A classical particle is moving in an external potential field V(x, y, z) which is invariant under the following infinitesimal transformations
\[\begin{array}{*{20}{c}}
{x \to x'}& = &{x + \delta x} \\
{y \to y'}& = &{y + \delta y} \\
{\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right] \to \left[ {\begin{array}{*{20}{c}}
{x'} \\
{y'}
\end{array}} \right]}& = &{{R_z}\left[ {\begin{array}{*{20}{c}}
x \\
y
\end{array}} \right]}
\end{array}\]
where, Rz is the matrix corresponding to rotation about the Z-axis. The conserved quantities are (the symbols have their usual meaning)
A. px, pz, Lz
B. px, py, Lz, E
C. py, Lz, E
D. py, pz, Lx, E
Answer: Option B
This Question Belongs to Engineering Physics >> Quantum Mechanics
A. In the ground state, the probability of finding the particle in the interval $$\left( {\frac{L}{4},\,\frac{{3L}}{4}} \right)$$ is half
B. In the first excited state, the probability of finding the particle in the interval $$\left( {\frac{L}{4},\,\frac{{3L}}{4}} \right)$$ is half This also holds for states with n = 4, 6, 8, . . . .
C. For an arbitrary state $$\left| \psi \right\rangle ,$$ the probability of finding the particle in the left half of the well is half
D. In the ground state, the particle has a definite momentum
A. (e-ax1 - e-ax2)
B. a(e-ax1 - e-ax2)
C. e-ax2 (e-ax1 - e-ax2)
D. None of the above
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