# Quantum particle's state in momentum eigenfunctions basis

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• cianfa72
cianfa72
TL;DR Summary
How to express quantum particle's state in momentum eigenfunctions basis considering the fact that momentum eigenfunctions are not square-integrable
Hi, as discussed in this recent thread, for a particle without spin the quantum state of the particle is described by a "point" in the Hilbert space of the (equivalence classes) of ##L^2## square-integrable functions ##|{\psi} \rangle## defined on ##\mathbb R^3##.

The square-integrable condition for complex-valued function ##f## means that its square ##|f|^2## has finite Lebesgue integral on the measurable space ##(\mathbb R^3, \mathcal A)## where ##\mathcal A## is the sigma-algebra of Lebesgue measurable sets (or perhaps simply the Borel sigma algebra ##\mathcal B(\mathbb R^3)##).

That said, consider the eigenfunctions ##|{\psi_k} \rangle## of momentum operator ##\vec{P}##. Now the Lebesgue integral of each of their equivalence classes is not finite.

How do we cope with this ? Thanks.

Last edited:
jbergman and PeroK
cianfa72 said:
The square-integrable condition for complex-valued function ##f## means that its square ##|f|^2## has finite Lebesgue integral on the measurable space ##(\mathbb R^3, \mathcal A)## where ##\mathcal A## is the sigma-algebra of Lebesgue measurable sets (or perhaps simply the Borel sigma algebra ##\mathcal B(\mathbb R^3)##).
From a theoretical point of view which type of sigma-algebra is implicitly implied (Borel or Lebesgue) ?

Lebesgue on the domain and Borel on the image i.e. the functions are measurable functions ##f: (\mathbb R^3, \mathcal L) \rightarrow (\mathbb C, \mathcal B)##.

martinbn said:
Lebesgue on the domain and Borel on the image i.e. the functions are measurable functions ##f: (\mathbb R^3, \mathcal L) \rightarrow (\mathbb C, \mathcal B)##.
Actually not ##f## itself but complex-valued functions ##f## such that the square ##g=|f|^2## is a measurable function ##g: (\mathbb R^3, \mathcal L) \rightarrow (\mathbb C, \mathcal B)##

cianfa72 said:
Actually not ##f## itself but complex-valued functions ##f## such that the square ##g=|f|^2## is a measurable function ##g: (\mathbb R^3, \mathcal L) \rightarrow (\mathbb C, \mathcal B)##
It is the same. Composing a continuous function and a measurable function gives a measurable one.

PeroK and cianfa72
martinbn said:
It is the same. Composing a continuous function and a measurable function gives a measurable one.
Ah ok, this because the Borel ##\sigma##-algebra on ##\mathbb C## is generated by open sets in ##\mathbb C## and the map ##|\, .|^2 : (\mathbb C, \mathcal B) \rightarrow (\mathbb C, \mathcal B)## is continuous (i.e. the preimage of borel sets are borel sets as well).

martinbn
What about a particle with spin (like the qbit) ? Its quantum state is defined by a point in the projective Hilbert abstract space of dimension 2.

Does exist in this case the concept of wave function, hence the requirement to work with ##L^2## square-integrable functions ?

Edit: perhaps the point is that the Hilbert space for a spin ##1/2## particle is (or it is isomorphic to) the ##\mathbb C^2## Hilbert space (actually the projective line). Hence it is complete under the norm derivated from the standard inner product in ##\mathbb C^2##.

Last edited:
cianfa72 said:
TL;DR Summary: How to express quantum particle's state in momentum eigenfunctions basis considering the fact that momentum eigenfunctions are not square-integrable

Hi, as discussed in this recent thread, for a particle without spin the quantum state of the particle is described by a "point" in the Hilbert space of the (equivalence classes) of ##L^2## square-integrable functions ##|{\psi} \rangle## defined on ##\mathbb R^3##.

The square-integrable condition for complex-valued function ##f## means that its square ##|f|^2## has finite Lebesgue integral on the measurable space ##(\mathbb R^3, \mathcal A)## where ##\mathcal A## is the sigma-algebra of Lebesgue measurable sets (or perhaps simply the Borel sigma algebra ##\mathcal B(\mathbb R^3)##).

That said, consider the eigenfunctions ##|{\psi_k} \rangle## of momentum operator ##\vec{P}##. Now the Lebesgue integral of each of their equivalence classes is not finite.

How do we cope with this ? Thanks.
Hall's book Quantum Theory for Mathematicians has the most rigorous analysis of this I have seen if you are really interested.

martinbn
In the Hall's book section 3.3 he claims that ##X\psi(x) = x\psi(x)## might fail to be in ##L^2(\mathbb R)##.

However we know that the function ##x## is continuous and ##\psi(x)## is integrable (w.r.t. the Lebesgue integral over the Lebesgue ##\sigma##-algebra over ##\mathbb R##) by hypothesis. Hence ##x\psi(x)## is measurable too.

Perhaps the point is that we know the above Lebesgue integral exists, however we can not say for sure that it is bounded (not ##+\infty##), right?

Is the equality in your first paragraph an equation, or a definition of operator action?

dextercioby said:
Is the equality in your first paragraph an equation, or a definition of operator action?
It is a definition (it defines how the operator ##X## acts on state/vector ##\psi(x)##).

cianfa72 said:
In the Hall's book section 3.3 he claims that ##X\psi(x) = x\psi(x)## might fail to be in ##L^2(\mathbb R)##.

However we know that the function ##x## is continuous and ##\psi(x)## is integrable (w.r.t. the Lebesgue integral over the Lebesgue ##\sigma##-algebra over ##\mathbb R##) by hypothesis. Hence ##x\psi(x)## is measurable too.

Perhaps the point is that we know the above Lebesgue integral exists, however we can not say for sure that it is bounded (not ##+\infty##), right?
Yes, integrable means finite integral. For example ##\frac1{x^2}## is integrable, but when multiplied by ##x## gives you a non integrable one.

mattt and cianfa72
martinbn said:
Yes, integrable means finite integral. For example ##\frac1{x^2}## is integrable, but when multiplied by ##x## gives you a non integrable one.
I was thinking about what happens at infinity. My example isnt great. One should change the function around zero.

cianfa72 said:
In the Hall's book section 3.3 he claims that ##X\psi(x) = x\psi(x)## might fail to be in ##L^2(\mathbb R)##.

However we know that the function ##x## is continuous and ##\psi(x)## is integrable (w.r.t. the Lebesgue integral over the Lebesgue ##\sigma##-algebra over ##\mathbb R##) by hypothesis. Hence ##x\psi(x)## is measurable too.

Perhaps the point is that we know the above Lebesgue integral exists, however we can not say for sure that it is bounded (not ##+\infty##), right?
##\psi \in L^2(\mathbb R) \rightarrow \int |\psi(x)|^2dx < \infty##. Hall's statement is merely that there exists ##\psi \in L^2(\mathbb R)## such that ##\int |x\psi(x)|^2dx## is not bounded.

cianfa72
Suppose one deal with a quantum system with position/momentum and spin not entangled. Then the state of system is in the form $$\ket{\psi} \otimes \ket{\chi}$$ and this is a unit vector in the Hilbert tensor product space (i.e. it has norm 1). It follows that the norms of ##\ket{\psi}## and ##\ket{\chi}## considered separately are such that their product is 1, right?

cianfa72 said:
Suppose one deal with a quantum system with position/momentum and spin not entangled. Then the state of system is in the form $$\ket{\psi} \otimes \ket{\chi}$$ and this is a unit vector in the Hilbert tensor product space (i.e. it has norm 1). It follows that the norms of ##\ket{\psi}## and ##\ket{\chi}## considered separately are such that their product is 1, right?
That's my understanding, yes.

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