A peculiarity in uncertainty principle

In summary, the author thinks that the uncertainty principle should be right even in eigenstates, but there are still some issues with the way the operators are defined.
  • #36
That's very good Goerge,thanks.
But sth that still bothers me is that you somewhere assumed one of the operators to be P.Doesn't that restrict the domain of your argument?I mean,that doesn't tell us that for every A and B for which [A,B]=cI,we can't switch the order,but just about P and X.And also that doesn't tell that there is no function that we can do such calculations on,but that there are some functions that are not suitable.So I think you should generalize your argument.(Or I'm wrong somewhere?)
Thanks again
 
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  • #37
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
Shyan said:
Ok,but can you prove [itex] \langle a | B | a \rangle=\infty [/itex] and [itex] \langle a | a \rangle=\infty [/itex] for all A,B and [itex] |a \rangle [/itex] satisfying the conditions?

Well, you certainly know that [itex]\langle a \vert a' \rangle = \delta(a - a')[/itex], so, from the commutator it would follow that:
[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]
 
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  • #39
Dickfore said:
Well, you certainly know that [itex]\langle a \vert a' \rangle = \delta(a - a')[/itex], so, from the commutator it would follow that:
[tex]
\langle a \vert B \vert a' \rangle = \frac{c}{a - a'} \, \delta(0) \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)
[/tex]

I mean can you tell that when [A,B]=cI,then one the operators has Continuous spectrum?
 
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  • #40
Shyan said:
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, [itex]\left[ J_x, J_y \right] = i \, J_z[/itex], so the commutator between two operators with a discrete spectrum is not the identity operator.

EDIT:
By the Baker-Hausdorf 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.
 
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  • #41
Dickfore said:
This is a hard question, but it may be true. I have not seen a proof. But, notice that, for angular momentum, [itex]\left[ J_x, J_y \right] = i \, J_z[/itex], 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?
 
  • #42
Some general results:

George Jones said:
Set [itex]\hbar = 1[/itex] and assume

[tex]AB - BA = iI.[/tex]

Multiplying the commutation relation by [itex]B[/itex] and reaaranging gives

[tex]
\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*}
[/tex]

By induction,

[tex]AB^n - B^n A = niB^{n-1}[/tex]

for every positive integer [itex]n[/itex]. Consequently,

[tex]
\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*}
[/tex]

Because this is true for every [itex]n[/itex], at least one of [itex]A[/itex] and [itex]B[/itex] must be unbounded. Say it is [itex]A[/itex]. Then, by the Hellinger-Toeplitz theorem, if [itex]A[/itex] is self-adjoint, the domain of physical observable [itex]A[/itex] cannot be all of Hilbert space!
 
  • #43
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(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
[/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
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 <Weyl unitaries>, 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 <Stone theorem> 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.
 
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  • #45
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
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(a-a')
[/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} (a-a)\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 |B|a \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.
 
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  • #47
There is another proof for [itex] \langle a |B|a \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 \psi|AB|\psi \rangle=\lambda \langle \psi |B|\psi\rangle \\
\langle \psi|BA|\psi \rangle=\lambda \langle \psi |B|\psi\rangle
[/itex]
But:
[itex]
\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
[/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 self-adjoint operators and [itex] A|\psi \rangle=\lambda |\psi\rangle [/itex],then [itex] \langle \psi |B|\psi \rangle=\infty [/itex].
 
  • #48
Shyan said:
I don't know how,but looks like whenever we have [A,B]=cI for two self-adjoint operators and [itex] A|\psi \rangle=\lambda |\psi\rangle [/itex],then [itex] \langle \psi |B|\psi \rangle=\infty [/itex].

[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).
 
  • #49
Jorriss said:
[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!
 
  • #50
Jorriss said:
[...]A trace is not defined on an infinite dimensional vector space (as far as I know).

It is. Just think of von Neumann's density operator (trace class). Tr(rho) is perfectly well defined.
 
<h2>1. What is the uncertainty principle?</h2><p>The uncertainty principle is a fundamental concept in quantum mechanics that states that it is impossible to know both the position and momentum of a particle with absolute certainty at the same time. This means that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa.</p><h2>2. How does the uncertainty principle affect our understanding of the physical world?</h2><p>The uncertainty principle challenges our traditional understanding of the physical world, as it suggests that there are inherent limitations to our ability to measure and predict the behavior of particles at a subatomic level. It also implies that there is a level of randomness and unpredictability in the universe.</p><h2>3. What is the significance of Heisenberg's uncertainty principle?</h2><p>Heisenberg's uncertainty principle, named after German physicist Werner Heisenberg, is one of the most important principles in quantum mechanics. It has led to the development of new theories and technologies, and has fundamentally changed our understanding of the nature of reality.</p><h2>4. Can the uncertainty principle be violated or overcome?</h2><p>No, the uncertainty principle is a fundamental law of nature and cannot be violated or overcome. It is a consequence of the wave-like behavior of particles at a subatomic level and is supported by numerous experimental observations.</p><h2>5. How does the uncertainty principle relate to other principles in physics?</h2><p>The uncertainty principle is closely related to other principles in physics, such as the wave-particle duality and the observer effect. It also has implications for other areas of physics, including thermodynamics and information theory, and has been linked to concepts such as entanglement and quantum tunneling.</p>

1. What is the uncertainty principle?

The uncertainty principle is a fundamental concept in quantum mechanics that states that it is impossible to know both the position and momentum of a particle with absolute certainty at the same time. This means that the more precisely we know the position of a particle, the less precisely we can know its momentum, and vice versa.

2. How does the uncertainty principle affect our understanding of the physical world?

The uncertainty principle challenges our traditional understanding of the physical world, as it suggests that there are inherent limitations to our ability to measure and predict the behavior of particles at a subatomic level. It also implies that there is a level of randomness and unpredictability in the universe.

3. What is the significance of Heisenberg's uncertainty principle?

Heisenberg's uncertainty principle, named after German physicist Werner Heisenberg, is one of the most important principles in quantum mechanics. It has led to the development of new theories and technologies, and has fundamentally changed our understanding of the nature of reality.

4. Can the uncertainty principle be violated or overcome?

No, the uncertainty principle is a fundamental law of nature and cannot be violated or overcome. It is a consequence of the wave-like behavior of particles at a subatomic level and is supported by numerous experimental observations.

5. How does the uncertainty principle relate to other principles in physics?

The uncertainty principle is closely related to other principles in physics, such as the wave-particle duality and the observer effect. It also has implications for other areas of physics, including thermodynamics and information theory, and has been linked to concepts such as entanglement and quantum tunneling.

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