Chapter 7, Problem 7.54P

(a)

To determine

The discussion based on Equation 7.118 when $\Omega =H\text{'}$ and explain why this is consistent with Equation 7.15.

Answer to Problem 7.54P

Form Equation 7.118 when $\Omega =H\text{'}$, $\langle H\text{'}\rangle =2E_n^0$ and this is consistent with Equation 7.15 as the second-order correction of the $n^{th}$ energy is $E_n^0$.

Explanation of Solution

From Equation 7.118,

$\langle H\text{'}\rangle =2\mathrm{Re}\frac{{\left|\langle {\psi }_m^0|H\text{'}|{\psi }_n^0\rangle \right|}^2}{E_n^0-E_m^0}$

The sum of the manifestly real and then the above equation is precisely Equation 7.15.

Write Equation 7.15,

$E_n^2={\displaystyle \sum _{m\ne n}\frac{{\left|\langle {\psi }_m^0|H\text{'}|{\psi }_n^0\rangle \right|}^2}{E_n^0-E_m^0}}$

Therefore,

$\begin{array}{c}\langle H\text{'}\rangle =2{\displaystyle \sum _{m\ne n}\frac{{\left|\langle {\psi }_m^0|H\text{'}|{\psi }_n^0\rangle \right|}^2}{E_n^0-E_m^0}}\\ =2E_n^0\end{array}$

The second-order correction to the energy is the term of order ${\lambda }^2$ in

$\frac{\langle {\psi }_n|H_0+\lambda H\text{'}|{\psi }_n\rangle }{\langle {\psi }_n|{\psi }_n\rangle }=\frac{\langle {\psi }_n|H_0|{\psi }_n\rangle }{\langle {\psi }_n|{\psi }_n\rangle }+\lambda \frac{\langle {\psi }_n|H\text{'}|{\psi }_n\rangle }{\langle {\psi }_n|{\psi }_n\rangle }$        (I)

Where, ${\psi }_n$ is the exact $n^{th}$ eigenstate.

Solving the terms separately from equation (I),

$\begin{array}{c}\langle {\psi }_n|H_0|{\psi }_n\rangle =\langle \left({\psi }_n^0+\lambda {\psi }_0^1+{\lambda }^2{\psi }_n^2+\mathrm{...}\right)|H_0|\left({\psi }_n^0+\lambda {\psi }_0^1+{\lambda }^2{\psi }_n^2+\mathrm{...}\right)\rangle \\ =\{\begin{array}{l}\langle {\psi }_n^0|H_0|{\psi }_n^0\rangle +\lambda \left[\langle {\psi }_n^0|H_0|{\psi }_n^1\rangle +\langle {\psi }_n^1|H_0|{\psi }_n^0\rangle \right]\\ +{\lambda }^2\left[\langle {\psi }_n^0|H_0|{\psi }_n^2\rangle +\langle {\psi }_n^1|H_0|{\psi }_n^1\rangle +\langle {\psi }_n^2|H_0|{\psi }_n^0\rangle \right]+\mathrm{...}\end{array}\\ \approx \{\begin{array}{l}{E}_n^0+\lambda \left[E_n^0\left(\langle {\psi }_n^0|{\psi }_n^1\rangle +\langle {\psi }_n^1|{\psi }_n^0\rangle \right)\right]\\ +{\lambda }^2\left[E_n^0\left(\langle {\psi }_n^0|{\psi }_n^2\rangle +\langle {\psi }_n^2|{\psi }_n^0\rangle \right)+\langle {\psi }_n^1|H_0|{\psi }_n^1\rangle \right]\end{array}\\ =E_n^0+{\lambda }^2\langle {\psi }_n^1|H_0|{\psi }_n^1\rangle \end{array}$        (II)

Solving for $\langle {\psi }_n|{\psi }_n\rangle$,

$\begin{array}{c}\langle {\psi }_n|{\psi }_n\rangle =\{\begin{array}{l}\langle {\psi }_n^0|{\psi }_n^0\rangle +\lambda \left[\langle {\psi }_n^0|{\psi }_n^1\rangle +\langle {\psi }_n^1|{\psi }_n^0\rangle \right]\\ +{\lambda }^2\left[\langle {\psi }_n^0|{\psi }_n^2\rangle +\langle {\psi }_n^1|{\psi }_n^1\rangle +\langle {\psi }_n^2|{\psi }_n^0\rangle \right]+\cdots \end{array}\\ =1+{\lambda }^2\langle {\psi }_n^1|{\psi }_n^1\rangle \end{array}$        (III)

Divide equation (III) in (II) to solve for $\frac{\langle {\psi }_n|H_0|{\psi }_n\rangle }{\langle {\psi }_n|{\psi }_n\rangle }$

$\begin{array}{c}\frac{\langle {\psi }_n|H_0|{\psi }_n\rangle }{\langle {\psi }_n|{\psi }_n\rangle }=\frac{E_n^0+{\lambda }^2\langle {\psi }_n^1|H_0|{\psi }_n^1\rangle }{1+{\lambda }^2\langle {\psi }_n^1|{\psi }_n^1\rangle }\\ =\left[E_n^0+{\lambda }^2\langle {\psi }_n^1|H_0|{\psi }_n^1\rangle \right]{\left[1+{\lambda }^2\langle {\psi }_n^1|{\psi }_n^1\rangle \right]}^{-1}\\ \approx \left[E_n^0+{\lambda }^2\langle {\psi }_n^1|H_0|{\psi }_n^1\rangle \right]\left[1-{\lambda }^2\langle {\psi }_n^1|{\psi }_n^1\rangle \right]\\ =\left[E_n^0+{\lambda }^2\langle {\psi }_n^1|H_0|{\psi }_n^1\rangle -E_n^2\langle {\psi }_n^1|{\psi }_n^1\rangle \right]\end{array}$

Solving further,

$\frac{\langle {\psi }_n|H_0|{\psi }_n\rangle }{\langle {\psi }_n|{\psi }_n\rangle }=\left[E_n^0+{\lambda }^2\langle {\psi }_n^1|\left(H_0-E_n^0\right)|{\psi }_n^1\rangle \right]$

Using Equation 7.13,

$\begin{array}{c}\langle {\psi }_n^1|\left(H_0-E_n^0\right)|{\psi }_n^1\rangle =\langle {\psi }_n^1|\left(H_0-E_n^0\right){\displaystyle \sum _{m\ne n}\frac{H_{mn}^{\text{'}}}{\left(E_n^0-E_m^0\right)}}|{\psi }_n^1\rangle \\ =-\langle {\psi }_n^1|{\displaystyle \sum _{m\ne n}{H}_{mn}^{\text{'}}}|{\psi }_m^0\rangle \\ =-{\displaystyle \sum _{p\ne n}{\displaystyle \sum _{m\ne n}\frac{H_{pn}^{\text{'}}*}{E_n^0-E_p^0}}}{H}_{mn}^{\text{'}}\langle {\psi }_p^0|{\psi }_m^0\rangle \\ =-{\displaystyle \sum _{m\ne n}\frac{{\left|H_{mn}^{\text{'}}\right|}^2}{\left(E_n^0-E_m^0\right)}}\end{array}$

Therefore,

$\langle {\psi }_n^1|\left(H_0-E_n^0\right)|{\psi }_n^1\rangle =-E_n^2$

Substituting in Equation (II), the second-order correction of the $n^{th}$ energy is

$-E_n^2+2E_n^2=E_n^2$

Conclusion:

Form Equation 7.118 when $\Omega =H\text{'}$, $\langle H\text{'}\rangle =2E_n^0$ and this is consistent with Equation 7.15 as the second-order correction of the $n^{th}$ energy is $E_n^0$.

(b)

To determine

Show that $\alpha =-2q^2{\displaystyle \sum _{m\ne n}\frac{{\left|\langle {\psi }_m^0|H\text{'}|{\psi }_n^0\rangle \right|}^2}{E_n^0-E_m^0}}$ and determine the polarizability of the ground state of a one-dimensional harmonic oscillator. Compare the result to the classical answer.

Answer to Problem 7.54P

It has been prove that $\alpha =-2q^2{\displaystyle \sum _{m\ne n}\frac{{\left|\langle {\psi }_m^0|H\text{'}|{\psi }_n^0\rangle \right|}^2}{E_n^0-E_m^0}}$ and the polarizability of the ground state of a one-dimensional harmonic oscillator is $\alpha =\frac{q^2}{m{\omega }^2}$. The result is same as the classical answer.

Explanation of Solution

The first-order expectation value of $p_e$ in the $n^{th}$ state is

$\begin{array}{c}\langle p_e\rangle =2\mathrm{Re}{\displaystyle \sum _{m\ne n}\frac{\langle {\psi }_n^0|qx|{\psi }_m^0\rangle \langle {\psi }_n^0|-q{E}_{ext}x|{\psi }_m^0\rangle }{E_n^0-E_m^0}}\\ =-2q^2{E}_{ext}{\displaystyle \sum _{m\ne n}\frac{{\left|\langle {\psi }_n^0|x|{\psi }_m^0\rangle \right|}^2}{E_n^0-E_m^0}}\end{array}$

Given, the expectation value of $p_e$ is proportional to the applied field and the proportionality factor called polarizabiltiy $\alpha$.

Therefore,

$\alpha =-2q^2{\displaystyle \sum _{m\ne n}\frac{{\left|\langle {\psi }_n^0|x|{\psi }_m^0\rangle \right|}^2}{E_n^0-E_m^0}}$        (IV)

Hence proved.

For one-dimensional oscillator, Equation 3.114

$\langle {\psi }_n^0|x|{\psi }_m^0\rangle =\sqrt{\frac{\hslash }{2m\omega }}\left(\sqrt{m}{\delta }_{n,m-1}+\sqrt{n}{\delta }_{n,m-1}\right)$

Solving for $\langle {\psi }_0^0|x|{\psi }_m^0\rangle$ using above equation,

$\begin{array}{c}\langle {\psi }_0^0|x|{\psi }_m^0\rangle =\sqrt{\frac{\hslash }{2m\omega }}\sqrt{m}{\delta }_{n,m-1}\\ =\sqrt{\frac{\hslash }{2\omega }}{\delta }_{n,m-1}\end{array}$

Substitute the above equation in equation (IV)

$\begin{array}{c}\alpha =-2q^2\frac{\hslash /2m\omega }{\left(\hslash \omega /2\right)-\left(3\hslash \omega /2\right)}\\ =\frac{q^2}{m{\omega }^2}\end{array}$

The classical answer for the polarizability is also $\alpha =\frac{q^2}{m{\omega }^2}$. Equating electric force $q{E}_{ext}$ and the spring for $kx$, $p_e=qx=q^2{E}_{ext}/k$. Therefore, $\alpha =q^2/k=q^2/m{\omega }^2$.

Conclusion:

It has been prove that $\alpha =-2q^2{\displaystyle \sum _{m\ne n}\frac{{\left|\langle {\psi }_m^0|H\text{'}|{\psi }_n^0\rangle \right|}^2}{E_n^0-E_m^0}}$ and the polarizability of the ground state of a one-dimensional harmonic oscillator is $\alpha =\frac{q^2}{m{\omega }^2}$. The result is same as the classical answer.

(c)

To determine

The expectation value of $\langle x\rangle$ to the first order, in the $n^{th}$ energy eigenstate.

Answer to Problem 7.54P

The expectation value of $\langle x\rangle$ to the first order, in the $n^{th}$ energy eigenstate is ${\langle x\rangle }^1=\left(2n+1\right)\frac{\hslash \kappa }{4m^2{\omega }^3}$.

Explanation of Solution

The first order expectation value of $\langle x\rangle$ is

$\begin{array}{c}{\langle x\rangle }^1=2\mathrm{Re}{\displaystyle \sum _{m\ne n}\frac{\langle {\psi }_n^0|x|{\psi }_m^0\rangle \langle {\psi }_m^0|\left(-\frac{1}{6}\kappa x^3\right)|{\psi }_n^0\rangle }{\left(n+\frac{1}{2}\right)\hslash \omega -\left(m+\frac{1}{2}\right)\hslash \omega }}\\ =-\frac{\kappa }{3\hslash \omega }\mathrm{Re}{\displaystyle \sum _{m\ne n}\frac{\langle {\psi }_n^0|x|{\psi }_m^0\rangle \langle {\psi }_m^0|x^3|{\psi }_n^0\rangle }{n-m}}\\ =-\frac{\kappa }{3\hslash \omega }\sqrt{\frac{\hslash }{2m\omega }}\mathrm{Re}{\displaystyle \sum _{m\ne n}\frac{\left(\sqrt{m}{\delta }_{n,m-1}+\sqrt{n}{\delta }_{n,m-1}\right)}{n-m}\langle {\psi }_m^0|x^3|{\psi }_n^0\rangle }\\ =-\frac{\kappa }{3\hslash \omega }\sqrt{\frac{\hslash }{2m\omega }}\mathrm{Re}\left[-\sqrt{n+1}\langle {\psi }_{n+1}^0|x^3|{\psi }_n^0\rangle +\sqrt{n}\langle {\psi }_{n-1}^0|x^3|{\psi }_n^0\rangle \right]\end{array}$        (V)

Solving the terms in the above equation separately for simplicity, using Equation 2.70,

$\begin{array}{c}{x}^3={\left(\frac{\hslash }{2m\omega }\right)}^{3/2}{\left(a_{+}+a_{-}\right)}^3\\ ={\left(\frac{\hslash }{2m\omega }\right)}^{3/2}\left(a_{+}^3+a_{+}^2{a}_{-}+a_{+}{a}_{-}{a}_{+}+a_{+}{a}_{-}^2+a_{-}{a}_{+}^2+a_{-}{a}_{+}{a}_{-}+a_{-}^2{a}_{+}+a_{-}^3\right)\end{array}$

Therefore,

$\langle n+1|x^3|n\rangle ={\left(\frac{\hslash }{2m\omega }\right)}^{3/2}\langle n+1|\left(a_{+}^2{a}_{-}+a_{+}{a}_{-}{a}_{+}+a_{-}{a}_{+}^2\right)|n\rangle$        (VI)

And,

$\langle n-1|x^3|n\rangle ={\left(\frac{\hslash }{2m\omega }\right)}^{3/2}\langle n-1|\left(a_{+}{a}_{-}^2+a_{-}{a}_{+}{a}_{-}+a_{-}^2{a}_{+}\right)|n\rangle$        (VII)

Solving the terms inside the bracket in equation (VI) by using Equation 2.67 repeatedly,

$\begin{array}{c}{a}_{+}^2{a}_{-}|n\rangle =\sqrt{n}{a}_{+}^2|n-1\rangle \\ =\sqrt{n}\sqrt{n}{a}_{+}|n\rangle \\ =n\sqrt{n+1}|n+1\rangle \end{array}$,

$\begin{array}{c}{a}_{+}{a}_{-}{a}_{+}|n\rangle =\sqrt{n+1}{a}_{+}{a}_{-}|n+1\rangle \\ =\sqrt{n+1}\sqrt{n+1}{a}_{+}|n\rangle \\ =\left(n+1\right)\sqrt{n+1}|n+1\rangle \end{array}$,

And

$\begin{array}{c}{a}_{-}{a}_{+}^2|n\rangle =\sqrt{n+1}{a}_{-}{a}_{+}|n+1\rangle \\ =\sqrt{n+1}\sqrt{n+2}{a}_{-}|n+2\rangle \\ =\left(n+2\right)\sqrt{n+1}|n+1\rangle \end{array}$

Therefore, $\langle n+1|\left(a_{+}^2{a}_{-}+a_{+}{a}_{-}{a}_{+}+a_{-}{a}_{+}^2\right)|n\rangle$ in equation (VI) becomes,

$\begin{array}{c}\langle n+1|\left(a_{+}^2{a}_{-}+a_{+}{a}_{-}{a}_{+}+a_{-}{a}_{+}^2\right)|n\rangle =\langle n+1|\left(n\sqrt{n+1}+\left(n+1\right)\sqrt{n+1}+\left(n+2\right)\sqrt{n+1}\right)|n+1\rangle \\ =3\left(n+1\right)\sqrt{n+1}\end{array}$

Similarly solving for equation (VII),

$\begin{array}{c}{a}_{+}{a}_{-}^2|n\rangle =\sqrt{n}{a}_{+}{a}_{-}|n-1\rangle \\ =\sqrt{n}\sqrt{n+1}{a}_{+}|n-2\rangle \\ =\left(n-1\right)\sqrt{n}|n-1\rangle \end{array}$,

$\begin{array}{c}{a}_{-}{a}_{+}{a}_{-}|n\rangle =\sqrt{n}{a}_{-}{a}_{+}|n-1\rangle \\ =\sqrt{n}\sqrt{n}{a}_{-}|n\rangle \\ =n\sqrt{n}|n-1\rangle \end{array}$,

And

$\begin{array}{c}{a}_{-}^2{a}_{+}|n\rangle =\sqrt{n+1}{a}_{-}^2|n+1\rangle \\ =\sqrt{n+1}\sqrt{n+1}{a}_{-}|n\rangle \\ =\left(n+1\right)\sqrt{n}|n-1\rangle \end{array}$

Therefore, $\langle n-1|\left(a_{+}{a}_{-}^2+a_{-}{a}_{+}{a}_{-}+a_{-}^2{a}_{+}\right)|n\rangle$ in equation (VII) becomes,

$\begin{array}{c}\langle n-1|\left(a_{+}{a}_{-}^2+a_{-}{a}_{+}{a}_{-}+a_{-}^2{a}_{+}\right)|n\rangle =\langle n-1|\left(\left(n-1\right)\sqrt{n}+n\sqrt{n}+\left(n+1\right)\sqrt{n}\right)|n-1\rangle \\ =3n\sqrt{n}\end{array}$

Therefore, equation (VI) and (VII) becomes,

$\langle n+1|x^3|n\rangle ={\left(\frac{\hslash }{2m\omega }\right)}^{3/2}3\left(n+1\right)\sqrt{n+1}$

And,

$\langle n-1|x^3|n\rangle ={\left(\frac{\hslash }{2m\omega }\right)}^{3/2}3n\sqrt{n}$

Substitute the above equations in equation (V)

$\begin{array}{c}{\langle x\rangle }^1=-\frac{\kappa }{3\hslash \omega }{\left(\frac{\hslash }{2m\omega }\right)}^2\left[-\sqrt{n+1}\left(3\right)\left(n+1\right)\sqrt{n+1}+\sqrt{n}\left(3n\right)\sqrt{n}\right]\\ =-\frac{\kappa }{3\hslash \omega }{\left(\frac{\hslash }{2m\omega }\right)}^2\left[-{\left(n+1\right)}^2+n^2\right]\\ =\left(2n+1\right)\frac{\hslash \kappa }{4m^2{\omega }^3}\end{array}$

Conclusion:

Thus, the expectation value of $\langle x\rangle$ to the first order, in the $n^{th}$ energy eigenstate is ${\langle x\rangle }^1=\left(2n+1\right)\frac{\hslash \kappa }{4m^2{\omega }^3}$.