X measurement modeled in non-separable Hilbert space

[normvcr]

TL;DR

Following a QP textbook for modeling a particle in Euclidean space, but the action of the symmetry group (Galilei group) is not continuous. This seems wrong.

Am reading a book (Ballentine, "Quantum Mechanics: A modern development) which I have found very helpful. Am now puzzled by section 3.4, where the position operator satisfies Q|x> = x |x> (I have simplified from 3 dims to 1 dim). Here, x is any real number. There are, thus, uncountably many eigenvalues for Q, and the eigenvectors are orthogonal: <x|y> = 0 for x != y, as they have distinct eigenvalues, and Q is self-adjoint (though not bounded). So far, so good. The group of translation symmetries acts as follows:

Pa |x> = |x+a>​

Herein lies the problem. The group action is supposed to be continuous, for the sake of being physically reasonable. But, we cannot have

Pa |x> --> |x> as a --> 0​

since

| Pa |x> - |x> |^2 = 2 for a != 0 (this follows from <x | Pa |x> = 0 for a != 0 )​

​

??
Thank you.

 

[Demystifier]

The norm that you consider, as a function of $a$, has a jump at $a=0$. Therefore you cannot find the value at $a=0$ by considering the limit $a\to 0$.

 

[normvcr]

I agree, and that is precisely my point: The group action is not continuous. Is this not a problem? Certainly conceptually, but also technically, as it is the continuity of the group action that guarantees the group generators have a common, invariant, dense domain in the Hilbert space.

 

[jbergman]

I agree, and that is precisely my point: The group action is not continuous. Is this not a problem? Certainly conceptually, but also technically, as it is the continuity of the group action that guarantees the group generators have a common, invariant, dense domain in the Hilbert space.

What Hilbert space does $\ket{x}$ live in? Technically none as I understand it as it is a delta function.

 

[strangerep]

Am reading a book (Ballentine, "Quantum Mechanics: A modern development) which I have found very helpful. Am now puzzled by section 3.4, [...]

I get the feeling you didn't read (or don't remember) section 1.4 "Hilbert Space and Rigged Hilbert Space". (?)

We've had extensive threads on PF about this. Try doing some searches.

 

[normvcr]

I did read the RHS section 1.4, but did not think that RHS was adopted by the text as the place to model QP -- only that this is one of the directions people are thinking about. For example, the postulates of section 2 assume Hermitian operators i.e. Hilbert spaces. Unitary operators are discussed only in the context of Hilbert spaces (section 3.1).

Indeed, there are many interesting discussions about RHS on this forum. I need to resolve for myself how to move forward -- use RHS or stay within Hilbert space itself, or ... . Thanks for the pointers.

 

[jbergman]

I did read the RHS section 1.4, but did not think that RHS was adopted by the text as the place to model QP -- only that this is one of the directions people are thinking about. For example, the postulates of section 2 assume Hermitian operators i.e. Hilbert spaces. Unitary operators are discussed only in the context of Hilbert spaces (section 3.1).

Indeed, there are many interesting discussions about RHS on this forum. I need to resolve for myself how to move forward -- use RHS or stay within Hilbert space itself, or ... . Thanks for the pointers.

You could also just look at $\|P_a\ket{\psi} - \ket{\psi}\|^2$, where $\ket{\psi}$ is in the Hilbert space with unit norm. Then, I believe, you will find that it is continuous in a. You can just imagine two wave functions that are overlapping more and more.

I've found Hall's book Quantum Theory for Mathematicians to be the clearest exposition of ways to deal with these kind of issues.

 

[normvcr]

Agreed, if you are referring to the action taking place in the separable Hilbert space $L^2(R)$. In this case, $P_a\Psi (x) = \Psi(x-a)$, and things are continuous. The problem, of course, is that the x-operator is defined as $Q\Psi (x) = x\Psi(x)$, whose eigen states are not in the Hilbert space . Like everyone else, I want to have my cake and eat it, too! I like Hall's book, and have a confession to make, that I am using the copy from his web site, and have not bought the book. Is this a no-no?
PS I did not find the preview button for previewing a reply. Where is it located, please?

 

[jbergman]

Agreed, if you are referring to the action taking place in the separable Hilbert space $L^2(R)$. In this case, $P_a\Psi (x) = \Psi(x-a)$, and things are continuous. The problem, of course, is that the x-operator is defined as $Q\Psi (x) = x\Psi(x)$, whose eigen states are not in the Hilbert space . Like everyone else, I want to have my cake and eat it, too! I like Hall's book, and have a confession to make, that I am using the copy from his web site, and have not bought the book. Is this a no-no?
PS I did not find the preview button for previewing a reply. Where is it located, please?

I think it's fine to use his book if its on his site.

As to your original question, I would say that eigen state like $\ket{x}$ isn't even a physical state so you really should limit yourself to states like I described. Also, I don't think separable or not has anything to do with whether $\ket{x}$ is in it or not, but I could be wrong.

 

[normvcr]

Are you saying $\ket{x}$ is not a physical state, because no measuring device can be so accurate? That makes sense. I like the example of the Hilbert space of band-limited functions, whose spectral domain is contained in the interval $[-1/2, 1/2]$. An orthonormal basis is $$sinc(t-N), N = 0, \pm1, \pm 2, ...$$. Each of these makes sense as a physical state.