8

Perhaps the following comment is helpful: If $M=M^{\prime}\times M^{\prime\prime}$ is a product manifold then the exterior differential $\mathrm{d}=\mathrm{d}^{\prime}+\mathrm{d}^{\prime\prime}$ on $M$ is a sum of the exterior differentials on $M^{\prime}$ and $M^{\prime\prime}$. We can assign degree $(1,0)$ to $\mathrm{d}^{\prime}$ and degree $(0,1)$ to $\...


4

On one hand, $$ 0~=~[0,0]~=~[f(x),\vec{p}\cdot\vec{\nabla}f]~=~i\hbar (\vec{\nabla}f)^2.$$ On the other hand, a constraint function $f$ typically satisfies a regularity condition $$ \left .\vec{\nabla}f \right|_{f=0}~\neq~\vec{0}.$$


4

Would an example suffice? If so, consider the case $f(\vec x) = x_1$. Then (1) says $x_1=0$, which is already inconsistent with the commutation relation, and (3) says $p_1=0$, which is again inconsistent with the commutation relation. If $x_1$ or $p_1$ is zero, then we can't have $[x_1,p_1]\neq 0$.


3

OP's Lagrangian density is up to a total divergence term equal to $${\cal L}~=~\bar{\psi}(\frac{i}{2}\gamma^{\mu}\underbrace{\stackrel{\leftrightarrow}{\partial}_{\!\mu}}_{=\stackrel{\rightarrow}{\partial}_{\mu}-\stackrel{\leftarrow}{\partial}_{\mu}} - m)\psi - \frac{1}{4}F_{\mu\nu}F^{\mu\nu}-e\bar{\psi}\gamma^{\mu}A_\mu\psi,\tag{A}$$ which in turn is real. ...


3

The terms $K$ and $W$ are not kinetic and potential terms, rather $K$ is a K"ahler potential and $W$ is a superpotential. Neither term enters the Lagrangian directly, but they are used to construct it. For details see, for example, Cyril Closset's lecture notes on supersymmetry.


3

To elaborate on Qmechanic's answer to show why anticommutation in a bigraded differential algebra is natural, consider a manifold $X$ and its exterior algebra $\Omega(X)$. Suppose that there is a bigrading on $\Omega(X)$ such that $$ \Omega(X)=\bigoplus_{(r,s)\in\mathbb Z^2}\Omega^{r,s}(X), $$ where the sum is a direct sum. Suppose furthermore that when ...


2

Your calculation is correct, with one more step you would have got the book result $$-\frac{m}{2}\int dt_1\,y(t_1)\frac{d^2y(t_1)}{d t_1^2}=-\frac{m}{2}\int dt_1\,\Bigg[\frac{d}{dt_1}\Bigg(y(t_1)\frac{dy(t_1)}{d t_1}\Bigg)-\Bigg(\frac{dy(t_1)}{dt_1}\Bigg)^2\Bigg]=\\=\frac{m}{2}\int dt_1\Bigg(\frac{dy(t_1)}{dt_1}\Bigg)^2$$


1

Assuming no explicit time dependence, the Heisenberg equation of motion relates the time derivative of an operator to its commutator with the Hamiltonian as follows: $$\frac{d}{dt}\mathcal{O} = \frac{i}{\hbar} [H,\mathcal{O}]$$ From this we see that, in such cases, an operator that commutes with the Hamiltonian corresponds to a conserved quantity. The ...


1

Suppose you have an arbitrary tensor $T_{\mu\nu}$. It can be separated into the symmetric part and the antisymmetric part, i.e. $T_{\mu\nu}=T_{(\mu\nu)}+T_{[\mu\nu]}$. If you contract its indices with an antisymmetric tensor $B^{\mu\nu}$, you will have $$T_{\mu\nu}B^{\mu\nu}=T_{[\mu\nu]}B^{\mu\nu}\tag{1},$$ but this does not mean that $T_{\mu\nu}=T_{[\mu\nu]}...


1

Dividing the equation of motion (EoM) that you find by $ml^2$, we obtain $$\ddot\theta+\frac{g+\ddot y_s}{l}\sin\theta=0.$$ Compare this EoM with the one for the pendulum with the pivot fixed: $$\ddot\theta+\frac{g}{l}\sin\theta=0.$$ Renormalize the acceleration just means that now the pendulum does not feel a value $g$ of the acceleration of gravity, but ...


1

According to these lecture notes, the combination of the following Hamiltonian density and constraint gives rise to the Maxwell equations: $$\mathcal{H} = \frac{\varepsilon_0}{2}\mathbf{E}^2 + \frac{1}{2\mu_0}\mathbf{B}^2 - j_\mu A^\mu$$ and $$\nabla\cdot\mathbf{E}=\frac{\rho}{\varepsilon_0}$$


1

For each path, we can assign the following quantity known as action: $$ S[\vec{r}]= \int_{t_1}^{t_2} dt( L)$$ Where, $$ L = \text{ total kinetic energy} - \text{total potential energy}$$ The quantity 'S' is called a functional. A functional can be stated loosely as a function of functions. A concrete example of such objects being used to solve problems can ...


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