Related Questions

The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$        where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$   is the angular momentum. The Hamiltonian does not commute with
In a cubic system with cell edge a, two phonons with wave vectors $${\overrightarrow {\bf{q}} _1}$$ and $${\overrightarrow {\bf{q}} _2}$$ collide and produce a third phonon with a wave. vector $${\overrightarrow {\bf{q}} _3}$$ such that $${\overrightarrow {\bf{q}} _1} + {\overrightarrow {\bf{q}} _2} = {\overrightarrow {\bf{q}} _3} + \overrightarrow {\bf{R}} $$    where, $$\overrightarrow {\bf{R}} $$ is a lattice vector. Such a collision process will lead to(a)
If for a system of N particles of different masses m1, m2, . . . mN with position vectors $${\overrightarrow {\bf{r}} _1},\,{\overrightarrow {\bf{r}} _2},\,.\,.\,.\,{\overrightarrow {\bf{r}} _N}$$    and corresponding velocities $${\overrightarrow {\bf{v}} _1},\,{\overrightarrow {\bf{v}} _2},\,.\,.\,.\,{\overrightarrow {\bf{v}} _N}$$    respectively such that $$\sum\limits_i {\overrightarrow {{{\bf{v}}_i}} = 0,} $$   then
Let $$\nabla \cdot \left( {{\text{f}}\overrightarrow {\text{v}} } \right) = {{\text{x}}^2}{\text{y}} + {{\text{y}}^2}{\text{z}} + {{\text{z}}^2}{\text{x}},$$      where f and v are scalar and vector fields respectively. If $$\overrightarrow {\text{v}} = {\text{y}}\overrightarrow {\text{i}} + {\text{z}}\overrightarrow {\text{j}} + {\text{x}}\overrightarrow {\text{k}} ,$$     then $$\overrightarrow {\text{v}} \cdot \nabla {\text{f}}$$   is
The primitive translation vectors of the body centred cubic lattice are $$\overrightarrow {\bf{a}} = \frac{a}{2}\left( {{\bf{\hat x}} + {\bf{\hat y}} - {\bf{\hat z}}} \right),\,\overrightarrow {\bf{b}} = \frac{a}{2}\left( { - {\bf{\hat x}} + {\bf{\hat y}} + {\bf{\hat z}}} \right)$$        and $$\overrightarrow {\bf{c}} = \frac{a}{2}\left( {{\bf{\hat x}} - {\bf{\hat y}} + {\bf{\hat z}}} \right)$$    . The primitive translation vectors $$\overrightarrow {\bf{A}} ,\,\overrightarrow {\bf{B}} $$  and $$\overrightarrow {\bf{C}} $$ of the reciprocal lattice are
Given $$\overrightarrow {\text{F}} = \left( {{{\text{x}}^2} - 2{\text{y}}} \right)\overrightarrow {\text{i}} - 4{\text{yz}}\overrightarrow {\text{j}} + 4{\text{x}}{{\text{z}}^2}\overrightarrow {\text{k}} ,$$       the value of the line integral $$\int\limits_{\text{c}} {\overrightarrow {\text{F}} \cdot d\overrightarrow l } $$   along the straight line c from (0, 0, 0) to (1,1,1) is
An atom with net magnetic moment $$\overrightarrow \mu $$ and net angular momentum $$\overrightarrow {\bf{L}} \left( {\overrightarrow \mu = \gamma \overrightarrow {\bf{L}} } \right)$$   is kept in a uniform magnetic induction $$\overrightarrow {\bf{B}} = {B_0}{\bf{\hat k}}.$$  The magnetic moment $$\overrightarrow \mu \left( { = {\mu _x}} \right)$$  is
For a vector potential $$\overrightarrow {\bf{A}} ,$$ the divergence of $$\overrightarrow {\bf{A}} $$ is $$\overrightarrow \nabla \cdot \overrightarrow {\bf{A}} = - \frac{{{\mu _0}}}{{4\pi }} \cdot \frac{Q}{{{r^2}}},$$    where Q is a constant of appropriate dimension. The corresponding scalar potential $$\phi \left( {r,\,t} \right)$$  that makes $$\overrightarrow {\bf{A}} $$ and $$\phi $$ Lorentz gauge invariant is
For two non-zero vectors $$\overrightarrow {\text{A}} $$ and $$\overrightarrow {\text{B}} $$, if $$\overrightarrow {\text{A}} $$ + $$\overrightarrow {\text{B}} $$ is perpendicular to $$\overrightarrow {\text{A}} $$ - $$\overrightarrow {\text{B}} $$ then,
Consider two particles with position vectors $$\overrightarrow {{{\bf{r}}_1}} $$ and $$\overrightarrow {{{\bf{r}}_2}} $$ . The force exerted by particle 2 on particle 1 is $$\overrightarrow {\bf{F}} \left( {\overrightarrow {{{\bf{r}}_1}} ,\,\overrightarrow {{{\bf{r}}_2}} } \right) = \left( {{{{\bf{\dot r}}}_2} - {{{\bf{\dot r}}}_1}} \right)\left( {{r_2} - {r_1}} \right).$$       The force is