The discussion centers on the behavior of linear operators in quantum mechanics, specifically whether the result of an operator acting on a ket vector remains within the same Hilbert space. It is established that while an operator maps a ket vector to another ket vector, the resulting vector may not belong to the original Hilbert space, as demonstrated with the momentum operator acting on the ground state wave function of an infinite square well. The momentum operator, defined as ##\hat{p} = -i\hbar\partial_x##, does not satisfy the boundary conditions required for the wave functions in this context, indicating that it is not a valid linear operator on this vector space.
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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.gaiussheh said: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?
That's very much a mathematical issue!Haborix said: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.
Technically, I assume it should be "That would be mathematical matter."PeroK said:That's very much a mathematical issue!
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 said: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.
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.Albertus Magnus said: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?
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 said: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.
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.gaiussheh said: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.gaiussheh said: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?"
A definition of an operator involvesgaiussheh said: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?
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 said: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.
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 said: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.
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.Demystifier said: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.Demystifier said: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.