1. **Problem Statement:** We are given the 2p orbital wavefunctions of the hydrogen atom: $$\psi_{2p,x}(r) = A x e^{-r/a_0}, \quad \psi_{2p,y}(r) = A y e^{-r/a_0}, \quad \psi_{2p,z}(r) = A z e^{-r/a_0}$$ where $$A = \frac{1}{4 \sqrt{2} a_0^{5/2}}$$. We need to: (a) Write down the rotation operators about the x-, y-, and z-axes by angle $$\pi/2$$ (anti-clockwise). (b) Show the sequence of transformations: $$\psi_{2p,x} \xrightarrow{R_z} \psi_{2p,y} \xrightarrow{R_x} \psi_{2p,z} \xrightarrow{R_y} \psi_{2p,x}$$ (c) Find the combined operator $$R_y R_x R_z$$ and discuss its geometric meaning. --- 2. **Rotation Operators:** The rotation operators act on the coordinates $$(x,y,z)$$ as follows for an anti-clockwise rotation by angle $$\theta$$: - About the x-axis: $$R_x(\theta): \begin{cases} x' = x \\ y' = y \cos\theta - z \sin\theta \\ z' = y \sin\theta + z \cos\theta \end{cases}$$ - About the y-axis: $$R_y(\theta): \begin{cases} x' = x \cos\theta + z \sin\theta \\ y' = y \\ z' = -x \sin\theta + z \cos\theta \end{cases}$$ - About the z-axis: $$R_z(\theta): \begin{cases} x' = x \cos\theta - y \sin\theta \\ y' = x \sin\theta + y \cos\theta \\ z' = z \end{cases}$$ For $$\theta = \pi/2$$, $$\cos(\pi/2) = 0$$ and $$\sin(\pi/2) = 1$$, so: - $$R_x(\pi/2): (x,y,z) \to (x, -z, y)$$ - $$R_y(\pi/2): (x,y,z) \to (z, y, -x)$$ - $$R_z(\pi/2): (x,y,z) \to (-y, x, z)$$ 3. **Applying Rotations to Wavefunctions:** Since $$\psi_{2p,x} = A x e^{-r/a_0}$$, under rotation the coordinates transform and the wavefunction changes accordingly. - (1) Apply $$R_z(\pi/2)$$ to $$\psi_{2p,x}$$: $$x \to -y, \quad y \to x, \quad z \to z$$ So, $$\psi_{2p,x} \xrightarrow{R_z} A (-y) e^{-r/a_0} = -\psi_{2p,y}$$ Ignoring the overall sign (phase), this is $$\psi_{2p,y}$$. - (2) Apply $$R_x(\pi/2)$$ to $$\psi_{2p,y} = A y e^{-r/a_0}$$: Coordinates transform as $$y \to -z, z \to y$$, so $$y \to -z$$ Thus, $$\psi_{2p,y} \xrightarrow{R_x} A (-z) e^{-r/a_0} = -\psi_{2p,z}$$ Again, ignoring sign, this is $$\psi_{2p,z}$$. - (3) Apply $$R_y(\pi/2)$$ to $$\psi_{2p,z} = A z e^{-r/a_0}$$: Coordinates transform as $$z \to -x$$, so $$\psi_{2p,z} \xrightarrow{R_y} A (-x) e^{-r/a_0} = -\psi_{2p,x}$$ Ignoring sign, this returns to $$\psi_{2p,x}$$. Thus, the sequence holds: $$\psi_{2p,x} \to \psi_{2p,y} \to \psi_{2p,z} \to \psi_{2p,x}$$ 4. **Combined Rotation $$R_y R_x R_z$$:** Apply the rotations in order: - $$R_z: (x,y,z) \to (-y, x, z)$$ - $$R_x: (x,y,z) \to (x, -z, y)$$ - $$R_y: (x,y,z) \to (z, y, -x)$$ Calculate combined effect on $$\mathbf{r} = (x,y,z)$$: First $$R_z$$: $$\mathbf{r}_1 = (-y, x, z)$$ Then $$R_x$$ on $$\mathbf{r}_1$$: $$x' = -y$$ $$y' = -z$$ $$z' = x$$ Then $$R_y$$ on result: $$x'' = z' \cos(\pi/2) + y' \sin(\pi/2) = 0 + (-z) \times 1 = -z$$ $$y'' = y' = -z$$ (But this conflicts with previous step; re-check carefully) Wait, carefully: Apply $$R_y(\pi/2)$$ to $$\mathbf{r}_2 = (x', y', z') = (-y, -z, x)$$: $$x'' = x' \cos(\pi/2) + z' \sin(\pi/2) = (-y) \times 0 + x \times 1 = x$$ $$y'' = y' = -z$$ $$z'' = -x' \sin(\pi/2) + z' \cos(\pi/2) = -(-y) \times 1 + x \times 0 = y$$ So final coordinates: $$\mathbf{r}'' = (x, -z, y)$$ This is exactly the effect of $$R_x(\pi/2)$$. **Geometrical meaning:** The combined rotation $$R_y R_x R_z$$ equals a single rotation about the x-axis by $$\pi/2$$. This shows that the product of these three rotations is equivalent to one rotation, illustrating the non-commutative nature of 3D rotations and how sequences of rotations combine. --- **Final answers:** - (a) Rotations about axes by $$\pi/2$$: $$R_x: (x,y,z) \to (x,-z,y), \quad R_y: (x,y,z) \to (z,y,-x), \quad R_z: (x,y,z) \to (-y,x,z)$$ - (b) The sequence of rotations transforms the wavefunctions cyclically: $$\psi_{2p,x} \to \psi_{2p,y} \to \psi_{2p,z} \to \psi_{2p,x}$$ - (c) The combined rotation $$R_y R_x R_z = R_x$$ (rotation about x-axis by $$\pi/2$$). This completes the problem.