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7A: Particles in Two-Dimensional Boxes

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    16973
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    A quantum particle of mass \(m\) in a two-dimensional square box by a potential energy \(V(x,y)\) that is zero if \(x \; \epsilon \; [0,L]\) and \(y\; \epsilon \;[0,L]\) and infinite otherwise. Inside the box, the energy is entirely kinetic because \(V(x,y)=0\), so the classical energy is

    \[\frac{p_{x}^{2}}{2m} +\dfrac{p_{y}^{2}}{2m}=E\]

    where \(p_x\) and \(p_y\) are the two components of the particle's momentum. We see that the energy naturally is expressible as a sum of kinetic energies associated with motion in the \(x\) and \(y\) directions:

    \[E=\varepsilon_x +\varepsilon_y\]

    Because the energy is a simple sum, the solutions of the Schrödinger equation can be expressed as simple products of the solutions of the one-dimensional Schrödinger equation for this problem. Note that it is only when the energy is expressible in this way that simple product solutions are rigorously correct. Thus, the wavefunction \(\psi (x,y)\) is of the form

    \[\psi (x,y)=A\sin\left ( \sqrt{\frac{2m\varepsilon_x}{\hbar^2}}x \right ) \sin \left ( \sqrt{\frac{2m\varepsilon_y}{\hbar^2}}y \right )\]

    which satisfies the boundary conditions at \(x=0\) and \(y=0\), namely \(\psi (0,y)=0\) and \(\psi (x,0)=0\). In order to satisfy the remaining boundary conditions \(\psi (L,y)=0\) and \(\psi (x,L)=0\), we have two conditions:

    \[A\sin\left ( \sqrt{\frac{2m\varepsilon_x}{\hbar^2}}L \right ) \sin \left ( \sqrt{\frac{2m\varepsilon_y}{\hbar^2}}y \right ) = 0 \label{cond1}\]

    and

    \[A\sin\left ( \sqrt{\frac{2m\varepsilon_x}{\hbar^2}}x \right ) \sin \left ( \sqrt{\frac{2m\varepsilon_y}{\hbar^2}}L \right ) = 0 \label{cond2}\]

    Equation \ref{cond1} can be satisfied if

    \[\sin\left ( \sqrt{\dfrac{2m\varepsilon_x}{\hbar^2}}L \right ) =0\]

    independent of \(y\), while Equation \ref{cond2} can be satisfied if

    \[\sin\left ( \sqrt{\dfrac{2m\varepsilon_y}{\hbar^2}}L \right ) =0​\]

    independent of \(x\). These are the same conditions that we encountered for the one-dimensional box, hence we already know the \(\sin\) function in each case can be zero in many places. In fact, these two conditions are satisfied if

    \[\sqrt{\frac{2m\varepsilon_x}{\hbar^2}}L = n_x \pi\]

    and

    \[ \sqrt{\frac{2m\varepsilon_y}{\hbar^2}}L = n_y \]

    which yield the allowed values of \(\varepsilon_x\) and \(\varepsilon_y\) as

    \[\varepsilon_{n_x}=\frac{\hbar^2 \pi^2}{2mL^2}n_{x}^{2}\;\;\;\; \varepsilon_{n_y}=\frac{\hbar^2 \pi^2}{2mL^2}n_{y}^{2}\]

    We need two different integers \(n_x\) and \(n_y\) because the conditions are completely independent and can be satisfied by any two different (or similar) values of these integers. The allowed values of the total energy are now given by

    \[E_{n_x n_y}=\frac{\hbar^2 \pi^2}{2mL^2}(n_{x}^{2}+n_{y}^{2})\]

    Note that the allowed energies now depend on two integers \(n_x\) and \(n_y\) rather than one. These arise from the two independent boundary conditions in the \(x\) and \(y\) directions. As in the one-dimensional box, the values of \(n_x\) and \(n_y\) are both restricted to the natural numbers \(1,2,3,...\) Note, therefore, that the ground state energy \(E_{11}\) is

    \[E_{11}=\frac{\hbar^2 \pi^2}{mL^2}\]

    is larger than for the one-dimensional box because of the contributions from kinetic energy in the \(x\) and \(y\) directions.

    Once the conditions on \(\varepsilon_{n_x}\) and \(\varepsilon_{n_y}\) are substituted in, the wavefunctions become

    \[\psi_{n_x n_y}(x,y)=A\sin\left ( \frac{n_x \pi x}{L} \right ) \sin \left ( \frac{n_y \pi y}{L} \right )\]

    The constant \(A\) is now determined by the normalization condition

    \[\begin{align*}\int_{0}^{L} \int_{0}^{L}|\psi_{n_x n_y}(x,y)|^2 dxdy &= 1\\ A^2 \int_{0}^{L}\sin^2\left ( \frac{n_x \pi x}{L} \right ) dx\int_{0}^{L}\sin^2 \left ( \frac{n_y \pi y}{L} \right )dy &= 1\\ A^2\frac{L}{2}\cdot \frac{L}{2} &= 1\\ A &= \frac{2}{L}\end{align*}\]

    so that

    \[\psi_{n_x n_y}(x,y)=\frac{2}{L}\sin\left ( \frac{n_x \pi x}{L} \right ) \sin\left ( \frac{n_y \pi y}{L} \right )\]

    The wavefunctions are somewhat more difficult to visualize because of they are two dimensional. Nevertheless, we can still visualize them, and the figure below shows the following wavefunctions: \(\psi_{11}(x,y)\), \(\psi_{21}(x,y)\) and \(\psi_{22}(x,y)\).

    Figure: First few wavefunctions for a particle in a two-dimensional square box
    \includegraphics[scale=0.7]{twoD_box.eps}

    Note that the figure makes use of what we call dimensionless coordinates \(\bar{x}\) and \(\bar{y}\), which are defined to be the coordinates of a particle in a unit box, i.e. a box with \(L=1\). Thus, the dimensionless coordinates are defined by

    \[\bar{x}=\dfrac{x}{L}\;\;\;\; \bar{y}=\frac{y}{L}\]

    It is important to note that the wavefunctions can be either positive or negative, even though the associated probability density \(p_{n_x n_y}(x,y)=|\psi_{n_x n_y}(x,y)|^2\) is strictly positive. In the figure, the red part of the function is positive, and the blue part is negative. Unlike in the one-dimensional case, where nodes in the wavefunction are points where \(\psi_{n}(x)=0\), here entire lines can be nodal. These are called nodal lines. For example, in the state \(\psi_{21}(x,y)\), there is a nodal line at \(\psi_{21}(L/2,y)\). Along the entire line \(x=L/2\), the wavefunction is \(0\) independent of the value of \(y\). The wavefunction \(\psi_{22}(x,y)\) has two nodal lines when \(x=L/2\) and when \(y=L/2\). The sign of the wavefunction and its nodal structure will play central roles later when we consider chemical bonding.

    Note that if the box were rectangular rather than square, then instead of having a length of \(L\) on both sides, there would be two different lengths \(L_x\) and \(L_y\). The formulas for the energies and wavefunctions become only slightly more complicated:

    \[E_{n_x n_y}=\frac{\hbar^2 \pi^2}{2m}\left ( \dfrac{n_{x}^{2}}{L_{x}^{2}}+\frac{n_{y}^{2}}{L_{y}^{2}} \right )\] \[\psi_{n_x n_y}(x,y)=\frac{2}{\sqrt{L_x L_y}}\sin\left ( \frac{n_x \pi x}{L_x} \right ) \sin \left ( \frac{n_y \pi y}{L_y} \right )\]

    The definition of the dimensionless variables is also altered slightly

    \[\bar{x}=\frac{x}{L_x}\;\;\;\; \bar{y}=\frac{y}{L_y}\]

    The fact that the wavefunction \(\psi_{n_x n_y}(x,y)\) is a product of one-dimensional wavefunctions:

    \[\psi_{n_x n_y}(x,y)=\psi_{n_x}(x)\psi_{n_y}(y)\]

    makes the calculation of probabilities rather easy. The probability that a measurement the particle's position will yield a value of \(x\epsilon [a,b]\) and \(y\epsilon [c,d]\) is

    \[\begin{align*}P(x\epsilon [a,b] \ and \ y\epsilon [c,d]) &= \int_{a}^{b}dx\int_{c}^{d}dy|\psi_{n_x n_y}(x,y)|^2\\ &= \int_{a}^{b}dx\int_{c}^{d}dy\psi_{n_x}^{2}(x)\psi_{n_y}^{2}(y)\\ &= \left [ \int_{a}^{b}\psi_{n_x}^{2}(x)dx \right ] \left [ \int_{c}^{d}\psi_{n_y}^{2}(y)dy \right ]\end{align*}\]

    Example \(\PageIndex{1}\)

    For a particle in a two-dimensional box, if the particle is in the state \(\psi_{12}(x,y)\), what is the probability that a measurement of the particle's position will yield \(x\epsilon [0,L/2]\) and \(y\epsilon [0,L/2]\)?

    Solution

    Substituting into the above formula, we have

    \[\begin{align*}P(x\epsilon [0,L/2] \ and \ y\epsilon [0,L/2] &= \left [ \int_{0}^{L/2}\psi_{1}^{2}(x)dx \right ] \left [ \int_{0}^{L/2}\psi_{2}^{2}(y)dy \right ]\\ &= \left [ \frac{2}{L}\int_{0}^{L/2}\sin^2 \left ( \frac{\pi x}{L} \right ) dx \right ] \left [ \frac{2}{L}\int_{0}^{L/2}\sin^2 \left ( \frac{2\pi y}{L} \right ) dy \right ]\\ &= \left [ \frac{2}{L} \frac{L}{4} \right ] \left [ \frac{2}{L}\frac{L}{4} \right ]\\ &= \frac{1}{4}\end{align*}\]

    Contributors and Attributions


    This page titled 7A: Particles in Two-Dimensional Boxes is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Mark E. Tuckerman.