91影视

The Hellmann鈥揊eynman theorem connects the derivative of total energy with respect to a parameter with the expectation value of the Hamiltonian's derivative with respect to the same parameter. All the forces in the system can be estimated using classical electrostatics once the spatial distribution of the electrons has been known by solving the Schr枚dinger equation, according to the theorem.

(a)

Show the following relationship:

$\frac{鈭�E_n}{鈭�位}={蠄}_n\left|\frac{鈭�H}{鈭�位}\right||{蠄}_n$

Using Equation 6.9, and get $E_n^1={蠄}_n^0\left|H^{\text{'}}\right|{蠄}_n^0$. Inserting this in the first expression, and get

$=\frac{鈭�{蠄}_n^0}{鈭�位}\left|H^{\text{'}}\right|{蠄}_n^0+{蠄}_n^0\frac{鈭�H^{\text{'}}}{鈭�位}{蠄}_n^0鈥�+{蠄}_n^0\left|H^{\text{'}}\right|\frac{鈭�{蠄}_n^0}{鈭�位}$

But, that,$H^{\text{'}}|{蠄}_n^0>=E_n|{蠄}_n^0>$and ${蠄}_n^0鈭�{蠄}_n^0=1$It follows:

$$\begin{gathered} \frac{鈭�}{鈭�位}{蠄}_n^0鈭�{蠄}_n^0=0 \\ 鈬�鈥�\frac{鈭�{蠄}_n^0}{鈭�位}|{蠄}_n^0+{蠄}_n^0|\frac{鈭�{蠄}_n^0}{鈭�位}=0 \end{gathered}$$

Returning to expression the following:

$$\begin{gathered} \frac{鈭�E_n^1}{鈭�位}=E_n\frac{鈭�{蠄}_n^0}{鈭�位}|{蠄}_n^0+{蠄}_n^0|\frac{鈭�H^{\text{'}}}{鈭�位}|{蠄}_n^0+E_n鈥�鈥�鈥�{蠄}_n^0|\frac{鈭�{蠄}_n^0}{鈭�位} \\ ={蠄}_n^0\left|\frac{鈭�H^{\text{'}}}{鈭�位}\right||{蠄}_n^0+E_n\frac{鈭�{蠄}_n^0}{鈭�位}|{蠄}_n^0+{蠄}_n^0|\frac{鈭�{蠄}_n^0}{鈭�位} \\ 鈬�E_n={蠄}_n^0|\frac{(鈭�H^{\text{'}})}{鈭�位}|{蠄}_n^0 \end{gathered}$$

To prove that the provided equation is correct$E_n={蠄}_n^0\left|\frac{(鈭�H^{\text{'}})}{鈭�位}\right|{蠄}_n^0$

b) Hamiltonian for 1D a harmonic oscillator is:

$$\begin{gathered} H=\frac{p^2}{2m}+\frac{m{蝇}^2{x}^2}{2}.x \\ \sqrt{\frac{魔}{2m蝇}}{a}^{-}+a^{+}鈥�鈥�鈥�p=i\sqrt{\frac{m鈩�蝇}{2}}{a}^{+}-a^{-} \\ {a}^{-}鈭�n>=鈭�n|n-1>鈥�鈥�鈥�a^{+}|n>=鈭�(n+1)鈭�n+1> \end{gathered}$$

(i) $位=蝇$

localid="1658214254502" $$\begin{gathered} \frac{鈭�H}{鈭�蝇}=m蝇x^2 \\ \frac{鈭�E_n}{鈭�蝇}=n\left|m蝇x^2\right|n \\ =m蝇鈰�\frac{鈩�}{2m蝇n}|a^{-}+a^{+}{a}^{-}+a^{+}|n \\ =\frac{鈩�}{2}n|a^{-}{a}^{-}+a^{-}{a}^{+}+a^{+}{a}^{-}+a^{+}{a}^{+}|n \end{gathered}$$

$$\begin{gathered} =\frac{鈩�}{2n}|a^{-}{a}^{+}+a^{+}{a}^{-}|n,鈥�鈥�鈥�n\left|a^{+}{a}^{-}\right|n=n \\ =\frac{鈩�}{2n}(n+n+1)=鈩�鈥�鈥�鈥�n+\frac{1}{2} \\ 鈬�鉄�V鉄�=\frac{1}{2}\left(n+\frac{1}{2}\right)鈩�蝇 \end{gathered}$$

(ii) $位=魔$Rewrite Hamiltonian as:

$$\begin{gathered} H=-\frac{{鈩�}^2}{2m{鈭�}^2}\frac{{鈭�}^2}{鈭�x^2}+\frac{m{蝇}^2{x}^2}{2}鈥�鈥�鈥�\frac{{鈭�}^2}{鈭�x^2}=-p^2/{鈩�}^2 \\ 鈬�\overline{)\frac{鈭�E_n}{鈭�鈩�}=n}鈥�鈥�-\overline{)\frac{鈩�}{m}\frac{{鈭�}^2}{鈭�x^2}}n \\ =\frac{1}{m鈩�}n\left|p^2\right|n=-\frac{1}{m鈩�}\frac{m鈩�蝇}{2}鈥�鈥�鈥�n\left|\right(a^{+}-a^{-}\left)\right(a^{+}-a^{-}\left)\right|n \\ =-\frac{蝇}{2}n\left|\right(a^{+}{a}^{+}-a^{+}{a}^{-}-a^{-}{a}^{+}+a^{-}{a}^{-}\left)\right|n=\frac{蝇}{2}(2n+1) \\ 鈬�鉄�T鉄�=\frac{1}{2}\left(n+\frac{1}{2}\right)鈩�蝇鈥� \end{gathered}$$

(iii) $位=m$Hamiltonian is:$H=\frac{p^2}{2m}+\frac{m{蝇}^2{x}^2}{2}$It follows:

$$\begin{gathered} \frac{鈭�H}{鈭�m}=-\frac{p^2}{2m^2}+\frac{{蝇}^2{x}^2}{2}\frac{鈭�E_n}{鈭�m} \\ =n\left|\frac{-p^2}{2m^2}+\frac{{蝇}^2{x}^2}{2}\right|n \\ =\frac{{蝇}^2}{2}\frac{鈩�}{2m蝇}(2n+1)-\frac{1}{2m^2}\frac{m蝇鈩�}{2}(2n+1) \\ =\frac{鈩�蝇}{4m}(2n+1)-\frac{鈩�蝇}{4m}(2n+1)=0 \end{gathered}$$

Hamiltonian is$\left(T\right)=\left(V\right)$$\left(T\right)=\left(V\right)$