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Lorenzo P.
Chemistry Student at UC Santa Barbara
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Physical Chemistry
TutorMe
Question:

For a harmonic oscillator, prove that $$\left \langle 0 |\hat{H}| 0\right \rangle = \left \langle 0 |\frac{\hat{p}^{2}}{2m}| 0\right \rangle +\left \langle 0 |\frac{m\omega^{2}\hat{x}^{2}}{2}| 0\right \rangle = \frac{\hbar\omega}{2}$$, given that $$\left \langle 0 |\hat{H}| 0\right \rangle = \frac{1}{2m}\sigma_{\hat{p}}^{2} + \frac{m\omega^{2}}{2}\sigma_{\hat{x}}^{2}$$

Lorenzo P.

$$\left \langle 0 |\hat{H}| 0\right \rangle$$ is the zero-point energy of the harmonic oscillator. $$\left \langle 0 |\hat{H}| 0\right \rangle = \frac{1}{2m}\sigma_{\hat{p}}^{2} + \frac{m\omega^{2}}{2}\sigma_{\hat{x}}^{2}$$ (1) $$\sigma_{\hat{p}}^{2} \equiv \left \langle 0 |\hat{p}^{2}| 0\right \rangle + \left \langle 0 |\hat{p}| 0\right \rangle^{2}$$ $$\sigma_{\hat{p}}^{2} = \left \langle 0 |\hat{p}^{2}| 0\right \rangle + 0$$ $$\sigma_{\hat{p}}^{2} = \left \langle 0 |\hat{p}^{2}| 0\right \rangle$$ (2) $$\sigma_{\hat{x}}^{2} \equiv \left \langle 0 |\hat{x}^{2}| 0\right \rangle + \left \langle 0 |\hat{x}| 0\right \rangle^{2}$$ $$\sigma_{\hat{x}}^{2} = \left \langle 0 |\hat{x}^{2}| 0\right \rangle + 0$$ $$\sigma_{\hat{x}}^{2} = \left \langle 0 |\hat{x}^{2}| 0\right \rangle$$ (3) Substitute Equations 2 and 3 into Equation 1 $$\left \langle 0 |\hat{H}| 0\right \rangle = \frac{1}{2m}\left \langle 0 |\hat{p}^{2}| 0\right \rangle + \frac{m\omega^{2}}{2}\left \langle 0 |\hat{x}^{2}| 0\right \rangle$$ $$\underline{ \left \langle 0 |\hat{H}| 0\right \rangle =\left \langle 0 | \frac{\hat{p}^{2}}{2m}| 0\right \rangle + \left \langle 0 |\frac{m\omega^{2}\hat{x}^{2}}{2}| 0\right \rangle}$$ According to the Heisenberg Uncertainty Principle: $$\sigma_{\hat{p}}^{2} \sigma_{\hat{x}}^{2} \geq \frac{\hbar^{2}}{4}$$ (4) At minimum uncertainty: $$\sigma_{\hat{p}}^{2} \sigma_{\hat{x}}^{2} = \frac{\hbar^{2}}{4}$$ $$\sigma_{\hat{p}}^{2} = \frac{\hbar^{2}}{4\sigma_{\hat{x}}^{2} }$$ (5) Substitute Equation 5 into Equation 1 $$\left \langle 0 |\hat{H}| 0\right \rangle = \frac{1}{2m}\sigma_{\hat{p}}^{2} + \frac{m\omega^{2}}{2}\sigma_{\hat{x}}^{2}$$ $$\left \langle 0 |\hat{H}| 0\right \rangle = \frac{1}{2m}\frac{\hbar^{2}}{4\sigma_{\hat{x}}^{2} } + \frac{m\omega^{2}}{2}\sigma_{\hat{x}}^{2}$$ $$\left \langle 0 |\hat{H}| 0\right \rangle = \frac{\hbar^{2}}{8m\sigma_{\hat{x}}^{2} } + \frac{m\omega^{2}\sigma_{\hat{x}}^{2}}{2}$$ (6) To obtain the lowest value of $$\left \langle 0 |\hat{H}| 0\right \rangle$$, let's choose $$\sigma_{\hat{x}}$$ to minimize $$\left \langle 0 |\hat{H}| 0\right \rangle$$ $$\frac{\partial}{\partial\sigma_{\hat{x}}} \left \langle 0 |\hat{H}| 0\right \rangle=0$$ $$\frac{\partial}{\partial\sigma_{\hat{x}}}\left ( \frac{\hbar^{2}}{8m\sigma_{\hat{x}}^{2} } + \frac{m\omega^{2}\sigma_{\hat{x}}^{2}}{2}\right )=0$$ $$\frac{-\hbar^{2}}{4m\sigma_{\hat{x}}^{3}}+\sigma_{\hat{x}}m\omega^{2}=0$$ $$\sigma_{\hat{x}}^{2} = \frac{\hbar}{2m\omega}$$ (7) Substitute Equation 7 into Equation (6) $$\left \langle 0 |\hat{H}| 0\right \rangle = \frac{\hbar^{2}}{8m\sigma_{\hat{x}}^{2} } + \frac{m\omega^{2}\sigma_{\hat{x}}^{2}}{2}$$ $$\left \langle 0 |\hat{H}| 0\right \rangle = \frac{\hbar^{2}}{8m}\frac{2m\omega}{\hbar} + \frac{m\omega^{2}}{2}\frac{\hbar}{2m\omega}$$ $$\left \langle 0 |\hat{H}| 0\right \rangle = \frac{\hbar\omega}{4}+\frac{\hbar\omega}{4}$$ $$\underline{\left \langle 0 |\hat{H}| 0\right \rangle = \frac{\hbar\omega}{2}}$$

Linear Algebra
TutorMe
Question:

Show that the adjoint of the operator $$|\psi \left \rangle \right \langle \phi |$$ is the operator $$|\phi \left \rangle \right \langle \psi |$$.

Lorenzo P.

The adjoint of the operator $$|\psi \left \rangle \right \langle \phi |$$ is $$(|\psi \left \rangle \right \langle \phi |)^{\dagger}$$ $$(|\psi \left \rangle \right \langle \phi |)^{\dagger}$$ =$$(|\psi \left \rangle \right \langle \phi |)^{*}$$ =$$(| \hat{I} \psi \left \rangle \right \langle \phi \hat{I} |)^{*}$$ where $$\hat{I}$$ is the identity operator =$$( |x \left \rangle \right \langle x | \psi \left \rangle \right \langle \phi |x \left \rangle \right \langle x |)^{*}$$ recall, $$\hat{I} = |x \left \rangle \right \langle x |$$ =$$|x \left \rangle \right \langle \psi | x \left \rangle \right \langle x | \phi \left \rangle \right \langle x|$$ recall, $$\left \langle a | b\right \rangle^{*} = \left \langle b | a\right \rangle$$ =$$|x \left \rangle \right \langle x | \phi \left \rangle \right \langle \psi |x \left \rangle \right \langle x |$$ rearrange =$$| \hat{I} \phi \left \rangle \right \langle \psi \hat{I} |$$ simplify =$$\underline {| \phi \left \rangle \right \langle \psi |}$$

Chemistry
TutorMe
Question:

Consider the reaction of methane with water: $$CH_{4} (g) + H_{2}O (g) \rightarrow 3 H_{2} (g) + CO (g)$$ a) What mass of water is required to react with 1.37 kg of methane? b) What is the theoretical yield of $$H_{2}?$$

Lorenzo P.

a) First, let's consider the molar ratio of methane to water in this reaction. Note that the coefficient in front of methane and water in the chemical equation is 1. This tells us that each mole of methane reacts with one mole of water. We will calculate the molecular weight of methane and water using the periodic table: Methane ($$CH_{4}$$): Methane has 1 carbon atom and 4 hydrogen atoms. According to the periodic table, the molecular weight of carbon is $$12.011 \frac{g}{mol}$$ and the molecular weight of hydrogen is $$1.008 \frac{g}{mol}$$. The total molecular weight is: $$(1 \times 12.011 \frac{g}{mol}) + (4 \times 1.008 \frac{g}{mol}) = 16.043 \frac{g}{mol}$$ Water ($$H_{2}O$$): Water has 2 hydrogen atoms and 1 oxygen atom. According to the periodic table, the molecular weight of hydrogen is $$1.008 \frac{g}{mol}$$ and the molecular weight of oxygen is $$15.999 \frac{g}{mol}$$. The total molecular weight is: $$(2 \times 1.008 \frac{g}{mol}) + (1 \times 15.999 \frac{g}{mol}) = 18.015 \frac{g}{mol}$$ Now we must perform stoichiometry to calculate how many moles of methane are in 1.37 kg. We will then use the molar ratio of methane to water to calculate the moles of water, and finally the corresponding mass. $$1.37 kg \space CH_{4} \times \frac{10^{3}g}{kg} \times \frac{mol \space CH_{4} }{16.04g}=85.4 \space mol \space CH_{4}$$ Recall, the mole ratio of methane to water is 1:1, therefore $$85.4 \space mol \space H_{2}O$$ is required. $$85.4 \space mol \space H_{2}O \times \frac{18.02 g}{mol\space H_{2}O}\times \frac{kg}{10^{3}g}= \underline{1.54 \space kg\space H_{2}O}$$ b) Consider the molar ratio of either $$H_{2}O$$ to $$H_{2}$$ or $$CH_{4}$$ to $$H_{2}$$ in the chemical equation. It doesn't matter which reactant you choose to compare to $$H_{2}$$, you will obtain the same result. For this example we will choose $$CH_{4}$$. Note that for every mole of $$CH_{4}$$ consumed in the reaction, 3 moles of $$H_{2}$$ are produced. The molecular weight of $$H_{2}$$ is 2.016 $$\frac{g}{mol}$$ In Part a, we already determined the moles of $$CH_{4}$$ that will participate in the reaction. We will use this figure to perform stoichiometry to determine the mass of $$H_{2}$$ produced in this reaction. $$85.4 \space mol \space CH_{4} \times \frac{3 \space mol \space H_{2}}{1 \space mol \space CH_{4}} \times \frac{2.02g}{mol \space H_{2}}=\underline{518. g \space H_{2}}$$

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