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Next: Particle in a Finite Up: Wave Mechanics Previous: Stationary States

Three-Dimensional Wave Mechanics

Up to now, we have only discussed wave mechanics for a particle moving in one dimension. However, the generalization to a particle moving in three dimensions is fairly straightforward. A massive particle moving in three dimensions has a complex wavefunction of the form [cf., (785)]
\begin{displaymath}
\psi(x,y,z,t) = \psi_0\,{\rm e}^{\,{\rm i}\,({\bf k}\cdot{\bf r}-\omega\,t)},
\end{displaymath} (844)

where $\psi_0$ is a complex constant, and ${\bf r}=(x,\,y,\,z)$. Here, the wavevector, ${\bf k}$, and the angular frequency, $\omega $, are related to the particle momentum, ${\bf p}$, and energy, $E$, according to [cf., (773)]
\begin{displaymath}
{\bf p} = \hbar\,{\bf k},
\end{displaymath} (845)

and [cf., (771)]
\begin{displaymath}
E = \hbar\,\omega,
\end{displaymath} (846)

respectively. Generalizing the analysis of Section 12.5, the three-dimensional version of Schrödinger's equation is easily shown to take the form [cf., (793)]
\begin{displaymath}
{\rm i}\,\hbar\,\frac{\partial\psi}{\partial t} = - \frac{\hbar^2}{2\,m}\,\nabla^2\psi + U({\bf r})\,\psi,
\end{displaymath} (847)

where the differential operator
\begin{displaymath}
\nabla^2 \equiv \frac{\partial^2}{\partial x^2} + \frac{\partial^2 }{\partial y^2} + \frac{\partial^2}{\partial z^2}
\end{displaymath} (848)

is known as the Laplacian. The interpretation of a three-dimensional wavefunction is that the probability of finding the particle between $x$ and $x+dx$, between $y$ and $y+dy$, and between $z$ and $z+dz$, at time $t$ is [cf., (796)]
\begin{displaymath}
P(x,y,z,t) = \vert\psi(x,y,z,t)\vert^2\,dx\,dy\,dz.
\end{displaymath} (849)

Moreover, the normalization condition for the wavefunction becomes [cf., (798)]
\begin{displaymath}
\int_{-\infty}^\infty\int_{-\infty}^\infty\int_{-\infty}^\infty\vert\psi(x,y,z,t)\vert^2\,dx\,dy\,dz =1.
\end{displaymath} (850)

Incidentally, it is easily demonstrated that Schrödinger's equation, (848), preserves the normalization condition, (851), of a localized wavefunction. Heisenberg's uncertainty principle generalizes to [cf., (826)]
$\displaystyle \Delta x\,\Delta p_x$ $\textstyle \stackrel {_{\normalsize >}}{_{\normalsize\sim}}$ $\displaystyle \frac{\hbar}{2},$ (851)
$\displaystyle \Delta y\,\Delta p_y$ $\textstyle \stackrel {_{\normalsize >}}{_{\normalsize\sim}}$ $\displaystyle \frac{\hbar}{2},$ (852)
$\displaystyle \Delta z\,\Delta p_z$ $\textstyle \stackrel {_{\normalsize >}}{_{\normalsize\sim}}$ $\displaystyle \frac{\hbar}{2}.$ (853)

Finally, a stationary state of energy $E$ is written [cf., (830)]
\begin{displaymath}
\psi(x,y,z,t) = \psi(x,y,z)\,{\rm e}^{-{\rm i}\,(E/\hbar)\,t},
\end{displaymath} (854)

where the stationary wavefunction, $\psi(x,y,z)$, satisfies [cf., (832)]
\begin{displaymath}
- \frac{\hbar^2}{2\,m}\,\nabla^2\psi + U({\bf r})\,\psi = E\,\psi.
\end{displaymath} (855)

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

\begin{displaymath}
U(x,y,z) = \left\{
\begin{array}{lll}
0&\mbox{\hspace{0.5cm}...
...\leq a\\ [0.5ex]
\infty &&\mbox{otherwise}
\end{array}\right..
\end{displaymath} (856)

Within the well, the stationary wavefunction, $\psi(x,y,z)$, satisfies
\begin{displaymath}
- \frac{\hbar^2}{2\,m}\,\nabla^2\psi = E\,\psi,
\end{displaymath} (857)

subject to the boundary conditions
\begin{displaymath}
\psi(0,y,z) = \psi(x,0,z)=\psi(x,y,0) =0,
\end{displaymath} (858)

and
\begin{displaymath}
\psi(a,y,z) = \psi(x,a,z)=\psi(x,y,a) =0,
\end{displaymath} (859)

since $\psi=0$ outside the well. Let us try a seperable wavefunction of the form
\begin{displaymath}
\psi(x,y,z) =\psi_0\,\sin(k_x\,x)\,\sin(k_y\,y)\,\sin(k_z\,z).
\end{displaymath} (860)

This expression automatically satisfies the boundary conditions (859). The remaining boundary conditions, (860), are satisfied provided
$\displaystyle k_x$ $\textstyle =$ $\displaystyle n_x\,\frac{\pi}{a},$ (861)
$\displaystyle k_y$ $\textstyle =$ $\displaystyle n_y\,\frac{\pi}{a},$ (862)
$\displaystyle k_z$ $\textstyle =$ $\displaystyle n_z\,\frac{\pi}{a},$ (863)

where $n_x$, $n_y$, and $n_z$ are (independent) positive integers. Substitution of the wavefunction (861) into Equation (858) yields
\begin{displaymath}
E = \frac{\hbar^2}{2\,m}\,(k_x^{\,2} + k_y^{\,2}+k_z^{\,2}).
\end{displaymath} (864)

Thus, it follows from Equations (862)-(864) that the particle energy is quantized, and that the allowed energy levels are
\begin{displaymath}
E_{n_x,n_y,n_z} = \frac{\hbar^2}{2\,m\,a^2}\,(n_x^{\,2}+n_y^{\,2}+n_z^{\,2}).
\end{displaymath} (865)

The properly normalized [see Equation (851)] stationary wavefunctions corresponding to these energy levels are
\begin{displaymath}
\psi_{n_x,n_y,n_z}(x,y,z) = \left(\frac{2}{a}\right)^{3/2}\,...
...i\,\frac{y}{a}\right)\,\sin\left(n_z\,\pi\,\frac{z}{a}\right).
\end{displaymath} (866)

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

\begin{displaymath}
E_{1,1,1} = 3\,E_1.
\end{displaymath} (867)

Here,
\begin{displaymath}
E_1 = \frac{\hbar^2}{2\,m\,a^2}.
\end{displaymath} (868)

is the ground state (i.e., the lowest energy state) energy in the one-dimensional case. Now, it is clear, from (866), that distinct permutations of $n_x$, $n_y$, and $n_z$ which do not alter the value of $n_x^{\,2}+n_y^{\,2}+ n_z^{\,2}$ 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, since the only combination of ($n_x$, $n_y$, $n_z$) which gives this energy is ($1$, $1$, $1$). However, the next highest energy level, $6\,E_1$, is degenerate, since it is obtained when ($n_x$, $n_y$, $n_y$) take the values ($2$, $1$, $1$), or ($1$, $2$, $1$), or ($1$, $1$, $2$). In fact, it is not difficult to see that a non-degenerate energy level corresponds to a case where the three mode numbers (i.e., $n_x$, $n_y$, and $n_z$) all have the same value, whereas a three-fold degenerate energy level corresponds to a case where only two of the mode numbers have the same value, and, finally, a six-fold degenerate energy level corresponds to a case where the mode numbers are all different.


next up previous
Next: Particle in a Finite Up: Wave Mechanics Previous: Stationary States
Richard Fitzpatrick 2010-10-11