Like people, particles trapped in claustrophobic boxes tend to behave strangely. Since the behavior of particles in quantum mechanics is defined by wave functions, let's try to derive the wave functions governing particles trapped in infinite square and box wells and observe the effects that these potential wells have on the particles' energies.
Writing the Schrodinger Equation
Before we start playing with wave functions, we need to define the scenario first by setting up our Schrodinger equation:\begin{equation}
E\psi(\mathbf{r}) = -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r}) + V(\mathbf{r})\psi(\mathbf{r})
\end{equation}
where $\mathbf{r}$ is the position vector and $\psi(\mathbf{r})$ is the wave function of the particle. $\nabla^2$ is the del squared operator which can be written as $\nabla^2 = \frac{\partial^2}{{\partial x}^2} + \frac{\partial^2}{{\partial y}^2} + \frac{\partial^2}{{\partial z}^2}$ and $V(\mathbf{r})$ is the particle's potential energy which can be split up like $V(\mathbf{r})= V_x(x)+V_y(y)+V_z(z)$. Because both the kinetic and potential energies of the particle can be split into dimensional components, the total energy, $E$, can also be split: $E=E_x+E_y+E_z$. Furthermore, cutting the Hamiltonian up like this means that $\psi(\mathbf{r})=\psi_x(x)\psi_y(y)\psi_z(z)$, where $\psi_x(x)$, $\psi_y(y)$, and $\psi_z(z)$ are respectively the $x$, $y$, and $z$ components of the wave function.E\psi(\mathbf{r}) = -\frac{\hbar^2}{2m}\nabla^2\psi(\mathbf{r}) + V(\mathbf{r})\psi(\mathbf{r})
\end{equation}
\begin{align}
E_x\psi_x(x) & = -\frac{\hbar^2}{2m}\frac{\partial^2}{{\partial x}^2}\psi_x(x) + V_x(x)\psi_x(x) \\
E_y\psi_y(y) & = -\frac{\hbar^2}{2m}\frac{\partial^2}{{\partial y}^2}\psi_y(y) + V_y(y)\psi_y(y) \\
E_z\psi_z(z) & = -\frac{\hbar^2}{2m}\frac{\partial^2}{{\partial z}^2}\psi_z(z) + V_z(z)\psi_z(z) \\
\end{align}
Now, let's define our potential well like this:E_x\psi_x(x) & = -\frac{\hbar^2}{2m}\frac{\partial^2}{{\partial x}^2}\psi_x(x) + V_x(x)\psi_x(x) \\
E_y\psi_y(y) & = -\frac{\hbar^2}{2m}\frac{\partial^2}{{\partial y}^2}\psi_y(y) + V_y(y)\psi_y(y) \\
E_z\psi_z(z) & = -\frac{\hbar^2}{2m}\frac{\partial^2}{{\partial z}^2}\psi_z(z) + V_z(z)\psi_z(z) \\
\end{align}
$$V_x(x)= \begin{cases} 0, &0 < x < L_x \\ \infty, &\text{otherwise} \end{cases}$$
$$V_y(y)= \begin{cases} 0, &0 < y < L_y \\ \infty, &\text{otherwise} \end{cases}$$
$$V_z(z)= \begin{cases} 0, &0 < z < L_z \\ \infty, &\text{otherwise} \end{cases}$$
where $L_x$, $L_y$, and $L_z$ are the dimensions of the infinite box well. And so, for the $x$ dimension, inside the infinite square well, the Schrodinger equation is:
\begin{equation}
E_x\psi_x(x) = -\frac{\hbar^2}{2m} \frac{\partial^2}{{\partial x}^2} \psi_x(x)
\end{equation}
(For the sake of not having huge long equations, let's work only in the $x$ dimension for now and put everything back into three dimensions later.) E_x\psi_x(x) = -\frac{\hbar^2}{2m} \frac{\partial^2}{{\partial x}^2} \psi_x(x)
\end{equation}
Solving for the Wave Function
It's time to solve for $\psi_x(x)$. Equation $(5)$ is a homogeneous second-order linear differential equation and has the general solution:\begin{equation}
\psi_x(x)=ae^{r_ax}+be^{r_bx}
\end{equation}
where $a$, $r_a$, $b$, and $r_b$ are constants and $r_a$ and $r_b$ are the two solutions to the equation $\hbar^2r^2/2m + E_x=0$. Interestingly enough, we find that $r_a=i\sqrt{2mE_x}/\hbar$ and $r_b=-i\sqrt{2mE_x}/\hbar$. Remember that the particle's momentum in the $x$ direction is given by $\hbar k_x = \pm\sqrt{2m[E_x-V_x(x)]}$, where $k_x$ is the $x$-component wavenumber of the particle's wave function. In the case of $V_x(x)=0$ inside the box well, \psi_x(x)=ae^{r_ax}+be^{r_bx}
\end{equation}
\begin{equation}
k_x=\pm\frac{\sqrt{2mE_x}}{\hbar}
\end{equation}
And so,k_x=\pm\frac{\sqrt{2mE_x}}{\hbar}
\end{equation}
\begin{equation}
\psi_x(x)=ae^{ik_xx} + be^{-ik_xx}
\end{equation}
Using Euler's formula, we can expand this to:\psi_x(x)=ae^{ik_xx} + be^{-ik_xx}
\end{equation}
\begin{equation}
\psi_x(x)=(a+b)\cos{k_xx} + i(a-b)\sin{k_xx}
\end{equation}
Now, since this particle is trapped in the infinite box well, its wave function must be zero for $x \leq 0$ and $x \geq L_x$, so $\psi_x(0) = \psi_x(L_x) = 0$. Therefore, $b=-a$ and $k_x=n_x\pi / L_x$ where $n_x\in \{1,2,3,...\}$. (Although $n_x$ mathematically could be negative or zero, only the positive values are used by convention. The negative values of $n_x$ give wave function shapes identical to those produced by positive values of $n_x$ except for insignificant sign changes which have no physical effect. Meanwhile, $n_x=0$ just makes $\psi_x(x)=0$, meaning there's no particle in the well at all).\psi_x(x)=(a+b)\cos{k_xx} + i(a-b)\sin{k_xx}
\end{equation}
$$\psi_x(x)=2ia\sin{\frac{n_x \pi x}{L_x}}$$
Let $A=2ia$.\begin{equation}
\psi_x(x)=A\sin{\frac{n_x \pi x}{L_x}}, \qquad n_x\in\{1,2,3,...\}
\end{equation}
\psi_x(x)=A\sin{\frac{n_x \pi x}{L_x}}, \qquad n_x\in\{1,2,3,...\}
\end{equation}
Normalizing the Wave Function
In the case of our potential well, $\int_{0}^{L_x} |\psi_x(x)|^2 dx = 1$ must be true for our $x$-component wave function to be normalized. We're going to use this to solve for $A$.\begin{equation}
|A|^2\int_{0}^{L_x} \sin^2{\frac{n_x\pi x}{L_x}} dx = 1
\end{equation}
\begin{equation}
|A|^2\frac{L_x}{n_x\pi} \int_{0}^{n_x\pi} \sin^2{u} du = 1
\end{equation}
Using a double angle identity of cosine, we can rewrite the above equation like this:|A|^2\int_{0}^{L_x} \sin^2{\frac{n_x\pi x}{L_x}} dx = 1
\end{equation}
\begin{equation}
|A|^2\frac{L_x}{n_x\pi} \int_{0}^{n_x\pi} \sin^2{u} du = 1
\end{equation}
\begin{equation}
|A|^2\frac{L_x}{n_x\pi} \int_{0}^{n_x\pi} \left(\frac{1}{2} - \frac{1}{2}\cos{2u} \right) du = 1
\end{equation}
\begin{equation}
|A|^2\frac{L_x}{n_x\pi} \left(\frac{n_x\pi}{2} - \frac{1}{4}\sin{2n_x\pi} \right) = 1
\end{equation}
Since $n_x\in\{1,2,3,...\}$, $\sin{2n_x\pi}=0$.|A|^2\frac{L_x}{n_x\pi} \int_{0}^{n_x\pi} \left(\frac{1}{2} - \frac{1}{2}\cos{2u} \right) du = 1
\end{equation}
\begin{equation}
|A|^2\frac{L_x}{n_x\pi} \left(\frac{n_x\pi}{2} - \frac{1}{4}\sin{2n_x\pi} \right) = 1
\end{equation}
\begin{equation}
|A|^2\frac{L_x}{2} = 1
\end{equation}
\begin{equation}
|A| = \sqrt{\frac{2}{L_x}}
\end{equation}
Note that, as long as the above equation is satisfied, $A$ can be any complex number without causing any physical difference. However by convention, $A=\sqrt{2/L_x}$. Thus, we've derived the $x$-component of the wave function to be:|A|^2\frac{L_x}{2} = 1
\end{equation}
\begin{equation}
|A| = \sqrt{\frac{2}{L_x}}
\end{equation}
\begin{equation}
\psi_x(x)=\sqrt{\frac{2}{L_x}}\sin{\frac{n_x \pi x}{L_x}}, \qquad n_x\in \{1,2,3,...\}
\end{equation}
Similarly, \psi_x(x)=\sqrt{\frac{2}{L_x}}\sin{\frac{n_x \pi x}{L_x}}, \qquad n_x\in \{1,2,3,...\}
\end{equation}
\begin{align}
\psi_y(y) & = \sqrt{\frac{2}{L_y}}\sin{\frac{n_y \pi y}{L_y}}, & n_y\in \{1,2,3,...\} \\
\psi_z(z) & = \sqrt{\frac{2}{L_z}}\sin{\frac{n_z \pi z}{L_z}}, & n_z\in \{1,2,3,...\} \\
\end{align}
Putting all this together for a particle trapped in a three-dimensional infinite box well,\psi_y(y) & = \sqrt{\frac{2}{L_y}}\sin{\frac{n_y \pi y}{L_y}}, & n_y\in \{1,2,3,...\} \\
\psi_z(z) & = \sqrt{\frac{2}{L_z}}\sin{\frac{n_z \pi z}{L_z}}, & n_z\in \{1,2,3,...\} \\
\end{align}
\begin{equation}
\psi(\mathbf{r}) = \sqrt{\frac{8}{L_xL_yL_z}} \sin\left(\frac{n_x \pi x}{L_x}\right) \sin\left(\frac{n_y \pi y}{L_y}\right) \sin\left(\frac{n_z \pi z}{L_z}\right)
\end{equation}
where $n_x\in \{1,2,3,...\}$, $n_y\in \{1,2,3,...\}$, and $n_z\in \{1,2,3,...\}$.\psi(\mathbf{r}) = \sqrt{\frac{8}{L_xL_yL_z}} \sin\left(\frac{n_x \pi x}{L_x}\right) \sin\left(\frac{n_y \pi y}{L_y}\right) \sin\left(\frac{n_z \pi z}{L_z}\right)
\end{equation}
Adding Time Dependence
To add time dependence to our wave function, consider the time-dependent Schrodinger equation, $i\hbar \frac{\partial}{\partial t}\psi= \hat{H}\psi $.\begin{equation}
i\hbar \frac{\partial}{\partial t}\psi(\mathbf{r}, t) = E\psi (\mathbf{r}, t)
\end{equation}
The energy of a particle can be written as $E=\hbar\omega$ where $\omega$ is the angular frequency of the particle's wave function.i\hbar \frac{\partial}{\partial t}\psi(\mathbf{r}, t) = E\psi (\mathbf{r}, t)
\end{equation}
\begin{equation}
\int\frac{1}{\psi (\mathbf{r}, t)} d\psi(\mathbf{r}, t) = -i\int\omega dt
\end{equation}
\begin{equation}
\exp{\text{Log}{\psi (\mathbf{r}, t)} } = \exp(-i\omega t+C_1)
\end{equation}
($\text{Log}z$ is the complex logarithm of $z$.) Let $C=e^{C_1}.$\int\frac{1}{\psi (\mathbf{r}, t)} d\psi(\mathbf{r}, t) = -i\int\omega dt
\end{equation}
\begin{equation}
\exp{\text{Log}{\psi (\mathbf{r}, t)} } = \exp(-i\omega t+C_1)
\end{equation}
\begin{equation}
\psi (\mathbf{r}, t) = Ce^{-i\omega t}
\end{equation}
Thus, the time-dependent component of $\psi(\mathbf{r}, t)$ is $e^{-i\omega t}$ and the time-independent component is denoted as $C$ in the above equation. Since $\psi(\mathbf{r}, t)$ still needs to be an eigenfunction of the Hamiltonian, which in this scenario isn't time-dependent but rather position-dependent, $C$ must be the component of $\psi(\mathbf{r}, t)$ that is the normalized eigenfunction of the Hamiltonian. Remember that $\psi(\mathbf{r})$, which we found in equation $(20)$, is the normalized eigenfunction of the Hamiltonian, so $C=\psi(\mathbf{r})$. Therefore, the time-dependent normalized wave function for a particle trapped in an infinite box well is:\psi (\mathbf{r}, t) = Ce^{-i\omega t}
\end{equation}
\begin{equation}
\psi (\mathbf{r}, t) = \sqrt{\frac{8}{L_xL_yL_z}} \sin\left(\frac{n_x \pi x}{L_x}\right) \sin\left(\frac{n_y \pi y}{L_y}\right) \sin\left(\frac{n_z \pi z}{L_z}\right) e^{-i\omega t}
\end{equation}
where $n_x\in \{1,2,3,...\}$, $n_y\in \{1,2,3,...\}$, and $n_z\in \{1,2,3,...\}$.\psi (\mathbf{r}, t) = \sqrt{\frac{8}{L_xL_yL_z}} \sin\left(\frac{n_x \pi x}{L_x}\right) \sin\left(\frac{n_y \pi y}{L_y}\right) \sin\left(\frac{n_z \pi z}{L_z}\right) e^{-i\omega t}
\end{equation}
Symmetric Square and Box Wells
So far, we've only been dealing with non-symmetric infinite wells and that's perfectly fine, but say we want to work with potential wells that are symmetric like this:$$V_x(x)= \begin{cases} 0, &-\frac{L_x}{2} < x < \frac{L_x}{2} \\ \infty, &\text{otherwise} \end{cases}$$
$$V_y(y)= \begin{cases} 0, &-\frac{L_y}{2} < y < \frac{L_y}{2} \\ \infty, &\text{otherwise} \end{cases}$$
$$V_z(z)= \begin{cases} 0, &-\frac{L_z}{2} < z < \frac{L_z}{2} \\ \infty, &\text{otherwise} \end{cases}$$
Since all we've done is shift the system $L_x/2$, $L_y/2$, and $L_z/2$ in the negative $x$, $y$, and $z$ directions respectively, we just need to do the same for our wave function.
\begin{equation}
\psi (\mathbf{r}, t) = \sqrt{\frac{8}{L_xL_yL_z}} \sin\left[\frac{n_x \pi}{L_x}\left(x + \frac{L_x}{2}\right)\right] \sin\left[\frac{n_y \pi}{L_y} \left(y + \frac{L_y}{2}\right)\right] \sin\left[\frac{n_z \pi}{L_z} \left(z + \frac{L_z}{2}\right) \right] e^{-i\omega t}
\end{equation}
That's about as far as we'll get if we try to simplify the three-dimensional wave function for a particle in a symmetric infinite box well, but if we were considering a one-dimensional system with a particle trapped in a symmetric infinite square well, we can go a little further.\psi (\mathbf{r}, t) = \sqrt{\frac{8}{L_xL_yL_z}} \sin\left[\frac{n_x \pi}{L_x}\left(x + \frac{L_x}{2}\right)\right] \sin\left[\frac{n_y \pi}{L_y} \left(y + \frac{L_y}{2}\right)\right] \sin\left[\frac{n_z \pi}{L_z} \left(z + \frac{L_z}{2}\right) \right] e^{-i\omega t}
\end{equation}
\begin{equation}
\psi (x, t) = \sqrt{\frac{2}{L_x}} \sin\left(\frac{n_x \pi x}{L_x} + \frac{n_x \pi }{2}\right) e^{-i\omega t}
\end{equation}
\begin{equation}
\psi (x, t) = \begin{cases} \sqrt{\frac{2}{L_x}} \cos\left(\frac{n_x \pi x}{L_x}\right) e^{-i\omega t}, & n_x \in \{1,3,5,...\} \\ \sqrt{\frac{2}{L_x}} \sin\left(\frac{n_x \pi x}{L_x}\right) e^{-i\omega t}, & n_x\in \{2,4,6,...\} \end{cases}
\end{equation}
(Sign changes for $\psi (x, t)$ are disregarded since sign changes only invert the shape of the wave function and have no physical significance.)\psi (x, t) = \sqrt{\frac{2}{L_x}} \sin\left(\frac{n_x \pi x}{L_x} + \frac{n_x \pi }{2}\right) e^{-i\omega t}
\end{equation}
\begin{equation}
\psi (x, t) = \begin{cases} \sqrt{\frac{2}{L_x}} \cos\left(\frac{n_x \pi x}{L_x}\right) e^{-i\omega t}, & n_x \in \{1,3,5,...\} \\ \sqrt{\frac{2}{L_x}} \sin\left(\frac{n_x \pi x}{L_x}\right) e^{-i\omega t}, & n_x\in \{2,4,6,...\} \end{cases}
\end{equation}
Energy Levels
Remember how earlier on, we found that $k_x=n_x\pi / L_x$, where $n_x\in \{1,2,3,...\}$, and that $k_x=\pm\sqrt{2mE_x}/ \hbar$ for a particle in a box well? This means that the energy of the particle must be quantized to specific energy levels.\begin{equation}
E_x = \frac{\hbar^2 k_x^2}{2m}
\end{equation}
\begin{equation}
E_x = \frac{{n_x}^2 \pi^2 \hbar^2 }{2m{L_x}^2}, \qquad n_x\in \{1,2,3,...\}
\end{equation}
Similarly, E_x = \frac{\hbar^2 k_x^2}{2m}
\end{equation}
\begin{equation}
E_x = \frac{{n_x}^2 \pi^2 \hbar^2 }{2m{L_x}^2}, \qquad n_x\in \{1,2,3,...\}
\end{equation}
\begin{align}
E_y & = \frac{{n_y}^2 \pi^2 \hbar^2 }{2m{L_y}^2}, & n_y\in \{1,2,3,...\} \\
E_z & = \frac{{n_z}^2 \pi^2 \hbar^2 }{2m{L_z}^2}, & n_z\in \{1,2,3,...\}
\end{align}
Putting all this together, for a particle trapped in an infinite box potential well, the energy levels are defined by:E_y & = \frac{{n_y}^2 \pi^2 \hbar^2 }{2m{L_y}^2}, & n_y\in \{1,2,3,...\} \\
E_z & = \frac{{n_z}^2 \pi^2 \hbar^2 }{2m{L_z}^2}, & n_z\in \{1,2,3,...\}
\end{align}
\begin{equation}
E = \frac{\pi^2 \hbar^2 }{2m}\left(\frac{{n_x}^2}{{L_x}^2} + \frac{{n_y}^2}{{L_y}^2} + \frac{{n_z}^2}{{L_z}^2}\right)
\end{equation}
where $n_x\in \{1,2,3,...\}$, $n_y\in \{1,2,3,...\}$, and $n_z\in \{1,2,3,...\}$ are quantum numbers.E = \frac{\pi^2 \hbar^2 }{2m}\left(\frac{{n_x}^2}{{L_x}^2} + \frac{{n_y}^2}{{L_y}^2} + \frac{{n_z}^2}{{L_z}^2}\right)
\end{equation}
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