Question

(a) Generalize Problem $8.2,$ using the trial wave function $^{18}$ $$\psi(x)=\frac{A}{\left(x^{2}+b^{2}\right)^{n}}$$ for arbitrary $n .$ Partial answer: The best value of $b$ is given by $$b^{2}=\frac{\hbar}{m \omega}\left[\frac{n(4 n-1)(4 n-3)}{2(2 n+1)}\right]^{1 / 2}$$ (b) Find the least upper bound on the first excited state of the harmonic oscillator using a trial function of the form $$\begin{aligned} &\psi(x)=\frac{B x}{\left(x^{2}+b^{2}\right)^{n}}\\ &\text {Partial answer: The best value of } b \text { is given by }\\ &b^{2}=\frac{\hbar}{m \omega}\left[\frac{n(4 n-5)(4 n-3)}{2(2 n+1)}\right]^{1 / 2} \end{aligned}$$ (c) Notice that the bounds approach the exact energies as $n \rightarrow \infty$. Why is that? Hint: Plot the trial wave functions for $n=2, n=3,$ and $n=4$ and compare them with the true wave functions (Equations 2.60 and 2.63 ). To do it analytically, start with the identity $$e^{z}=\lim _{n \rightarrow \infty}\left(1+\frac{z}{n}\right)^{n}$$

          (a) Generalize Problem $8.2,$ using the trial wave function $^{18}$
$$\psi(x)=\frac{A}{\left(x^{2}+b^{2}\right)^{n}}$$
for arbitrary $n .$ Partial answer: The best value of $b$ is given by
$$b^{2}=\frac{\hbar}{m \omega}\left[\frac{n(4 n-1)(4 n-3)}{2(2 n+1)}\right]^{1 / 2}$$
(b) Find the least upper bound on the first excited state of the harmonic oscillator using a trial function of the form
$$\begin{aligned}
&\psi(x)=\frac{B x}{\left(x^{2}+b^{2}\right)^{n}}\\
&\text {Partial answer: The best value of } b \text { is given by }\\
&b^{2}=\frac{\hbar}{m \omega}\left[\frac{n(4 n-5)(4 n-3)}{2(2 n+1)}\right]^{1 / 2}
\end{aligned}$$
(c) Notice that the bounds approach the exact energies as $n \rightarrow \infty$. Why is that? Hint: Plot the trial wave functions for $n=2, n=3,$ and $n=4$ and compare them with the true wave functions (Equations 2.60 and 2.63 ). To do it analytically, start with the identity
$$e^{z}=\lim _{n \rightarrow \infty}\left(1+\frac{z}{n}\right)^{n}$$

Added by Joseph S.

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