MY-ASSIGNMENTEXPERT™可以为您提供my.uq.edu PHYS4040 Quantum field theory量子场论课程的代写代考和辅导服务！
这是昆士兰大学 量子场论课程的代写成功案例。
PHYS4040课程简介
Course Description: This course covers advanced topics from single-particle quantum mechanics, such as advanced quantum dynamics, path-integrals, and quantum decoherence. It will also cover introductory non-relativistic many-body quantum theory, relativistic quantum mechanics, and relativistic quantum field theory. We will look at selected applications from atom optics, condensed matter physics, and particle physics.
Assumed Background:
Students should have a sound understanding of quantum physics at the level of PHYS3040 (or equivalent). Background in special relativity (e.g. PHYS2100), electromagnetic fields (e.g. PHYS3051), condensed matter physics (e.g. PHYS2020) and statistical mechanics (e.g. PHYS3020) is strongly recommended.
Prerequisites
This course aims at a first introduction to relativistic quantum field theory. This plays a fundamental role in our understanding of particle physics but arises also in the study of statistical mechanics and second quantised condensed matter systems.
Course Changes in Response to Previous Student Feedback
The course has been modified to be more interactive, with more chance for students to work through problems with assistance.
PHYS4040 Quantum field theory HELP（EXAM HELP， ONLINE TUTOR）
Bjorken and Drell, Ch. 19, problem 1, which is:
Prove the generalized Ward identity [the Ward-Takahashi identity] by forming the three-fold vacuum expectation value
$$
\left\langle 0\left|T\left(\psi(x) \bar{\psi}(y) j_\mu(z)\right)\right| 0\right\rangle
$$
and using current conservation and the field equations.
(The current $j_\mu(z)$ is as usual $\bar{\psi}(z) \gamma_\mu \psi(z)$.)
The basic object is
$$
\left\langle 0\left|T \psi(x) \bar{\psi}(y) j_\mu(z)\right| 0\right\rangle=\left\langle 0\left|T \psi(x) \bar{\psi}(y) \bar{\psi}(z) \gamma_\mu \psi(z)\right| 0\right\rangle
$$
Be aware that the fields are fields with all interactions included. Also be aware that the equal time canonical commutation relations of the fields only depend on the rules of quantum mechanics, and so work for the interacting fields. There is some choice regarding what steps to take first.
I will start with using current conservation.
$$
\begin{aligned}
\frac{\partial}{\partial z_\mu}\left\langle 0\left|T \psi(x) \bar{\psi}(y) j_\mu(z)\right| 0\right\rangle=\frac{\partial}{\partial z_\mu}\langle 0| & \theta\left(x_0-y_0\right) \theta\left(y_0-z_0\right) \psi(x) \bar{\psi}(y) j_\mu(z) \
& +\theta\left(x_0-z_0\right) \theta\left(z_0-z_0\right) \psi(x) j_\mu(z) \bar{\psi}(y) \
& +4 \text { more terms } \quad|0\rangle .
\end{aligned}
$$
The derivatives will act on the theta-functions, as well as on the current. On the current it will give $\partial_z^\mu j_\mu=0$, and on the theta-functions it will give delta-functions. One will find,
$$
\begin{aligned}
\partial_z^\mu\left\langle 0\left|T \psi(x) \bar{\psi}(y) j_\mu(z)\right| 0\right\rangle & =\delta\left(y_0-z_0\right)\left\langle 0\left|T \psi(x)\left[j_0(z), \bar{\psi}(y)\right]\right| 0\right\rangle \
& -\delta\left(x_0-z_0\right)\left\langle 0\left|T\left[\psi(x), j_0(z)\right] \bar{\psi}(y)\right| 0\right\rangle
\end{aligned}
$$
Then, with “ET” standing for “equal time,”
$$
\begin{aligned}
& {\left[j_0(z), \psi(y)^{\dagger}\right]{\mathrm{ET}}=\left[\psi^{\dagger}(z) \psi(z), \psi(y)^{\dagger}\right]{\mathrm{ET}}=\psi^{\dagger}(z)\left{\psi(z), \psi(y)^{\dagger}\right}_{\mathrm{ET}}=\psi^{\dagger}(z) \delta^3(z-y)} \
& {\left[\psi(x), j_0(z)\right]_{\mathrm{ET}}=\psi(z) \delta^3(x-z)}
\end{aligned}
$$
Hence,
$$
\partial_z^\mu\left\langle 0\left|T \psi(x) \bar{\psi}(y) j_\mu(z)\right| 0\right\rangle=\delta^4(y-z)\langle 0|T \psi(x) \bar{\psi}(y)| 0\rangle-\delta^4(x-z)\langle 0|T \psi(x) \bar{\psi}(y)| 0\rangle,
$$
or
$$
\partial_z^\mu\left\langle 0\left|T \psi(x) \bar{\psi}(y) j_\mu(z)\right| 0\right\rangle=i S_F^{\prime}(x-z) \delta^4(y-z)-i S_F^{\prime}(z-y) \delta^4(x-z) .
$$
To get factors of “i”, etc., work out 3-point function in LO.
$$
\begin{aligned}
& \left\langle 0\left|T \psi(x) \bar{\psi}(y) \bar{\psi}(z) \gamma_\mu \psi(z)\right| 0\right\rangle=\left\langle 0\left|T \psi(x) \bar{\psi}(z) \gamma_\mu \psi(z) \bar{\psi}(y)\right| 0\right\rangle \
& \quad=\langle 0|T \psi(x) \bar{\psi}(z)| 0\rangle \gamma_\mu\langle 0|\psi(z) \bar{\psi}(y)| 0\rangle=i S_F(x-z) \gamma_\mu i S_F(z-y)
\end{aligned}
$$
Using momentum space,
$$
\begin{aligned}
S_F(x-z) & =\int\left(d^4 k\right) e^{-i k(x-z)} S_F(k), \
S_F\left(p^{\prime}\right) & =\int\left(d^4 x\right) e^{i p^{\prime}(x-z)} S_F(x-z), \
S_F(p) & =\int\left(d^4 x\right) e^{i p(z-y)} S_F(z-y) .
\end{aligned}
$$
Hence,
$$
\int\left(d^4 x\right)\left(d^4 y\right) e^{i p^{\prime}(x-z)} e^{i p(z-y)}\left\langle 0\left|T \psi(x) \bar{\psi}(y) j_\mu(z)\right| 0\right\rangle=i S_F\left(p^{\prime}\right) \gamma_\mu i S_F(p)
$$
General case definition,
$$
\int\left(d^4 x\right)\left(d^4 y\right) e^{i p^{\prime}(x-z)} e^{i p(z-y)}\left\langle 0\left|T \psi(x) \bar{\psi}(y) j_\mu(z)\right| 0\right\rangle=i S_F^{\prime}\left(p^{\prime}\right) \Gamma_\mu i S_F^{\prime}(p) .
$$
Finally,
$$
\begin{aligned}
& \int\left(d^4 x\right)\left(d^4 y\right) e^{i p^{\prime}(x-z)} e^{i p(z-y)} \partial_z^\mu\left\langle 0\left|T \psi(x) \bar{\psi}(y) j_\mu(z)\right| 0\right\rangle=i\left(p^{\prime}-p\right)^\mu \int \ldots \
& =i^3 S_F^{\prime}\left(p^{\prime}\right)\left(p^{\prime}-p\right)^\mu \Gamma_\mu S_F^{\prime}(p) \
& =\int\left(d^4 x\right)\left(d^4 y\right) e^{i p^{\prime}(x-z)} e^{i p(z-y)}\left[i S_F^{\prime}(x-z) \delta^4(y-z)-i S_F^{\prime}(z-y) \delta^4(x-z)\right] \
& =i S_F^{\prime}\left(p^{\prime}\right)-i S_F^{\prime}(p) .
\end{aligned}
$$
Thus,
$$
S_F^{\prime}\left(p^{\prime}\right)\left(p^{\prime}-p\right)^\mu \Gamma_\mu S_F^{\prime}(p)=S_F^{\prime}(p)-S_F^{\prime}\left(p^{\prime}\right) .
$$
Thus,
$$
\left(p^{\prime}-p\right)^\mu \Gamma_\mu=S_F^{\prime-1}\left(p^{\prime}\right)-S_F^{\prime-1}(p) .
$$
Suppose the QED Lagrangian contained an actual magnetic moment interaction, as
$$
\mathcal{L}=\bar{\psi}\left(i \not \partial-m_0\right) \psi-e_0 \bar{\psi} \gamma_\mu \psi A^\mu-\frac{\kappa_0}{2 m_0} \bar{\psi} \sigma_{\mu \nu} \psi F^{\mu \nu} .
$$
Give the superficial degree of divergence for Feynman diagrams in this theory in $4 \mathrm{D}$ in terms of the numbers of external legs and, if necessary, the number of vertices in the diagram. How many primitively divergent diagrams are there? For the sake of simplicity, you may let $e_0=0$.
Look up again, or recall, the meaning of renormalizable, superrenormalizable, and nonrenormalizable, and state which of these QED with an intrinsic magnetic moment term is.
What is the degree of divergence of the box diagram,
where each vertex is the $\kappa_0$ vertex above?
The $\kappa_0$ vertex has a derivative (within $F^{\mu \nu}$ ), and hence a power of momentum in momentum space. Say all the vertices are $\kappa_0$ vertices.
$$
D=4 L-2 P_\gamma-P_e+p\left(V-N_\gamma\right)
$$
where $p=1$ for the $\kappa_0$ vertex, and the standard result can be checked using $p=0 . P_\gamma$ and $P_e$ are the number of internal photon and electron propagators, and $N_\gamma$ and $N_e$ are the number of external photon and electron lines.
$$
\begin{aligned}
L & =P_\gamma+P_e-V+1 \
V & =2 P_\gamma+N_\gamma \
2 V & =2 P_e+N_e
\end{aligned}
$$
Substituting,
$$
D=4-N_\gamma-\frac{3}{2} N_e+p\left(V-N_\gamma\right)
$$
The are an infinite number of primitive divergences.
The theory is non-renormalizable.
The box diagram has
$$
D=4-0-6+(4-0)=2 \text {. }
$$
(For ordinary QED it would have $D=-2$.)
MY-ASSIGNMENTEXPERT™可以为您提供MY.UQ.EDU PHYS4040 QUANTUM FIELD THEORY量子场论课程的代写代考和辅导服务！
