Suppose the Hamiltonian H, for a particular quantum system, is a function of some parameter $\mathit{\lambda }$let ${\mathit{E}}_{\mathbf{n}}\left(\lambda \right)$and ${\mathit{\psi }}_{\mathbf{n}}\left(\lambda \right)$be the eigen values and

Eigen functions of. The Feynman-Hellmann theorem${}^{\mathbf{22}}$states that

$\frac{\mathbf{\partial }{\mathbf{E}}_{\mathbf{n}}}{\mathbf{\partial }\mathbf{\lambda }}\mathbf{=}\left({\psi }_n\left|\frac{\partial H}{\partial \lambda }\right|{\psi }_n\right)$

(Assuming either that ${\mathit{E}}_{\mathbf{n}}$is nondegenerate, or-if degenerate-that the ${\mathit{\psi }}_{\mathbf{n}}$'s are the "good" linear combinations of the degenerate Eigen functions).

(a) Prove the Feynman-Hellmann theorem. Hint: Use Equation 6.9.

(b) Apply it to the one-dimensional harmonic oscillator,(i)using $\mathit{\lambda }\mathbf{=}\mathit{\omega }$(this yields a formula for the expectation value of V), (II)using $\mathit{\lambda }\mathbf{=}\mathit{ħ}$(this yields (T)),and (iii)using $\mathit{\lambda }\mathbf{=}\mathit{m}$(this yields a relation between (T)and (V)). Compare your answers to Problem 2.12, and the virial theorem predictions (Problem 3.31).

The Hellmann–Feynman 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 Schrödinger equation, according to the theorem.

(a)

Show the following relationship:

$\frac{\partial E_n}{\partial \lambda }={\psi }_n\left|\frac{\partial H}{\partial \lambda }\right||{\psi }_n$

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

$=\frac{\partial {\psi }_n^0}{\partial \lambda }\left|H^{\text{'}}\right|{\psi }_n^0+{\psi }_n^0\frac{\partial H^{\text{'}}}{\partial \lambda }{\psi }_n^0\hspace{1em}+{\psi }_n^0\left|H^{\text{'}}\right|\frac{\partial {\psi }_n^0}{\partial \lambda }$

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

$$\begin{gathered} \frac{\partial }{\partial \lambda }{\psi }_n^0\mid {\psi }_n^0=0 \\ \Rightarrow \hspace{1em}\frac{\partial {\psi }_n^0}{\partial \lambda }|{\psi }_n^0+{\psi }_n^0|\frac{\partial {\psi }_n^0}{\partial \lambda }=0 \end{gathered}$$

Returning to expression the following:

$$\begin{gathered} \frac{\partial E_n^1}{\partial \lambda }=E_n\frac{\partial {\psi }_n^0}{\partial \lambda }|{\psi }_n^0+{\psi }_n^0|\frac{\partial H^{\text{'}}}{\partial \lambda }|{\psi }_n^0+E_n\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\psi }_n^0|\frac{\partial {\psi }_n^0}{\partial \lambda } \\ ={\psi }_n^0\left|\frac{\partial H^{\text{'}}}{\partial \lambda }\right||{\psi }_n^0+E_n\frac{\partial {\psi }_n^0}{\partial \lambda }|{\psi }_n^0+{\psi }_n^0|\frac{\partial {\psi }_n^0}{\partial \lambda } \\ \Rightarrow E_n={\psi }_n^0|\frac{(\partial H^{\text{'}})}{\partial \lambda }|{\psi }_n^0 \end{gathered}$$

To prove that the provided equation is correct$E_n={\psi }_n^0\left|\frac{(\partial H^{\text{'}})}{\partial \lambda }\right|{\psi }_n^0$

b) Hamiltonian for 1D a harmonic oscillator is:

$$\begin{gathered} H=\frac{p^2}{2m}+\frac{m{\omega }^2{x}^2}{2}.x \\ \sqrt{\frac{ħ}{2m\omega }}{a}^{-}+a^{+}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}p=i\sqrt{\frac{m\hslash \omega }{2}}{a}^{+}-a^{-} \\ {a}^{-}\mid n>=\surd n|n-1>\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{a}^{+}|n>=\surd (n+1)\mid n+1> \end{gathered}$$

(i) $\lambda =\omega$

localid="1658214254502" $$\begin{gathered} \frac{\partial H}{\partial \omega }=m\omega x^2 \\ \frac{\partial E_n}{\partial \omega }=n\left|m\omega x^2\right|n \\ =m\omega \cdot \frac{\hslash }{2m\omega n}|a^{-}+a^{+}{a}^{-}+a^{+}|n \\ =\frac{\hslash }{2}n|a^{-}{a}^{-}+a^{-}{a}^{+}+a^{+}{a}^{-}+a^{+}{a}^{+}|n \end{gathered}$$

$$\begin{gathered} =\frac{\hslash }{2n}|a^{-}{a}^{+}+a^{+}{a}^{-}|n,\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}n\left|a^{+}{a}^{-}\right|n=n \\ =\frac{\hslash }{2n}(n+n+1)=\hslash \hspace{0.33em}\hspace{0.33em}\hspace{0.33em}n+\frac{1}{2} \\ \Rightarrow ⟨V⟩=\frac{1}{2}\left(n+\frac{1}{2}\right)\hslash \omega \end{gathered}$$

(ii) $\lambda =ħ$Rewrite Hamiltonian as:

$$\begin{gathered} H=-\frac{{\hslash }^2}{2m{\partial }^2}\frac{{\partial }^2}{\partial x^2}+\frac{m{\omega }^2{x}^2}{2}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\frac{{\partial }^2}{\partial x^2}=-p^2/{\hslash }^2 \\ \Rightarrow \overline{)\frac{\partial E_n}{\partial \hslash }=n}\hspace{0.33em}\hspace{0.33em}-\overline{)\frac{\hslash }{m}\frac{{\partial }^2}{\partial x^2}}n \\ =\frac{1}{m\hslash }n\left|p^2\right|n=-\frac{1}{m\hslash }\frac{m\hslash \omega }{2}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}n\left|\right(a^{+}-a^{-}\left)\right(a^{+}-a^{-}\left)\right|n \\ =-\frac{\omega }{2}n\left|\right(a^{+}{a}^{+}-a^{+}{a}^{-}-a^{-}{a}^{+}+a^{-}{a}^{-}\left)\right|n=\frac{\omega }{2}(2n+1) \\ \Rightarrow ⟨T⟩=\frac{1}{2}\left(n+\frac{1}{2}\right)\hslash \omega \hspace{0.33em} \end{gathered}$$

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

$$\begin{gathered} \frac{\partial H}{\partial m}=-\frac{p^2}{2m^2}+\frac{{\omega }^2{x}^2}{2}\frac{\partial E_n}{\partial m} \\ =n\left|\frac{-p^2}{2m^2}+\frac{{\omega }^2{x}^2}{2}\right|n \\ =\frac{{\omega }^2}{2}\frac{\hslash }{2m\omega }(2n+1)-\frac{1}{2m^2}\frac{m\omega \hslash }{2}(2n+1) \\ =\frac{\hslash \omega }{4m}(2n+1)-\frac{\hslash \omega }{4m}(2n+1)=0 \end{gathered}$$

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