This question comes to mind while reading the quote in different books which is "Hilbert space is infinite dimensional". While the electron spin is 2 Dimensional problem. Secondly, we assume that spin wave function is square integrable which is a property of a function in Hilbert space. THen how do we know if spin wavefunctions belongs to Hilbert Space or not? Please help me clear this concept of Hilbert space.
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1Honestly, I find the usage of "Hilbert space" in physics needlessly confusing. It's frequently used as a synonym for "the state space of the system under consideration," when all it means is a particular kind of vector space. It makes no more sense to say that something "lives in Hilbert space" (which we see all the time) than to say it "lives in vector space." For the same reason, "Hilbert space is infinite dimensional" makes as little sense as "vector space is infinite dimensional." – A_P Sep 17 '21 at 16:04
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A Hilbert space is a complex vector space with an inner product that is complete w.r.t. the norm induced by that inner product. It may be finite-dimensional, it may be infinite-dimensional - the dimensionality is not part of the definition. Neither is square-integrability. Either your sources are wrong - or you are misinterpreting them, not paying attention to a difference between the generic definition of a Hilbert space and the definition of a specific Hilbert space as the space of states of a physical system.
The space of states of position (or momentum) wavefunctions of objects without spin in quantum mechanics is the infinite-dimensional Hilbert space of square-integrable functions $L^2(\mathbb{R}^n)$.
The space of (spin) states for a particle with spin $s$ is the finite-dimensional Hilbert space $\mathbb{C}^{2s+1}$.
Since spaces of states are combined via the tensor product, the space of states of position (or momentum) wavefunctions for an object with spin $s$ is the Hilbert space $L^2(\mathbb{R}^n)\otimes \mathbb{C}^{2s+1}$, which is the space of $\mathbb{C}^{2s+1}$-valued functions that are square-integrable in each component.
ACuriousMind
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1Thanks for this excellent clarification. I was reading Griffiths on this topic and through his whole discussion he assumed wavefunctions as a function of position, probably for reader's convenience. Later I found in Zettili the exact definition of Hilbert Space. Your comment helped me a lot. Thanks again. – AB1998 Sep 16 '21 at 12:45
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For what it's worth, "wave function" (almost?) always refers to the Hilbert space of position / momentum (which, as you know, occupy the same space), and confusingly sometimes only to the position representation of that space. Nobody calls a spin state a "wave function" (though again, confusingly, sometimes the position $\otimes$ spin state is called that). – A_P Sep 17 '21 at 16:10
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A Hilbert space is a vector space $\mathcal H$ with an inner product which satisfies a particular metric condition with respect to the sense of distance induced by this inner product. (Specifically, the space must be complete, i.e., every Cauchy sequence in $\mathcal H$ converges to a limit in $\mathcal H$.)
Every finite-dimensional inner-product space is complete (because it is isomorphic to $\mathbb C^n$ and $\mathbb C^n$ is complete), so therefore every finite-dimensional inner-product space is a Hilbert space.
As such, anybody who claims that "Hilbert space [must be] infinite dimensional" is basically dead wrong about that. Hilbert spaces can be infinite-dimensional, but they don't need to be.
Emilio Pisanty
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"anybody who claims that "Hilbert space [must be] infinite dimensional" is basically dead wrong" - this statement may be too strong, as this depends on the definition. Sometimes Hilbert space is defined as an infinite-dimensional one (https://www.britannica.com/science/Hilbert-space) – akhmeteli Sep 16 '21 at 12:52
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1Emilio, you answer is excellent as usual, but completeness is not a topological condition, as, roughy speaking, it is not invariant under homeomorphisms. – Valter Moretti Sep 16 '21 at 15:46
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1@ValterMoretti Many thanks! This one got away from me, and I still find it surprising. Indeed, completeness is not topological. Who knew! (apart from, you know, everybody...). – Emilio Pisanty Sep 16 '21 at 20:04
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1$(-\infty,\infty)$ and $(-1,1)$ are homeomorphic but the former is complete whereas the latter is not, using the natural metric topology. (Ah yes you quoted a link on that fact). – Valter Moretti Sep 16 '21 at 20:13