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Q14CQ

Expert-verifiedFound in: Page 278

Book edition
2nd Edition

Author(s)
Randy Harris

Pages
633 pages

ISBN
9780805303087

**At heart, momentum conservation is related to the universe being "translationally invariant," meaning that it is the same if you shift your coordinates to the right or left. Angular momentum relates to rotational invariance. Use these ideas to explain at least some of the differences between the physical properties quantized in the cubic three-dimensional box versus the hydrogen atom.**

The use of the Cartesian coordinates is appropriate for the symmetry of the three-dimensional cube and it leads to the quantization of the linear momentum of the particles trapped which is in an analogous way. In case of the hydrogen atom the symmetry is also appropriate spherical coordinates to boundary condition leads to quantization of the angular momentum other than the linear momentum.

The case when the particle is trapped in the three-dimensional cube the three-dimensional coordinate is used to describe the symmetry of the system. The linear momentum of the system is in the direction of the symmetry of the system and is quantized by the boundary condition.

The energy level of the hydrogen atoms is labelled by the number *n* which is the largest value of the angular momentum for a given energy. The allowed energy in the three-dimensional plane is named as quantum numbers $({n}_{x},{n}_{y},{n}_{z})$

Consider the case when the Cartesian axis are normal to the face of the cube and are formed by the edge of the potential well. The equation for the Schrödinger equation before the boundary condition are applied is given by,

$\Psi \left(x\right)=A\mathrm{sin}\left({k}_{x}x\right)\mathrm{sin}\left({k}_{y}x\right)\mathrm{sin}\left({k}_{z}x\right)$

The expression for the time dependent Schrödinger equation inside the region of the well where the potential U is taken as zero and the equation is given by,

$\frac{-{h}^{2}{\nabla}^{2}}{2m}\psi =E\psi $

Solve further as

$\begin{array}{rcl}\frac{-{h}^{2}{\nabla}^{2}}{2m}x\left(A\mathrm{sin}\left({k}_{x}x\right)\mathrm{sin}\left({k}_{y}x\right)\mathrm{sin}\left({k}_{z}x\right)\right)& =& EA\mathrm{sin}\left({k}_{x}x\right)\mathrm{sin}\left({k}_{y}x\right)\mathrm{sin}\left({k}_{z}x\right)\\ \frac{{h}^{2}\left({k}_{x}^{2}+{k}_{y}^{2}+{k}_{z}^{2}\right)}{2m}\psi & =& E\psi \\ E& =& \frac{{p}^{2}}{2m}\\ & & \end{array}$

The quantization of the boundary condition of the wave function is zero at the barriers allow only the specific of the three-wave number $({k}_{x},{k}_{y},{k}_{z})$. The expression for the linear momentum is given by,

$\begin{array}{rcl}p& =& hk\\ \u200a\u200a\u200a\u200a& =& h({k}_{x},{k}_{y},{k}_{z})\\ \u200a\u200a\u200a\u200a\u200a& =& h\left(\frac{\pi h{n}_{x}}{L},\frac{\pi h{n}_{y}}{L},\frac{\pi h{n}_{y}}{L}\right)\\ & & \end{array}$

In case of the hydrogen atom, the spherical coordinate fit the symmetry of the system and the angular momentum replace the linear momentum. The wave function that vanishes at the barrier but approaches to zero with large radial distances. This is sufficient to determine the energy level. The remaining quantum numbers are determined by the boundary conditions on the angular variables, thus when the complete rotation occurs through $2\pi $ it returns wave function to its original value. The quantum number results label the possible values of the magnitude of angular momentum which is consistent with the uncertainty principle.

Therefore, the use of the cartesian coordinates is appropriate for the symmetry of the three-dimensional cube and it leads to the quantization of the linear momentum of the particles trapped which is in an analogous way. In case of the hydrogen atom the symmetry is also appropriate spherical coordinates to boundary condition leads to quantization of the angular momentum other than the linear momentum.

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