A peculiarity in uncertainty principle

[ShayanJ]

In Quantum mechanics:Concepts and applications by Nouredine Zettili (2nd edition),one can find the following:
$| \psi \rangle = \int d^3 r | \vec{r} \rangle \langle \vec{r} | \psi \rangle = \int d^3 r \psi(\vec{r}) | \vec{r} \rangle$
With $\langle \vec{r} | \psi \rangle = \psi(\vec{r})$
I tried to write the paradox with this convention:
$<br /> 1=\langle a | a \rangle=\frac{1}{c} \langle a | cI | a \rangle=\frac{1}{c} [ \int d^3 r a^*(\vec{r}) (cI) \int d^3 r a(\vec{r}) ] | \vec{r} \rangle=\frac{1}{c} [ \int d^3 r a^*(\vec{r}) [A,B] \int d^3 r a(\vec{r}) ] | \vec{r} \rangle=\frac{1}{c} [ \int d^3 r a^*(\vec{r}) AB \int d^3 r a(\vec{r})-\int d^3 r a^*(\vec{r}) BA \int d^3 r a(\vec{r}) ] | \vec{r} \rangle<br />$
I'm not sure how to continue but I think it may help.So I give it to more capable hands.

 

[ShayanJ]

Yeah,again me.
This time i asked a professor of pure mathematics.
He said the problem is that $[A,B] | \psi \rangle =c |\psi \rangle$ doesn't mean $[A,B]=cI$
I think the reason is that the domain of I is the whole space but the domain of [A,B] is more restricted so the two operators only have the same prescription but different domains so they're not equal.

 

[Dickfore]

Yeah,again me.
This time i asked a professor of pure mathematics.
He said the problem is that $[A,B] | \psi \rangle =c |\psi \rangle$ doesn't mean $[A,B]=cI$
I think the reason is that the domain of I is the whole space but the domain of [A,B] is more restricted so the two operators only have the same prescription but different domains so they're not equal.

I think the simplest way to put it is that $\langle a \vert a \rangle$ and $\langle a \vert B \vert a \rangle$, where $\vert a \rangle$ is an eigenvector of A formally do not exist (they diverge, are infinite)!
Calculate the M.E. of this commutation relation between two different eigenstates:
$$\langle a \vert [A, B] \vert a' \rangle = (a - a') \, \langle a \vert B \vert a' \rangle = c \langle a \vert a' \rangle$$
When $a' \neq a$, the r.h.s. is zero, so we conclude $\langle a \vert B \vert a' \rangle = 0, \ a' \neq a$. But, if we take the limit $a' \rightarrow a$, the l.h.s. is an indeterminate form of the type $0 \cdot \infty$, whereas the r.h.s. is $c \, \delta(a - a')$, according to the orthonormalization condition for continuous spectra.

This was the motivation behind making the eigenvalues of at least one of the operators A or B, discrete (in my case, the momentum operator p), because then you may normalize everything to a finite norm.

 

[ShayanJ]

I think the simplest way to put it is that $\langle a \vert a \rangle$ and $\langle a \vert B \vert a \rangle$, where $\vert a \rangle$ is an eigenvector of A formally do not exist (they diverge, are infinite)!
Calculate the M.E. of this commutation relation between two different eigenstates:
$$\langle a \vert [A, B] \vert a' \rangle = (a - a') \, \langle a \vert B \vert a' \rangle = c \langle a \vert a' \rangle$$
When $a' \neq a$, the r.h.s. is zero, so we conclude $\langle a \vert B \vert a' \rangle = 0, \ a' \neq a$. But, if we take the limit $a' \rightarrow a$, the l.h.s. is an indeterminate form of the type $0 \cdot \infty$, whereas the r.h.s. is $c \, \delta(a - a')$, according to the orthonormalization condition for continuous spectra.

This was the motivation behind making the eigenvalues of at least one of the operators A or B, discrete (in my case, the momentum operator p), because then you may normalize everything to a finite norm.

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?

 

[George Jones]

Consider two hermitian operators A and B and a system in state $|a\rangle$ which is an eigenstate of A with eigenvalue $\lambda$
So we have:
$\langle a|[A,B]|a\rangle=\langle a|AB|a\rangle-\langle a |BA|a\rangle=(A^{\dagger}|a\rangle)^{\dagger}B|a \rangle-\lambda \langle a |B|a\rangle=(A|a \rangle)^{\dagger} B | a \rangle-\lambda \langle a |B|a \rangle=\lambda \langle a|B|a \rangle - \lambda \langle a|B|a \rangle=0$

There is still a point here.
We can't tell that the assumption that A has an eigenstate,causes the paradox because even if P doesn't have an eigenstate,one still can find an operator that does and do the calculation for that operator and its eigenstate and again arrive at the paradox.So I think sth else should be
wrong.
I've been trying to understand it via reading the thread that Goerge suggested but I can't follow the discussion because I don't know enough math.
Can somebody present the final result of that thread in a simple language?
thanks

There is a problem with the step

$$\left\langle a \vert AB \vert a\right\rangle =\left\langle Aa \vert Ba\right\rangle$$
To see the problem in terms of domains, let's work through in detail a fairly elementary example for which the eigenstates and eigenvalues above actually exist. In this example, subtleties with domains definitely come into play.

First, consider something even more elementary, real-valued functions of a single real variable. The domain of such a function $f$ is the (sub)set of all real numbers $x$ on which $f$ is allowed to act. Suppose $f$ is defined by $f\left(x\right) = 1/x$. The domain of $f$ cannot be the set of all real numbers $\mathbb{R}$, but it can be any subset of $\mathbb{R}$ that doesn't contain zero. Take the domain of $f$ to be the set of all non-zero real numbers. Define $g$ by $g\left(x\right) = 1/x$ with domain the set of all positive real numbers. As functions, $f \ne g$, because $f$ and $g$ have different domains, i.e., it takes both a domain and an action to specify a function. As functions, $f = g$ only when $f$ and $g$ have the same actions and the same domains.

Let the action of the momentum operator be given by $P=-id/dx$ (for convenience, set $\hbar =1$). On what wave functions can $P$ act, i.e, what is the domain, $D_{P}$, of $P$? Since $P$ operates the Hilbert space $H$ of square-integrable functions, $D_{P}$ must be a subset of the set of square-integrable functions. The action of $P$ has to give as output something that lives in the Hilbert space $H$, i.e., the output has to be square-integrable, and thus $D_{P}$ must be subset of the set of square-integrable functions whose derivatives are also square-integrable. Already, we see that the domain of $P$ cannot be all of the Hilbert space $H$.

As an observable, we want $P$ to be self-adjoint, i.e, we want $P=P^{\dagger }$. As in the case of functions above, this means that the actions of $P$ and $P^{\dagger }$ must be the same, and that the domains (the states on which $P$ and $P^{\dagger }$ act) $D_{P}$ and $D_{P^{\dagger }}$ must be the same. For concreteness, take wave functions on the interval with endpoints $x=0$ and $x=1$. The adjoint of the momentum operator is defined by

$$\begin{align}<br /> \left\langle P^{\dagger }g \vert f\right\rangle &=\left\langle g \vert Pf\right\rangle \\<br /> &=-i\int_{0}^{1}g* \frac{df}{dx}dx \\<br /> &=-i\left( \left[ g* f\right] _{0}^{1}-\int_{0}^{1}\frac{dg}{dx}* fdx\right) \\<br /> & =-i\left[ g* f\right] _{0}^{1}+\left\langle Pg \vert f\right\rangle ,<br /> \end{align}$$
where integration by parts has been used.

Consequently, the actions of $P$ and $P^{\dagger }$ are the same as long as the first term in the last line vanishes, i.e., as long as

$$\begin{align}<br /> 0 &= g* \left( 1\right) f\left( 1\right) -g* \left( 0\right) f\left( 0\right) \\<br /> \frac{f\left( 1\right) }{f\left( 0\right) } &= \frac{g* \left( 0\right) }{g* \left( 1\right) }<br /> \end{align}$$
for non zero $f\left( 0\right)$ and $g\left( 1\right)$. Now, $f\left( 1\right) /f\left( 0\right)$ is some complex number, say $\lambda$, so $f\left( 1\right) = \lambda f\left( 0\right)$. Hence,

$$\begin{align}<br /> \lambda & =\frac{g* \left( 0\right) }{g* \left( 1\right) }\\<br /> \lambda * & =\frac{g\left( 0\right) }{g\left( 1\right)} \\<br /> g\left( 1\right) & =\frac{1}{\lambda * }g\left( 0\right) .<br /> \end{align}$$
From the relation $\left\langle P^{\dagger }g \vert f\right\rangle =\left\langle g \vert Pf\right\rangle$, we see that $g\in D_{P^{\dagger }}$ and $f\in D_{P}$. These domains can be made to be the same if the same $\lambda$ restrictions are placed on $f$ and $g$, i.e, if $\lambda =1/\lambda*$, or $\lambda* \lambda =1$. This means that $\lambda$ can be written as $\lambda =e^{i\theta }$ with $\theta$ real. Different choices of $\theta$ correspond to different boundary conditions, with $\theta =0$ corresponding to the periodic boundary condition used by Dickfore above (with $L=1$). Let's use this choice, so that $f$ is in $D_{P^{\dagger }}=D_{P}$ if $f$ is square-integrable, $f'$ is square-integrable, and $f\left( 1\right) =f\left( 0\right)$. Note that this works even if $0=f\left( 0\right)$. With this choice, $P=P^{\dagger }$, and we can write $\left\langle Pg \vert f\right\rangle =\left\langle g \vert Pf\right\rangle$. We cannot use this when $P$ (on either side) acts on a wave function $h$ that is not in $D_{P^{\dagger }}=D_{P}$. This is the problem with the "proof" above.

In $\left\langle a \vert AB \vert a\right\rangle$, take $A=P$ and $B=X$. Take $f\left( x\right) =e^{2\pi ix}$. Then, $\left( Pf\right) \left( x\right) =2\pi f\left( x\right)$ and $\left( PXf\right) \left( x\right) =P\left( xf\left( x\right) \right)$. However, $h\left( x\right) =xf\left( x\right) =xe^{2\pi ix}$ does not satisfy the boundary condition $h\left( 1\right) =h\left( 0\right)$, so $h$ is not in the domain $D_{P}$, and we cannot just slide [/itex]A=P[/itex] to the left.

 

[ShayanJ]

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

 

[ShayanJ]

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?
$\lambda=\frac{f(1)}{f(0)}=\frac{g^*(0)}{g^*(1)}= \frac{f^*(0)}{f^*(1)}=\frac{1}{\lambda^*}$

 

[Dickfore]

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:
$$\langle p \vert x \vert p' \rangle = \frac{1}{2\pi \hbar} \, \int_{-\infty}^{\infty} x \, e^{-\frac{i}{\hbar}(p - p') x} \, dx$$
$$= i \, \frac{\partial}{\partial p} \, \left( \frac{1}{2\pi} \, \int_{-\infty}^{\infty} e^{-\frac{i}{\hbar}(p - p') x} \, dx \right) = i \, \hbar \, \delta'(p - p') \rightarrow i \, \hbar \ \delta'(0), \ p \rightarrow p'$$

 

[ShayanJ]

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?

 

[Dickfore]

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.

 

[ShayanJ]

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?

 

[George Jones]

Some general results:

Set $\hbar = 1$ and assume

$$AB - BA = iI.$$

Multiplying the commutation relation by $B$ and reaaranging gives

$$\begin{equation*}<br /> \begin{split}<br /> AB - BA &= iI \\<br /> AB^2 - BAB &= iB \\<br /> AB^2 - B \left( BA + iI \right) &= iB \\<br /> AB^2 - B^2 A &= 2iB.<br /> \end{split}<br /> \end{equation*}$$

By induction,

$$AB^n - B^n A = niB^{n-1}$$

for every positive integer $n$. Consequently,

$$\begin{equation*}<br /> \begin{split}<br /> n\left\| B \right\|^{n-1} &=\left\| AB^n - B^n A \right\| \\<br /> &\leq 2\left\| A \right\| \left\| B \right\|^n\\<br /> n &\leq 2\left\| A \right\| \left\| B \right\| .<br /> \end{split}<br /> \end{equation*}$$

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!

 

[ShayanJ]

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?)

 

[dextercioby]

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.

 

[dextercioby]

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).

 

[ShayanJ]

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.

 

[ShayanJ]

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 $A | \psi \rangle=\lambda | \psi \rangle$:
$\langle \psi|AB|\psi \rangle=\lambda \langle \psi |B|\psi\rangle \\<br /> \langle \psi|BA|\psi \rangle=\lambda \langle \psi |B|\psi\rangle$
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 $\langle \psi |B|\psi \rangle=\infty$.

 

[Jorriss]

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).

 

[ShayanJ]

[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!

 

[dextercioby]

[...]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.