Quantum particle's state in momentum eigenfunctions basis

[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.

 

[martinbn]

How do we cope with this ? Thanks.

Most often people cope with this by ignoring the subtlety, which is probably the best thing to do at first. Otherwise you need to find a text that treats it in detail. Look into rigged Hilbert spaces or Gelfand's triples.

https://en.wikipedia.org/wiki/Rigged_Hilbert_space
https://ncatlab.org/nlab/show/rigged+Hilbert+space
https://ncatlab.org/nlab/show/Gelfand+triple
https://arxiv.org/abs/quant-ph/0502053

 

[cianfa72]

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) ?

 

[martinbn]

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

 

[cianfa72]

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 $g: (\mathbb R^3, \mathcal L) \rightarrow (\mathbb C, \mathcal B)$

 

[martinbn]

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.

 

[cianfa72]

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).

 

[cianfa72]

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 $\mathbb C^2$.

 

[jbergman]

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.

 

[cianfa72]

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?

 

[dextercioby]

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

 

[cianfa72]

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 $\psi(x)$).

 

[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?

Yes, integrable means finite integral. For example $\frac1{x^2}$ is integrable, but when multiplied by $x$ gives you a non integrable one.

 

[martinbn]

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.

 

[jbergman]

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?

 

[jbergman]

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.