Symmetries of a differential equation, its solutions and hydrogen atom

Question

A symmetry of a differential equation need not be shared by its solutions. However, under that symmetry, the one solution goes to another. For example, consider the time-independent Schrodinger equation (TISE) $H\phi=E\phi$ of an one-dimensional SHO. The TISE is invariant under reflections i.e. $x\to -x$ but the solutions $\phi(x)$ under reflections behave as $$\phi(x)\to\phi(x)~~~~~~~~ {\rm or}~~~~~~~ \phi(x)\to -\phi(x).$$

If the above statements are generically true, then the rotational symmetry of the TISE of the hydrogen atom must also exhibit the same feature. For instance, the wavefunction of the $2p_x$ (or $2p_y$) orbital, under some rotation, must be be transformable to $2p_z$ orbital. However, I don't think that is possible or is it? If possible, what is that operator which takes, for example, $\psi_{2p_x}\to\psi_{2p_z}$?

Answer 1

The hydrogen hamiltonian can be written as $$ H = \frac{p_r^2}{2m} + \frac{L^2}{2mr^2} - \frac{e^2}{r} $$ where $p_r$ is the radial component of the momentum. Since $H$ depends on $L^2$, $p_r$ and $r$, the hamiltonian commutes with every component of $\mathbf L$: $$ [H, \mathbf L] = 0, $$ which in turn means we can simulteneously diagonalize $H$, $L^2$ and one component of $\mathbf L$ (say, $L_z$). We can now introduce the rotation operator: $$ D(\phi, \mathbf n) = \exp \left (- \frac{i \mathbf L \cdot \mathbf n \phi}{\hbar} \right ), $$ which rotates the states by an angle $\phi$ around the unit vector $\mathbf n$. We can transform the hamiltonian under a rotation in the following manner: $$ H \to H' = D(\phi, \mathbf n)^\dagger H D(\phi, \mathbf n). $$ But $H$ commutes with $\mathbf L$, so $H$ also commutes with $D$, and thus $H' = D^\dagger D H = H$. We proved the hamiltonian is invariant under rotations. This means that some degeneracies will occur, i.e., some states will share the same energy. Note however that unlike I said previously (which was a mistake) you cannot rotate one state $|nlm\rangle$ to create one with different $m$ around the $z$-axis. Since the eigenstates of the hamiltonian are also eigenstates of $L_z$, the rotation operator will only introduce a phase factor on the state: $$ D(\phi, \mathbf{\hat z})|nlm\rangle = \exp\left (-\frac{iL_z \phi}{\hbar} \right ) |nlm\rangle = e^{-im\phi}|nlm\rangle $$ But states with different $m$ are orthogonal (their inner product is zero), which is clearly not the case for the rotated state above: $$ \langle nlm | \big(D |nlm\rangle\big) = e^{-im\phi} \neq 0. $$ Now, rotations around the $x$ or $y$-axis will change the value of $m$ since $|nlm\rangle$ is not an eigenstate of $L_x$ or $L_y$. Indeed, you can write $L_x$ and $L_y$ in terms of the ladder operators $L_\pm = L_x \pm i L_y$ which increase/decrease the value of $m$ to calculate the action of the rotation operator. The only thing I can guarantee you is in the end you will arrive at least in a linear combination of states with the same $n$ and $l$ but with different $m$'s.

Answer 2

One cannot transform a $p$ orbital into a $d$ orbital by a rotation. This is because the generators of rotations are the angular momentum operators, and their action can only connect states with the same value $\ell$ of the angular momentum quantum number. Thus, rotations can only connect states with the same $\ell$.

In the case of the parity example that you give, the parity $P$ and the identity form an Abelian group and all the representations (there are two of them) are of dimension one so if the TISE Schrodinger equation is also $P$-invariants then solutions can have either even or odd parity. This is the case for instance for symmetric potentials for which $V(-x)=V(x)$.