### 物理代写|量子场论代写Quantum field theory代考|PHYC90008

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• Advanced Mathematical Statistics 高等数理统计学
• (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|量子场论代写Quantum field theory代考|Measuring Two Different Observables on the Same System

Suppose now that we consider a second observable $\mathcal{O}^{\prime}$ with a corresponding Hermitian operator $B$. Then if a system in state $|\alpha\rangle$ is such that it will always yield the same value
14 One should say that the problem here is formulated in a rather outrageous way to make the point clear. The universe seems to have done fine prior to our existing! The problem is nonetheless real. On the one hand, one may argue that Quantum Mechanics applies only to the microscopic world. Then a measurement process will be anything that interacts with a macroscopic object such as a photosensitive chemical in a photographic emulsion and has nothing to do with a conscious observer. On the other hand, if one refuses this arbitrary and ill-defined boundary between macroscopic and microscopic worlds, the quantum realm extends all the way to the consciousness of the observer. It is then very difficult to escape the conclusion that this consciousness plays a role, and matters become very murky. The theory of decoherence tries to address these issues.
15 Including cases where the mathematical justification is not ironclad.
16 The measured value of the observable $O$ in state $|\alpha\rangle$ is a random variable, and as such has an expected value. When we repeat the experiment many times and average the corresponding measurements, the quantity we obtain is near this expected value. This is why both names are used.

when $\mathcal{O}$ is measured, and also when $\mathcal{O}^{\prime}$ is measured, then $|\alpha\rangle$ must be an eigenvector of both $A$ and $B$. Then $[A, B]|\alpha\rangle=0$. When no such $|\alpha\rangle$ exists, ${ }^{17}$ there does not exist a state for which both $\mathcal{O}$ and $\mathcal{O}^{\prime}$ can be measured with certainty. It is simply impossible to ever know at the same time the values of both $\mathcal{O}$ and $\mathcal{O}^{\prime}$ for any state of the system. It is fallacious to think that to know them both one just has to measure $\mathcal{O}$ and then $\mathcal{O}^{\prime}$. After the measurement of $\mathcal{O}^{\prime}$ has taken place you no longer know the value of $\mathcal{O}$. If you measure $\mathcal{O}$, then measure $\mathcal{O}^{\prime}$, and “immediately after” measure $\mathcal{O}$ again, the result of this second measurement of $\mathcal{O}$ will sometimes be different from the result of the first measurement. This is because, as we explained earlier, the measurement of $\mathcal{O}^{\prime}$ has changed the state of the system. Right after the measurement of $\mathcal{O}^{\prime}$, the state of the system is an eigenvector of $B$, and by hypothesis, this eigenvector of $B$ is not an eigenvector of $A$, so that in this state of the system, the result of the measurement of $A$ cannot be predicted with certainty.

## 物理代写|量子场论代写Quantum field theory代考|Uncertainty

Heisenberg’s uncertainty principle, which we will study in the present section, is related to but different from the phenomenon of the previous section. If the system is in state $x$ with $(x,[A, B] x) \neq 0$, one cannot measure both observables $\mathcal{O}$ and $\mathcal{O}^{\prime}$ with arbitrary accuracy. We are not talking here of successive measurements on the same experiment, where the first measurement changes the state of the system. We are talking of measurements on different experiments. One repeats the experiment many times, each time measuring either $\mathcal{O}$ or $\mathcal{O}^{\prime}$. If the results of measuring $\mathcal{O}$ are concentrated in a small interval, then the results of measuring $\mathcal{O}^{\prime}$ must spread out. In the important special case where $[A, B]$ is a multiple of the

identity, whatever the state of the system, you can never measure both $\mathcal{O}$ and $\mathcal{O}^{\prime}$ with arbitrary accuracy. A quantitative version of this statement is given in Proposition 2.3.2.
Definition 2.3.1 Consider an observable $\mathcal{O}$ with associated Hermitian operator $A$. The uncertainty $\Delta_{x} A \geq 0$ of $\mathcal{O}$ in the state $x \in \mathcal{H}$ is given by
$$\left(\Delta_{x} A\right)^{2}:=\left(x, A^{2} x\right)-(x, A x)^{2} .$$
For instance, in the case of the basic example, the uncertainty of the position in state $x$ is $\sqrt{\sum_{i \leq n} \lambda_{i}^{2}\left|x_{i}\right|^{2}-\left(\sum_{i \leq n} \lambda_{i}\left|x_{i}\right|^{2}\right)^{2}}$. To make sense of $(2.9)$, one may observe that $\left(\Delta_{x} A\right)^{2}$ is just the variance of the probability distribution of Principle 4 , or in other words, $\Delta_{x} A$ is the standard deviation of this probability distribution. The physical content of this definition should be stressed. When you make a measurement of the observable $\mathcal{O}$ for a system in state $x$, you get a random result, and $\Delta_{x} A$ is the standard deviation of this random result. ${ }^{18}$
To explain (2.9) in more mathematical terms, this quantity measures “the squaredeviation of $\mathcal{O}$ from its average in state $x “$. Indeed, denoting by 1 the identity operator of $\mathcal{H}$ (and since $x$ is of norm 1 ),
$$\left(\Delta_{x} A\right)^{2}=\left(x, A^{\prime 2} x\right)$$
where the Hermitian operator $A^{\prime}:=A-(x, A x) 1$ is “the deviation of $\mathcal{O}$ from its average in state $x^{\prime \prime}$. When $x$ is an eigenvector of $A$, obviously $\left(\Delta_{x} A\right)^{2}=0$. Conversely, when $\left(\Delta_{x} A\right)^{2}=0$, since
$$\left(x, A^{\prime 2} x\right)=\left(A^{\prime} x, A^{\prime} x\right)=\left|A^{\prime} x\right|^{2},$$
then $A^{\prime} x=0$ so that $x$ is an eigenvector of $A$. Therefore, $\left(\Delta_{x} A\right)^{2}=0$ if and only if $x$ is an eigenvector of $A$, i.e. if and only if the measurement of $\mathcal{O}$ in state $x$ offers no uncertainty, in accord with our calling $\Delta_{x} A$ the “uncertainty of $\mathcal{O}$ in state $x$ “.
Observe finally that $\left(x, A^{\prime} x\right)=0$, so that $(2.10)$ means
$$\left(\Delta_{x} A\right)^{2}=\left(\Delta_{x} A^{\prime}\right)^{2} .$$

## 物理代写|量子场论代写Quantum field theory代考|Finite versus Continuous Models

Mathematicians are trained to think of physical space as $\mathbb{R}^{3}$. But our continuous model of physical space as $\mathbb{R}^{3}$ is of course an idealization, both at the scale of the very large and at the scale of the very small. This idealization has proved to be very powerful, but in the case of Quantum Field Theory, it creates multiple problems, and in particular the infamous infinities (in the form of diverging integrals).

Can we dispense with continuous models and their analytical problems? A physical measurement is made through a device with finite accuracy, and this measurement is no different from the same measurement rounded to the last significant digit. The result of the measurement is also bounded, ${ }^{19}$ so it may yield only finitely many possible values, and we might be able to study physics using only finite-dimensional Hilbert spaces (of huge dimension).
There is a fundamental reason why we stubbornly keep infinite models. Probably the most important guiding principle in finding good models is that a proper theory should be Lorentz invariant, ${ }^{20}$ reflecting the fact that physics should be the same for all inertial observers ${ }^{21}$ (who undergo no acceleration). There is no way this can be implemented in a finite model, say one which replaces the continuous model of physical space by a finite grid. Lorentz invariance can be recovered from a finite model only “in the infinite limit”. Further, there is no canonical choice for such a finite model, so that one has to show that the results obtained by a finite approximation are indeed essentially independent of how this approximation is performed. Heuristically, this is plausible but it is quite another matter to really prove independence. In fact, it can be argued that settling this question is of the same order of difficulty as constructing a continuous model which in some sense would be the limit of the finite model as the grid becomes finer. In the case of Quantum Field Theory, this is a highly non-trivial task. Most importantly, considering finite models does not really solve anything. The infinities reappear in the guise of quantities that blow up as the grid becomes finer, and it is very hard to make sense of this behavior.

## 物理代写|量子场论代写Quantum field theory代考|Measuring Two Different Observables on the Same System

14 应该说，这里的问题是以一种相当离谱的方式表述的，以明确这一点。在我们存在之前，宇宙似乎做得很好！问题仍然是真实的。一方面，有人可能会争辩说，量子力学只适用于微观世界。那么测量过程将是与宏观物体相互作用的任何东西，例如照相乳剂中的感光化学物质，与有意识的观察者无关。另一方面，如果拒绝这种在宏观和微观世界之间任意且不明确的界限，那么量子领域就会一直延伸到观察者的意识中。那么很难摆脱这个意识起作用的结论，事情变得非常模糊。退相干理论试图解决这些问题。
15 包括数学论证不明确的情况。
16 可观测的测量值○处于状态|一个⟩是一个随机变量，因此具有期望值。当我们多次重复实验并平均相应的测量值时，我们得到的数量接近这个预期值。这就是使用这两个名称的原因。

## 物理代写|量子场论代写Quantum field theory代考|Uncertainty

(ΔX一个)2:=(X,一个2X)−(X,一个X)2.

(ΔX一个)2=(X,一个′2X)
Hermitian 算子在哪里一个′:=一个−(X,一个X)1是“的偏差○从它的平均状态X′′. 什么时候X是一个特征向量一个， 明显地(ΔX一个)2=0. 相反，当(ΔX一个)2=0， 自从

(X,一个′2X)=(一个′X,一个′X)=|一个′X|2,

(ΔX一个)2=(ΔX一个′)2.

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## MATLAB代写

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