物理代写|量子力学代写quantum mechanics代考|PHYS3034

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

物理代写|量子力学代写quantum mechanics代考|Quantum Lagrangian

For each $\Psi \in \sec (\boldsymbol{E}, \boldsymbol{Q})$, we obtain the gauge independent and observer independent quantum lagrangian (see Theorem 17.5.2)
$$\left.\mathrm{L}[\Psi]:=-d t \wedge\left(\mathrm{imh}\eta(\Psi, \text { д }\lrcorner \nabla^{\uparrow} \Psi\right)+\frac{1}{2}\left(\bar{G} \otimes \mathrm{h}\eta\right)\left(\check{\nabla}^{\uparrow} \Psi, \check{\nabla}^{\dagger} \Psi\right)\right): \boldsymbol{E} \rightarrow \Lambda^4 T^{\dagger} \boldsymbol{E},$$
with observed and coordinate expressions
\begin{aligned} & \mathrm{L}[\Psi]=-d t \wedge\left(\mathrm{imh}\eta\left(\Psi, \nabla{\nexists[o]}[o] \Psi\right)+\frac{1}{2}\left(\bar{G} \otimes \mathrm{h}_\eta\right)(\nabla[o] \Psi, \nabla[o] \Psi)\right), \ & \mathrm{L}[\Psi]=\frac{1}{2}\left(-G_0^{i j} \partial_i \bar{\psi} \partial_j \psi+\mathfrak{i}\left(\bar{\psi} \partial_0 \psi-\psi \partial_0 \bar{\psi}\right)-\mathfrak{i} A_0^j\left(\bar{\psi} \partial_j \psi-\psi \partial_j \bar{\psi}\right)+2 \alpha_0 \bar{\psi} \psi\right) v^0 ; \ & \end{aligned}

With reference to the distinguished quantum basis $b_{\Psi}$, the distinguished observer $o_{\Psi}$, and the potential $A[\Psi]$ “seen by” $\Psi$, the above expression can be written in the following remarkable way (see Theorem 15.2.31 and Corollary 17.5.3)
$$\mathrm{L}[\Psi]=\left(\frac{1}{2} \bar{G}(d|\Psi|, d|\Psi|)+A[\Psi]|\Psi|^2\right) \otimes v .$$
We stress that, here again, the explicit mention of the phase polar degree of freedom of the quantum particle $((\Psi))$ disappears; however, it is implicitly encoded in $A[\Psi]$.

According to the standard lagrangian formalism, for each $\Psi \in \sec (\boldsymbol{E}, \boldsymbol{Q})$, we obtain the “quantum momentum form” (see Proposition 17.5.7)
$$\mathrm{P}:=\vartheta_Q \bar{\wedge} V_Q \mathrm{~L}: J_1 \boldsymbol{Q} \rightarrow \Lambda^4 T^* \boldsymbol{Q},$$
with coordinate expression
\begin{aligned} \mathrm{P}= & \frac{1}{2} \mathfrak{i}(\bar{z} d z-z d \bar{z}) \wedge v_0^0-\frac{1}{2}\left(G_0^{i j}\left(\bar{z}_i d z+z_i d \bar{z}\right)+\mathfrak{i} A_0^j(z d \bar{z}-\bar{z} d z)\right) \wedge v_j^0 \ & +\left(-\frac{1}{2} \mathfrak{i}\left(\bar{z} z_0-z \bar{z}_0\right)+\frac{1}{2}\left(G_0^{i j}\left(\bar{z}_i z_j+z_i \bar{z}_j\right)+\mathfrak{i} A_0^j\left(z \bar{z}_j-\bar{z} z_j\right)\right)\right) v^0 . \end{aligned}

物理代写|量子力学代写quantum mechanics代考|Schrödinger Operator

For each $\Psi \in \sec (\boldsymbol{E}, \boldsymbol{Q})$, we obtain the gauge independent and observer independent Schrödinger operator (see Theorem 17.6.5 and [219])
$$\left.\mathrm{S}[\Psi]:=\frac{1}{2}(\text { д }\lrcorner \nabla^{\uparrow} \Psi+\delta^{\uparrow}(Q[\Psi])\right) \in \sec \left(\boldsymbol{E}, \mathbb{T}^* \otimes \boldsymbol{Q}\right),$$
with observed and coordinate expression

\begin{aligned} & \mathrm{S}[\Psi]=\nabla[o]{\mu[o]} \Psi+\frac{1}{2} \operatorname{div}\eta \mu[o] \Psi-\mathfrak{i} \frac{1}{2} \Delta[G, o] \Psi, \ & \mathrm{S}[\Psi]=\left(\partial_0 \psi-\frac{1}{2} \mathrm{i} G_0^{i j} \partial_{i j} \psi-\left(A_0^j+\frac{1}{2} \mathrm{i} \frac{\partial_i\left(G_0^{i j} \sqrt{|g|}\right)}{\sqrt{|g|}}\right) \partial_j \psi\right. \ & \left.+\frac{1}{2}\left(\frac{\partial_0 \sqrt{|g|}}{\sqrt{|g|}}-\frac{\partial_i\left(A_0^i \sqrt{|g|}\right)}{\sqrt{|g|}}-\mathfrak{i} 2 \alpha_0\right) \psi\right) u^0 \otimes \mathrm{b} . \ & \end{aligned}
Several authors say that the Schrödinger equation is observer dependent; this fact happens if we consider an arbitrary phenomenological potential. But, if we deal with the joined gravitational and electromagnetic potential, then the Schrödinger equation turns out to be observer equivariant. In fact, the above joined potential fulfills a distinguished transition law (determined by the upper quantum connection), which turns out to be responsible for the observer equivariance of the Schrödinger equation. Of course, a possible additional phenomenological potential might be added by hand to our Schrödinger equation, but so doing we would break the covariance of the equation.

With reference to the distinguished quantum basis $\mathrm{b}{\Psi}$, the distinguished observer $o{\Psi}$, and the potential $A[\Psi]$ “seen by” $\Psi$, the above expression can be written in the following remarkable way (see Corollary 17.6.9)
$$\mathrm{S}[\Psi]=\left(\text { д }\left[o_{\Psi}\right] \cdot|\Psi|+\frac{1}{2}|\Psi| \operatorname{div}\eta \text { д }\left[o{\Psi}\right]-\mathfrak{i}\left(\frac{1}{2} \Delta[G]|\Psi|+A[\Psi]\right)|\Psi|\right) \otimes \mathrm{b}_{\Psi} .$$

物理代写|量子力学代写quantum mechanics代考|Quantum Lagrangian

$$\left.\mathrm{L}[\Psi]:=-d t \wedge\left(\operatorname{imh} \eta(\Psi, \Omega\lrcorner \nabla^{\uparrow} \Psi\right)+\frac{1}{2}(\bar{G} \otimes \mathrm{h} \eta)\left(\check{\nabla}^{\uparrow} \Psi, \check{\nabla}^{\dagger} \Psi\right)\right): \boldsymbol{E} \rightarrow \Lambda^4 T^{\dagger} \boldsymbol{E},$$

$$\mathrm{L}[\Psi]=-d t \wedge\left(\operatorname{imh} \eta(\Psi, \nabla \nexists[o][o] \Psi)+\frac{1}{2}\left(\bar{G} \otimes \mathrm{h}_\eta\right)(\nabla[o] \Psi, \nabla[o] \Psi)\right), \quad \mathrm{L}[\Psi]=\frac{1}{2}\left(-G_0^{i j}\right.$$

$$\mathrm{L}[\Psi]=\left(\frac{1}{2} \bar{G}(d|\Psi|, d|\Psi|)+A[\Psi]|\Psi|^2\right) \otimes v$$

$$\mathrm{P}:=\vartheta_Q \bar{\wedge} V_Q \mathrm{~L}: J_1 \boldsymbol{Q} \rightarrow \Lambda^4 T^* \boldsymbol{Q},$$

$$\mathrm{P}=\frac{1}{2} \mathrm{i}(\bar{z} d z-z d \bar{z}) \wedge v_0^0-\frac{1}{2}\left(G_0^{i j}\left(\bar{z}_i d z+z_i d \bar{z}\right)+\mathfrak{i} A_0^j(z d \bar{z}-\bar{z} d z)\right) \wedge v_j^0 \quad+\left(-\frac{1}{2} \mathrm{i}\left(\bar{z} z_0\right.\right.$$

物理代写|量子力学代写quantum mechanics代考|Schrödinger Operator

$$\left.\mathrm{S}[\Psi]:=\frac{1}{2}(\text { ㅍ. }\lrcorner \nabla^{\uparrow} \Psi+\delta^{\uparrow}(Q[\Psi])\right) \in \sec \left(\boldsymbol{E}, \mathbb{T}^* \otimes \boldsymbol{Q}\right),$$

$$\mathrm{S}[\Psi]=\nabla[o] \mu[o] \Psi+\frac{1}{2} \operatorname{div} \eta \mu[o] \Psi-\mathrm{i} \frac{1}{2} \Delta[G, o] \Psi, \quad \mathrm{S}[\Psi]=\left(\partial_0 \psi-\frac{1}{2} \mathrm{i} G_0^{i j} \partial_{i j} \psi-\left(A_0^j\right.\right.$$

$$\mathrm{S}[\Psi]=\left(\text { д }\left[o_{\Psi}\right] \cdot|\Psi|+\frac{1}{2}|\Psi| \operatorname{div} \eta \text { д }[o \Psi]-\mathrm{i}\left(\frac{1}{2} \Delta[G]|\Psi|+A[\Psi]\right)|\Psi|\right) \otimes \mathrm{b}_{\Psi} .$$

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