Physical Chemistry , 1st ed.

Despite the separated wavefunctions in one dimension each, it is important to understand that the operator must operate on the e ...

This product of three integrals is relatively easy to evaluate, despite its length. The xand zintegrals are exactly the same as ...

numbers themselves exchange values, the energies would be exactly the same even though the wavefunctions are different. This con ...

Here is an example of degeneracy by accident. The corresponding wavefunc- tions have no common quantum numbers, but their energy ...

 511  a 8 3 sin^ 5 a x sin 1 a y sin 1 a z  151  a 8 3 sin^ 1 a x sin 5 a y sin 1 a z  115  a 8 3 sin ...

The orthogonality and normality properties of wavefunctions are usually combined into a single expression termed orthonormality: ...

10.14 The Time-Dependent Schrödinger Equation Although the time-independent Schrödinger equation is heavily utilized in this cha ...

On the right,cancels, and the minus sign cancels i^2. Since the Hamiltonian operator does not include time, the exponential on ...

Although no particle truly exists in a box having infinite walls, the particle- in-a-box illustrates all of the important aspect ...

10.1.State the postulates of quantum mechanics introduced throughout the chapter in your own words. 10.2 Wavefunctions 10.2.What ...

10.4 Uncertainty Principle 10.14.Calculate the uncertainty in position, x, of a baseball having mass 250 g going at 1602 km/hr ...

10.31.In exercise 10.30a, the wavefunction is not normal- ized. Normalize the wavefunction and verify that it still satis- fies ...

10.57.Verify that  1  2 dx 2  1 dx0 for the 1-D particle-in-a-box, showing that the order of the wavefunctions inside th ...

11 315315 T HE PREVIOUS CHAPTER INTRODUCED the basic postulates of quan- tum mechanics, illustrated key points, and applied the ...

Two-dimensional rotational motion, which describes motion in a circu- lar path Three-dimensional rotational motion, which descr ...

absolute position of the mass at the starting time, when t0). We get this equation by solving the appropriate equations of moti ...

The expression is rearranged to solve for k, and the result is 197 N/m. There are 10^5 dynes per newton, 1000 mdyn per dyne, and ...

or  d d 2 x   2  2  m 2 E ^2 x^2  0 (11.6) where equation 11.6 has been rearranged from the previous expression to ...

Every term in equation 11.10 has the exponential ex (^2) /2 in it, so it can be al- gebraically divided out. Its residual infl ...

Mathematically, the function fhas not changed.What has changed is the index, which has shifted by 2. It is the same second deri ...