# A peculiarity in uncertainty principle

by Shyan
Tags: peculiarity, principle, uncertainty
 P: 697 And a question,Why by having same $\lambda$ 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? $\lambda=\frac{f(1)}{f(0)}=\frac{g^*(0)}{g^*(1)}= \frac{f^*(0)}{f^*(1)}=\frac{1}{\lambda^*}$
P: 3,015
 Quote by Shyan Ok,but can you prove $\langle a | B | a \rangle=\infty$ and $\langle a | a \rangle=\infty$ for all A,B and $|a \rangle$ satisfying the conditions?
Well, you certainly know that $\langle a \vert a' \rangle = \delta(a - a')$, so, from the commutator it would follow that:
$$\langle a \vert B \vert a' \rangle = \frac{c}{a - a'} \, \delta(a - a') \sim c \, \delta'(a - a'), \ a \rightarrow a'$$

EDIT:
Example:
$$\langle p \vert x \vert p' \rangle = \frac{1}{2\pi \hbar} \, \int_{-\infty}^{\infty} x \, e^{-\frac{i}{\hbar}(p - p') x} \, dx$$
$$= i \, \frac{\partial}{\partial p} \, \left( \frac{1}{2\pi} \, \int_{-\infty}^{\infty} e^{-\frac{i}{\hbar}(p - p') x} \, dx \right) = i \, \hbar \, \delta'(p - p') \rightarrow i \, \hbar \ \delta'(0), \ p \rightarrow p'$$
P: 697
 Quote by Dickfore Well, you certainly know that $\langle a \vert a' \rangle = \delta(a - a')$, so, from the commutator it would follow that: $$\langle a \vert B \vert a' \rangle = \frac{c}{a - a'} \, \delta(0) \sim c \, \delta'(a - a'), \ a \rightarrow a'$$ EDIT: Example: $$\langle p \vert x \vert p' \rangle = \frac{1}{2\pi \hbar} \, \int_{-\infty}^{\infty} x \, e^{-\frac{i}{\hbar}(p - p') x} \, dx$$ $$= i \, \frac{\partial}{\partial p} \, \left( \frac{1}{2\pi} \, \int_{-\infty}^{\infty} e^{-\frac{i}{\hbar}(p - p') x} \, dx \right) = i \, \hbar \, \delta'(p - p') \rightarrow i \, \hbar \ \delta'(0)$$
I mean can you tell that when [A,B]=cI,then one the operators has Continuous spectrum?
P: 3,015
 Quote by Shyan I mean can you tell that when [A,B]=cI,then one the operators has continues spectrum?
This is a hard question, but it may be true. I have not seen a proof. But, notice that, for angular momentum, $\left[ J_x, J_y \right] = i \, J_z$, so the commutator between two operators with a discrete spectrum is not the identity operator.

EDIT:
By the Baker-Hausdorf lemma:

$$f(\alpha) \equiv e^{\alpha A} B e^{-\alpha A}$$
$$f(\alpha) = B + \alpha [A, B] + \frac{\alpha^2}{2} \, [A, [A, B]] + \ldots = B + \alpha \, c \, I$$
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.
P: 697
 Quote by Dickfore This is a hard question, but it may be true. I have not seen a proof. But, notice that, for angular momentum, $\left[ J_x, J_y \right] = i \, J_z$, so the commutator between two operators with a discrete spectrum is not the identity operator.
So I guess because the observables in QM are not so many,its possible to check all conjugate observables and see whether they have Continuous spectrum or not.
You agree?
Can someone list all conjugate observables?
Mentor
P: 6,009
Some general results:

 Quote by George Jones Set $\hbar = 1$ and assume $$AB - BA = iI.$$ Multiplying the commutation relation by $B$ and reaaranging gives $$\begin{equation*} \begin{split} AB - BA &= iI \\ AB^2 - BAB &= iB \\ AB^2 - B \left( BA + iI \right) &= iB \\ AB^2 - B^2 A &= 2iB. \end{split} \end{equation*}$$ By induction, $$AB^n - B^n A = niB^{n-1}$$ for every positive integer $n$. Consequently, $$\begin{equation*} \begin{split} n\left\| B \right\|^{n-1} &=\left\| AB^n - B^n A \right\| \\ &\leq 2\left\| A \right\| \left\| B \right\|^n\\ n &\leq 2\left\| A \right\| \left\| B \right\| . \end{split} \end{equation*}$$ Because this is true for every $n$, at least one of $A$ and $B$ must be unbounded. Say it is $A$. Then, by the Hellinger-Toeplitz theorem, if $A$ is self-adjoint, the domain of physical observable $A$ cannot be all of Hilbert space!
 P: 697 Ok Goerge,So let's say $D(A)=S \subset H \mbox{ and } D(B)=H$ Then we have: $D([A,B])=D(AB-BA)=D(AB) \cap D(BA)=\{ \psi \epsilon H | B \psi \epsilon S\} \cap \{ \psi \epsilon S | A \psi \epsilon H\}=\{\psi|B \psi \epsilon S\} \cap S$ I don't think this helps because we have $D([A,B]) \subset D(A) \subset D(B)$ 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?)
 Sci Advisor HW Helper P: 11,833 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 re-wrote 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 self-adjoint operators on a complex separable Hilbert space. In 1927 in his ground-breaking paper on group theory in quantum mechanics, H. Weyl re-wrote the commutation relations in terms of what we call today , i.e. semigroups of unitary operators, having the position and momentum coordinates as generators. 1928-1929: In his famous series of papers in the Proc. Nat. Acad. Sci., M. Stone in the US proves the famous which sets Weyl's work on rigorous mathematical terms. Stone even formulates the first version of the Stone-von 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 Born-Jordan 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.
 Sci Advisor HW Helper P: 11,833 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).
 P: 697 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: $\langle a | a' \rangle = \delta(a-a')$ So we have: $\infty=\langle a | a \rangle=\frac{1}{c}\langle a | cI | a \rangle =\frac{1}{c}\langle a |c[A,B]| a \rangle =\frac{1}{c} (a-a)\langle a |B| a \rangle \Rightarrow \langle a |B| a \rangle =\frac{\infty}{0}=\infty$ We just have to prove that the conditions stated,imply that the spectrum of the operators is purely continuous. And about $\langle a |B|a \rangle$ being infinite.consider: $\int _{-\infty} ^{\infty} e^{-ikx} x e^{ikx} dx= \frac{1}{2} x^2 ] _{-\infty} ^{\infty}= \infty - \infty$ which is not infinite.
 P: 697 There is another proof for $\langle a |B|a \rangle$ being infinite. $[A,B] |\psi \rangle=c | \psi \rangle \Rightarrow \langle \psi | [A,B] | \psi \rangle=c$ Now if $A | \psi \rangle=\lambda | \psi \rangle$: $\langle \psi|AB|\psi \rangle=\lambda \langle \psi |B|\psi\rangle \\ \langle \psi|BA|\psi \rangle=\lambda \langle \psi |B|\psi\rangle$ But: $\langle \psi|AB|\psi \rangle-\langle \psi |BA|\psi\rangle=\langle \psi|[A,B]|\psi \rangle \Rightarrow (\lambda-\lambda) \langle \psi|B|\psi \rangle=c \Rightarrow \langle \psi|B|\psi\rangle=\frac{c}{0} \Rightarrow \langle \psi |B|\psi \rangle=\infty$ 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 self-adjoint operators and $A|\psi \rangle=\lambda |\psi\rangle$,then $\langle \psi |B|\psi \rangle=\infty$.
P: 1,007
 Quote by Shyan I don't know how,but looks like whenever we have [A,B]=cI for two self-adjoint operators and $A|\psi \rangle=\lambda |\psi\rangle$,then $\langle \psi |B|\psi \rangle=\infty$.
[A,B]=cI can only be realized if A, B are infinite dimensional. A trace is not defined on an infinite dimensional vector space (as far as I know).
P: 697
 Quote by Jorriss [A,B]=cI can only be realized if A, B are infinite dimensional. A trace is not defined on an infinite dimensional vector space (as far as I know).
Well,we're OK with that,They ARE infinite dimensional!