Unbounded operators

Stream: PhysLean

[Izan Beltran]

Regarding #Is there code for X? > Unbounded operators (Quantum Mechanics).

I have seen PhysLean has some definitions like UnboundedOperator, but for example the Momentum operator is defined as

def momentumOperator (ψ : ℝ → ℂ) : ℝ → ℂ := fun x ↦ - Complex.I * ℏ * deriv ψ x

why is it like this, and not trying to frame it with some operators more general definition?

I see later on, some theorems like

@[fun_prop]
lemma continuous_momentumOperator (ψ : ℝ → ℂ) (hψ : ContDiff ℝ 1 ψ) :
    Continuous (momentumOperator ψ) := by
  rw [momentumOperator_eq_smul]
  fun_prop

So I guess it is handled in this more naive/case-specific way. I'd like to understand why it was formalized like this. Thanks!

[Joseph Tooby-Smith]

So further down that same file, are two variations of the momentum operator:

def momentumOperatorSchwartz : 𝓢(ℝ, ℂ) →L[ℂ] 𝓢(ℝ, ℂ) where ....

and

def momentumOperatorUnbounded : UnboundedOperator schwartzIncl schwartzIncl_injective :=
  UnboundedOperator.ofSelfCLM momentumOperatorSchwartz

The former is the momentum operator defined as a continuous linear map between Schwartz spaces, and the second is the bonafide momentum operator as an unbounded operator - with the domain equal to the Schwartz space.

momentumOperator is really just there to be available in proving theorems or stating results initially without getting too involved with unbounded operators etc.

[Joseph Tooby-Smith]

In addition to this @Afiq Hatta is working on the QM reflectionless potential defining things properly in terms of unbounded operators, and the related discussion to this is #PhysLean > QM reflectionless potential .

[Joseph Tooby-Smith]

Any improvements to this API would be very much welcome :).

[Izan Beltran]

Hi! Thanks for this explanation. Now I have checked more carefully the definitions, and I have another question: why is UnboundedOperator defined as

/-- An unbounded operator on the one-dimensional Hilbert space,
  corresponds to a subobject `ι : S →L[ℂ] HilbertSpace` of the Hilbert
  space along with the operator `op : S →L[ℂ] HilbertSpace` -/
@[nolint unusedArguments]
def UnboundedOperator {S : Type} [AddCommGroup S] [Module ℂ S]
    [TopologicalSpace S] (ι : S →L[ℂ] HilbertSpace)
    (_ : Function.Injective ι) := S →L[ℂ] HilbertSpace

I was curious about this since →L[ℂ] apparently denotes ContinuousLinearMap, which is equivalent to a "Bounded Linear Map", so this seems to be conflicting with the unbounded case. I saw it is used with Schwartz space defined operators mainly, and it makes sense in that case. My question is: why did you choose to call it Unbounded Operator?

[Joseph Tooby-Smith]

I'm slightly confused by your question here. An unbounded operator from $X$ to $Y$ is a (continuous) linear map $T$ from a linear subspace $D(T)$ to $Y$.

Here we are defining an unbounded operator from HilbertSpace to HilbertSpace, and S is what corresponds to $D(T)$ (it is a linear embedding into HilbertSpace). Admittedly ι and S should probably be part of the definition rather than an input.

[Izan Beltran]

I see, I was using a different definition apparently. In places like

• https://en.wikipedia.org/wiki/Unbounded_operator#Definitions_and_basic_properties

• https://en.wikipedia.org/wiki/Operator_(physics)#Operators_in_quantum_mechanics

• Mathematical Methods in Quantum Mechanics, Gerald Teschl, Section 0.5

it is usually defined just as a linear map from a subset $D(T) \subseteq X$ to $Y$ but not bounded/continuous (which is equivalent for linear maps), and they develop further the theory of the unbounded operators. In the $L^2(\mathbb{R})$ space, the position operator or the momentum operator are not bounded/continuous in the way I found them defined.

You seem to be working fully on the Schwartz space, using the Schwartz norm instead of the $L^2(\mathbb{R})$ norm and considering Schwartz a subset of $L^2(\mathbb{R})$, is that right? The issue is then how to lift things to $L^2(\mathbb{R})$ to use the inner product?

[Izan Beltran]

Also theorem 0.25 in that book claims this well-known theorem, just to clarify.
theorem 0.25 screenshot

[Joseph Tooby-Smith]

How would you define the position and momentum operators?

TBH I am now thoroughly confused, and the literature doesn't seem to be particularly clear.
The definition of unbounded operators which appeared in PhysLean originated from this conversation: #PhysLean > Turning kets to bras is surjective which I think may answer your question about norms.

[Joseph Tooby-Smith]

I think what I have defined is probably more properly called a linear operator on a rigged Hilbert space (or something like that), and from that one can get an unbounded linear operator in the sense you which to define it (which is not continuous on the induced topology).

[Moritz Doll]

I haven't read everything here, but let me just remark that the spaces under consideration are very important: for instance the Laplacian (and similarly for position and momentum operator) is an unbounded operator from $L^2 \to L^2$ with domain $H^2$ (that is the self-adjoint domain for $\mathbb{R}^n$), it is obviously also a well-defined operator with domain $C^\infty_c$, but then it is only essentially self-adjoint. Considered as an operator $\Delta : H^2 \to L^2$, the Laplacian is bounded (that is almost trivial depending on how you define the Sobolev spaces). Similarly, $\Delta : \mathcal{S} \to \mathcal{S}$ is bounded.
So basically, there are two concurrent stories: the abstract functional analytic one with unbounded operators and the spectral theorem, etc and the concrete one, where the operators are bounded on some smaller space.

[Izan Beltran]

Joseph Tooby-Smith said:

How would you define the position and momentum operators?

TBH I am now thoroughly confused, and the literature doesn't seem to be particularly clear.
The definition of unbounded operators which appeared in PhysLean originated from this conversation: #PhysLean > Turning kets to bras is surjective which I think may answer your question about norms.

Thanks for clarifying. Now I'm reading about Rigged Hilbert Spaces. This is a new concept to me, I think this was the root of discrepancy :)

I was thinking about operators defined in some dense subspace such that they are essentially self-adjoint, so that we can consider the unique self-adjoint extension of them. But I found some papers claiming that RHS are more suitable to describe QM, so I am going through them to understand better why.