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Stream: deprecated: physics

Topic: densely defined operators in quantum mechanics

Owen Lynch (Mar 04 2023 at 00:34):

This is a very basic question, but I don't have much QM background. When we do the 1D wave equation, the underlying Hilbert space is H = L^2(R), correct? Then for any finite time, we have a unitary operator on H which is the evolution of the wave for that time. However, the wave equation uses a second partial derivative. So it seems like the Hamiltonian for the Schrodinger equation here is only defined on a dense subset of H?

Owen Lynch (Mar 04 2023 at 00:38):

I'm familiar with the theory for stochastic processes with the Hille Yosida theorem, is there an analogous theorem for quantum mechanics?

John Baez (Mar 04 2023 at 01:26):

Owen Lynch said:

So it seems like the Hamiltonian for the Schrodinger equation here is only defined on a dense subset of H?

Yup. Whenever you have a quantum system where energies are arbitrarily large, the Hamiltonian is only densely defined - since an everywhere defined self-adjoint operator is necessarily bounded. So, most of the work involved in understanding self-adjoint operators involves figuring out how to handle those that are densely defined and unbounded. The bounded ones are trivial by comparison.

John Baez (Mar 04 2023 at 01:28):

Owen Lynch said:

I'm familiar with the theory for stochastic processes with the Hille Yosida theorem, is there an analogous theorem for quantum mechanics?

You bet! Stone's theorem on one-parameter unitary groups. This is one of the most important theorems in quantum mechanics. Since it relates observables to one-parameter groups of symmetries, it's philosophically connected to Noether's theorem, which does the same thing in a different context.

John Baez (Mar 04 2023 at 01:29):

I recommend Reed and Simon's Functional Analysis for the basics of this, and then their book Fourier Analysis and Self-Adjointness for how to actually prove operators are self-adjoint.

Owen Lynch (Mar 04 2023 at 01:46):

Is there an algebraic connection between unbounded operators and bounded operators? I.e. bounded operators form a C*-algebra; what sort of algebraic structure do unbounded self-adjoint operators form?

Owen Lynch (Mar 04 2023 at 02:19):

I'm now reading about "affiliated operators"

Owen Lynch (Mar 04 2023 at 02:41):

(I'm also reading the classical treatment; I'm not going too crazy with trying to categorify everything)

John Baez (Mar 04 2023 at 04:44):

Owen Lynch said:

Is there an algebraic connection between unbounded operators and bounded operators?

Yes.

I.e. bounded operators form a C*-algebra; what sort of algebraic structure do unbounded self-adjoint operators form?

Nothing good. What you need to do is take an unbounded self-adjoint operator $A$, form all the C*-algebra of bounded operators $f(A)$ that you can define by applying $L^\infty$ functions $f \colon \mathbb{R} \to \mathbb{C}$ to it using the [[functional calculus]], and work with them.

John Baez (Mar 04 2023 at 04:45):

In particular, Stone's theorem says knowing $A$ is the same as knowing all the unitary operators $\exp(it A)$.

John Baez (Mar 04 2023 at 04:49):

There's a huge amount to say about this subject, but the first set of tricks you need to learn are the functional calculus for unbounded self-adjoint operators, and the various ways to work with such operators $A$ indirectly by using the operators $f(A)$ instead.

Owen Lynch (Mar 04 2023 at 05:59):

OK, I like that picture! That gives me a good direction to go in.

John Baez (Mar 04 2023 at 06:14):

Good!

Owen Lynch (Mar 05 2023 at 08:08):

I finally understand the connection between the conditions on the Hille Yosida theorem and the conditions on the Stone theorem. Self adjoint means the spectrum is purely real, and so the spectrum of $e^{itA}$ is all in the unit circle, which is necessary to be unitary The condition for the Hille Yosida theorem is saying that the spectrum contains no positive values, and because we are working in a real Banach space, this means the spectrum is contained in the negative values. So the spectrum for $e^{tA}$ is contained in [0,1], which is necessary to be a contraction.

John Baez (Mar 05 2023 at 18:23):

Yes! I'm kind of surprised that you learned Hille-Yosida before Stone, because I'm pretty sure it came later and was developed as a kind of generalization of Stone's theorem. I learned Stone's theorem first. But I guess you ran into the math of Markov processes before the math of quantum mechanics... which makes some sense: probability theory is less weird than quantum mechanics.

John Baez (Mar 05 2023 at 18:27):

There aren't as many norm-preserving 1-parameter groups in Banach spaces that aren't Hilbert spaces, because the unit ball in a general Hilbert space isn't 'round'.

For example, if you have a finite-dimensional real Banach space that has enough norm-preserving linear transformations to map any point on the unit sphere to any other point on the unit sphere, the norm needs to be a Hilbert space norm!

John Baez (Mar 05 2023 at 18:28):

So the theory of 1-parameter norm-preserving groups on a Banach space is much less important than the theory of 1-parameter contractive semigroups, unless the Banach space is a Hilbert space.

Owen Lynch (Mar 05 2023 at 21:00):

Is there a more general spectral theorem for operators satisfying the conditions for Hille-Yosida? I learned about Hille-Yosida in a class on interacting particle systems that didn't have a functional analysis prerequisite; I took functional analysis the next year and I'm only now trying to put all the pieces together in my head.

Owen Lynch (Mar 05 2023 at 21:00):

I.e., the spectral theorems I've seen have been for self-adjoint operators

Owen Lynch (Mar 06 2023 at 05:05):

OK, I'm now seeing how Hille-Yosida is more general:

image.png

Owen Lynch (Mar 06 2023 at 05:07):

Because for a self-adjoint operator $A$, $\rho(iA) \subset \mathbb{C} \setminus i \mathbb{R}$, which then satisfies the conditions of this theorem.

Owen Lynch (Mar 06 2023 at 05:07):

The surprising thing is that $(iii)$ is equivalent to $(ii)$!

John Baez (Mar 06 2023 at 18:22):

Yes! I guess it's related to how changing the imaginary part of $c \in \mathbb{C}$ doesn't change the growth (or decay) rate of $\exp(ct)$ as $t \to \infty$ along the real axis.