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Chapter 5: Problem 19
Prove that the derivative of an even (odd) function is odd (even).
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The Schr枚dinger equation for a one-dimensional harmonic oscillator is $$ \hat{H} \psi(x)=E \psi(x) $$ where the Hamiltonian operator is given by $$ \hat{H}=-\frac{\hbar^{2}}{2 \mu} \frac{d^{2}}{d x^{2}}+\frac{1}{2} k x^{2} $$ where \(k=\mu \omega^{2}\) is the force constant. Let \(\hat{P}\) and \(\hat{X}\) be the operators for momentum and position, respectively. If we define \(\hat{p}=(\mu \hbar \omega)^{-1 / 2} \hat{P}\) and \(\hat{x}=(\mu \omega / \hbar)^{1 / 2} \hat{X}\), show that $$ \hat{H}=\frac{\hat{P}^{2}}{2 \mu}+\frac{k}{2} \hat{X}^{2}=\frac{\hbar \omega}{2}\left(\hat{p}^{2}+\hat{x}^{2}\right) $$ Use the definitions of \(\hat{p}\) and \(\hat{x}\) to show that $$ \hat{p}=-i \frac{d}{d x} $$ and $$ \hat{p} \hat{x}-\hat{x} \hat{p}=[\hat{p}, \hat{x}]=-i $$
Consider the transformation from Cartesian coordinates to plane polar coordinates where $$\begin{array}{ll} x=r \cos \theta & r=\left(x^{2}+y^{2}\right)^{1 / 2} \\ y=r \sin \theta & \theta=\tan ^{-1}\left(\frac{y}{x}\right) \end{array}$$ If a function \(f(r, \theta)\) depends upon the polar coordinates \(r\) and \(\theta,\) then the chain rule of partial differentiation says that \\[ \left(\frac{\partial f}{\partial x}\right)_{y}=\left(\frac{\partial f}{\partial r}\right)_{\theta}\left(\frac{\partial r}{\partial x}\right)_{y}+\left(\frac{\partial f}{\partial \theta}\right)_{r}\left(\frac{\partial \theta}{\partial x}\right)_{y} \\] and that \\[ \left(\frac{\partial f}{\partial y}\right)_{x}=\left(\frac{\partial f}{\partial r}\right)_{\theta}\left(\frac{\partial r}{\partial y}\right)_{x}+\left(\frac{\partial f}{\partial \theta}\right)_{r}\left(\frac{\partial \theta}{\partial y}\right)_{x} \\] For simplicity, we will assume \(r\) is constant so that we can ignore terms involving derivatives with respect to \(r .\) In other words, we will consider a particle that is constrained to move on the circumference of a circle. This system is sometimes called a particle on a ring. Using Equations 1 and \(2,\) show that \\[ \left(\frac{\partial f}{\partial x}\right)_{y}=-\frac{\sin \theta}{r}\left(\frac{\partial f}{\partial \theta}\right)_{r} \quad \text { and } \quad\left(\frac{\partial f}{\partial y}\right)_{x}=\frac{\cos \theta}{r}\left(\frac{\partial f}{\partial \theta}\right)_{r} \quad(r \text { fixed }) \\] Now apply Equation 2 again to show that \\[ \begin{aligned} \left(\frac{\partial^{2} f}{\partial x^{2}}\right)_{y} &=\left[\frac{\partial}{\partial x}\left(\frac{\partial f}{\partial x}\right)_{y}\right]=\left[\frac{\partial}{\partial \theta}\left(\frac{\partial f}{\partial x}\right)_{y}\right]_{r}\left(\frac{\partial \theta}{\partial x}\right)_{y} \\\ &=\left\\{\frac{\partial}{\partial \theta}\left[-\frac{\sin \theta}{r}\left(\frac{\partial f}{\partial \theta}\right)_{r}\right]\right\\}\left(-\frac{\sin \theta}{r}\right) \\ &=\frac{\sin \theta \cos \theta}{r^{2}}\left(\frac{\partial f}{\partial \theta}\right)_{r}+\frac{\sin ^{2} \theta}{r^{2}}\left(\frac{\partial^{2} f}{\partial \theta^{2}}\right)_{r}(r \text { fixed }) \end{aligned} \\] Similarly, show that \\[ \left(\frac{\partial^{2} f}{\partial y^{2}}\right)_{x}=-\frac{\sin \theta \cos \theta}{r^{2}}\left(\frac{\partial f}{\partial \theta}\right)_{r}+\frac{\cos ^{2} \theta}{r^{2}}\left(\frac{\partial^{2} f}{\partial \theta^{2}}\right)_{r}(r \text { fixed }) \\] and that \\[ \nabla^{2} f=\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}} \rightarrow \frac{1}{r^{2}}\left(\frac{\partial^{2} f}{\partial \theta^{2}}\right)_{r} \quad(r \text { fixed }) \\] Now show that the Schr枚dinger equation for a particle of mass \(m\) constrained to move on a circle of radius \(r\) is (see Problem \(3-28\) ) \\[ -\frac{\hbar^{2}}{2 I} \frac{\partial^{2} \psi(\theta)}{\partial \theta^{2}}=E \psi(\theta) \quad 0 \leq \theta \leq 2 \pi \\] where \(I=m r^{2}\) is the moment of inertia.
To normalize the harmonic-oscillator wave functions and calculate various expectation values, we must be able to evaluate integrals of the form \\[ I_{v}(a)=\int_{-\infty}^{\infty} x^{2 v} e^{-a x^{2}} d x \quad v=0,1,2, \dots \\] We can simply either look them up in a table of integrals or continue this problem. First, show that \\[ I_{v}(a)=2 \int_{0}^{\infty} x^{2 v} e^{-a x^{2}} d x \\] The case \(v=0\) can be handled by the following trick. Show that the square of \(I_{0}(a)\) can be written in the form \\[ I_{0}^{2}(a)=4 \int_{0}^{\infty} \int_{0}^{\infty} d x d y e^{-a\left(x^{2}+y^{2}\right)} \\] Now convert to plane polar coordinates, letting \\[ r^{2}=x^{2}+y^{2} \quad \text { and } \quad d x d y=r d r d \theta \\] Show that the appropriate limits of integration are \(0 \leq r<\infty\) and \(0 \leq \theta \leq \pi / 2\) and that \\[ I_{0}^{2}(a)=4 \int_{0}^{\pi / 2} d \theta \int_{0}^{\infty} d r r e^{-a r^{2}} \\] which is elementary and gives \\[ I_{0}^{2}(a)=4 \cdot \frac{\pi}{2} \cdot \frac{1}{2 a}=\frac{\pi}{a} \\] or that \\[ I_{0}(a)=\left(\frac{\pi}{a}\right)^{1 / 2} \\] Now prove that the \(I_{v}(a)\) may be obtained by repeated differentiation of \(I_{0}(a)\) with respect to \(a\) and, in particular, that \\[ \frac{d^{v} I_{0}(a)}{d a^{v}}=(-1)^{v} I_{v}(a) \\] Use this result and the fact that \(I_{0}(a)=(\pi / a)^{1 / 2}\) to generate \(I_{1}(a), I_{2}(a),\) and so forth.
The general solution for the classical harmonic oscillator is \(x(t)=C \sin (\omega t+\phi) .\) Show that the displacement oscillates between \(+C\) and \(-C\) with a frequency \(\omega\) radian \(\cdot \mathrm{s}^{-1}\) or \(v=\omega / 2 \pi\) cycle \(\cdot \mathrm{s}^{-1} .\) What is the period of the oscillations; that is, how long does it take to undergo one cycle?
\text { Show that the reduced mass of two equal masses, } m \text {, is } m / 2 \text {. }
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