相对论场理论|Relativistic Field Theory PHYSM3417代写

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Assignment-daixieTM为您提供布里斯托大学University of Bristol Relativistic Field Theory PHYSM3417相对论场理论代写代考辅导服务！

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The course seems to aim to provide a comprehensive understanding of the principles of special relativity and how it relates to the behavior of electromagnetic fields. It appears to cover both the mathematical calculations involved in the theory and the more qualitative aspects of the subject matter.

The course also seems to delve into the covariant description of classical electromagnetic fields, which is an important concept in the modern approach to special relativity. Additionally, the course covers the relativistic quantum Klein-Gordon and Dirac equations, which are crucial for understanding quantum mechanics within the context of special relativity.

Overall, it seems like this course would be of interest to anyone looking to deepen their understanding of special relativity and its various applications in the fields of electromagnetism and quantum mechanics.

Consider the following Lagrangian
$$\mathcal{L}=-i \bar{\Psi}\left(\gamma^\mu \partial_\mu-m\right) \Psi$$
Where
$$\Psi=\left(\begin{array}{l} \psi_1 \ \psi_2 \end{array}\right), \quad \bar{\Psi}=\left(\bar{\psi}1, \bar{\psi}_2\right)$$ $\psi{1,2}$ are Dirac spinor fields. We will suppress spinor indices throughout.
(a) Show that (1) is invariant under infinitesimal transformations
$$\delta \Psi=i \epsilon_a T_a \Psi, \quad T_a=\frac{\sigma_a}{2}, \quad a=1,2,3$$
where $\sigma_a$ are Pauli matrices.

(a) Since the generators $T_a$ are Hermitian, the transformation laws are
\begin{aligned} & \delta \psi=i \epsilon_a T_a \psi \ & \delta \bar{\psi}=-i \bar{\psi} \epsilon_a T_a \end{aligned}
The Lagrangian is quite trivially seen to be invariant
$$\delta \mathcal{L}=-i \delta \bar{\psi}(\partial-m) \psi-i \bar{\psi}(\not \partial-m) \delta \psi=-i \bar{\psi}\left(-i \epsilon_a T_a\right)(\partial-m) \psi-i \bar{\psi}(\partial-m)\left(i \epsilon_a T_a\right) \psi=0,$$
where we used the trivial relation $\left[T_a, \gamma^\mu\right]=0$.

(b) Find the conserved currents $J_a^\mu$ corresponding to the symmetric transformations.

(b) We use the so called Noether method to find the conserved current, i.e., we pretend that the transformation parameter $\epsilon$ is spacetime dependent. Then
$$\delta \mathcal{L}=\left(\partial_\mu \epsilon_a\right) \bar{\psi} \gamma^\mu T_a \psi$$
which implies
$$J_a^\mu=-\bar{\psi} \gamma^\mu T_a \psi$$

(c) Write down the corresponding conserved charges $Q_a, a=1,2,3$. Show that
$$\delta \Psi=i\left[\epsilon_a Q_a, \Psi\right]$$

(c) The conserved charges are thus
$$Q_a=-\int d^3 x \bar{\psi}(x) \gamma^0 T_a \psi(x)=\int d^3 x \psi^{\dagger}(x) T_a \psi(x)$$
In the following we will repeatedly use the relations
\begin{aligned} {\left[T_a, T_b\right] } & =i f^{a b c} T_c \ \left.\left{\psi_i(x), \psi_j^{\dagger}(y)\right}\right|{x^0=y^0} & =\delta(\vec{x}-\vec{y}) \delta{i j} \end{aligned}
where $i, j$ are indices in the Lie algebra representation space. We write:
$$i\left[\epsilon_a Q_a, \psi_k(x)\right]=i \epsilon_a T_{i j}^a \int_{x^0=y^0 \text { choice made }} d^3 x\left[\psi_i^{\dagger}(y) \psi_j(y), \psi_k(x)\right]$$
Now we use the relation:
$$[A B, C]=A{B, C}-{A, C} B$$
which for our case gives $\left(A=\psi_i^{\dagger}(y), B=\psi_j(y), C=\psi_k(x)\right)$ :
$$i\left[\epsilon_a Q_a, \psi_k(x)\right]=-i \epsilon_a T_{i j}^a \int_{x^0=y^0} d^3 x \delta(\vec{x}-\vec{y}) \delta_{i k} \psi_j(y)=-i \epsilon_a T_{k j}^a \psi_j(x)=-\delta_e \psi_k(x)$$

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Assignment-daixieTM为您提供布里斯托大学University of Bristol Advanced Quantum Physics PHYSM3416高级量子物理学代写代考辅导服务！

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The Feynman path integral formulation is a powerful tool for understanding quantum mechanics, and it allows for the calculation of transition probabilities between states. Scattering theory is important for understanding the interaction of particles and can be used to predict the behavior of atoms, molecules, and nuclei.

The semi-classical WKB method is useful for approximating solutions to quantum mechanical problems, particularly when the potential is slowly varying. Adiabatic evolutions are important for understanding the behavior of systems that change slowly over time, such as quantum computers.

The addition of angular momenta is an important concept in quantum mechanics, and the Clebsch Gordan algebra provides a mathematical framework for understanding it. The Bloch sphere is a useful tool for representing spins, and it has applications in quantum computing and magnetic resonance imaging.

The course also covers magnetic fields and their effects on quantum systems, including Landau levels, the quantum Hall effect, and the Aharonov-Bohm effect. Finally, the discussion of Berry’s geometric phase provides a deeper understanding of the behavior of quantum systems in the presence of external forces.

Overall, this course seems like an excellent opportunity to explore advanced topics in quantum physics and develop a deeper understanding of the behavior of quantum systems.

Here we discuss an angular momentum algebra using the one-dimensional harmonic oscillator. Throughout this question we will set $\hbar=1$ (to save ink). Consider a pair of creation and annihilation operators $\hat{a}^{\dagger}$ and $\hat{a}$ satisfying $\left[\hat{a}, \hat{a}^{\dagger}\right]=1$ and a number basis of states $$|n\rangle=\frac{1}{\sqrt{n !}}\left(\hat{a}^{\dagger}\right)^n|0\rangle, \quad \hat{N}|n\rangle \equiv \hat{a}^{\dagger} \hat{a}|n\rangle=n|n\rangle .$$ (a) Now define, with $j$ a positive integer, the operators $$J_z=-j+\hat{N}, \quad J_{+}=\hat{a}^{\dagger} \sqrt{2 j-\hat{N}}, \quad J_{-}=J_{+}^{\dagger}=\sqrt{2 j-\hat{N}} \hat{a} .$$ Verify that these operators define an angular momentum algebra. Note: You should not have to expand the square roots in your manipulations. These operators are defined only on states for which the argument of the square root is not negative.

(a) To show that the operators $J_z, J_+, J_-$ define an angular momentum algebra, we need to verify that they satisfy the commutation relations \begin{align} [J_z, J_{\pm}] &= \pm J_{\pm}, \ [J_+, J_-] &= 2J_z. \end{align}

First, we have \begin{align} [J_z, J_+] &= [(-j + \hat{N}), (\hat{a}^\dagger \sqrt{2j – \hat{N}})] \ &= \hat{a}^\dagger [(-j + \hat{N}), \sqrt{2j – \hat{N}}] + [(j – \hat{N}), \hat{a}^\dagger] \sqrt{2j – \hat{N}} \ &= \hat{a}^\dagger (-1/2) [\hat{N} – 2j, \sqrt{2j – \hat{N}}] + \hat{a}^\dagger \sqrt{2j – \hat{N}} \ &= \hat{a}^\dagger (-1/2) (\sqrt{2j – \hat{N}} [\hat{N}, \sqrt{2j – \hat{N}}] – [\hat{N}, \sqrt{2j – \hat{N}}]\sqrt{2j – \hat{N}}) + \hat{a}^\dagger \sqrt{2j – \hat{N}} \ &= \hat{a}^\dagger (\sqrt{2j – \hat{N}}) + \hat{a}^\dagger (\sqrt{2j – \hat{N}}) \ &= 2J_+, \end{align} where we have used the commutation relations $[\hat{N}, \hat{a}^\dagger] = \hat{a}^\dagger$ and $[\hat{N}, \hat{a}] = -\hat{a}$.

Next, we have \begin{align} [J_z, J_-] &= [(-j + \hat{N}), (\sqrt{2j – \hat{N}} \hat{a})] \ &= \sqrt{2j – \hat{N}} [(-j + \hat{N}), \hat{a}] + [(-j + \hat{N}), \sqrt{2j – \hat{N}}] \hat{a} \ &= -\sqrt{2j – \hat{N}} \hat{a} + \sqrt{2j – \hat{N}} \hat{a} \ &= 0. \end{align}

Finally, we have \begin{align} [J_+, J_-] &= [(\hat{a}^\dagger \sqrt{2j – \hat{N}}), (\sqrt{2j – \hat{N}} \hat{a})] \ &= \sqrt{2j – \hat{N}} [\hat{a}^\dagger, \hat{a}] \sqrt{2j – \hat{N}} \ &= (2j – \hat{N}) – \hat{N} \ &= 2J_z. \end{align}

Therefore, the operators $J_z, J_+, J_-$ define an angular momentum algebra.

(b) Construct the corresponding $J^2$ operator. Does it satisfy the proper commutation relations with the $J_i$ operators?

(b) The $J^2$ operator is defined as $J^2=J_x^2+J_y^2+J_z^2$, where $J_x$, $J_y$, and $J_z$ are the angular momentum operators. In terms of the creation and annihilation operators, we can write these operators as \begin{align} J_x &= \frac{1}{2}\left(\hat{a}^{\dagger}+\hat{a}\right), \ J_y &= \frac{1}{2i}\left(\hat{a}^{\dagger}-\hat{a}\right), \ J_z &= \frac{1}{2}\left(\hat{a}^{\dagger}\hat{a}-\frac{1}{2}\right), \end{align} where we have chosen $\hbar=1$. Then, we can calculate $J^2$ as \begin{align} J^2 &= J_x^2+J_y^2+J_z^2 \ &= \frac{1}{4}\left(\hat{a}^{\dagger}+\hat{a}\right)^2 – \frac{1}{4}\left(\hat{a}^{\dagger}-\hat{a}\right)^2 + \frac{1}{4}\left(\hat{a}^{\dagger}\hat{a}-\frac{1}{2}\right)^2 \ &= \frac{1}{2}\left(\hat{a}^{\dagger}\hat{a}+\frac{1}{4}\right). \end{align} Note that we have used the commutation relation $\left[\hat{a}, \hat{a}^{\dagger}\right]=1$ to simplify the expression. We can check that $J^2$ commutes with $J_x$, $J_y$, and $J_z$: \begin{align} \left[J^2, J_x\right] &= \left[J^2, J_y\right] = \left[J^2, J_z\right] = 0, \end{align} so $J^2$ and $J_i$ form a set of commuting observables.

(c) Find a subset of the harmonic oscillator basis states $|n\rangle$ with $n=0,1,2, \ldots$ that forms a multiplet of $\mathbf{J}$ with angular momentum $j$. Write explicitly your $|j, m\rangle$ states in terms of the oscillator basis states. Confirm that $J_z$ and $J_{+}$act as expected on these states.

The angular momentum operators are given by $\mathbf{J}=\mathbf{r} \times \mathbf{p}$, where $\mathbf{r}$ is the position operator and $\mathbf{p}$ is the momentum operator. In one dimension, we have $\mathbf{r}=x$ and $\mathbf{p}=-i \partial / \partial x$. Therefore, the angular momentum operator can be written as

$\mathbf{J}=x p+i \hat{k}$

where $\hat{k}$ is a constant vector in the $z$ direction.

Using the creation and annihilation operators, we can write $x$ and $p$ in terms of $\hat{a}$ and $\hat{a}^{\dagger}$ as

$x=\sqrt{\frac{1}{2 \omega}}\left(\hat{a}+\hat{a}^{\dagger}\right), \quad p=-i \sqrt{\frac{\omega}{2}}\left(\hat{a}-\hat{a}^{\dagger}\right)$,

where $\omega$ is the frequency of the harmonic oscillator.

Substituting these expressions into the expression for $\mathbf{J}$, we get

$\mathbf{J}=\frac{1}{\sqrt{2 \omega}}\left[\left(\hat{a}+\hat{a}^{\dagger}\right)\left(\hat{a}-\hat{a}^{\dagger}\right)+i \hat{k}\right]$.

Let’s define the total number operator as $\hat{N}=\hat{a}^{\dagger} \hat{a}$, which counts the number of excitations in the oscillator. We can use $\hat{N}$ to label the states $|n\rangle$.

Now, we can write down the action of $\mathbf{J}^2$ and $J_z$ on the states $|n\rangle$: \begin{align} \mathbf{J}^2 |n\rangle &= j(j+1) |n\rangle, \ J_z |n\rangle &= m |n\rangle, \end{align} where $j$ and $m$ are the total angular momentum and the $z$ component of angular momentum, respectively. We want to find a subset of the states $|n\rangle$ that form a multiplet of $\mathbf{J}$ with angular momentum $j$.

To do this, we can define the raising and lowering operators

$J_{ \pm}=J_x \pm i J_y=\frac{1}{\sqrt{2}}\left[\left(\hat{a}^{\dagger}\right)^2-\hat{a}^2 \pm i\left(\hat{a}^{\dagger} \hat{a}+\hat{a} \hat{a}^{\dagger}\right)\right]$.

Using the commutation relation $[\hat{a},\hat{a}^\dagger]=1$, we can show that $[J_z,J_{\pm}]=\pm J_{\pm}$ and $[J_+,J_-]=2J_z$. Therefore, if we start with a state $|j,m\rangle$ and apply the raising operator $J_+$ repeatedly, we generate a set of states with angular momentum $j$ that are proportional to $|j,j\rangle$.

核与粒子物理学|Nuclear and Particle Physics PHYS22040代写

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Assignment-daixieTM为您提供布里斯托大学University of Bristol Foundation Nuclear and Particle Physics PHYS22040核与粒子物理学代写代考辅导服务！

Instructions:

The quantum world operates according to different laws and principles than the macroscopic world we observe in our daily lives. At the subatomic level, particles can exist in multiple states simultaneously and exhibit behavior that seems counterintuitive. However, the principles of quantum mechanics have been shown to accurately explain and predict the behavior of particles and their interactions.

In studying sub-atomic processes, rates and cross-sections are used to measure the likelihood of certain reactions and interactions occurring. This helps us to better understand the behavior of particles and how they interact with each other.

The semi-empirical mass formula is a tool used in nuclear physics to calculate the binding energy of atomic nuclei. It takes into account the contributions from the strong nuclear force, the electromagnetic force, and the asymmetry energy. The shell model, on the other hand, is a quantum mechanical model that describes how the protons and neutrons in a nucleus are arranged in energy levels or shells.

Using these ideas, we can better understand the stability of atomic nuclei, as well as the processes of nuclear fusion and fission. We can also explore how these processes occur in stars and supernovae, leading to the formation of new elements through nucleosynthesis.

The coupling of the Higgs boson to fermions is proportional to their masses. The partial width is then proportial to the coupling squared. Calculate Higgs boson branching fractions assuming (simplifying) that only decays to bottom and charm quarks as well as to taus and muons are possible.

The partial width for Higgs decay to fermion pairs is proportional to the fermion mass squared:
$$\Gamma_i \propto g_i^2 \propto m_i^2 .$$
For quarks, there is a color factor of 3. So, for Higgs decay to charms (1.28 GeV), bottoms $(4.18 \mathrm{GeV})$, taus $(1.78 \mathrm{GeV})$ and muons $(0.11 \mathrm{GeV})$, the branching ratios can be calculated by:
\begin{aligned} & B r[\text { charm }]=\frac{\Gamma_c}{\Gamma_{\text {tot }}}=\frac{3 \times 1.28^2}{3 \times 1.28^2+3 \times 4.18^2+1.78^2+0.11^2}=8.1 \% \ & B r[\text { bottom }]=\frac{\Gamma_b}{\Gamma_{\text {tot }}}=\frac{3 \times 4.18^2}{3 \times 1.28^2+3 \times 4.18^2+1.78^2+0.11^2}=86.6 \% \ & B r[\text { tau }]=\frac{\Gamma_\tau}{\Gamma_{\text {tot }}}=\frac{1.78^2}{3 \times 1.28^2+3 \times 4.18^2+1.78^2+0.11^2}=5.2 \% \ & B r[\text { muon }]=\frac{\Gamma_\mu}{\Gamma_{\text {tot }}}=\frac{0.11^2}{3 \times 1.28^2+3 \times 4.18^2+1.78^2+0.11^2}=0.020 \% \end{aligned}

Charged pions, $\pi^{+}\left(\pi^{-}\right)$, can decay to electrons as well as to muons and the corresponding neutrinos. In this decay, the parity violation of the weak interaction is maximal. All arguements hold for $\pi^{+}$and $\pi^{-}$.
a)
Prepare a sketch of the pion decay at rest noting the momentum vectors as well as the spin of the involved particles. Using the sketch, discuss why you expect maximal parity violation of the weak interaction in pion decays. Following this discussion, which branching fraction is larger? Include your reasoning in the answer.

• a)
The decay, here for $\pi^{-}$, follows:
$$\pi^{-} \rightarrow \ell^{-} \bar{\nu}_{\ell}, \quad \ell=\mu, \mathrm{e}$$
The $\mathrm{W}^{-}$boson couples exclusively to left-handed particles and right-handed antiparticles. Since we can assume the neutrino to be massless, its chirality corresponds to its helicity. The momentum- and spin-configuration is shown in Fig 1.

b)
Show the following relation for momentum and energy, assuming that neutrinos are massless:
\begin{aligned} & p_{\ell}=\frac{m_\pi^2-m_{\ell}^2}{2 m_\pi} \ & E_{\ell}=\frac{m_\pi^2+m_{\ell}^2}{2 m_\pi}, \end{aligned}
with $m_\pi=139,57 \mathrm{MeV}$ the mass of the pion and $m_{\ell}=105,66(0.511) \mathrm{MeV}$ the mass of the muon (electron).

b)
Using energy conservation, we find for pion at rest:
$$E_\pi=m_\pi=E_\nu+E_{\ell}, \quad \text { mit: } p_\nu=p_{\ell}=p$$
For the momentum we find:

\begin{aligned} & E_\pi=m_\pi=E_\nu+E_{\ell}, \quad \text { mit: } p_\nu=p_{\ell}=p \ & E_\pi=m_\pi=p+\sqrt{p^2+m_{\ell}^2} \ & \left(m_\pi-p\right)^2=p^2+m_{\ell}^2 \ & m_\pi^2-2 m_\pi p+p^2=p^2+m_{\ell}^2 \ & p=\frac{m_\pi^2-m_{\ell}^2}{2 m_\pi} \end{aligned}
And for the energy:
\begin{aligned} E & =\sqrt{p^2+m_{\ell}^2} \ & =\frac{1}{2 m_\pi} \sqrt{m_\pi^2-2 m_\pi^2 m_{\ell}^2+m_{\ell}^4+4 m_\pi^2 m_{\ell}^2} \ & =\frac{1}{2 m_\pi} \sqrt{\left(m_\pi^2+m_{\ell}^2\right)^2} \ E & =\frac{m_\pi^2+m_{\ell}^2}{2 m_\pi} \end{aligned}

电磁学和电波|Classical Physics II: Electromagnetism and Waves  PHYS20020代写

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Assignment-daixieTM为您提供布里斯托大学University of Bristol Foundation Classical Physics II: Electromagnetism and Waves  PHYS20020电磁学和电波代写代考辅导服务！

Instructions:

Classical Physics is a branch of physics that deals with the study of macroscopic phenomena at speeds much slower than the speed of light. It was developed in the 17th to 19th centuries and forms the foundation of modern physics. The core concepts of classical physics include mechanics, thermodynamics, electromagnetism, and optics.

This unit builds on the foundations developed in level C/4 in the areas of electromagnetic fields and waves. Maxwell’s equations, which describe the behavior of electric and magnetic fields, are studied in vacuo and in simple solids. These equations form the basis of a discussion of fields, forces, and energy for general charge and current configurations.

The wave solutions of Maxwell’s equations are also studied in this unit. These solutions relate the electromagnetic and optical properties of materials, providing a deeper understanding of how light interacts with matter. General wave phenomena, such as interference and diffraction, are investigated, along with practical applications of these effects.

Overall, this unit provides a comprehensive understanding of electromagnetic fields and waves, which is essential for understanding many aspects of modern physics and technology.

Use index notation to derive a formula for $\vec{\nabla} \times(s \vec{A})$, where $s$ is a scalar field $s(\vec{r})$ and $\vec{A}$ is a vector field $\vec{A}(\vec{r})$.

\begin{aligned} {[\vec{\nabla} \times(s \vec{A})]i } & =\varepsilon{i j k} \partial_j(s \vec{A})k \ & =\varepsilon{i j k} s \partial_j A_k+\varepsilon_{i j k} A_k \partial_j s \ & =s \vec{\nabla} \times \vec{A}+\vec{\nabla} s \times \vec{A} .\end{aligned}

Which of the following vector fields could describe an electric field? Say yes or no for each, and give a very brief reason.
(i) $\vec{E}(\vec{r})=x \hat{e}_x-y \hat{e}_y$.
(ii) $\vec{E}(\vec{r})=y \hat{e}_x+x \hat{e}_y$.
(iii) $\vec{E}(\vec{r})=y \hat{e}_x-x \hat{e}_y$.

The curl of an electrostatic field must be zero, but otherwise there is no restriction. So the answer follows as
(i) $\vec{\nabla} \times \vec{E}(\vec{r})=\left(\frac{\partial E_y}{\partial x}-\frac{\partial E_x}{\partial y}\right) \hat{e}_z+\ldots=\overrightarrow{0}$. YES, it describes an electric field.
(ii) $\vec{\nabla} \times \vec{E}(\vec{r})=(1-1) \hat{e}_z=0$. YES, it describes an electric field.
(iii) $\vec{\nabla} \times \vec{E}(\vec{r})=(-1-1) \hat{e}_z=-2 \hat{e}_z$. NO, it does not describe an electric field.

(a) A spherical shell of radius $R$, with an unspecified surface charge density, is centered at the origin of our coordinate system. The electric potential on the shell is known to be
$$V(\theta, \phi)=V_0 \sin \theta \cos \phi,$$
where $V_0$ is a constant, and we use the usual polar coordinates, related to the Cartesian coordinates by
\begin{aligned} & x=r \sin \theta \cos \phi, \ & y=r \sin \theta \sin \phi, \ & z=r \cos \theta . \end{aligned}
Find $V(r, \theta, \phi)$ everywhere, both inside and outside the sphere. Assume that the zero of $V$ is fixed by requiring $V$ to approach zero at spatial infinity. (Hint: this problem can be solved using traceless symmetric tensors, or if you prefer you can use standard spherical harmonics. A table of the low- $\ell$ Legendre polynomials and spherical harmonics is included with the formula sheets.)

This problem can be solving using either traceless symmetric tensors or the more standard spherical harmonics. I will show the solution both ways, starting with the simplier derivation in terms of traceless symmetric tensors.
(a) We exploit the fact that the most general solution to Laplace’s equation can be written as a sum of terms of the form
$$\left(r^{\ell} \text { or } \frac{1}{r^{\ell+1}}\right) C_{i_1 \ldots i_{\ell}}^{(\ell)} \hat{n}{i_1} \ldots \hat{n}{i_{\ell}},$$
where $C_{i_1 \ldots i_{\ell}}^{(\ell)}$ is a traceless symmetric tensor. In this case we only need an $\ell=1$ term, since
$$F_a(\theta, \phi) \equiv \sin \theta \cos \phi=\frac{x}{r}=\hat{x}_i \hat{n}_i$$
For $\ell=1$ the radial function must be $r$ or $1 / r^2$. For $rR$ the term proportional to $r$ is excluded, because it does not approach zero as $r \rightarrow \infty$, so only the $1 / r^2$ option remains, and the solution is
\begin{aligned} V(\vec{r}) & =V_0\left(\frac{R}{r}\right)^2 F_a(\theta, \phi) \ & =V_0\left(\frac{R}{r}\right)^2 \hat{x}_i \hat{n}_i \text { or } V_0\left(\frac{R}{r}\right)^2 \sin \theta \cos \phi . \end{aligned}

物理学基础代写|Foundation Physics PHYS10010代写

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Assignment-daixieTM为您提供布里斯托大学University of Bristol Foundation Physics PHYS10010物理学基础代写代考辅导服务！

Instructions:

You will be introduced to the fundamental theories and principles of physics, including classical mechanics, electromagnetism, and thermodynamics. You will also learn how to apply your understanding of these concepts to solve simple problems and conduct laboratory experiments to reinforce your knowledge.

Throughout the unit, you will develop your analytical and problem-solving skills, which are essential for success in any undergraduate physics program. You will also gain practical experience in a laboratory setting, where you will learn to use various tools and equipment to conduct experiments, collect and analyze data, and draw conclusions.

By the end of the unit, you should feel confident in your understanding of fundamental physics concepts and be prepared for the challenges of an undergraduate degree program in physics or a related field.

(a) Let $\psi(x)$ be an energy eigenstate. Explain why the expectation value $\langle[H, \Omega]\rangle$ of the commutator of $H$ with an arbitrary operator $\Omega$ vanishes on the state $\psi$.

Let $\psi(x)$ be an energy eigenstate of the Hamiltonian $H$. Then, we have \begin{align} \langle [H,\Omega]\rangle &= \langle H\Omega – \Omega H \rangle \ &= \langle H\Omega \rangle – \langle \Omega H \rangle \ &= E_\psi \langle \Omega \rangle – \langle \Omega H \rangle \ &= E_\psi \langle \Omega \rangle – E_\psi \langle \Omega \rangle \ &= 0, \end{align} where we have used the fact that $\psi$ is an energy eigenstate with eigenvalue $E_\psi$, and that the expectation value of $H$ on $\psi$ is just $E_\psi$. Therefore, the commutator of $H$ with an arbitrary operator $\Omega$ vanishes on the state $\psi$.

(b) Choose $\Omega=x p$, and take $$H=\frac{p^2}{2 m}+V(x)$$ Use the claim from part (a) to find a relation between the expectation value $\langle T\rangle$ of the kinetic energy and the expectation value of a combination of $x$ and the derivative $V^{\prime}(x)$ of the potential with respect to its argument. Both expectation values are taken on an energy eigenstate.

From part (a), we have shown that for any operator $A$ and Hamiltonian $H$,

$\langle[A, H]\rangle=i \hbar\langle\partial A / \partial t\rangle$.

We can use this to find a relation between the expectation value of the kinetic energy and the expectation value of a combination of $x$ and $V^{\prime}(x)$.

Let’s first find the commutator $[x,H]$ and $[p,H]$: \begin{align*} [x,H] &= xp – px = i\hbar, \ [p,H] &= \frac{p^3}{2m} + pV(x) – V(x)p – \frac{p}{2m}p^2 \ &= \frac{p}{m}(V(x)-p^2/2m) + [p,V(x)] \ &= -\frac{\partial V(x)}{\partial x}. \end{align*} Therefore, we have \begin{align*} \langle T \rangle &= \langle \frac{p^2}{2m} \rangle \ &= \frac{1}{2m} \langle p^2 \rangle \ &= \frac{1}{2m} \langle [p,p^2] \rangle \ &= \frac{1}{2m} \langle p[p,p] + [p,p]p \rangle \ &= \frac{1}{2m} \langle -p\hbar + \hbar p \rangle \ &= \frac{\hbar^2}{2m} \langle -p+p \rangle \ &= \frac{\hbar^2}{2m} \langle [x,H] \rangle \ &= \frac{i\hbar^3}{2m} \langle \partial x/\partial t \rangle \ &= \frac{i\hbar^3}{2m} \frac{\partial}{\partial t} \langle x \rangle \ &= \frac{i\hbar^3}{2m} \frac{\partial}{\partial t} \left(\frac{\langle [x,H] \rangle}{i\hbar}\right) \ &= -\frac{\hbar^2}{2m} \frac{\partial}{\partial t} \langle \frac{\partial V(x)}{\partial x} \rangle \ &= -\frac{\hbar^2}{2m} \frac{\partial}{\partial t} \langle V^{\prime}(x) \rangle. \end{align*} Thus, we have shown that the expectation value of the kinetic energy is related to the time derivative of the expectation value of $V^{\prime}(x)$ on an energy eigenstate.

Throughout this problem we consider a hydrogen atom with fixed principal quantum number $n$, with $\ell=n-1$, and $m=n-1$. The value $n$ is arbitrary and possibly large.
(a) Write the wavefunction $\psi_{n, \ell, m}(r, \theta, \phi)$ in terms of the relevant spherical harmonic and a radial factor fully determined except for an overall unit-free normalization constant $N$.

The wavefunction for a hydrogen atom with principal quantum number $n$, $\ell=n-1$, and $m=n-1$ can be expressed as:

$\psi_{n, \ell, m}(r, \theta, \phi)=R_{n, \ell}(r) Y_{\ell}^m(\theta, \phi)$

where $R_{n,\ell}(r)$ is the radial part of the wavefunction and $Y_{\ell}^{m}(\theta,\phi)$ is the spherical harmonic. Since $\ell=n-1$ and $m=n-1$, we have $\ell=m=n-1$. Therefore, the spherical harmonic reduces to:

$Y_{n-1}^{n-1}(\theta, \phi)=(-1)^{n-1} \sqrt{\frac{(2 n-1) ! !}{4 \pi(n-1) !}} P_{n-1}^{n-1}(\cos \theta) e^{i(n-1) \phi}$

where $P_{n-1}^{n-1}(\cos\theta)$ is the associated Legendre polynomial of degree $n-1$. The radial part of the wavefunction $R_{n,\ell}(r)$ can be obtained from the radial Schrödinger equation for hydrogen atom:

$\frac{d^2 R_{n, \ell}(r)}{d r^2}+\frac{2}{r} \frac{d R_{n, \ell}(r)}{d r}-\frac{\ell(\ell+1)}{r^2} R_{n, \ell}(r)+\frac{-\frac{1}{2}\left(\frac{1}{n}\right)^2}{-\frac{1}{2}} R_{n, \ell}(r)=0$

Simplifying the equation, we get:

$\frac{d^2 R_{n, \ell}(r)}{d r^2}+\frac{2}{r} \frac{d R_{n, \ell}(r)}{d r}-\frac{\ell(\ell+1)}{r^2} R_{n, \ell}(r)+\frac{1}{n^2} R_{n, \ell}(r)=0$

This equation has the standard form of a differential equation for the radial part of a spherically symmetric potential. The solution can be written in terms of the associated Laguerre polynomial $L_{n-\ell-1}^{2\ell+1}(2r/n)$ as:

$R_{n, \ell}(r)=\sqrt{\frac{(n-\ell-1) !}{2 n[(n+\ell) ! 3}}\left(\frac{2 r}{n}\right)^{\ell} e^{-r / n} L_{n-\ell-1}^{2 \ell+1}\left(\frac{2 r}{n}\right)$

Therefore, the wavefunction for a hydrogen atom with principal quantum number $n$, $\ell=n-1$, and $m=n-1$ is:

$\psi_{n, n-1, n-1}(r, \theta, \phi)=N \sqrt{\frac{(2 n-1) ! !}{4 \pi(n-1) !}}\left(\frac{2 r}{n}\right)^{n-1} e^{-r / n} L_1^{2 n-1}\left(\frac{2 r}{n}\right) e^{i(n-1) \phi}$

where $N$ is an overall unit-free normalization constant.

物理学的基本数学代写|Essential Maths for Physics PHYS11400代写

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Assignment-daixieTM为您提供布里斯托大学University of Bristol Essential Maths for Physics PHYS11400物理学的基本数学代写代考辅导服务！

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Mathematics plays a crucial role in understanding and describing physical phenomena. Here are some essential math topics for physics:

1. Calculus: Calculus is the branch of mathematics that deals with the study of continuous change. It is essential for physics because it helps in understanding how things change over time. Calculus concepts like differentiation and integration are widely used in physics, for instance, in the description of motion and the calculation of rates of change.
2. Linear Algebra: Linear algebra deals with the study of linear equations, matrices, and vector spaces. It is used to solve systems of equations, which frequently arise in physics problems. Linear algebra is also used to describe transformations and rotations, which are essential in understanding the behavior of objects in motion.
3. Differential Equations: Differential equations are equations that involve derivatives or rates of change. They are widely used in physics to describe the behavior of systems that change over time. Examples include the motion of particles under the influence of forces, the behavior of circuits, and the decay of radioactive materials.
4. Complex Analysis: Complex analysis deals with the study of complex numbers and their properties. It is used to analyze functions that are defined in the complex plane, such as those that describe the behavior of electromagnetic waves. Complex analysis is also essential for understanding quantum mechanics.
5. Statistics and Probability: Statistics and probability are essential for making predictions and inferences in physics. They are used to analyze data and to make quantitative predictions about the behavior of physical systems. Examples include the probability of a particle being found in a certain location and the probability of a system undergoing a particular change.
6. Fourier Analysis: Fourier analysis is used to describe periodic functions and their properties. It is used extensively in physics to analyze signals and to understand the behavior of waves, including sound waves, electromagnetic waves, and quantum waves.
7. Group Theory: Group theory is the study of symmetry and structure. It is used in physics to describe the behavior of particles, including their interactions with each other. Group theory is also essential for understanding the behavior of crystals, which have a regular, repeating structure.

These are just some of the essential math topics for physics. A strong foundation in these areas is crucial for understanding the mathematical underpinnings of physical phenomena.

Reduce the following ordinary differential equation to a first-order vector differential equations, which you should write out completely, in vector format. $$\left(\frac{d^3 y}{d x^3}\right)^3–\frac{d^2 y}{d x^2}-y^2=0$$

Let $y_1=y$, $y_2=\frac{d y}{d x}$, and $y_3=\frac{d^2 y}{d x^2}$. Then, we can write the given third-order differential equation as a system of three first-order differential equations:

\begin{align*} \frac{d y_1}{d x} &= y_2 \ \frac{d y_2}{d x} &= y_3 \ \frac{d y_3}{d x} &= \sqrt[3]{y_3^3+y_2+y_1^2} \end{align*}

Thus, we have reduced the given third-order differential equation to a first-order vector differential equation in vector format:

$\frac{d}{d x}\left(\begin{array}{l}y_1 \ y_2 \ y_3\end{array}\right)=\left(\begin{array}{c}y_2 \ y_3 \ \sqrt[3]{y_3^3+y_2+y_1^2}\end{array}\right)$

Divergence of acceleration in phase space. (a) Prove that particles of charge $q$ moving in a magnetic field $\mathbf{B}$ and hence subject to a force $q \mathbf{v} \times \mathbf{B}$, nevertheless have $\nabla_v \cdot \mathbf{a}=0$.

(a) The acceleration $\mathbf{a}$ of a charged particle moving in a magnetic field is given by $\mathbf{a} = (q/\gamma)m \mathbf{v} \times \mathbf{B}$, where $\gamma$ is the relativistic factor and $m$ is the mass of the particle. Using the vector identity $\nabla \cdot (\mathbf{a} \times \mathbf{b}) = \mathbf{b} \cdot (\nabla \times \mathbf{a}) – \mathbf{a} \cdot (\nabla \times \mathbf{b})$, we can write:

\begin{align} \nabla_v \cdot \mathbf{a} &= \frac{\partial}{\partial v_x} \left(\frac{q}{\gamma} m B_y v_z\right) + \frac{\partial}{\partial v_y} \left(\frac{q}{\gamma} m B_z v_x\right) + \frac{\partial}{\partial v_z} \left(\frac{q}{\gamma} m B_x v_y\right) \ &= \frac{q}{\gamma} m B_y \frac{\partial}{\partial v_x}(v_z) + \frac{q}{\gamma} m B_z \frac{\partial}{\partial v_y}(v_x) + \frac{q}{\gamma} m B_x \frac{\partial}{\partial v_z}(v_y) \ &= 0, \end{align}

where we have used the fact that $\partial v_i/\partial v_j = \delta_{ij}$, the Kronecker delta.

Therefore, the divergence of acceleration for a charged particle in a magnetic field is zero.

(b) Consider a frictional force that slows particles down in accordance with $\mathbf{a}=-K \mathbf{v}$, where $K$ is a constant. What is the “velocity-divergence”, of this acceleration, $\nabla_v \cdot \mathbf{a}$ ? Does this cause the distribution function $f$ to increase or decrease as a function of time?

For a frictional force that slows particles down according to $\mathbf{a} = -K \mathbf{v}$, the velocity-divergence is given by:

\begin{align} \nabla_v \cdot \mathbf{a} &= \frac{\partial}{\partial v_x}(-Kv_x) + \frac{\partial}{\partial v_y}(-Kv_y) + \frac{\partial}{\partial v_z}(-Kv_z) \ &= -3K. \end{align}

The velocity-divergence is negative, indicating that the distribution function $f$ will decrease as a function of time. This is because the frictional force removes kinetic energy from the particles, causing them to slow down and reduce their probability of occupying higher velocity states.

金融风险管理|FINANCIAL RISK MANAGEMENT ACFI342代写

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Assignment-daixieTM为您提供利物浦大学University of Liverpool FINANCIAL RISK MANAGEMENT ACFI342金融风险管理代写代考辅导服务！

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Financial Risk Management is an essential aspect of any financial institution. It involves identifying, measuring, and mitigating risks that may arise in financial markets. Internal aspects of financial institutions include identifying risks in credit and operational risk, as well as managing liquidity, market, and interest rate risk.

External factors that affect the investment arena and modern financial markets include macroeconomic events such as changes in interest rates, inflation rates, and political instability. Financial risk management involves understanding how these factors impact investments and developing strategies to manage these risks.

Theoretical foundations that underpin modern investment and risk management techniques include portfolio theory, option pricing theory, and capital asset pricing models. Understanding these theories can help in developing investment strategies and managing risks.

The practical dimension of financial risk management involves applying theoretical knowledge to real-world scenarios. This includes developing investment strategies that incorporate risk management techniques, analyzing financial data, and using financial models to make informed decisions.

Overall, a degree in Financial Risk Management can provide you with a solid understanding of the theoretical foundations of modern investment and risk management techniques, as well as practical skills that can be applied to equity and credit markets. This can help in developing effective investment strategies and managing risks in financial markets.

Your friend is celebrating her 35 th birthday today and wants to start saving for her anticipated retirement at age 65 (she will retire on her 65 th birthday). She would like to be able to withdraw $\$ 80,000$from her savings account on each birthday for at least 20 years following her retirement (the first withdrawal will be on her 66th birthday). Your friend intends to invest her money in the local savings bank which offers 4 percent per year. She wants to make equal annual deposits on each birthday in a new savings account she will establish for her retirement fund. If she starts making these deposits on her 36 th birthday and continues to make deposits until she is 65 (the last deposit will be on her 65 th birthday), what amount must she deposit annually to be able to make the desired withdrawals upon retirement? 证明 . To determine the amount your friend must deposit annually, we can use the concept of present value, which is the current value of a future sum of money after accounting for the time value of money. We know that your friend wants to withdraw$80,000 per year for at least 20 years after her retirement, and the first withdrawal will be on her 66th birthday. Assuming that she lives for 20 years after the first withdrawal, the total number of withdrawals she will make is:

20years+1=21withdrawals

We can use the formula for present value of an annuity to determine how much she needs to deposit each year to be able to make these withdrawals. The formula is:

$P V=\frac{P M T}{r}\left[1-\frac{1}{(1+r)^n}\right]$

where PV is the present value of the annuity, PMT is the annual payment, r is the interest rate, and n is the number of periods.

In this case, we want to find the PMT, which represents the annual deposit your friend must make. We know that:

• The interest rate is 4% per year
• The number of periods is 21 (from her 45th birthday to her 65th birthday)
• The future value of the annuity is the total amount she will withdraw over 20 years, which is:$F V=20 \times \$ 80,000=\$1,600,000$

Substituting these values into the formula, we get:

$P V=\frac{\mathrm{PMT}}{0.04}\left[1-\frac{1}{(1+0.04)^{21}}\right]=\$ 1,600,000$Solving for PMT, we get:$\mathrm{PMT}=\frac{\$1,600,000 \times 0.04}{1-\frac{1}{(1+0.04)^{21}}} \approx \$ 36,044.68$Therefore, your friend must deposit approximately $$36,044.68 each year from her 36th birthday to her 65th birthday to be able to withdraw$$80,000$ per year for at least 20 years after her retirement.

The current level of the S\&P 500 is 1040 . The risk-free interest rate per year is $2 \%$. Assume negligible dividends. The 6 month futures contract is trading at 1060 . (a)Is there an arbitrage opportunity? Briefly explain.

(a) To determine if there is an arbitrage opportunity, we can calculate the theoretical futures price using the cost-of-carry model:

$\mathrm{F}=\mathrm{S}^* \mathrm{e}^{\wedge}\left(r^{\star} \mathrm{t}\right)$

where F is the theoretical futures price, S is the current spot price, r is the risk-free interest rate, and t is the time to expiration in years.

In this case, the time to expiration is 0.5 years, and the risk-free interest rate is 2%. Plugging in the values, we get:

$F=1040 * e^{\wedge}(0.02 * 0.5)=1051.2$

Since the current futures price is 1060, there is an arbitrage opportunity. The theoretical futures price is lower than the current futures price, which means we can make a riskless profit by using the cost-of-carry model.

(b)If there is an arbitrage opportunity, what strategy would you use to exploit it without using any funds of your own?

(b) To exploit the arbitrage opportunity without using any funds of our own, we can use a cash-and-carry arbitrage strategy. The steps involved are:

1. Borrow money to buy the underlying asset (in this case, the S&P 500 index) at the spot price.
2. Sell a futures contract for the same asset, and receive the current futures price.
3. Invest the proceeds from the futures sale at the risk-free rate.
4. When the futures contract expires, take delivery of the underlying asset and use it to repay the loan.

Since the theoretical futures price is lower than the current futures price, we can use the cash-and-carry arbitrage strategy to lock in a profit. Here’s how it works:

1. Borrow $1040 to buy the S&P 500 index at the current spot price. 2. Sell a futures contract for the S&P 500 index at the current futures price of$1060.
3. Invest the $1060 at the risk-free rate of 2% for 6 months, which gives us: 4.$ 1060 * \mathrm{e}^{\wedge}(0.02 * 0.5)=\$1071.21$
5. When the futures contract expires in 6 months, take delivery of the S&P 500 index and use it to repay the loan of $1040.The profit from this strategy is: 6. Profit$=\$1071.21-\$ 1040=\$31.21$
7. Since we didn’t use any of our own funds to execute this strategy, the profit is risk-free.

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Assignment-daixieTM为您提供利物浦大学University of Liverpool ADVANCED MACROECONOMICS ECON343进阶宏观经济学代写代考辅导服务！

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Countries grow due to a combination of factors such as technological progress, investment, human capital accumulation, and favorable institutions and policies.
Recessions and booms are caused by fluctuations in aggregate demand and supply shocks.
Unemployment is the result of a mismatch between the number of workers and the available jobs, and it can be influenced by factors such as labor market policies, technology, and economic growth.
Inflation can be caused by increases in aggregate demand, cost-push factors such as higher oil prices, or monetary factors such as excessive money supply growth.
Government policies can affect macroeconomic variables such as output, unemployment, inflation, and growth through fiscal and monetary policies, and by influencing institutions and regulations that affect the economy.

In this problem you will replicate Figures on pages 12 and 14 of the lecture notes (demand shocks, part I). Consider a stochastic growth model with preferences and technology given by \begin{aligned} U\left(C_t, N_t\right) & =\frac{1}{1-\sigma} C_t^{1-\sigma}-\frac{1}{1+\eta} N_t^{1+\eta} \\ A_t F\left(K_{t-1}, N_t\right) & =A_t K_{t-1}^\alpha N_t^{1-\alpha} . \end{aligned} The process for $A_t$ is as follows $$\begin{gathered} A_t=e^{a_t}, \\ a_t=\rho a_{t-1}+\epsilon_t . \end{gathered}$$ Use parameters \begin{aligned} \beta & =0.99, \quad \delta=0.025 \\ \eta & =1, \quad \sigma=1, \\ \alpha & =0.36, \quad \rho=0.95 . \end{aligned} You can use the Matlab package Dynare. (i) Setup the planner problem and derive the first order conditions.

(i) The planner problem is to maximize the household’s utility subject to the resource constraint:

$\begin{array}{r}\max {\left{C_t, N_t, K_t\right}} \sum{t=0}^{\infty} \beta^t U\left(C_t, N_t\right) \ \text { s.t. } C_t+K_{t+1}-(1-\delta) K_t=A_t K_t^\alpha N_t^{1-\alpha},\end{array}$

where $K_t$ is the capital stock, $C_t$ is consumption, $N_t$ is labor, and $A_t$ is the level of technology. The first-order conditions are:

\begin{aligned} C_t: & C_t^{-\sigma}=\beta \mathbb{E}t\left{C{t+1}^{-\sigma}\left[(1-\delta)+\alpha A_{t+1} K_{t+1}^{\alpha-1} N_{t+1}^{1-\alpha}\right]\right} \ N_t: & N_t^\eta=\beta \mathbb{E}t\left{N{t+1}^\eta\left[(1-\alpha) A_{t+1} K_{t+1}^\alpha N_{t+1}^{-\alpha}-\frac{1}{1+\eta}\right]\right} \ K_t: & C_t+(1-\delta) K_t=\alpha A_t K_t^\alpha N_t^{1-\alpha}+K_{t+1}\end{aligned}

Consider the version of the Lucas (1972) model derived in class. (i) Derive an expression for the constant $\xi$ or (which is the same) for the average price level $\bar{p}$. (Hint: you can take unconditional expectations on both sides of the labor supply equation to get $$E\left[N_{i, t}\right]=E\left[\frac{P_{i, t}}{P_{j, t+1}}\left(1+x_{t+1}\right)\right],$$

(i) In the Lucas (1972) model, the aggregate labor supply is given by

$N_t=\sum_{i=1}^N n_{i, t}$,

where $n_{i,t}$ is the labor supply of the $i$th individual. From the individual’s problem, we have

$n_{i, t}=\frac{w_t}{P_{i, t}}\left(\frac{u_{c, i, t}}{u_{l, i, t}}\right)$,

where $u_{c, i,t}$ and $u_{l, i,t}$ are the partial derivatives of the individual’s utility function with respect to consumption and leisure, respectively.

Taking the unconditional expectation on both sides of the labor supply equation and using the law of iterated expectations, we get \begin{align*} E[N_t]&=E\left[\sum_{i=1}^{N}n_{i,t}\right]\ &=\sum_{i=1}^{N}E\left[\frac{w_t}{P_{i,t}}\left(\frac{u_{c, i,t}}{u_{l, i,t}}\right)\right]\ &=\frac{w_t}{\bar{p}}\left(\frac{\bar{u}c}{\bar{u}l}\right), \end{align*} where $\bar{p}$ is the average price level defined by $\bar{p}=\left(\frac{1}{N}\sum{i=1}^{N}P{i,t}\right)$, and $\bar{u}_c$ and $\bar{u}_l$ are the average partial derivatives of the utility function with respect to consumption and leisure, respectively.

Since the model assumes that there is no aggregate demand for money, we have $M_t=P_{t}Y_t$, where $M_t$ is the nominal money supply, $P_t$ is the aggregate price level, and $Y_t$ is the level of real output. Therefore, we have

$P_t=\frac{M_t}{Y_t}$

which implies that $P_t$ is a function of $Y_t$ and $M_t$. Using this expression for $P_t$ and the fact that the real money supply $M_t/P_{t}$ is constant, we get

$\bar{p}=\frac{1}{N} \sum_{i=1}^N \frac{M_t}{Y_{i, t}}$.

Substituting this expression for $\bar{p}$ into the expression for $E[N_t]$, we get

$E\left[N_t\right]=\frac{w_t}{\left(\frac{1}{N} \sum_{i=1}^N \frac{M_t}{Y_{i, t}}\right)}\left(\frac{\bar{u}_c}{\bar{u}_l}\right)$.

Solving for $\bar{p}$, we get

$\bar{p}=\frac{1}{N} \sum_{i=1}^N \frac{M_t}{Y_{i, t}}=\frac{w_t}{E\left[N_t\right]}\left(\frac{\bar{u}_c}{\bar{u}_l}\right)$.

(ii) Study the effect of changing $\sigma_\epsilon^2$ on average labor supply and average output, interpret.

(ii) The effect of changing $\sigma_\epsilon^2$ on average labor supply and average output can be studied by analyzing the impact of changes in this parameter on the average marginal utility of consumption $\bar{u}_c$ and the average marginal disutility of labor $\bar{u}_l$.

From the individual’s

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Assignment-daixieTM为您提供利物浦大学University of Liverpool BUSINESS FINANCE ECON304商业经济学代写代考辅导服务！

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The module covers a range of topics, including bond and stock valuation, capital structure, dividend policy, and leasing contracts.

The module will be delivered through a combination of small group face-to-face sessions and asynchronous online lectures. This approach allows for more personalized interaction and engagement in small group sessions, while also providing flexibility for students to access the course content at their own pace through online lectures.

Overall, this module seems to provide a comprehensive introduction to corporate finance, equipping students with the knowledge and skills needed to understand and navigate the financial decision-making processes of businesses.

Your car dealer offers you a loan for part of the purchase price of a new car, citing an annual percentage rate (APR) of 8.5%. What is the effective annual rate of such a loan (recall that an auto loan typically requires monthly payments)?

To calculate the effective annual rate of the loan, we need to take into account the compounding effect of the monthly payments. Here’s how to do it:

1. First, we need to find the monthly interest rate. We can do this by dividing the APR by 12 (the number of months in a year):Monthly interest rate = 8.5% / 12 = 0.7083%
2. Next, we can use the following formula to calculate the effective annual rate (EAR) of the loan:EAR = (1 + r/n)^n – 1Where: r = the monthly interest rate (0.7083%) n = the number of compounding periods in a year (12 for monthly payments)Plugging in the numbers, we get:EAR = (1 + 0.7083/100)^12 – 1 = 9.0715%

Therefore, the effective annual rate of the loan is 9.0715%. This means that if you were to borrow money from the car dealer for a year, with monthly payments, the interest rate would be equivalent to 9.0715% per year.

Company $\mathrm{ABC}$ has just paid a dividend of 50 cents per share. Because of its growth potential, its dividend is forecasted to grow at a rate of 7 percent per year indefinitely. If the company’s appropriate cost of capital (given its risk) is 11 percent, what was ABC’s share price immediately before it paid its 50 cent dividend, i.e the stock price right before the ex-dividend date?

We can use the dividend discount model to find the price of the stock before the dividend was paid. The formula for the price of a stock using the dividend discount model is:

P = D / (r – g)

Where:

• P is the price of the stock
• D is the dividend paid per share
• r is the required rate of return or cost of capital
• g is the expected growth rate of the dividend

In this case, we are looking for the price of the stock right before the dividend was paid, so we can use the current dividend of 50 cents per share. We are also given that the expected growth rate of the dividend is 7% per year and the cost of capital is 11%.

Plugging in the values, we get:

P = 0.5 / (0.11 – 0.07) P = 0.5 / 0.04 P = 12.50

Therefore, the share price of ABC right before it paid its 50 cent dividend was $12.50. 问题 3. The current market price of a two-year 25 percent coupon bond with a$\$1,000$ face value is $\$ 1,219.71$(recall that such a bond pays coupons of$\$250$ at the end of years 1 and 2 , and the principal of $\$ 1,000$at the end of year 2). The current market price of a one-year pure discount bond with a$\$50$ face value is $\$ 44.64$. (a) (15 points) What must the price of a two-year pure discount bond with a$\$2,500$ face value be in order to avoid arbitrage?

To avoid arbitrage, the principle of no-arbitrage states that if two securities have the same cash flows in the future, they must have the same price today. We can use this principle to find the price of a two-year pure discount bond with a face value of $2,500$.

First, we need to calculate the present value of the cash flows from the two-year 25% coupon bond. We know that the bond pays coupons of $250$ at the end of years $1$ and $2$, and the principal of $$1,000 at the end of year 2. Using a discount rate of r, the present value of these cash flows is: PV = \frac{250}{(1 + r)} + \frac{250 + 1,000}{(1 + r)^2} We can use the current market price of the bond to solve for r:$$1,219.71 = \frac{250}{(1 + r)} + \frac{250 + 1,000}{(1 + r)^2}$Solving this equation, we find that$r = 8.211%$. Next, we need to find the present value of the face value of the bond, which is $$2,500 and will be received at the end of year 2. Using the same discount rate of 8.211%, the present value of this cash flow is: PV = \frac{2,500}{(1 + r)^2} PV = \frac{2,500}{(1 + 0.08211)^2} = 2,079.35 Therefore, the price of a two-year pure discount bond with a face value of$$2,500$ must be 2,079.35\$ to avoid arbitrage.

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Assignment-daixieTM为您提供利物浦大学University of Liverpool ADVANCED MICROECONOMICS ECON342进阶微观经济学代写代考辅导服务！

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Advanced microeconomics is a branch of economics that deals with the behavior of individuals, firms, and markets. It builds upon the principles of microeconomics and incorporates more complex theoretical and mathematical models to analyze the decisions made by economic agents in various situations.

Advanced microeconomics covers topics such as game theory, asymmetric information, general equilibrium theory, welfare economics, and mechanism design. It also involves the study of topics like consumer behavior, producer behavior, market structure, and competition.

Advanced microeconomics is important because it provides a deeper understanding of economic phenomena, helps to identify potential inefficiencies in markets, and guides the design of policies to improve economic outcomes. It is used in many fields, including finance, business, public policy, and academia.

)Give a formal definition of what it means for a multistage game with observed
actions to be continuous at infinity? Why do we care whether games are continuous at
infinity?

See the lecture notes or FT for the formal definition. We care about continuity at infinity because if this property is satisfied, then checking a proposed SPE is equivalent to checking the single-deviationproperty (ie that no player can achieve a higher payoff by deviating from their equilibrium strategy at a single node, and playing according to their equilibrium strategy at all other nodes).

State Kakutani’s theorem. What correspondence is it applied to in the proof that
any finite game has a Nash equilibrium? Where does the argument break down if you try
to use Kakutani’s theorem in the same way to prove the existence of an equilibrium in the
”Name the Largest Number” game?

See the notes for the definitions. The assumption that breaks down in the “name the largest
number” game is that C, the space of strategy profiles, is not compact (because there is no limit on how
large the numbers named by the players can be).

Given an example of a game in which you could argue that the subgame perfect
equilibrium concept is too restrictive and rules out a reasonable outcome. Give an example
of a game in which you could argue that the subgame perfect equilibrium concept is not
restrictive enough and fails to rule out an unreasonable outcome. (Explain briefly what
you would argue about each example.)

The centipede game that Glenn discussed in class is an example where SPE/backward induction
rules out a reasonable outcome. If you remember, the SPE is to play down immediately, giving the players a
low payoff. But it is probably more reasonable from an empirical point of view to expect players to cooperate
for a while before they play down.
An example where SPE gives an unreasonable result – Glenn gives several of these in the notes. He
discusses games in which SPE allows equilibria that involve players choosing actions at an information set
that are inconsistent with any possible set of beliefs the player could hold at that information set. Signalling
games will often have unreasonable SPE, because there are no subgames (you can’t separate parts of the
game tree without cutting information sets).