📂Quantum Mechanics

Energy Levels in an Infinite Potential Well

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To determine the wave functions (eigenfunctions) and energies (eigenvalues) in an infinite potential well, refer to here. Now, let’s bring in the results and examine their significance.

Eigenfunction $\displaystyle \psi_{(x)} =\sqrt{\frac{2}{a}}\sin \frac{n\pi}{a}x$ Eigenvalue $\displaystyle E_{n}=\frac{n^2\pi^2\hbar^2}{2ma^2}$

For the wave function in an infinite potential well, the expectation value of momentum is $0$, but the expectation value of the square of momentum is not $0$. $1.$ Expectation value of momentum \begin{align*} \langle p \rangle &= \int_{0}^a {\psi_{n}(x)}^{\ast} p {\psi_{n}(x)} dx \\ &= \int_{0}^a \psi_{n}(x) \left( \frac{\hbar}{i}\frac{\partial}{\partial x} \right) {\psi_{n}(x)} dx \\ &= \frac{\hbar}{i} \int_{0}^a \psi_{n}(x) \frac{\partial}{\partial x}{\psi_{n}(x)} dx \end{align*}. In this case, since $\displaystyle \frac{\partial \psi ^2}{\partial x}=2\psi\frac{\partial \psi}{\partial x}$, it is $\displaystyle \psi \frac{\partial \psi}{\partial x}=\frac{1}{2}\frac{\partial \psi^2}{\partial x}$. $\displaystyle{ \therefore \frac{\hbar}{i} \int_{0}^a \psi_{n}(x) \frac{\partial}{\partial x}{\psi_{n}(x)} dx \\ = \frac{\hbar}{i} \int_{0}^a \frac{1}{2} \frac{\partial}{\partial x}{\psi_{n}(x)}^2 dx \\ = \frac{\hbar}{i}\frac{1}{2} \left[ \frac{2}{a} \sin^2 \frac{n\pi x}{a} \right]_{0}^a \\ = \frac{\hbar}{i}\frac{1}{2} \frac{2}{a} (\sin ^2 n\pi – \sin^2 0) \\ = 0 }$ The expectation value of momentum over the entire range is $0$. Does this mean the particle does not exist? No, it does not mean that. Just because the expectation value is $0$ does not mean the momentum is $0$. Let’s verify below that the expectation value of the square of the momentum is not $0$. $2.$ Expectation value of the square of momentum $\displaystyle \langle p^2 \rangle = \int_{0}^a \psi_{n}(x) \left( -\hbar^2 \frac{\partial ^2}{\partial x^2} \right) \psi_{n}(x) dx$. In this case, since the Schrödinger equation is $\displaystyle \frac{-\hbar^2}{2m}\frac{\partial^2 }{\partial x^2}\psi = E_{n}\psi$, it is $\displaystyle -\hbar^2 \frac{\partial^2 }{\partial x^2}\psi = 2m E_{n}\psi$. Therefore, \begin{align*} \int_{0}^a \psi_{n}(x) \left( -\hbar^2 \frac{\partial ^2}{\partial x^2} \right) \psi_{n}(x) dx &= \int_{0}^a \psi_{n}(x) 2m E_{n} \psi_{n}(x)dx \\ &= 2mE_{n}\int_{0}^a {\psi_{n}(x)}^2 dx \\ &= 2mE_{n} \\ &= 2m\frac{n^2 \hbar ^2 \pi^2}{2ma^2} \\ &= \left( \frac{2\hbar\pi}{a} \right)^2 \end{align*} . Now, using the result of $2$, we can find the expectation value of kinetic energy. $3.$ Expectation value of kinetic energy \begin{align*} \langle K \rangle &= \langle \frac{1}{2m} p^2 \rangle \\ &= \frac{1}{2m} \langle p^2 \rangle \\ &= \frac{1}{2m} (2mE_{n}) \\ &= E_{n}=\frac{n^2 \hbar^2 \pi^2}{2ma^2} \end{align*} . At this point, let’s inspect the wavelength of the wave function. Since the wave function is $\displaystyle \sin \frac{n\pi x}{a}$, the wavelength is $\displaystyle 2\pi (\frac{a}{n\pi})=\frac{2a}{n}$. The same result can be obtained using De Broglie’s matter-wave formula. $\displaystyle \lambda = \frac{h}{p}=\frac{2\pi \hbar}{\hbar k}=\frac{2\pi}{k}=2\pi \frac{a}{n\pi}=\frac{2a}{n}$ For each $n$, the energy and wavelength can be written as follows: $\displaystyle E_{1}=\frac{\pi^2 \hbar^2}{2ma^2}$, $\lambda_{1} = 2a$ $E_2 = \frac{4\pi^2 \hbar^2}{2ma^2}=4E_{1}$, $\lambda_2=a$ $E_{3} = \frac{9\pi^2 \hbar^2}{2ma^2}=9E_{1}$, $\lambda_2=\frac{2a}{3}$. The results are visually represented below. What has physical significance here is not $\psi$ but $|\psi|^2$, so the phase is not important. This means that in the two figures below, although the blue waves are out of phase with each other, it does not matter which one. The wave function is symmetric about $\frac{a}{2}$, and energy is proportional to $n^2$. The more the wave function oscillates, the higher the energy level (the greater the energy), or the greater the energy, the more the wave function oscillates.