# Rigorous spectral theorem in quantum mechanics

As far as I am aware, ordinary (nonrelativistic) quantum mechanics can be put on a completely rigorous mathematical foundation (unlike, say, QFT).

With that said, having learned some functional analysis, I remember spectral theorems existing for

• Compact, self-adjoint operators on Hilbert spaces, which is very similar to the finite-dimensional case;

• Bounded, self-adjoint operators on Hilbert-spaces, where the spectral theorem involves a projection valued measure and is essentially given by $$A=\int_{\sigma(A)}\lambda dE_\lambda.$$

The problem is that when we represent the Hilbert-space of physical states as $L^2(\mu)$, then all relevant operators are differential operators, multiplication operators or some kind of combinations of the two. To the best of my knowledge, neither of these operators are bounded, and definitely not compact.

Moreover, as I recall from functional analysis, eigenvalues/eigenvectors can only be assigned to point spectra and only compact operators are guaranteed to have point spectrum.

Question: How to reconcile these claims with quantum mechanics? Is it possible to turn the position and momentum operators into bounded operators somehow? If so, is a satisfactory formulation of QM possible where projection operators of the form $\left|n\right\rangle\left\langle n\right|$ (essentially tensor product operators) are replaced with the projection valued measure?

Or how does one do this rigorously?

• You should look into the theory of unbounded operators. There is a spectral theorem for unbounded normal (in particular self-adjoint) operators. One could try to avoid dealing with unbounded operators by only considering spectral projections as observables (forming an algebra) but still needs to know the spectral theorem. A satisfactory formulation of QM with PVMs is certainly possible. For a broad overview of rigorous results I recomend the book by Peter Bongaarts. Mar 27, 2017 at 8:24

The spectral theorem holds for any self-adjoint (actually normal) operator on a separable Hilbert space. In particular, it holds for unbounded self-adjoint operators. In the formula $$A=\int_{\sigma(A)}\lambda \,\mathrm{d}E(\lambda)\; ,$$ the $\sigma(A)$ is simply an unbounded closed subset of the reals, and not a bounded one (usually, it is of the form $[M,\infty)$ for some $M\in\mathbb{R}$). Functional calculus is also available for unbounded operators and unbounded functions (such as taking the square root of the modulus, exponentiating, etc.).
In conclusion, it is perfectly possible to set non-relativistic quantum mechanics in the mathematical framework of operator theory in the Schrödinger representation of the canonical commutation relations (i.e. in $L^2(V)$, $V$ finite dimensional, where position is the multiplication operator by the canonical variable $x$ of $V$, and the momentum is $-i\nabla_x$).
• I don't think you need separability for the spectral theorem. Then again all sources I seem to find just write $H$ without specifying if it's assumed to be separable, so I'm not positive. Also I have trouble interpreting the integral expression. It's clear for bounded operators, but for unbounded operators I have only seen integral representations for expectation values. The expression suggests that the integral converges in some sense to an unbounded operator. Do you know sources where this is defined? Mar 27, 2017 at 16:18
• @Adomas: the best reference I know about these integrals $\int f(\lambda) dE(\lambda)$ introduced by von Neumann, is the book "Unbounded Self-adjoint Operators on Hilbert Space" by Konrad Schmudgen. BTW you only need separability if you want to prove the spectral without the axiom of choice (i.e. only with DC) when decomposing your space in terms of cyclic vectors. Apr 8, 2017 at 18:49