# How to write this down in a rigorous manner?

I'm studying Quantum Mechanics and there's something I'm in doubt in how to write it down in a rigorous way. The idea is: consider a Hilbert space $\mathcal{H}$ and one hermitian operator $A\in \mathcal{L}(\mathcal{H})$. Suppose $\lambda$ is na eigenvalue of $A$. In that case the eigenvector equation is

$$A\psi = \lambda \psi.$$

Let's call $\psi_\lambda$ the solution to that equation. Now, the eigenvalue $\lambda$ might be degenerate in the sense that there's not just one $\psi_\lambda$ but several.

In the book Quantum Mechanics by Cohen-Tannoudji, the author says that there might be one countable collection of eigenvectors associated to the eigenvalue $\lambda$, in the case that we might distinguish the different eigenvectors by a discrete index $k\in \mathbb{N}$. In that case the eigenvectors would be $\psi_{\lambda, k}$ where $k\in \mathbb{N}$.

On the other hand the author states that the collection of eigenvectors might not be countable. In that case the distinct eigenvectors associated to $\lambda$ are labelled by a continuous (i.e. real) index $\alpha\in \mathbb{R}$. In this case the eigenvectors are $\psi_{\lambda,\alpha}$.

My issue here is: several times before knowing if there are any degeneracies corresponding to the eigenvalue $\lambda$ and before knowing if the degeneracies are discrete or continuous the author simply says "we will label the eigenvectors with an additional index $k$ which can be discrete or continuous to account for possible degeneracies".

I find this to be not rigorous, because we don't not upfront if $k$ is a discrete or continuous index. In that case when we write $\psi_{\lambda,k}$ we don't know what $k$ is (is it a natural or a real number?). This makes me quite concerned because I really want to find a way to write this down in a rigorous manner.

In that case, when we have one hermitian operator, one eigenvalue and we want to label the eigenvectors associated to the eigenvalue in order to account for degeneracies, if we don't know upfront if the degeneracies are discrete or continuous, how can we label the eigenvectors in a rigorous manner? How can we do the same kind of thing the author of the book I'm reading does but in a rigorous way?

• I don't see anything non-rigorous about what the author says. We don't know in that context whether the set of eigenvectors is countable or not, so we say $k$ is either an integer index or a continuous index, depending. What's the problem? – David C. Ullrich Feb 11 '16 at 17:57
• I thought that not specifiying upfront whether $k\in \mathbb{N}$ or $k\in \mathbb{R}$ could be an issue. In the end $k$ becomes just an unknown that can be determined upon solving the equations then? By the way, since $\mathbb{N}\subset \mathbb{R}$ we could say that $k\in \mathbb{R}$ anyway, and the solution would just determine the allowed values (i.e if $k\in \mathbb{N}$ or if $k$ can be any real number). Is that the idea? – user1620696 Feb 11 '16 at 18:02
• Let $K$ be the set of possible values for $k$. Then for the $\lambda$'th eigenvalue, the eigenvectors are $\{\psi_{\lambda,k}:k\in K\}$. The set $K$ is all you need to identify individual eigenvectors. It just happens that $K$ is a subset of the real numbers. – grand_chat Feb 11 '16 at 18:03
• No, $k$ will never be determined. It's just a label. If there are two eigenvectors there's no way to "determine" that this one corresponds to $k=1$ instead of $k=2$. Yes, in a specific problem we eventually determine whether $k$ is discrete or continuous. (Yes, since the natural numbers are a subset of the reals it doesn't matter. But I don't like to say yes to that, because it's much more important for you to realize that the problem this solves simply doesn't exist!) – David C. Ullrich Feb 11 '16 at 18:16
• Thanks @DavidC.Ullrich, you're right about that. Labeling the eigenvectors is a matter of choice. My only issue (which I recognize that it's quite silly) is that if I have a set $S$ and a "labeling set" $\Lambda$ we can label the elements of $S$ by picking an injective function $f : \Lambda \to S$. What we call $\psi_k$ then is the image $f(k)$ for $k\in \Lambda$. What concerns me here is that we don't know upfront what is this labeling set $\Lambda$. We are thus talking about $\psi_k$ which would be the mapping $k\mapsto \psi_k$ without knowing the domain of this mapping. Isn't this an issue? – user1620696 Feb 11 '16 at 18:30

## 1 Answer

The Hilbert spaces of Quantum Mechanics must be separable. That's basically an axiom of QM that is a requirement of constructibility. That means that a selfadjoint operator can never have more than a countable number of eigenvectors. For example, $$L=\frac{1}{i}\frac{d}{dx} \\ \mathcal{D}(L)=\{ f \in \mathcal{AC}\subset L^2[0,2\pi] : f'\in L^2[0,2\pi],\;f(0)=f(2\pi) \}$$ is selfadjont and has a countable number of eigenvectors on $L^2[-\pi,\pi]$. The operator $$L=\frac{1}{i}\frac{d}{dx} \\ \mathcal{D}(L)=\{ f \in\mathcal{AC}\subset L^2(\mathbb{R}) : f'\in L^2(\mathbb{R}) \}$$ has no eigenvalues or eigenvectors. It has approximate eigenvectors, but no actual eigenvectors. That's because $e^{i\lambda x} \notin L^2(\mathbb{R})$. The function $e^{i\lambda x}$ is a classical eigenfunction, but not an eigenvector. Small packets of such things are in $L^2(\mathbb{R})$: $$\varphi_{\lambda,\epsilon}(x)=\frac{1}{\sqrt{2\pi\epsilon}}\int_{\lambda-\epsilon/2}^{\lambda+\epsilon/2}e^{i\mu x}d\mu.$$ Using Parseval's identity, $\varphi_{\lambda,\epsilon}\in L^2$ because it is the Fourier transform of $\frac{1}{\epsilon}\chi_{[\lambda-\epsilon/2,\lambda+\epsilon/2]}$, which has $L^2$ norm equal to $1$. So, $$\|\varphi_{\lambda,\epsilon}\|=\left\|\frac{1}{\epsilon}\chi_{[\lambda-\epsilon/2,\lambda+\epsilon/2]}\right\|=1.$$ This is an approximate eigenvector because Parseval's identity gives \begin{align} \|(L-\lambda I)\varphi_{\lambda,\epsilon}\| & =\left\|\frac{1}{\sqrt{2\pi\epsilon}}\int_{\lambda-\epsilon/2}^{\lambda+\epsilon/2}e^{i\mu x}(\mu-\lambda)dx\right\| \\ & \le \frac{\epsilon}{2}\|\varphi_{\lambda,\epsilon}\|=\frac{\epsilon}{2}\rightarrow 0\;\;\;\mbox{ as } \epsilon\rightarrow 0. \end{align} For a general selfadjoint operator, there are eigenvectors and approximate eigenvectors, but ideal objects such as $e^{i\lambda x}$ don't naturally exist. They do for ODEs because the classical eigenfunctions exist. There are been various attempts to append such objects, but IMHO, its seems much easier to deal with the problem in a direct head-on fashion rather than trying to conform the subject to fit the ideas of some early founders of the subject. The techniques of Hilbert space are sufficient, even without the presence of the ideal objects. The Spectral Theorem for selfadjoint linear operators on a Hilbert space has been around a long time, and provides a rigorous characterization of selfadjoint operators, and it deals well with any multiplicity.