# 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 $$|e_1\rangle = (1\:0\:\cdots)^T,|e_2\rangle = (0\:1\:0\:\cdots)^T,\cdots$$.

## Motivation

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)

Created: 2024-05-30 Thu 21:15

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