# Symmetries of a differential equation, its solutions and hydrogen atom

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, $$\psi_{2p_x}\to\psi_{2p_z}$$?

• I don't understand what the "this" in your question "Is this always true, for instance, in case of nontrivial symmetries such as rotation?" refers to. It sounds as if you are asking whether all symmetries act on solutions either as the identity or as reflection, but why would you think that? – ACuriousMind Feb 10 at 14:36
• Related: physics.stackexchange.com/q/174957/50583 | Also, the 1s wavefunction is rotationally symmetric itself (which you can see by either looking at its popular depictions or at its actual mathematical form), so I don't understand the second part of your question, either. – ACuriousMind Feb 10 at 14:39
• @ACuriousMind Does it make sense now? – SRS Feb 10 at 14:45
• It makes sense, except for the part where you think that it is not possible to rotate a $p_z$ into a $p_x$ orbital. Again, just looking at the shapes of these orbitals (the $p_x,p_y,p_z$ are the three to the right in the picture in the related question I linked) seems to suggest to me very strongly that you can rotate them into each other simply by a rotation that takes the x-axis to the z-axis. Why do you think this won't work? – ACuriousMind Feb 10 at 14:51
• I think what you're grasping towards is the ideas underlying representation theory and the so-called "irreducible representations". You might find it worthwhile to read up on the subject a bit; Peter Woit has a textbook draft (PDF) available on the subject. – Michael Seifert Feb 10 at 14:54

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.
• Is the TISE invariant under rotation for a fixed value of $n$ and $l$? @ErickShock – SRS Feb 10 at 15:53
• Well, of course $n$ and $l$ must be fixed. If they varied, then you be trying to match 2 different states. – Hanting Zhang Feb 10 at 17:12
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 $$V(-x)=V(x)$$.