Browse & Filter

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Show under an infinitesimal rotation of a vector \(\vec{A}\) about an axis \(\hat{n}\) has the form \(\Delta \vec{A}= c \hat{n}\times{\vec{A}}\), where \(c\) is an infinitesimal constant. }
\Hint{Use the fact that lengths and angles do not change under rotations} \QField{Comment}{ When ever you see an equation of the form \(\Delta \vec{A}= c \hat{n}\times{\vec{A}}\), you should realize that some rotation is taking place. Several examples can be given

1. A electric dipole placed in an electric field will experience a torque and its equation of motion can be written down as\[ \dd[\vec{p}]{t}\propto \vec{p}\times \vec{E}\] and it will tend to rotate.
2. A similar statement can be made about magnetic dipole in a magentic field.
3. A related example is that of spin in a uniform magnetic field. The Hamitonian is of the form \(H=-\vec{\mu}\cdot\vec{B}\) and the equation of motion takes the form \[ i\hbar\dd[\vec{S}]{t} = [\vec{S},H]_- \] With magnetic moment being proportional to the spin \(\vec{S}\), the right hand side becomes proportional to \(\vec{B}\times \vec{S}\).

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The Feynman propagator \(\Delta_F\) for a real scalar field is defined as the vacuum expectation value of time ordered product of fields: \[ \Delta_F(x-y) = \matrixelement{0}{T\big(\phi(x)\phi(y)\big)}{0}.\] We will now show that the Feynman propagator satisfies the equation \[ (\Box + M^2) \Delta_F(x-y)= - \delta^{(4)}(x-y).\] }Since \(\phi(x)\) obeys the Klein Gordon equation, it is obvious that \(\Delta^{(\pm)}(x-y)\) satisfy the Klein Gordon equation \begin{equation} (\Box + M^2)\Delta^{(\pm)}=0 . \end{equation} The time ordered product can be written as \begin{eqnarray} T(\phi(x)\phi(y)) &=&(\phi(x)\phi(y))\theta(x_0-y_0) + (\phi(y)\phi(x))\theta(y_0-x_0) \end{eqnarray} Now we compute action of the operator \(\displaystyle\Box =\frac{\partial^2}{x^{0\,2}}- \frac{\partial^2}{\partial x^{k\,2}}\) on the time ordered product. Note that the space derivatives in \(\displaystyle\Box = \PP{{x^0}}- \PP{{x^k}}\) will act only on the fields but the time derivative will act on theta functions also. So let us compute the time derivatives using \(\displaystyle\dd[\theta(x_0-y_0)]{x_0}= \delta(x_0-y_0)\). We will get \begin{eqnarray} { \pp{x_0}T(\phi(x)\phi(y))\nonumber }\nonumber &=&\Big\{\pp[\phi(x)]{x_0}\phi(y)\theta(x_0 - y_0) + (\phi(x)\phi(y))\pp{x_0}\theta(x_0 - y_0)\Big\} - \Big\{\phi(y)\pp[\phi(x)]{x_0}\theta(y_0 - x_0) - (\phi(y)\phi(x))\pp{x_0}\theta(y_0 - x_0)\Big\}\nonumber &=& \dot{\phi}(x)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\dot{\phi}(x)\theta(y_0 - x_0) + \big[\phi(x), \phi(y)\big] \delta(x_0 - y_0)\\ &=& \big(\partial_0{\phi}(x)\big)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\big(\partial_0{\phi}(x)\big)\theta(y_0 - x_0) \end{eqnarray} The last term in \eqRef{EQ19} becomes equal time commutator of the fields and hence it is zero. Differentiating \eqRef{EQ20} once again w.r.t. \(x_0\) we will get \begin{eqnarray}\nonumber { \frac{\partial^2}{\partial x^{02}}T(\phi(x)\phi(y)) }\nonumber &=& \ddot{\phi}(x)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\ddot{\phi}(x)\theta(y_0 - x_0) +\nonumber \dot{\phi}(x)\,\phi(y)\delta(x_0 - y_0) - \phi(y)\,\dot{\phi}(x)\delta(y_0 - x_0)\\\nonumber &=& \ddot{\phi}(x)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\ddot{\phi}(x)\theta(y_0 - x_0) \\\nonumber && \qquad + \big[\dot{\phi}(x)\,,\,\phi(y) \big] \delta(x_0-y_0) . \end{eqnarray} Substituting the value of equal time commutator \[\big[\dot{\phi}(x)\,,\,\phi(y)\big]\Big|_{x_0=y_0} = -\delta^{(3)}(\vec{x}-\vec{y}).\] we get\begin{eqnarray}\nonumber {\frac{\partial^2}{\partial x^{02}}\big\{T(\phi(x)\phi(y))\big\} }\ddot{\phi}(x)\,\phi(y)\theta(x_0 - y_0) - \phi(y)\,\ddot{\phi}(x)\theta(y_0 - x_0) - \delta^{(4)}(x-y). \end{eqnarray} Hence \begin{eqnarray}\nonumber {\big(\partial_0^2-\partial_k^2 + M^2\big)T(\phi(x)\phi(y))} &=& \big(\Box + M^2\big)\phi(x)\,\phi(y)\theta(x_0 - y_0)- \phi(y)\big(\Box + M^2\big)\phi(x)\theta(y_0- x_0) - \delta^{(4)}(x-y) - \delta^{(4)}(x-y) . \end{eqnarray} Remembering that the field satisfies Klein Gordon equation and taking vacuum expectation value, we get the desired answer \begin{equation} \big(\Box + M^2\big) \Delta_F(x) = - \delta^{(4)}(x-y) .\end{equation}

The matrix element for the decay process \(\pi \to \mu + \nu\) is given by
\[ m_{fi} =\frac{g}{\surd 2} \bar{u}^{(r)}(p) (\gamma_\mu iQ_\mu) (1-\gamma_5)v^{(s)}(k)\] for the life time computation we need to take absolute square, to average over initial spin and sum over final spin states. You may assume neutrino to be have a mass \(m_\nu\) and take limit 

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Assuming interactions of charged pions to be of the form \(\Lsc_\text{int}(x)= (g/4)(\pi(x)^+\pi(x)^-)^2\) find

• the \(S\) matrix element for \(\pi-\pi\) scattering \[\pi^+ + \pi^- \longrightarrow \pi^+ + \pi^-\]
• transition probability per unit time per unit volume for \(\pi - \pi\) scattering.
• Compute the total cross section for the scattering process and show that \[ \dd[\sigma]{\Omega}= \frac{g^2}{64\pi^2 E_\text{cm}^2}\]

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Assuming \(\frac{g}{4!} \pi^4(x)\) interaction for neutral pions of mass \(m\), find

• transition probability per unit time per unit volume for \(\pi - \pi\) scattering.
• Compute the total cross section for the scattering process and show that \[ \dd[\sigma]{\Omega}= \frac{g^2}{64\pi^2 E_\text{cm}^2}\]

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Write Dirac equation in external electromagentic potential \(A_\mu(x)\). Assume the potentials for Coulomb interactions with a nucleus of charge \(Ze\) to be \[ \vec{A}=0, \quad A_0 = \frac{Ze}{4\pi |x|}. \] Show that the differential cross section for electron scattering from nucleus is given by \[\dd[\sigma]{\Omega}= \left(\dd[\sigma]{\Omega}\right)_\text{R} \Big(1-v^2\sin^2(\theta/2)\Big)\] where \[\left(\dd[\sigma]{\Omega}\right)_\text{R} = \frac{Ze^2}{64\pi m^2v^4 \sin^4(\theta/2)}\] is the Rutherford cross section for nonrelativistic Coulomb scattering

The original four fermion interaction for beta decay of neutron \[n \longrightarrow p + e^- + \bar{\nu} \] is of the of form \[ \bar{\psi}_p(x)\gamma_\mu\psi_n(x) \bar{\psi}_\nu(x) \gamma^\mu \psi_e(x) + h.c.\] Now consider other processes given below. Which of these processes (real or virtual) are permitted and which ones are not permitted by the above interaction in the first order?

• \( \bar{p} \longrightarrow \bar{n} + e^- +\bar{\nu} \);
• \( \bar{p} \longrightarrow \bar{n} + e^- +\nu \);
• \( n \longrightarrow p + e^+ + \nu \);
• \( p \longrightarrow n + e^+ + \bar{\nu} \);
• \( \bar{n} \longrightarrow \bar{p} + e^+ + \nu \);
• \( \bar{n} \longrightarrow \bar{p} + e^+ + \bar{\nu} \).

Give brief reason in each case.