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The dimension of the Hilbert space of a free particle is countable. To see this, simply note that The Hilbert space of a free particle in three dimensions is $L^2(\mathbb{R}^3)$. The dimension theorem guarantees that any two bases of of a vector space have the same cardinality, which allows us to define the dimension of a vector space as the cardinality ...

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Although I think user Siva's argument is nice for intuition, I feel that the key mathetmatical point is being obscured; you just need to be careful about what you mean by the "size" of a vector space. The dimension theorem for vector spaces tells us that any two bases for a vector space must have the same cardinality. This allows us to define the dimension ...

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This question first posed to me by a friend of mine. For the subtleties involved, I love this question. :-) The "flaw" is that you're not counting the dimension carefully. As other answers have pointed out, $\delta$-functions are not valid $\mathcal{L}^2(\mathbb{R})$ functions, so we need to define a kosher function which gives the $\delta$-function as a ...

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The Hilbert space ${\cal H}$ of the one-dimensional harmonic oscillator in the position representation is the set $L^{2}(\mathbb{R})$ of square integrable functions $\psi:\mathbb{R}\to\mathbb{C}$ on the real line. The Dirac delta distribution $\delta(x-x_{0})$ is not a function and it is not square integrable. See also this Phys.SE post.

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Microstate:the number of distinct arrangements of partical in cells in phase space , Macrostate:the arrangements of partical in cells in phase space when partical are identical &the number of Microstate in a given Macrostate is called Thermodynamic probability

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The derivation by Sakurai is by no means mathematicaly rigorous, so you should expect something like your argument about the scalar product. Indeed, we have everything more or less fine until $$[x,\mathcal{T}(\epsilon)]|z\rangle=\epsilon|z+\epsilon\rangle$$ where we want to replace $|z+\epsilon\rangle$ by $|z\rangle$ and claim that it is ok in the first ...

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Here's the most logical way to proceed if you ask me. Given any $a\in\mathbb R$, we define the translation operator $T_a$ by its action on position basis vectors $$T_a|x\rangle = |x + a\rangle$$ One can prove the following properties: $T_a$ is unitary for each $a\in\mathbb R$. $T_aT_b = T_{a+b}$ for all $a,b\in\mathbb R$. It follows (by Stone's ...

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