# Spectral theorem/ exchanging limit of series and operator

I am currently learning quantum mechanics and there is one typical scenario i encounter in my physics books:

Suppose $$\mathcal{H}$$ is a Hilbert space and $$A: \operatorname{Dom}(A)\to \mathcal{H}$$ is a (linear) operator, $$Dom(A)\subseteq\mathcal{H}$$. In addition, let $$\mathcal{B}:=\{v_n: n \in\mathbb{N}\}$$ be a set of eigenvectors of $$A$$: $$A(v_n)=\lambda_nv_n$$.

If there is a $$v \in \operatorname{Dom}(A)$$ and a sequence of coefficients $$(c_n)_{n \in\mathbb{N}}$$ with $$v=\sum\limits_{n=1}^{\infty}c_nv_n$$, then the authors write $$A(v)=A(\sum\limits_{n=1}^{\infty}c_nv_n)=\sum\limits_{n=1}^{\infty}c_nA(v_n)=\sum\limits_{n=1}^{\infty}(c_n\lambda_n)v_n$$.

From what i know, A is not continuous in general, but this is a special case of the spectral theorem. I think that a physicist would write $$A=\sum\limits_{n=1}^{\infty}|v_n\rangle\langle v_n|$$ and call this the spectral theorem for the discrete case. The problem is that when looking into a math book about the spectral theorem, i see a lot of integrals and things get very complicated. So it would be nice if someone could explain this special case of the spectral theorem (its requirements and statements), or suggest a good source.

• I don't think anything weaker than continuity will be enough. Jul 29, 2019 at 8:47
• Potentially, in the case of a differential operator, it is genuinely continuous on some Sobolev-space and, accordingly, we might expect the above series to converge there, and not just in, say, $L^2$? Jul 29, 2019 at 9:10
• @WoolierThanThou I think that this would be too specific, since physicits often do that "trick" with the Hamiltonian operator. Jul 29, 2019 at 9:33
• You can do something like this for general operators. Equip Dom(A) with the norm $\sqrt{||x||^2+||A(x)||^2}$ (i.e. force $A$ to be continuous) and look at its completion with respect to this norm. Naturally, this is equivalent to saying that we expect the above sum to converge to the right thing before we start performing our calculation, so it's not useful for actually verifying anything unless you have a good idea of what the resulting Hilbert space looks like for other reasons. Jul 29, 2019 at 10:05
• @KaviRamaMurthy Physicits use this equation all the time, although a selfadjoint operator doesn't have to be continuous. This makes me wonder if there is some theorem that allows physicits to do so. By the way, i edited my post to make clearer why this might be related to the spectral theorem. Jul 29, 2019 at 11:12

The operator $$A$$ is typically not continuous, but since it is self-adjoint, it is closed (see below). One of the conclusions of the spectral theorem is that $$v\in \mathrm{Dom}(A)$$ is equivalent to the convergence of the series of nonnegative numbers $$\sum_n \lambda_n^2\lvert c_n\rvert^2.$$ This implies that the sequence $$S_N:= \sum_{n=1}^N \lambda_n c_n v_n$$ is convergent, because $$\lVert S_M-S_N\rVert=\left\lVert \sum_{n=N}^M \lambda_n c_n v_n\right\rVert=\sqrt{\sum_{n=N}^M \lvert c_n\rvert^2\lambda_n^2} ,$$ so $$S_N$$ satisfies the Cauchy condition.

The formula $$\tag{1} A\sum_n c_n v_n= \sum_n \lambda_n c_n v_n$$ is now proven by using the fact that $$A$$ is closed$$^{[1]}$$. Indeed, $$A(\sum_1^N c_n v_n)=\sum_{1}^N \lambda_n c_n v_n$$, and we just saw that the right-hand side converges. Since $$\sum_n \lvert c_n\rvert^2<\infty$$, by reasoning as above we see that $$\sum_1^N c_n v_n$$ converges, so by the closedness property we can pass to the limit and prove (1).

Remark. Here we supposed that $$A$$ has a discrete spectrum. This means that every spectral value is an eigenvalue and that there exists an orthonormal basis of $$\mathcal{H}$$ made of eigenvectors. If this is not the case, the series $$\sum_n$$ have to be replaced by integrals. This is why mathematics books explain the spectral theorem in terms of spectral integrals. The basic idea, however, is already fully contained in the discrete spectrum case.

$$[1]$$. This means that if $$f_n\in \mathrm{Dom}(A)$$ is such that $$f_n\to f$$ and $$Af_n \to g$$, then $$f\in \mathrm{Dom}(A)$$ and $$g=Af$$.

• First of all, thank you. However, i have two questions: 1. How is the limit with N and M going to infinity defined? 2. Is A having a discrete spectrum the only requirement or are there any extra requirements (for example the set of eigenvectors being a Hilbert basis)? Jul 29, 2019 at 13:27
• 1. It is just a shortcut for the Cauchy condition. This post proves that $S_N:=\sum_{n=1}^N c_n \lambda_n v_n$ is a Cauchy sequence in $\mathcal H$. 2. Here, we only used that $\{v_n\ :\ n\in\mathbb N\}$ is an orthonormal set; it actually is a Hilbert basis. For a general operator, such a Hilbert basis might have to be replaced with a set of generalized eigenfunctions; for example, the generalized eigenfunctions of the momentum operator on the line are complex exponentials $e^{ikx}$. This is technically more difficult, but the idea is the same. Jul 29, 2019 at 15:17
• Thank you again. Unfortunately, i don't see how $A\sum_{n=1}^{\infty} c_n v_n=\lim\limits_{N \to \infty}A\sum_{n=1}^{N} c_n v_n$ follows from $S_N:= \sum_{n=1}^N \lambda_n c_n v_n$ being convergent. Jul 29, 2019 at 17:36
• The formula is true for finite sums. Pass to the limit by using the same argument as before. Jul 29, 2019 at 17:38
• Well, pause for a moment. :) This is not a textbook, I just want to give some ideas. Anyway; yes, equation (1) only needs orthonormality. "Having discrete spectrum" is a way of saying I just invented. What I meant is what you said, which is also what I wrote. Maybe "a purely discrete spectrum" would have been a better terminology, but again, here we are just having an informal discussion. Jul 30, 2019 at 14:41