Infinite Dimensional Vector Spaces

Infinite Dimensional Vector Spaces

Given a basis, we can always represent \(|a\rangle\in\mathcal{V}\) as \(|a\rangle:=(a^1\:a^2\:a^3\:\cdots\:a^n)^T\) (with \(n\) to infinity). So one obvious way to get an infinite dimensional vector space is to allow \(\infty\) sequences. \(\Rightarrow\) \(|a\rangle:=(a^1\:\cdots)^T\in\mathbb{C}^\infty\). The standard basis is then \(|e_1\rangle = (1\:0\:\cdots)^T,|e_2\rangle = (0\:1\:0\:\cdots)^T,\cdots\).


Could we even tell the difference between dimension of \(\mathcal{V}\) being very large versus infinite (countable or uncountable).

Consider for two operators \([A,B]=\lambda \mathbb{I}\). Take the trace, \(\text{Tr}([A,B]) = \text{Tr}(AB)-\text{Tr}(BA) = \text{Tr}(AB)-\text{Tr}(AB) = 0 =\text{Tr}(\lambda \mathbb{I})=\lambda \text{Tr}(\mathbb{I})=\lambda \dim \mathcal{V} = \lambda n\). Thus, there are no operators on a finite dimensional vector space that has a commutator proportional to the Identity operator, Schor’s Lemma.

Consider \([x,p]=i\hbar\). Thus, they must be infinite dimensional. (If we were in a finite dimensional space then there would be no quantum mechanics)

Author: Christian Cunningham

Created: 2024-05-30 Thu 21:15