>[!summary] Zero point energy is finding the lowest possible energy level of a system. > **Key equation:** > General equation for zero-point energy: $E_n = (n+1/2)hf$ >[!Bug] Reason for Zero point From the classical harmonic oscillator ([[Quantum Harmonic Oscillator#Classical Harmonic Oscillator]]) $E = \frac{p^2}{2m}+\frac{1}{2}kx^2$ so for E = 0 > So for the quantum harmonic oscillator the lowest energy cannot be zero since that would mean that $\Delta x = 0$ and $\Delta p = 0$ which would violate the [[Uncertainty Principle]] meaning that the lowest possible energy level is never zero in a quantum sense # Finding the lowest energy level >[!warning] Assumptions We will use the example in [[Quantum Harmonic Oscillator]] where $E = \frac{p^2}{2m}+\frac{1}{2}kx^2$ to derive the lowest energy level. > Which we will generalize at the end. If we assume that distribution of momentum and position is this we can derive the following: $\begin{array}{c} \left< x \right> = 0 \\ \left< x^2 \right> = \Delta x^2 = A^2 \\\\ \left< p \right> = 0 \\ \left< p^2 \right> = \Delta p ^2 \geq \frac{\hbar^2}{4\Delta x ^2} = \Delta p ^2 \geq \frac{\hbar^2}{4\Delta A ^2} \end{array}$ ![[ZP_1.png]] [^1] >[!note] Explanation Example of the distribution of position on a axis. The average the postion is zero. From this and our early definition of energy we can assume that the average energy will be: $\left < E\right> = \frac{\hbar ^2}{8mA^2} + \frac{1}{2}kA^2$ From this we can draw the average energy with area. As well as the PE and KE individual ![[ZP_2.png]] [^1] >[!note] Explanation Average energy over area for the quantum harmonic oscillator. From finding $A_0$ and using the average energy we can solve for the lowest energy level ($E_0$) $ \begin{array}{c} E_0 = \frac{\hbar^2 \cdot 2\sqrt{km}}{8m\cdot \hbar} + \frac{1}{2} k \frac{\hbar}{2\sqrt{km}} \\ \text{Simplifying the first term first} \\ = \frac{\hbar^2 \cdot 2\sqrt{km}}{8m\cdot \hbar} = \frac{1}{4}\hbar \sqrt{\frac{k}{m}} \\ \text{Second term} \\ = \frac{1}{2} k \frac{\hbar}{2\sqrt{km}} = \frac{1}{4}\hbar \sqrt{\frac{k}{m}} \\ \\ \text{So our final equation is then:} \\ = \frac{1}{4}\hbar \sqrt{\frac{k}{m}} + \frac{1}{4}\hbar \sqrt{\frac{k}{m}} \quad \text{Let $\sqrt{\frac{k}{m}} = f$} \\ = \frac{1}{2 } hf \\ \\ \text{A generalized equation we can assume that following:} \\ \text{Energy has be quantized (cant have half energy)} \\ E_n = (n+1/2)hf \end{array}$ >[!note] Energy is quantized because of [[Quantization of Atomic Orbital's]] and due to Pauli Exclusion principle ([[Fermion, Bosons & Pauli Exclusion Principle]]) [^1]: Taken from R.Epp notes. --- > 📚 Like this note? [Star the GitHub repo](https://github.com/rajeevphysics/Obsidan-MathMatter) to support the project and help others discover it! ---