I Is the result of an operator acting on a ket vector always a ket？

[gaiussheh]

In quantum mechanics, is the result of a linear operator acting on a ket vector always a ket vector? I’ve seen many textbooks state this, but in the case of an infinite square well, when the momentum operator acts on the ground state wave function $\mathrm{sin}(πx/l)$, the resulting cosine function doesn’t satisfy the boundary conditions and thus doesn’t belong to this Hilbert space?

 

[PeroK]

In quantum mechanics, is the result of a linear operator acting on a ket vector always a ket vector? I’ve seen many textbooks state this, but in the case of an infinite square well, when the momentum operator acts on the ground state wave function $\mathrm{sin}(πx/l)$, the resulting cosine function doesn’t satisfy the boundary conditions and thus doesn’t belong to this Hilbert space?

Correct! The infinite square well is the limiting case of the finite square well with unbounded potential. Although presented in introductory texts, it is neither physically not mathematically valid, if you look closely enough.

 

[Haborix]

To clear up a semantic issue, I think it is more correct to say an operator always maps a ket vector to a ket vector, but the domain and range of the operator may be different vector spaces.

 

[PeroK]

To clear up a semantic issue, I think it is more correct to say an operator always maps a ket vector to a ket vector, but the domain and range of the operator may be different vector spaces.

That's very much a mathematical issue!

 

[Steve4Physics]

That's very much a mathematical issue!

Technically, I assume it should be "That would be mathematical matter."

(Link is here for anyone puzzled.)

 

[Albertus Magnus]

Correct! The infinite square well is the limiting case of the finite square well with unbounded potential. Although presented in introductory texts, it is neither physically not mathematically valid, if you look closely enough.

So what are the details? When is it the case that an operator defined on a Hilbert space has it's image also contained within the given Hilbert space? Further, why does the momentum operator not meet the criteria?

 

[PeroK]

So what are the details? When is it the case that an operator defined on a Hilbert space has it's image also contained within the given Hilbert space? Further, why does the momentum operator not meet the criteria?

In general, a linear operator is a mapping from one vector space to another. For any vector space, we can consider the special case of linear operators that map the vector space to itself. The starting point for the formalism of QM is that we have a vector space of valid states (or wave functions) for the system and each observable is represented by a linear operator on that vector space. I.e. the operator maps the vector space to itself.

For the infinite square well, the valid wave functions must be zero at the boundary. This means that the momentum operator is not a valid linear operator on this vector space of wave functions. This is because the derivative of a function is not necessarily zero at the end points.

This technicality doesn't stop the infinite square well being used as an interesting example of wave mechanics. If you think about it, a truly infinite potential is unphysical.

I would draw an analogy with the infinite uniformly charged plate in classical EM. The electric field is uniform, hence a particle in that field has infinite potential energy. All of these examples: infinite straight wire, infinite plate, infinite square well are all likely to run into a mathematical/physical problem at some point.

 

[gaiussheh]

In general, a linear operator is a mapping from one vector space to another. For any vector space, we can consider the special case of linear operators that map the vector space to itself. The starting point for the formalism of QM is that we have a vector space of valid states (or wave functions) for the system and each observable is represented by a linear operator on that vector space. I.e. the operator maps the vector space to itself.

For the infinite square well, the valid wave functions must be zero at the boundary. This means that the momentum operator is not a valid linear operator on this vector space of wave functions. This is because the derivative of a function is not necessarily zero at the end points.

This technicality doesn't stop the infinite square well being used as an interesting example of wave mechanics. If you think about it, a truly infinite potential is unphysical.

I would draw an analogy with the infinite uniformly charged plate in classical EM. The electric field is uniform, hence a particle in that field has infinite potential energy. All of these examples: infinite straight wire, infinite plate, infinite square well are all likely to run into a mathematical/physical problem at some point.

I think my question is more like "Okay you can say this problem is overly-simplified, it doesn't exist in the real world, blah blah blah, but really, how do you know that when you talk about abstract vector spaces representing quantum states, you can always write down $\hat{O}|\psi\rangle$? How do you know that it is not ill-defined?"

 

[PeroK]

I think my question is more like "Okay you can say this problem is overly-simplified, it doesn't exist in the real world, blah blah blah, but really, how do you know that when you talk about abstract vector spaces representing quantum states, you can always write down $\hat{O}|\psi\rangle$? How do you know that it is not ill-defined?"

That's an axiom of QM. Like how do you know that $F = ma$ is always satisfied? Newton;s law looks more natural because 3D vectors seem intuitively close to real forces. Whereas, abstract kets seem somewhat disconnected from our intuitive picture of reality. But, in their own ways, they are mathematical formulations that are taken as an axiom of the mathematical model in question.

And, in fact, just as Newtonian mechanics must eventually give way to Relativity and QM, so QM must eventually give way to QFT and an even more abstract and advanced set of axioms.

 

[Albertus Magnus]

I think my question is more like "Okay you can say this problem is overly-simplified, it doesn't exist in the real world, blah blah blah, but really, how do you know that when you talk about abstract vector spaces representing quantum states, you can always write down $\hat{O}|\psi\rangle$? How do you know that it is not ill-defined?"

You know you can write $\hat O|\psi\rangle$ for $|\psi\rangle$ in any abstract vector space if you are certain that $\hat O$ has been well defined for some domain in the given space containing $|\psi\rangle$. In the specific case of the momentum operator, it is not well defined on the Hilbert space of square well state functions. Thus, $\hat p|\psi\rangle$ is technically not a linear operator on the space but a linear transformation from one space to another.

 

[Demystifier]

In quantum mechanics, is the result of a linear operator acting on a ket vector always a ket vector? I’ve seen many textbooks state this, but in the case of an infinite square well, when the momentum operator acts on the ground state wave function $\mathrm{sin}(πx/l)$, the resulting cosine function doesn’t satisfy the boundary conditions and thus doesn’t belong to this Hilbert space?

A definition of an operator involves
(i) a rule of acting, in this case $\hat{p}\psi(x)=-i\hbar\partial_x\psi(x)$
and
(ii) a domain ${\cal D}$ of functions $\psi(x)$ on which this acts.
When domain is such that $\psi\in{\cal D}$ implies $\hat{p}\psi\in{\cal D}$, then we say that $\hat{p}$ is self-adjoint. In your case above, the ${\cal D}$ is spanned by $\mathrm{sin}(πx/l)$ alone, without the corresponding $\mathrm{cos}$-functions, so in this case $\hat{p}$ is not self-adjoint. To make it self-adjoint you have to consider a bigger domain, in which case we talk of self-adjoint extension of the operator.

For more details see e.g. the review https://arxiv.org/abs/2103.01080.

 

[PeroK]

A definition of an operator involves
(i) a rule of acting, in this case $\hat{p}\psi(x)=-i\hbar\partial_x\psi(x)$
and
(ii) a domain on which this acts, namely a Hilbert space ${\cal H}$ of functions $\psi(x)$.
When ${\cal H}$ is such that $\psi\in{\cal H}$ implies $\hat{p}\psi\in{\cal H}$, then we say that $\hat{p}$ is self-adjoint.

That's the definition of a function from ${\cal H}$ to ${\cal H}$. A self-adjoint operator is one where the dual of the dual operator is equivalent to the original operator. This is the most general definition, although often it is sufficient for an operator to be its own Hermitian conjugate.

 

[Demystifier]

That's the definition of a function from ${\cal H}$ to ${\cal H}$. A self-adjoint operator is one where the dual of the dual operator is equivalent to the original operator. This is the most general definition, although often it is sufficient for an operator to be its own Hermitian conjugate.

Yes, I was referring specifically to the $\hat{p}=-i\hbar\partial_x$ operator, so I took for granted that it is symmetric (hermitian) and wrote the additional condition needed for it to be self-adjoint.

 

[PeroK]

Yes, I was referring specifically to the $\hat{p}=-i\hbar\partial_x$ operator, so I took for granted that it is symmetric (hermitian) and wrote the additional condition needed for it to be self-adjoint.

That's very much a physicist's approach. Mathematically, it must be a well-defined operator on the Hilbert space to begin with. That's the first and foremost condition.

 

[gentzen]

Yes, I was referring specifically to the $\hat{p}=-i\hbar\partial_x$ operator, so I took for granted that it is symmetric (hermitian) and wrote the additional condition needed for it to be self-adjoint.

Your original condition (before the edit) was only sufficient, but not necessary. Your new condition (after the edit) fails to mention that the domain $\mathcal D$ should be dense in "the" Hilbert space $\mathcal H$. I have no idea whether it is sufficient or necessary.

 

[Morbert]

An interesting preprint from last month detailing the modelling of measurements on different intervals. I believe section A3 corresponds to OP's case.

https://arxiv.org/abs/2502.08494

 

[gaiussheh]

I came up with a more natural example that avoids the discussion of functions in a square well. See https://www.physicsforums.com/threa...ps-a-vector-out-of-the-hilbert-space.1078949/