# Relation between eigenvalue equation for an operator and for its square

Consider a time indipendent Schrodinger problem: $$\hat{H}\psi_E(p) = E \psi_E(p)$$ with suitable boundary conditions. We know that $$\psi_E$$ are the eigenfunctions of $$\hat{H}$$. If we now consider the operator $$\hat{H}^2$$ we know that the eigenfunctions of $$\hat{H}$$ are also eigenfunctions of $$\hat{H}^2$$ but in general the reverse is not true.

Consider the problem $$\hat{H}^2 \psi_\epsilon (p) = \epsilon \psi_\epsilon (p) \, .$$ What is the relation between the two solutions? My intuitive idea is that the eigenfunctions of $$\hat{H}$$ are eigenfunctions of $$\hat{H}^2$$ whenever $$E=+\sqrt{\epsilon}$$, so that starting from the "squared problem" we could, in principle, find the eigenfunctions and eigenvalues of the "non-squared problem" simply restricting the solutions to $$E = +\sqrt{\epsilon}$$ and $$\psi_E(p) = \psi_{\sqrt{\epsilon}}(p) \, .$$

Is this legit or are there inconsistencies when starting from the "squared problem"? Should I also consider the solutions with $$E=-\sqrt{\epsilon}$$?

I'll comment on only the case of finite dimensions, since in infinite dimensional Hilbert spaces, you have to be careful about domains: the domain of $$\hat{H}$$ might not be the same as the domain of $$\hat{H}^2$$. In addition, I'll assume that the operators are self-adjoint so that the eigenvalues are all real.

Since $$\hat{H}$$ and $$\hat{H}^2$$ commute, they share an eigenbasis, call it $$\lvert h\rangle$$. Then, $$\hat{H}\lvert h\rangle = h\lvert h\rangle\Longrightarrow \hat{H}^2\lvert h\rangle = h^2\lvert h\rangle\,.$$

From the other direction, let's start with the eigenbasis of $$\hat{H}^2$$, say, $$\hat{H}^2\lvert h, n\rangle=h\lvert h, n\rangle\,,$$ where we use the second label $$n$$ to indicate that some eigenvalues could be degenerate so that there are multiple eigenvectors of $$\hat{H}^2$$ with that same eigenvalue. (We should have done this above, too, but there's no danger in the implication, so we ignored it. This is the important direction anyway.) Then, for a particular eigenvalue $$h$$, there are two possibilities:

1. The eigenvalue $$h$$ is not degenerate. In that case, we just call $$\lvert h, 1\rangle=\lvert h\rangle$$, and it must be that $$\hat{H}\lvert h\rangle=\sqrt{h}\lvert h\rangle\,.$$

2. The eigenvalue $$h$$ is degenerate. For simplicity, we'll assume that it's doubly-degenerate; the general case is a straight-forward generalization. Then, $$\hat{H}^2\lvert h, 1\rangle=h\lvert h, 1\rangle$$ and $$\hat{H}^2\lvert h, 2\rangle=h\lvert h, 2\rangle$$. Since $$\hat{H}$$ and $$\hat{H}^2$$ commute, they share an eigenbasis. Furthermore, an eigenvector of $$\hat{H}^2$$ must be a linear combination of eigenvectors of $$\hat{H}^2$$ with the same eigenvalue. This is a general statement, but to see this for our specific case, note that $$\hat{H}^2\left(\hat{H}\lvert h,n\rangle\right) = \hat{H}\left(\hat{H}^2\lvert h,n\rangle\right)=\hat{H}\left(h\lvert h,n\rangle\right) = h\left(\hat{H}\lvert h,n\rangle\right)\,.$$ This shows that $$\hat{H}\lvert h,n\rangle$$ is an eigenvector of $$\hat{H}^2$$ with eigenvalue $$h$$, and hence it must be some linear combination of the $$\lvert h,n\rangle$$'s. In other words, \begin{align*}\hat{H}\lvert h,1\rangle &= \alpha_1 \lvert h,1\rangle + \alpha_2 \lvert h,2\rangle\,,\\\hat{H}\lvert h,2\rangle &= \beta_1 \lvert h,1\rangle + \beta_2 \lvert h,2\rangle\,.\end{align*} This implies that we can construct two eigenvectors of $$\hat{H}$$ out of $$\lvert h,1\rangle$$ and $$\lvert h,2\rangle$$. Thus, there are two eigenvectors of $$\hat{H}$$ that are eigenvectors of $$\hat{H}^2$$ with eigenvalue $$h$$. The corresponding eigenvalues can be $$\sqrt{h}$$ and $$-\sqrt{h}$$, and it depends on the details of $$\hat{H}$$ whether we get two of the same eigenvalue or one of each of $$\sqrt{h}$$ and $$-\sqrt{h}$$.

You should have put in some effort by considering some simple examples. An example that is impossible to avoid in the standard curriculum would be the momentum operator $$\hat p=-i\hslash\frac{\mathrm d\ }{\mathrm dx}$$ and the NR kinetic energy operator $$\hat T=\frac{\hat p^2}{2m}$$

We all know that the eigenfunctions of the momentum operator are $$e^{ipx/\hslash}$$ so that $$\hat pe^{ipx/\hslash}=-i\hslash\frac{\mathrm d\ }{\mathrm dx}e^{ipx/\hslash}=pe^{ipx/\hslash}$$ are also the eigenfunctions of the NR kinetic energy operator.

However, because $$-p$$ gives the same kinetic energy as $$+p$$, not only would we have $$e^{-ipx/\hslash}$$ to contend with, we actually would also have $$\cos\frac{px}\hslash$$ and $$\sin\frac{px}\hslash$$ as eigenfunctions of the NR kinetic energy operator to contend with. That is, the disentangling of the various eigenfunctions of the squared operator in order to get the eigenfunctions of the un-squared operator, is sometimes non-trivial. Needless to say, it gets progressively more complicated when the space of degeneracies increase. e.g. in 2D, the whole circle's worth of eigenfunctions get mapped to the same squared KE, and in 3D, it is a whole sphere's worth. This disentangling will then get much more difficult.

This is one of those questions that is not useful to cover; the above exposition makes it clear that even in the standard curriculum, if a person starts thinking about such things, they can easily work out a solution for themselves. On the other hand, nobody would be wanting to start from such things. Instead, we start with the basic quantities, the momenta, if you will, and compose them to get the complicated operators, like the squared operators. There is no foreseeable future whereby we would be starting with the squared operators and then be needing to work downwards. That is why it is afforded no space in the standard textbooks.