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I am new to quantum mechanics. I have been trying to understand why when we want to represent a function $$\psi(x)$$ as a ket in continuous basis |x> we us the integral:

$$\vert \psi(x)\rangle =\int\psi(x)\vert x\rangle dx$$

where in non-continuous basis it is :

$$\sum\psi(x)\vert x\rangle $$ clearly the $dx$ gives different units here so I am not sure if integrals make sense to use to expand the vector in these basis. Also, I have heard that continuous means uncountable which I am not sure how that is uncountable, can't we just index all the basis with natural numbers since last time I checked we have infinity of them?

ZeroTheHero
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NegativeTension
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1 Answers1

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You want to think of $$ \psi(x)=\langle x\vert\psi\rangle $$ as a (complex) number interpreted as the “component” of $\vert\psi\rangle$ on the basis vector $\vert x\rangle$, with $\langle x\vert\bar x\rangle=\delta(x-\bar x)$. This way \begin{align} \vert \psi\rangle &= \int\,dx\, \psi(x) \vert x\rangle \, ,\\ \psi(\bar x)=\langle \bar x\vert\psi\rangle &= \int \,dx\, \psi(x)\langle \bar x\vert x\rangle \end{align} is basically the (continuous) generalization of $$ \vec r = \sum_{i} {\hat \iota} \,\left({\hat \iota}\cdot \vec r \right). $$ where the resolution of the identity \begin{align} \hat I&=\sum_i {\hat \iota}\ {\hat \iota}\cdot\\ \hat I\vec r=\vec r&= \sum_i {\hat \iota}\ {\hat \iota}\cdot \vec r \end{align} is replaced by the continuous $$ \hat I= \int dx \vert x\rangle\langle x\vert\, . $$

ZeroTheHero
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  • see also: https://physics.stackexchange.com/q/364208/36194 – ZeroTheHero Nov 19 '17 at 02:05
  • so it is not actually component? I mean what kind of space is this is it still Hilbert space, if so is there a name for it?. how is the inner product defined here? also I am still confused of how the integral is considered an expansion why go through this path where you use the Dirac delta function to project. i.e why can't we just expand by the sum: $$\lim_{n\rightarrow \infty}(\sum_{i=1}^{ \infty} \psi(x+i(1/n))|x+i(1/n))>)$$ or is the integral just a matter of notation convenience? – NegativeTension Nov 19 '17 at 02:27
  • @NegativeTension I’m not sure I understand your comment. You can think of $\psi(x)$ as a component. The kets live in a Hilbert space, and the inner product $\langle \phi\vert\psi\rangle=\int dx \phi^*(x)\psi(x)$. – ZeroTheHero Nov 19 '17 at 02:36
  • But why go through the trouble of introducing a dx which scales the components which changes unites? why can't we still use the sum? I looked all over the web and most of what I read says that the basis is uncountable so we have to use an integral? which brings another problem what, do we mean by uncountable? – NegativeTension Nov 19 '17 at 02:46
  • @NegativeTension because $x$ is continuous rather than discrete? – ZeroTheHero Nov 19 '17 at 02:51
  • But wouldn't the sum above act like an integral without the dx? – NegativeTension Nov 19 '17 at 03:07
  • OP is asking about the dimensions. For example, what are the dimensions of $\psi(x)$? – DanielSank Nov 19 '17 at 05:20
  • I now get the countable vs uncountable part but I am still confused about using the integral to expand a vector. Specifically, what bugs me is the dx in the integral which I think multiplys every component. To make it clear what I am saying, if /psi(x) has unit of meter then the integral will have a unit of meter^2. I know in QM /psi has no units but still the dx is there. – NegativeTension Nov 19 '17 at 06:37
  • @NegativeTension I’m still not sure I understand why you have issues with dimensions. The dimensionless quantity is not $\psi(x)$ or $\vert\psi(x)\vert^2$ (the latter being a probability density) but $\vert\psi(x)\vert^2 dx$. An individual molecule of water has mass, but water has a continuous fluid as mass density, i.e. when you go from discrete to continuous the quantities do change units to become - say- $kg/m^3$. – ZeroTheHero Nov 19 '17 at 08:44