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15

$\lvert A\rangle \langle B \rvert$ is the tensor of a ket and a bra (well, duh). This means it is an element of the tensor product of a Hilbert space $H_1$ (that's where the kets live) and of a dual of a Hilbert space $H_2^\ast$, which is where the bras live. Although for Hilbert spaces their duals are isomorphic to the original space, this distinction ...


12

The notion of tensor product is independent from the Hilbert space structure, it is defined for vector spaces on the field $\mathbb K$ (usually $\mathbb R$ or $\mathbb C$). A formal definition is given below (there are many equivalent approaches). First, if $V$ is a vector space, $V^*$ denotes its algebraic dual space, namely the vector space of the linear ...


6

Take your first relation for the 3 Pauli matrices individually: $$\sigma_1(t)=U^\dagger\sigma_1(0)U$$ $$\sigma_1(t)=U^\dagger\sigma_2(0)U$$ $$\sigma_3(t)=U^\dagger\sigma_3(0)U$$ Now you define a "vector" for notational convenience like the OP says in the question. I will choose to rewrite it as a column vector to visually show the relation of the above 3 ...


4

Radioactivity occurs (with the exception of k-capture in which the nucleus captures an orbiting electron) inside the nucleus of atoms, not in the electron shells. Yes, the product has one extra proton, and thus is now a different element, jumping up one in the Periodic Table. No nucleus is electrically neutral. Whether or not an atom is an ion has nothing to ...


3

Radioactive materials are studied in bulk matter. The electron that left with beta decay will become part of the conduction band electrons for the lattice the atom is in, and another electron will be drawn in very fast from the conduction band.. To remain an ion the atom should be isolated from bulk matter, and usually experiments are done on enough bulk ...


2

So I'm not sure if this is in the spirit of the exercise and this is probably not the most elegant solution but one way to solve your first equation would be to write it as: $\left(k\delta_{ij}+\varepsilon_{ijk}P_k\right)X_j=Q_i,$ in which case we can call the term in parentheses $M_{ij}$: $M=\left( \begin{array}{ccc} k & p_3 & -p_2 \\ -p_3 ...


2

This symbol means that there's an error of aproximation, an example is the equation you mentioned: $f(r) \sim \frac{1}{4r} + \mathcal{O}(\frac{ln(r)}{r^2})$ Here, the function $f(r)$ is being aproximated by the first term, it means that you can aproximate the actual value of the function $f(r)$ by computing the term $\frac{1}{4r}$, but by doing so, you'll ...


2

These two formulae must give the same result. I think there are some mistakes in your calculation. The result a) is correct. a) $(ia_{\mu} + b_{\mu}) = i(a_{\mu} - ib_{\mu} )$ and then it is $$ i^2(a_{\mu}- ib_{\mu} )(a^{\mu}+ ib^{\mu} ) = i^2(a^2 + ia_{\mu}b^{\mu}-ia^{\mu}b_{\mu} - ib^{2}) = i^2(a^2+b^2) $$ . But result b) is wrong. The correct result is $$ ...


2

When one writes $A=c$, where $A$ is an operator and $c$ is a number, it is implicit that the r.h.s. actually denotes $c$ times the identity.


2

It's a LaTeX typo. The author meant to write $\mathbf{r}$, i.e. '\bf{r}' where they've defined \bf as a macro for \mathbf. They forgot the slash.


2

From the paper, which states fiber Bragg gratings (FBG) have been demonstrated to exhibit temperature dependent shifts in resonant wavelength of 10 pm/K it is fairly clear that the unit is picometer per kelvin. That is, you have some device with a resonance wavelength $\lambda_\mathrm{R}$ which depends on temperature, ...


1

If $\mathbf{x}=\left(x_{1},x_{2},x_{3}\right)$ is a 3-vector rotated to $\mathbf{x}^{\prime}=\left(x_{1}^{\prime},x_{2}^{\prime},x_{3}^{\prime}\right)$ then this rotation is expressed via special unitary matrices $U \in SU\left(2\right)$ as follows : \begin{equation} \mathbf{X}^{\prime}\equiv \begin{bmatrix} ...


1

I would assume that to mean the zeroth component of the energy-momentum four-vector, for which $p^\mu p_\mu=-m^2$ is an expression of the full version of the famous mass-energy equivalence formula (plus or minus, depending on a sign convention you can choose). $p^\mu p_\mu$ is Einstein notation for, in this case, $p^\mu p_\mu=\frac{1}{c^2}E^2 ...



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