# Applications of the Spectral Theorem to Quantum Mechanics

I'm currently learning some basic functional analysis. Yesterday I arrived at the spectral theorem of self-adjoint operators. I've heard that this theorem has lots of applications in Quantum Mechanics.

But let me first state the formulation of the theorem that I'm using:

Let $H$ be a Hilbert space. There's a 1-to-1-correspondence between self-adjoint operators $A$ on $H$ and spectral measures $P^{A}$ given by $$A~=~\int_{\mathbb{R}} \lambda ~dP^{A}.$$ ($\lambda$ denotes a constant, $\mathbb{R}$ denotes the real numbers.)

A corollary is:

Let $g:\mathbb{R}\to\mathbb{R}$ be a function. (Again: $\mathbb{R}$ denotes the set of real numbers.) Then: $$g(A)~:=~\int_{\mathbb{R}} g(\lambda)~ dP^{g(A)}$$
$$P^{g(A)}(\Delta) ~=~ P^{A}(g^{-1}(\Delta))$$ where $\Delta$ denotes a set in the $\sigma$-algebra of $\mathbb{R}$.

Okay. Now this is the theorem. First I don't really the application of the corollary in Quantum mechanics. I've heard that suppose you're given an operator $A$ this means that it's easy for you to define operators like $\exp(A)$, especially on infinite dimensional Hilbert spaces. This indeed could be useful in quantum mechanics. Especially when thinking about the "time-evolution operator" of a system.

However then I say: Why do you make things so complicated? Suppose you want to calculate $\exp(A)$. Why don't you define $$\exp(A)~:=~1+A+1/2 A^2 + \ldots$$ and require convergence with respect to the operator norm. An example: Consider the vectorspace spanned by the monomials $1,x,x^2,\ldots$ and let $A=d/dx$. Then you can perfectly define

$$\exp(d/dx)~:=~1+ d/dx + 1/2 d^2/dx^2 + \ldots$$

and require convergence with respect to the operator norm.

In addition to that I've heard that the spectral theorem gives a full description of all self-adjoint operators. Now why is that the case? I mean okay..there's a one to one correspondence between self-adjoint operators and spectral measures...but why does this give me any information about "the inner structure of the operator"? (And why is there this $\lambda$ in the integral? Looks somehow like an eigenvalue of $A$? But I'm just guessing)

I'd me more than happy, if you could provide me with some intuition and ideas of how the theorem can be used.

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Dear @Matt_Quantum. Two comments: 1) Would you be willing to restrict to separable Hilbert spaces only? 2) The two examples $A=\frac{d}{dx}$ and $\exp(\frac{d}{dx})$ are not self-adjoint operators, if $x$ is supposed to be a real variable. – Qmechanic♦ May 7 '11 at 8:52
1) Okay 2) Okay. Sorry for that. But assume A is some arbitrary self-adjoint operator, one could still define exp(A)=1+A+1/2 A^2 +... and require convergence with respect to the operator norm. – Matt_Quantum May 7 '11 at 12:36
Dear @Matt_Quantum. But the operator norm $||A||$ of an arbitrary self-adjoint operator $A$ could be $\infty$. In the question formulation (v1) you never restricted yourself to only bounded operators, which is actually good, because in quantum mechanics there are lots of unbounded operators. For instance, $|| \frac{d}{dx} ||=\infty$. – Qmechanic♦ May 7 '11 at 14:56
I may try to work this up into an answer, but for now, I'll just comment: in a way, the spectral measure determines the operator up to a change of variables. Every operator whose spectrum is the real line can be transformed to $id\over dx$ acting on some subspace of the functions on the real line, after a change of variable, for example. It's almost true that the spectrum determines the operator (up to a change of basis) but you do actually need a little more info: multiplicites of eigenvavlues, whether they are part of a continuous set of eigenvalues, etc., spectral measure type thingies. – joseph f. johnson Jan 7 '12 at 22:09

## 2 Answers

Meta:

It is true that a lot of quantum mechanics can be taught and understood without much knowledge of the mathematical foundations, and usually it is. Since QM is a mandatory class at many faculties that future experimental physicists have to attend, too, this also makes sense. But for future theoretical and mathematical physicists, it may pay off to learn a little bit about the math, too.

A little anecdote: John von Neumann once said to Werner Heisenberg that mathematicians should be grateful for QM, because it led to the invention of a lot of beautiful mathematics, but that mathematicians repaid this by clarifying, e.g., the difference between a selfadjoint and a symmetric operator. Heisenberg asked: "What is the difference?"

Suppose you want to calculte exp(A). Why don't you define exp(A):=1+A+1/2 A^2 + ... and require convergence with respect to the operator norm.

That's correct. The benefit of the spectral theorem is that you can define f(A) for any selfadjoint (or more generally, normal) operator for any bounded Borel function. This comes in handy in many proofs in operator theory.

In addition to that I've heard that the spectral theorem gives a full description of all self-adjoint operators. Now why is that the case? I mean okay..there's a one to one correspondence between self-adjoint operators and spectral measures..

That's correct, too. Spectral measures are much much simpler objects than selfadoint operators, that's why. Futhermore, you can use the spectral theorem to prove that every selfadjoint operator is unitarily equivalent to a multiplication operator (multiply f(x) by x). From an abstract viewpoint, this is a very satisfactory characterization. It does not help much for concrete calcuations in QM, though.

BTW: On a more advanced level, you'll need to understand the spectral theorem to understand what a mass gap is in Yang-Mills Theory (millenium problem).

Hint: In QFT in Minkowski-Spacetime, one usually assumes that there is a continuous representation of the Poincaré group, especially of the commutative subgroup of translations, on the Hilbert space that contains all physical states. The operators that form the representation have a common spectral measure, this is an application of the SNAG-theorem. The support of this spectral measure is bounded away from zero, that's the definition of the mass gap.

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Dear Constantin, the $A=\int_R \lambda\,dP^A$ is just a continuous version of spectral decomposition. Here $dP^A$ is a differential version of the projection operators that define the Hermitian operator.

For a discrete spectrum, the corresponding equation would be $$A = \sum_{i} \lambda_i P_{\lambda_{i}}$$ where the sum goes over the eigenvalues $\lambda_i$ and $P_{\lambda_i}$ are the projection operators on the subspace of the Hilbert space that contains the eigenvectors with the eigenvalue $\lambda_i$. Indeed, $\lambda$ is always meant to be a possible eigenvalue of the operator. And indeed, a Hermitian operator is fully determined by its spectrum and the corresponding eigenvectors (and multiplicities) for each eigenvalue, which is why the formula above is an equivalent way to rewrite a Hermitian operator.

When the spectrum of $A$ is continuous, the summation over $i$ has to be replaced by an integral, and the corresponding differential $d$ is added in front of $dP_{\lambda}$: it's really the differential of the projection operator on the space of eigenstates with eigenvalues in the interval $[-\infty, \lambda]$; $dP_\lambda = dP_\lambda / d\lambda \cdot d\lambda$, if you wish.

But it's really morally the same thing as in the case of the discrete spectrum (which produces delta functions in $dP_\lambda / d\lambda$ if we adopt this terminology). Also, some of your additional proofs involving $\Delta$ are just trivial substitutions under the integral sign. One would need to know lots of details of your mathematical axioms - lots of the particular "math culture" you're coming from - to figure out what could exactly be difficult for a mathematician about the substitutions under the integral sign. There are no difficulties from a physicist's perspective - it's high school maths.

http://en.wikipedia.org/wiki/Spectral_theorem#Hermitian_matrices

Mathematicians may worry about boundedness and well-definedness of all these things for most of their careers but these things are totally vacuous from the viewpoint of physics.

If a physicist finds out that the answer to a physics question requires him to calculate $f(A)$, a function of an operator - an observable - he just has to calculate it whether or not it looks hard or well-defined. In particular, the Taylor expansion for functions such as the exponential is always assumed to hold.

You discuss the function $g(A)$ of the operator as an example. The procedures you outline physically mean that one diagonalizes $A$ - which brings the projection operators to a simple form (only one number $1$ on the diagonal) - and then he simply applies the function $g$ to the eigenvalues. In other words, $$A = U D U^{-1} \quad \Rightarrow \quad g(A) = U g(D) U^{-1}$$ where $g(D)$ is simply a diagonal matrix with entries $g(D_{ii})$ on the diagonal. The formula above works because $U^{-1} U$ cancel everywhere in the middle if we write $g(A)$ e.g. as a Taylor expansion - and by generalization, we just declare the formula above to be right even if the Taylor expansion is not appropriate for a mathematician.

The Taylor expansion for the exponential is always OK from a physicist's viewpoint.

All these objects - Hilbert spaces, operators, their functions (especially exponentials), spectra, eigenvalues - and all these operations - exponentiation, search for projection operators etc. - are important in physics, indeed. And that's what mathematicians often declare when they want their students to listen. But it's equally true that all the points that mathematicians actually focus on most of their lives are totally uninteresting from a scientific viewpoint.

This is why the comments that the "material is important in physics" is morally wrong.

Mathematicians shouldn't try to support the attractiveness of their teachings by physical applications - especially because their real goal (and the real goal of pure maths) is to make them as independent of natural science as possible. One can't have it both ways. Doing physics or science means that one is allowed to "prove" all these things - such as $g(A)=g(A)$ which is really the content of the "difficult" proof you sketched) - much more elegantly and arguably naively than in maths. One is only worried about the lack of rigor if he can actually find a contradiction with the experiment or other calculations. If it doesn't exist, the science is just perfectly OK.

On the other hand, mathematicians are usually looking for problems even if no problems from a scientist's viewpoint exist. This implies a totally different set of priorities and it is unlikely that a student is going to be excited about both. One either prefers to measure the truth according to nitpicking based on predetermined sets of axioms - which is the mathematician's viewpoint - or one is ready to adjust his methods, axioms, and precise definitions of objects (and invent or learn totally new branches of physics) as needed to agree with the empirical data and other accurate calculations - which is the physicist's viewpoint.

They're different attitudes and this is why I think that your question should have been posted on a math forum because it's not really following physicist's way of thinking and it is not really motivated by the desire to understand Nature.

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