conditions, (11.166), are satisfied provided

$$k_x = l_x\,\frac{\pi}{a},$$ (11.168)

$$k_y =l_y\,\frac{\pi}{a},$$ (11.169)

$$k_z = l_z\,\frac{\pi}{a},$$ (11.170)

where $l_x$, $l_y$, and $l_z$ are (independent) positive integers. Substitution of the wavefunction (11.167) into Equation (11.164) yields

$$E = \frac{\hbar^{\,2}}{2\,m}\,(k_x^{\,2} + k_y^{\,2}+k_z^{\,2}).$$ (11.171)

Thus, it follows from Equations (11.168)–(11.170) that the particle energy is quantized, and that the allowed energy levels are

$$E_{l_x,l_y,l_z} = \frac{\hbar^{\,2}}{2\,m\,a^{\,2}}\,(l_x^{\,2}+l_y^{\,2}+l_z^{\,2}).$$ (11.172)

The properly normalized [see Equation (11.157)] stationary wavefunctions corresponding to these energy levels are

$$\psi_{l_x,l_y,l_z}(x,y,z) = \left(\frac{2}{a}\right)^{3/2}\,\sin\... ...\sin\left(l_y\,\pi\,\frac{y}{a}\right)\,\sin\left(l_z\,\pi\,\frac{z}{a}\right).$$ (11.173)

As is the case for a particle trapped in a one-dimensional potential well, the lowest energy level for a particle trapped in a three-dimensional well is not zero, but rather

$$E_{1,1,1} = 3\,E_1.$$ (11.174)

Here,

$$E_1 = \frac{\hbar^{\,2}}{2\,m\,a^{\,2}}.$$ (11.175)

is the ground state (i.e., the lowest energy state) energy in the one-dimensional case. It follows from Equation (11.172) that distinct permutations of $l_x$, $l_y$, and $l_z$ that do not alter the value of $l_x^{\,2}+l_y^{\,2}+ l_z^{\,2}$ also do not alter the energy. In other words, in three dimensions, it is possible for distinct wavefunctions to be associated with the same energy level. In this situation, the energy level is said to be degenerate. The ground-state energy level, $3\,E_1$, is non-degenerate, because the only combination of ($l_x$, $l_y$, $l_z$) that gives this energy is ($1$, $1$, $1$). However, the next highest energy level, $6\,E_1$, is degenerate, because it is obtained when ($l_x$, $l_y$, $l_y$) take the values ($2$, $1$, $1$), or ($1$, $2$, $1$), or ($1$, $1$, $2$). In fact, a non-degenerate energy level corresponds to a case where the three quantum numbers (i.e., $l_x$, $l_y$, and $l_z$) all have the same value, whereas a threefold degenerate energy level corresponds to a case where only two of the quantum numbers have the same value, and, finally, a sixfold degenerate energy level corresponds to a case where the quantum numbers are all different.