Question

From the diagrams in Figure 7.15and the qualitative behavior of the wave functions they represent, argue that a combination of the$\mathbf{2}{\mathbf{p}}_{\mathbf{Z}}$(i.e.,${\mathbf{m}}_{\mathbf{l}}\mathbf{=}\mathbf{0}$) and the negative of the 2swould produce a function that sticks out preferentially in the positive Zdirection. This is known as a hybridspstate.

Short answer

The answer is given below.

Step-by-step solution

Given data

In this problem, our task is to calculate how high the temperature must be so that the probability of a system of distinguishable harmonic oscillators to occupy the ground state is less than $p=\frac{1}{2}$.

Concept of the energies of a harmonic oscillator

Similar to the notation used in the book, we also assume that the potential energy is shifted by$\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathit{ħ}{\mathit{\omega }}_{\mathbf{0}}$so that the allowed energies of a harmonic oscillator is given by:

role="math" localid="1659762682190" ${\mathbf{E}}_{\mathbf{n}}\mathbf{=}\mathbf{n}\mathit{ħ}{\mathit{\omega }}_{\mathbf{0}}\mathbf{}\mathbf{where}\mathbf{}\mathbf{n}\mathbf{\ge }\mathbf{0}\mathbf{,}$

and its ground state energy is${\mathbf{E}}_{\mathbf{0}}\mathbf{=}\mathbf{0}$.

Argument on the given function 2pz 

The key idea here lies from the fact that we are now dealing with an enormous amount of distinguishable harmonic oscillators, that is to say, the probability at which an oscillator is in an energy level $n$ is now given by the Boltzmann probability distribution:

$\mathrm{P}\left({\mathrm{E}}_{\mathrm{n}}\right)={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{n}}/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}$

Where ${\mathrm{k}}_{\mathrm{B}}$Boltzmann is’s constant, and is temperature.

Before we could utilize eq. (2), we first have to find the value of the normalization constant A . To do so, we plug in eq. (1) into (2), set the whole expression to l, and evaluate the integral over an interval dn from 0 to $\infty$. This translates to:

$1={\int }_0^{\infty }{\mathrm{Ae}}^{-{\mathrm{nħ\omega }}_0/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\mathrm{dn}$

Evaluating the above integral, and solving for A we obtain:

role="math" localid="1659763564607" $$\begin{gathered} 1=\mathrm{A}{\int }_0^{\infty }{\mathrm{e}}^{-{\mathrm{nħ\omega }}_0/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}} \\ =\mathrm{A}{\left[-\frac{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{{\mathrm{ħ\omega }}_0}{\mathrm{e}}^{-{\mathrm{nħ\omega }}_0/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\right]}_0^{\infty } \\ =\mathrm{A}\left[-\frac{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{{\mathrm{ħ\omega }}_0}\left(0-1\right)\right] \\ =\mathrm{A}\frac{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{{\mathrm{ħ\omega }}_0}\to \mathrm{A}=\frac{{\mathrm{ħ\omega }}_0}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}} \\ \mathrm{Knowing},\mathrm{eq}.\left(2\right)\mathrm{now}\mathrm{becomes}: \\ \mathrm{P}\left(\mathrm{n}\right)=\frac{{\mathrm{ħ\omega }}_0}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{e}}^{-{\mathrm{nħ\omega }}_0/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}} \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{temperature}\mathrm{T}\mathrm{where}\mathrm{the}\mathrm{probability}\mathrm{in}\mathrm{the}\mathrm{ground}\mathrm{state}\mathrm{is}\mathrm{less}\mathrm{than}1/2, \\ \mathrm{we}\mathrm{set}\mathrm{n}=0\mathrm{in}\mathrm{eq}.\left(3\right)\mathrm{and}\mathrm{transform}\mathrm{the}\mathrm{expression}\mathrm{into}\mathrm{an}\mathrm{inequality}: \\ \frac{1}{2}>\frac{{\mathrm{ħ\omega }}_0}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}{\mathrm{e}}^{-{\mathrm{nħ\omega }}_0/{\mathrm{k}}_{\mathrm{B}}\mathrm{T}} \\ \mathrm{Solving}\mathrm{for}\mathrm{T}\mathrm{we}\mathrm{finally}\mathrm{get}: \\ \frac{1}{2}>\frac{{\mathrm{ħ\omega }}_0}{{\mathrm{k}}_{\mathrm{B}}\mathrm{T}}\to \frac{2{\mathrm{ħ\omega }}_0}{{\mathrm{k}}_{\mathrm{B}}} \end{gathered}$$