### 物理代写|量子力学代写quantum mechanics代考|PHYS3040

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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 Velocity

For each proper quantum section $\Psi \in \sec \left(\boldsymbol{E}, \boldsymbol{Q}{/ 0}\right)$, we obtain the gauge independent and observer independent “quantum velocity” (see Theorem 17.2.2) $$\mathrm{V}[\Psi]:=д+\vec{\nabla}^{\dagger \omega}((\Psi)) \in \sec \left(\boldsymbol{E}, \mathbb{T}^* \otimes T \boldsymbol{E}\right),$$ with observed and coordinate expressions \begin{aligned} \mathrm{V}[\Psi] & =\mu[o]+\vec{\nabla}^{\mathbb{E}}o) \ & =u^0 \otimes\left(\partial_0+\left(G_0^{i j} \partial_j \varphi-A_0^i\right) \partial_i\right) . \end{aligned} In particular, with reference to the distinguished observer $O{\Psi}$ associated with the proper quantum section $\Psi$, we obtain the equality $\mathrm{V}[\Psi]=д\left[o_{\Psi}\right]$.

The quantum velocity has a close relation with the kinetic quantum tensor, the kinetic quantum vector field and the quantum probability current (see Corollary 17.3.3 and Theorem 17.4.2). Moreover, the quantum velocity plays a key role in the context of the hydrodynamical picture of Quantum Mechanics (see Theorem 18.1.1).

The quantum velocity is a rather usual object of standard Quantum Mechanics. But, our 4-dimensional intrinsic presentation, the link with the “rest observer” and the related physical interpretation can be hardly achieved in standard Quantum Mechanics.

## 物理代写|量子力学代写quantum mechanics代考|Kinetic Quantum Tensor

For each $\Psi \in \sec (\boldsymbol{E}, \boldsymbol{Q})$, we obtain the gauge independent and observer independent “kinetic quantum tensor” (see Theorem 17.3.2)
$$\mathrm{Q}[\Psi]:=\text { д } \otimes \Psi-i \vec{\nabla}^{\dagger} \Psi \in \sec \left(\boldsymbol{E}, \mathbb{T}^* \otimes(T \boldsymbol{E} \otimes \boldsymbol{Q})\right),$$
with observed and coordinate expressions
\begin{aligned} \mathrm{Q}[\Psi] & =д[o] \otimes \Psi-\mathrm{i} \vec{\nabla}[o] \Psi \ & =\left(\psi \partial_0-\mathfrak{i} G_0^{i j}\left(\partial_j \psi-\mathfrak{i} A_j \psi\right) \partial_i\right) \otimes u^0 \otimes \mathrm{b} . \end{aligned}
In particular, for each $\Psi \in \sec \left(\boldsymbol{E}, \boldsymbol{Q}_{/ 0}\right)$, we have the splitting
$$\mathrm{Q}[\Psi]=(\mathrm{V}[\Psi]-\mathrm{i} \vec{d} \log |\Psi|) \otimes \Psi$$
The kinetic quantum tensor is a rather unusual object, with respect to standard Quantum Mechanics but this object plays a relevant role in our approach, as it is the source of the Schrödinger operator via the projectability criterion (see Theorem 17.6.5).

Indeed, the kinetic quantum tensor has a close relation with the quantum velocity (see Corollary 17.3.3). Furthermore, the kinetic quantum tensor has a close relation with the quantum momentum operator (see Example 20.1.12).

In this respect, we stress that the standard approach to Quantum Mechanics, the notion of quantum momentum is usually strictly linked with the Fourier formalism. However, in a curved spacetime, the Fourier methods can be hardly proposed in a covariant way. So, in our general curved framework, we are forced to follow a completely different geometric way, which, in the flat case, reproduces known objects and results.

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

$$\mathrm{V}[\Psi]:=\text { д }+\vec{\nabla}^{\dagger \omega}((\Psi)) \in \sec \left(\boldsymbol{E}, \mathbb{T}^* \otimes T \boldsymbol{E}\right),$$

## 物理代写|量子力学代写quantum mechanics代考|Kinetic Quantum Tensor

$$\mathrm{Q}[\Psi]:=\text { 刀 } \otimes \Psi-i \vec{\nabla}^{\dagger} \Psi \in \sec \left(\boldsymbol{E}, \mathbb{T}^* \otimes(T \boldsymbol{E} \otimes \boldsymbol{Q})\right)$$

$$\mathrm{Q}[\Psi]=\mathrm{I}[o] \otimes \Psi-\mathrm{i} \vec{\nabla}[o] \Psi \quad=\left(\psi \partial_0-\mathrm{i} G_0^{i j}\left(\partial_j \psi-\mathrm{i} A_j \psi\right) \partial_i\right) \otimes u^0 \otimes \mathrm{b} .$$

$$\mathrm{Q}[\Psi]=(\mathrm{V}[\Psi]-\mathrm{i} \vec{d} \log |\Psi|) \otimes \Psi$$

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