
#37
Dec2212, 01:42 PM

P: 738

And a question,Why by having same [itex] \lambda [/itex] restrictions on f and g,the domains become the same?
Can we tell that's because it makes the proportionality symmetrical between f and g?If yes,how this one affects the domains? [itex] \lambda=\frac{f(1)}{f(0)}=\frac{g^*(0)}{g^*(1)}= \frac{f^*(0)}{f^*(1)}=\frac{1}{\lambda^*} [/itex] 



#38
Dec2212, 02:11 PM

P: 3,015

[tex] \langle a \vert B \vert a' \rangle = \frac{c}{a  a'} \, \delta(a  a') \sim c \, \delta'(a  a'), \ a \rightarrow a' [/tex] EDIT: Example: [tex] \langle p \vert x \vert p' \rangle = \frac{1}{2\pi \hbar} \, \int_{\infty}^{\infty} x \, e^{\frac{i}{\hbar}(p  p') x} \, dx [/tex] [tex] = i \, \frac{\partial}{\partial p} \, \left( \frac{1}{2\pi} \, \int_{\infty}^{\infty} e^{\frac{i}{\hbar}(p  p') x} \, dx \right) = i \, \hbar \, \delta'(p  p') \rightarrow i \, \hbar \ \delta'(0), \ p \rightarrow p' [/tex] 



#39
Dec2212, 02:27 PM

P: 738





#40
Dec2212, 02:36 PM

P: 3,015

EDIT: By the BakerHausdorf lemma: [tex] f(\alpha) \equiv e^{\alpha A} B e^{\alpha A} [/tex] [tex] f(\alpha) = B + \alpha [A, B] + \frac{\alpha^2}{2} \, [A, [A, B]] + \ldots = B + \alpha \, c \, I [/tex] so any eigenstate of B, is also an eigenstate of f. Maybe this can be used to prove that B has a continuous spectrum, but I don't know how. 



#41
Dec2212, 02:40 PM

P: 738

You agree? Can someone list all conjugate observables? 



#43
Dec2212, 03:02 PM

P: 738

Ok Goerge,So let's say [itex]D(A)=S \subset H \mbox{ and } D(B)=H [/itex]
Then we have: [itex] D([A,B])=D(ABBA)=D(AB) \cap D(BA)=\{ \psi \epsilon H  B \psi \epsilon S\} \cap \{ \psi \epsilon S  A \psi \epsilon H\}=\{\psiB \psi \epsilon S\} \cap S [/itex] I don't think this helps because we have [itex] D([A,B]) \subset D(A) \subset D(B) [/itex] so we still can choose a state which is in the domain of all three operators.This helps only if you can prove there is no eigenstate of A in that subspace.(Is that what you mean?) 



#44
Dec2512, 05:20 AM

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P: 11,866

This thread really contains useful information so I'll add my bit (as usually from the historical perspective).
The commutation relations initially due to Born and Jordan 1925 written for infinite matrices (1925 was the year of matrix mechanics) were reinterpreted in terms of Hilbert spaces with the suitable example l^2 (R). In 1926 Schrödinger coined wave mechanics, a new form of quantum mechanics, based not on infinite sequences of numbers and infinite matrices, but on functions, a little different than 'ordinary functions. Without knowledge of Hilbert space methods (in 1926 most of Hilbert space methods did not exist !), Schrödinger proved the equivalence of these 2 new 'mechanics' in a formal manner. Today, one knows that matrix mechanics and wave mechanics are essentially the same, because l^2(R) and L^2(R) are isomorphic Hilbert spaces, a particular case of a general theorem which says that any 2 complex separable Hilbert spaces are isomorphic. In terms of wave functions, Schrödinger rewrote the fundamental commutation relations of Born and Jordan and 'found a solution', i.e. what we call today a representation of the Lie algebra of the Heisenberg group in terms of essentially selfadjoint operators on a complex separable Hilbert space. In 1927 in his groundbreaking paper on group theory in quantum mechanics, H. Weyl rewrote the commutation relations in terms of what we call today <Weyl unitaries>, i.e. semigroups of unitary operators, having the position and momentum coordinates as generators. 19281929: In his famous series of papers in the Proc. Nat. Acad. Sci., M. Stone in the US proves the famous <Stone theorem> which sets Weyl's work on rigorous mathematical terms. Stone even formulates the first version of the Stonevon Neumann's theorem and sketches a proof of it. 1929: In Germany John von Neumann was already working on the overall theory of Hilbert spaces and his interest in quantum mechanics through his friend and compatriot E. Wigner led him to apply his results in the theory of quantum mechanics and he particularly chose Weyl's 1927 paper and Stone's 1929 result. He formulated and proved a completely rigorous version of Stone's theorem in terms of Weyl's unitaries which said that the 1926 solution of the BornJordan commutation relations found by Erwin Schrödinger was essentially unique. In the 1950's von Neumann's work was obtained as a particular case of the general theory of unitary (and also projective unitary) representations of locally compact Lie grous developed by G.W. Mackey. Having gone through most of the work on the fundamental commutation relations for coordinate and momentum, I don't recall any rigorous mathematical result regarding the spectrum of the 2 operators in a general setting. I don't think there's a case in which one can imagine a physical situation in which both the coordinate and the momentum operators both have a completely discrete (perhaps unbounded) spectrum and obey the commutation relation on a dense everywhere subset of a complex separable Hilbert space. P.S. 0. Born and Jordan initially wrote the commutation relations in terms of frequencies, not position and momentum. 1. Position and momentum can be replaced by orbital angular momentum and azimuthal angle or by Hamiltonian and time operators. 2. If one can prove that at least 1 of the 2 operators in the commutator must have a part of its spectrum absolutely continuous, then one must bring in the theory of rigged Hilbert spaces. 



#45
Dec2612, 04:45 PM

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P: 11,866

This is a completion to post #42 by George above. The proof he quoted was similar to the one in
Wielandt, H.  Über die Unbeschränktheit der Operatoren der Quantenmechanik (Math. Annalen, 1949, S.21). 



#46
Dec3012, 01:39 PM

P: 738

I think this article helps.
For one thing,we have,for one of the eigenvectors of an operator having only continuous part in its spectrum: [itex] \langle a  a' \rangle = \delta(aa') [/itex] So we have: [itex] \infty=\langle a  a \rangle=\frac{1}{c}\langle a  cI  a \rangle =\frac{1}{c}\langle a c[A,B] a \rangle =\frac{1}{c} (aa)\langle a B a \rangle \Rightarrow \langle a B a \rangle =\frac{\infty}{0}=\infty [/itex] We just have to prove that the conditions stated,imply that the spectrum of the operators is purely continuous. And about [itex] \langle a Ba \rangle [/itex] being infinite.consider: [itex] \int _{\infty} ^{\infty} e^{ikx} x e^{ikx} dx= \frac{1}{2} x^2 ] _{\infty} ^{\infty}= \infty  \infty [/itex] which is not infinite. 



#47
Jan113, 10:32 PM

P: 738

There is another proof for [itex] \langle a Ba \rangle [/itex] being infinite.
[itex] [A,B] \psi \rangle=c  \psi \rangle \Rightarrow \langle \psi  [A,B]  \psi \rangle=c [/itex] Now if [itex] A  \psi \rangle=\lambda  \psi \rangle [/itex]: [itex] \langle \psiAB\psi \rangle=\lambda \langle \psi B\psi\rangle \\ \langle \psiBA\psi \rangle=\lambda \langle \psi B\psi\rangle [/itex] But: [itex] \langle \psiAB\psi \rangle\langle \psi BA\psi\rangle=\langle \psi[A,B]\psi \rangle \Rightarrow (\lambda\lambda) \langle \psiB\psi \rangle=c \Rightarrow \langle \psiB\psi\rangle=\frac{c}{0} \Rightarrow \langle \psi B\psi \rangle=\infty [/itex] So I think this should be the problem in the proof. I don't know how,but looks like whenever we have [A,B]=cI for two selfadjoint operators and [itex] A\psi \rangle=\lambda \psi\rangle [/itex],then [itex] \langle \psi B\psi \rangle=\infty [/itex]. 



#48
Jan113, 11:30 PM

P: 1,025





#50
Jan213, 04:17 AM

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