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4

As Kyle says, $\nu$ is just a (free) index. You can use any letter. More precisely, $x^\mu$ is the $\mu$-component of of the vector $\mathbf{x}=(x_1,x_2,\dots,x_n)$. And $x^\nu$ is the $\nu$-component of of the vector $\mathbf{x}=(x_1,x_2,\dots,x_n)$. So you can see that the vector is the same, $\mathbf{x}$, you just name the components with a different ...

5

Refer to the nice complement on coherent states in the book by Cohen-Tannoudji, Diu and Laloë, volume 1. It starts off defining coherent states as neither of the ones you mention, and then derives all properties. To answer the question, if you start with definition 2, you can easily show 1, and then from 2, 3. First expand the exponential using ...

1

No, not really. A gradient is the derivative of a scalar. It is not actually a vector, but a dual vector or 1-form. http://en.wikipedia.org/wiki/Gradient Vectors and 1-forms have different transformation properties, and used to be called contra-variant and co-variant vectors, but the language of exterior calculus makes this much cleaner. Intuitively, a ...

5

It's purely notation. Given a real-valued function $f(\mathbf r) = f(x^1, \dots, x^n)$ of $n$ real variables, one defines the derivative with respect to $\mathbf r$ as follows: \begin{align} \frac{\partial f}{\partial \mathbf r}(\mathbf r) = \left(\frac{\partial f}{\partial x^1}(\mathbf r), \dots, \frac{\partial f}{\partial x^n}(\mathbf r)\right) ...

2

The answer is no. Just as in the case without a gauge field, it is just a product of two derivatives of the field $\phi$. You might be interested in the chapter "Scalar Electrodynamics" in Srednicki's book.

3

First equation refers to the passive view of coordinate transformations while the second is the active view. Let $M$ be a manifold with metric $g$ which in local coordinates $x$ is written as $g_{\alpha\beta}(x)dx^{\alpha}\otimes dx^{\beta}$. Let $\phi:M\rightarrow M$ be a differentiable function and let $g'=\phi^*g$ be the pullback of the metric $g$, ...

3

Let's look at an example. Let's consider $0+1$ dimensions. Our manifold will be $M=\mathbb{R}$ and the coordinate system we will use will map the coordinate $x \in \mathbb{R}$ to the point $p \in M$ according to the rule $p(x^a) = x^0$. Now suppose the metric in this coordinate system has coordinates $g_{00}=x^0$ in this coordinate system. Now we would ...

4

The books are correct. The statement is a definite relation that is being imposed between the 'old' metric structure and the transformed one, for the transformation to be conformal. The equation you're unhappy about, $$g_{\mu \nu}'(x') = \Omega(x) g_{\mu \nu}(x)$$ states that the transformed metric $g_{\mu \nu}'$ at the transformed point $x'$ can be ...

2

When you have a matrix $\Phi = \begin{pmatrix} \phi_1\\ \phi_2\end{pmatrix}$, with one column and two rows, and its transpose matrix $\Phi^T = \begin{pmatrix} \phi_1 & \phi_2\end{pmatrix}$, with one row and two columns, the product of the two matrix $\Phi^T \Phi$ is a matrix $P$ with one column and one row: $P =\Phi^T \Phi = \begin{pmatrix} \phi_1 ... 5 There is no significance in the choice between upper- and lower-case$\psi$(or$\Psi\$) to denote a system's wavefunction. The two are used interchangeably and it is the author's discretion to use either symbol. (On the other hand, of course, one shouldn't use the two symbols interchangeably within the same text; if both are used they would refer to ...

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