Tuesday, 17 September 2019

Operator in quantum mechanics


I'm really confused by the definition and uses of operators in quantum mechanics. Usually we say that the state of a system is described by some vector $\lvert\psi\rangle$ in a Hilbert space $H$, and then we define operators acting on said vector, for example $\hat{p}: H\rightarrow H$. But often I read things like $$ \hat{p}\psi(x)=-i\hbar\frac{\partial}{\partial x}\psi(x)$$


I don't understand. $\psi(x)=\langle x\rvert\psi\rangle$ is a function in $L^2$ or some other space, not the same Hilbert space as $\lvert\psi\rangle$. More precisely $\psi(x)=\langle x\rvert\psi\rangle$ is an element of the field associated with $H$ for fixed $x$, I don't understand how can we apply $\hat{p}$ to this object.


How should I interpret this?



EDIT: I just realized that my question is a duplicate of this one, I must say that the "related" section is a much better search engine than the search engine. I have a question about ACuriousMind's answer. He writes that one can define a map $$\mathrm{Ket}: L^2(\mathbb{R},\mathbb{C})\rightarrow \mathcal{H}_{1D}, \psi \mapsto|\psi\rangle := \int_{-\infty}^\infty\psi(x)|x\rangle\mathrm{d}x $$


But I don't really understand how $$\int_{-\infty}^\infty\psi(x)|x\rangle\mathrm{d}x $$ is defined. How can one take an integral of a ket? The integral is a functional in $L^2$, not whatever space $\lvert x \rangle$ is in.



Answer



If we you want to know a rigorous formulation of quantum mechanics, please check the first chapter of the book Dirac Kets, Gamow Vectors and Gelfand Tripletes--The Rigged Hilbert Space formulation of Quantum Mechanics by A.Bohm and M.Gadella. This is a huge topic and cannot be answered in a few lines. I list some important facts below.


Complete system of commuting operators


$\{A_k\}$, $k=1,2,\cdots,N$ is a system of commuting operators on rigged Hilbert space $\Phi \subset H \subset \Phi^X$ iff



  1. $[A_i,A_k] = 0$ for all $i,k = 1,\cdots,N$

  2. $\sum A_k^2$ is essentially self adjoint



$\{A_k\}$ is a complete commuting system if there exists a vector $\phi \in \Phi$ such that $\{A\phi| A$ runs out the algebra generated by $\{A_k\}\}$ spans $H$.


An antilinear functional $F$ on $\Phi$ is a generalized eigenvector for the system $A_k$ if for any $k=1,\cdots,N$ $$(A_k)^X F = \lambda^{(k)}F$$ The set of numbers $\lambda = (\lambda^{(1)},\cdots,\lambda^{(N)})$ are called generalized eigenvalues $F_{\lambda} = |\lambda^{(1)},\cdots,\lambda^{(N)}\rangle$.


Nuclear Spectral Theorem


Let $\{A_k\}$, $k=1,2,\cdots,N$ be a complete system of commuting essentially $\tau_{\Phi}$-continuous operators on the rigged Hilbert space $\Phi \subset H \subset \Phi^X$. Then, there exists a set of generalized eigenvectors $$|\lambda^{(1)},\cdots,\lambda^{(N)}\rangle \in \Phi^X$$ $$(A_k)^X|\lambda^{(1)},\cdots,\lambda^{(N)}\rangle = \lambda^{(k)}|\lambda^{(1)},\cdots,\lambda^{(N)}\rangle$$ $$\lambda^{(k)} \in \Lambda^{(k)} = \mbox{ spectrum of } A_k$$ such that for every $\phi \in \Phi$ and some uniquely defined measure $\mu$ on $\Lambda = \Lambda^{(1)} \times \cdots \times \Lambda^{(N)}$, $$(\psi|\phi) = \int_{\Lambda} d\mu(\lambda) \langle \psi | \lambda^{(1)},\cdots,\lambda^{(N)} \rangle \langle \lambda^{(1)},\cdots,\lambda^{(N)} | \phi \rangle$$.


Comments


Roughly speaking, the equivalence of the $L^2(\mathbb{R},\mathbb{C})$ and $H$ is guaranteed by the fact that $X$ is a system of commuting operators on rigged Hilbert space. The demanded rigged Hilbert space should be constructed from the original Hilbert space if the algebra of operators are given. The notation of $|\psi\rangle = \int dx \langle x | \psi \rangle |x\rangle$ holds in the sense of performing inner product and is guaranteed by the nuclear spectrum theorem.


The whole construction is very complicated and subtle, and needs a lot of concepts of modern function analysis. Again, please check the book I recommended if you are really interested in this topic.


No comments:

Post a Comment

Understanding Stagnation point in pitot fluid

What is stagnation point in fluid mechanics. At the open end of the pitot tube the velocity of the fluid becomes zero.But that should result...