Prove that \mathbf{X} and \mathbf{D} are linear operators.
For \mathbf{X}()\equiv x:
\mathbf{X}\left(\alpha\,\psi(x) + \beta\,\phi(x)\right) =x\left(\alpha\,\psi(x) + \beta\,\phi(x)\right) = \alpha\,x\, \psi(x) + \beta\,x\,\phi(x)
So the operator is linear.
For \mathbf{D}()\equiv \frac{\mathrm d}{\mathrm dx}:
\mathbf{D}\left(\alpha\,\psi(x) + \beta\,\phi(x)\right) = \frac{\mathrm d}{\mathrm dx} \left(\alpha\,\psi(x) + \beta\,\phi(x)\right) = \alpha\,\frac{\mathrm d\psi(x)}{\mathrm dx} + \beta\,\frac{\mathrm d\phi(x)}{\mathrm dx}
So the operator is linear.
Derive Eq. 9.7 by plugging Eq. 9.6 into Eq. 9.5.
From:
-\frac{\hbar^2}{2m}\frac{\partial \psi(x)}{\partial x^2} = E\psi(x)
substituting:
\psi(x) = e^{\frac{ipx}{\hbar}}
Then:
-\frac{\hbar^2}{2m}\frac{\partial \psi(x)}{\partial x^2} = -\frac{\hbar^2}{2m}\frac{i^2p^2}{\hbar^2}e^{\frac{ipx}{\hbar}} =\frac{p^2}{2m} \psi(x) = E \psi(x)
which gives:
\frac{p^2}{2m} = E
Prove Eq. 9.10 by expanding each site and comparing the results.
Expanding both sides:
\begin{aligned} & [ \mathbf P^2, \mathbf X] = \mathbf P [ \mathbf P, \mathbf X] + [ \mathbf P, \mathbf X] \mathbf P \\ & \mathbf P^2 \mathbf X - \mathbf X \mathbf P^2 = \mathbf P \left(\mathbf P \mathbf X - \mathbf X \mathbf P \right) + \left(\mathbf P \mathbf X - \mathbf X \mathbf P \right)\mathbf P \\ & \mathbf P\mathbf P \mathbf X - \mathbf X \mathbf P \mathbf P = \mathbf P \mathbf P \mathbf X - \mathbf P \mathbf X \mathbf P + \mathbf P \mathbf X \mathbf P - \mathbf X \mathbf P \mathbf P \\ & \mathbf P\mathbf P \mathbf X - \mathbf X \mathbf P \mathbf P = \mathbf P \mathbf P \mathbf X - \mathbf X \mathbf P \mathbf P \end{aligned}
So the identity is proven.
Show that the right-hand side of Eq. 9.17 simplifies to the right-hand side of Eq. 9.16. Hint: First expand the second term by taking the derivative of the product. Then look for cancellations.
As suggested, let’s expand the right hand side of the equation taking the derivative of the product:
\begin{aligned} [\mathbf V(x), \mathbf P]\psi(x) & = V(x)\left(-i\hbar \frac{\mathrm d}{\mathrm dx}\right) \psi(x) - \left(-i\hbar \frac{\mathrm d}{\mathrm dx}\right)\left[V(x)\psi(x)\right] \\ & = V(x)\left(-i\hbar \frac{\mathrm d \psi(x)}{\mathrm dx}\right) - \left(-i\hbar \frac{\mathrm dV(x)}{\mathrm dx}\right)\psi(x) - \left(-i\hbar \frac{\mathrm d\psi(x)}{\mathrm dx}\right)V(x) \\ & = i\hbar \frac{\mathrm dV(x)}{\mathrm dx} \psi(x) \end{aligned}
as \psi(x) \ne 0:
[\mathbf V(x), \mathbf P] = i\hbar \frac{\mathrm dV(x)}{\mathrm dx}
Find the second time derivative of x in Eq. 10.9, and thereby show that it solves Eq. 10.8.
\ddot x = \frac{\mathrm d^2x(t)}{\mathrm dt^2} = -A\omega^2\cos(\omega t) - B\omega^2\sin(\omega t) = -\omega^2 x