Dirac Proves 0 =1

George Jones

Suppose [itex]A[/itex] is an observable, i.e., a self-adjoint operator, with real eigenvalue [itex]a[/itex] and normalized eigenket [itex] \left| a \right>[/itex]. In other words,

[tex]A \left| a \right> = a \left| a \right>, \hspace{.5 in} \left< a | a \right> = 1.[/tex]

Suppose further that [itex]A[/itex] and [itex]B[/itex] are canonically conjugate observables, so

[tex] \left[ A , B \right] = i \hbar I,[/tex]

where [itex]I[/itex] is the identity operator. Compute, with respect to [itex]\left| a \right>[/itex], the matrix elements of this equation divided by [itex]i \hbar[/itex]:

[tex]
\begin{equation*}
\begin{split}
\frac{1}{i \hbar} \left< a | \left[ A , B \right] | a \right> &= \left< a | I | a \right>\\
\frac{1}{i \hbar} \left( \left< a | AB | a \right> - \left<a | BA | a \right> \right) &= <a|a>.
\end{split}
\end{equation*}
[/tex]

In the first term, let [itex]A[/itex] act on the bra; in the second, let [itex]A[/itex] act on the ket:

[tex]\frac{1}{i \hbar} \left( a \left< a | B | a \right> - a \left<a | B | a \right> \right)= <a|a>.[/tex]

Thus,

[tex]0 = 1.[/tex]

This is my favourite "proof" of the well-known equation [itex]0 = 1[/itex].

What gives?

In order not spoil other people's fun, it might be best to put "spoiler" at the top of any post that explains what's happening.

Regards,
George

Super Nade

I don't think you can do that because A and B don't commute?

George Jones

That step is OK.

One way to see this is to take |b> = A|a> and |c> = B|a>, and then to consider <b|c>.


Any is to to look at (AB)^* = B^* A^* = B A, which takes care of the order of the operators.

Regards,
George

selfAdjoint

Isn't this the one about the domains of the operators?

George Jones

I don't think the problem is with domains. I think it is possible for the intersection of the domains of A, B, and [A , B] to be dense, and to still have the proof be "true".

Regards,
George

waht

Interesting, but the proof is based on an assumption that A and B are canonically conjugate observables. Therefore 0=1 is constrained to that condition.

how does <a|[A,B]|a> = <a|AB|a> - <a|BA|a>

btw?

Physics Monkey

What a wonderful proof! I have never seen this one before, George. My discussion is below.















***SPOILER***

Think about the real line where we can represent the algebra by the usual quantum mechanical operators X and P. The key is to realize that X and P have no normalizable eigenvectors! The usual "normalization" for position "eigenstates" (lots of scare quotes) is [tex] \langle x | x' \rangle = \delta(x-x')[/tex], so let's have some fun with this formula. Since X and P are canonically conjugate we have that [tex] [X,P] = i \hbar [/tex], and we can take matrix elements of both sides. The right side is [tex] \langle x | i \hbar | x' \rangle = i \hbar \delta(x-x') [/tex]. The left side is [tex] (x - x')( - i \hbar \frac{d}{dx} \delta(x-x')) [/tex] where I have used [tex] \langle x | P = - i \hbar \frac{d}{dx} \langle x | [/tex]. Thus we appear to have stumbled onto the rather cute identity [tex] - x \frac{d}{dx} \delta(x) = \delta(x) [/tex]. Go ahead, try it under an integral, it actually works! I love such silly little formulae between wildly singular objects.

A further amusing challenge:
It isn't always true that the derivative operator has no eigenstates. Suppose you look at the derivative operator on a finite interval. It turns out that the Neumann indices are (1,1), and thus self adjoint extensions exist which are parameterized by a phase (the boundary condition). One can now find proper eigenfunctions and eigenvalues for a given self adjoint extension of the derivative operator. Are we therefore back to proving that 0 = 1 or what?

mikeu

Except you can always find such A and B, so you can always find 0=1..... :)

It's just the definition of the commutator and linearity of the inner product:
[tex]\langle a | [A,B] | a\rangle = \langle a | (AB-BA) | a \rangle = \langle a | AB | a \rangle - \langle a | BA | a \rangle[/tex].

Physics Monkey, I've got a question about your spoiler below....







*** SPOILER cont. ***




I suspected (based on X and P ) that delta distributions would enter into it, since we end up with [tex]\frac{1}{i\hbar}(a-a)\langle a|B|a\rangle=1[/tex] so it is clear that [tex]\langle a|B|a\rangle[/tex] must be ill-defined (i.e. infinite) to get something like "[tex]0\cdot\infty=1[/tex]." Recovering the definition of the derivative of the delta was neat. What I still don't see though is what the flaw in the proof is in the case of discrete operators...?

George, I thought of another 'interpretation' of the 'proof' too: you could prove 0=ih => h=0 => things aren't quantized

George Jones

So, you want to take A = P and B = X for the Hilbert space of square-integrable functions on the closed interval [0 , 1], say.









SPOILER for Physic Monkey's Challenge.

It looks like, appropriately, selfAdjoint was right - domains are important. For the operator PX, operating by X on an eigenfunction of P results in a function that is not in the domain of selfadjointness for P, so P cannot be moved left while remaining to be P.

Easy direct calculations in this example reveal a lot.

As I said in another thread, if A and B satisfy [A , B] = ihbar, then at least one of A and B must be unbounded. In example of functions on the whole real line, both X and P are unbounded, while for functions on [0 ,1], X is bounded and P is unbounded.

Regards,
George

George Jones

Time to come clean!

I lifted (and addded a liitle elaboration) this example from the Chris Isham's nice little book Lectures on Quantum Theory: Mathematical and Structural Foundations.

Very interesting discussion!

Regards,
George

Physics Monkey

Very good, George. The commutator is indeed ill defined on the momentum eigenstates.

Well then, I think I might have to take a look at Isham's book.

Thanks for the interesting post!

P.S. To all you readers out there, I can't resist telling about some nice physical applications of such ideas. It turns out that the self adjoint extensions of the momentum operator on a finite interval describe physically the problem of a particle on a ring with a magnetic field through the ring. This is in turn equivalent to imposing a 'twisted' boundary condition [tex]\psi(x+L) = e^{i \alpha} \psi(x) [/tex] on the wavefunction for a particle on a ring with no magnetic field. But there's more! Impurities in a metal can localize electronic states and cause a metal to become an insulator. One way to tell if you have localized states is to look at how sensitive such states are to the boundary conditions of your sample. The above ideas can then be applied, and you can relate the question of localization to the behavior of the system under an applied magnetic field (a problem which can be attacked with perturbation theory). And you thought self adjoint extensions were dull! Shame on you.

reilly

First, if "0 = 1" is true then QM completely falls apart, sorta like proof by contradiction, and "0=1" is certainly a contradiction. That tells me that the various proofs must be incorrect, or most physicists have been living like Alice in Wonderland.

The problem is that P X | x> is not equal to P|x> x. As in, go to an x position representation in which P = -i d/dx. That is,

P X |x> = -i d/dx x |x> = (-i + {-i x d/dx})|x>

Delta functions and domaines are not at issue

Sometimes abstraction can lead even the best astray.

Think about Wick's Thrm, which would not hold if "0 =1" were true, nor would many standard manipulations of creation and destruction operators be legitimate. .

(For the abstract truth about momentum operators see Hille and Phillips, Functional Analysis and Semi Groups, Chap XIX, which discusses translation operators (d/dx) in great and highly rigorous detail. The authors demonstrate that there really is not a problem with such operators.

Again, if "0=1" then QM is inherently mathematically trustworthy, which seems to me to be a completely absurd idea.

Regards,
Reilly Atkinson

Hurkyl

Any notation can lead people astray. But abstraction has the advantage that there are fewer messy details, which means less opportunities to make mistakes, and less possibility for those mistakes to be obscured.

Avoiding abstraction certainly doesn't prevent one from making mistakes...


such as overworking your variables. The x in d/dx is not the same as the x as in |x>; the former is the coordinate variable of the position representation, and the latter is a constant denoting which position eigenstate we've selected.

If I relabel the variables so x is no longer being overworked, we're looking at -i d/dx x |a>. (And don't forget that x |a> = a |a>)


You could rewrite George's entire post in the A-representation (so that A = x, and B = -ih d/dx), but that doesn't resolve the paradox: you still wind up with 0 = 1.


That's not accurate: if 0=1 were true, then everything is true. (And simultaneously false)


I'm completely confused by this.

Gokul43201

Can we go over this again, slowly ? This is something that has bothered me for a little while. Is L the circumference of the ring ? Does this not destroy the single-valuedness of [itex]\psi(x)[/itex]? Or is that what is being probed ?

I think I've drunk too deep from the cup of Periodic BCs, what with all the goodies like flux quantization in SCs and Brillouin zones in crystals that it has thrown up like so many marshmallows!


Help me understand this, please.

Let's start with a simple case : the Anderson hamiltonian for non-interacting electrons in a cubic lattice.

The Hamiltonian consists of your favorite on-site disorder potential and the usual hopping term (nn, say). You then apply the above boundary condition to the single-particle eigenfunction in one or more directions. Ignoring what this means for now, this allows you to Taylor expand the eigenvalues [itex]E_i(\alpha) [/itex] and look at the coefficients of higher order terms in [itex]\alpha[/itex]. The deviations from 0 of these coefficients is what you call the phase sensitivity? If that's true, how exactly is this a "measure" of localization? Is the point to extract a dimensionless number (like T/U) and looking for a scaling law? And if not, what happens next?

lalbatros

George Jones,
Physics Monkey,

Would that "0=1" contradiction be a proof that no finite-dimentional matrix could satisfy the commutation relation [tex] \left[ A , B \right] = i \hbar I[/tex] ?

Would it possible to see that easily for two-dimentional matrices?

Michel

Haelfix

The problem is indeed one of domains of definition, its the last step in the sequence that is erroneous. Ask yourselves, what *is* the operator AB or BA and where and what are they defined on?

Most of this is easily demystified if you recall the spectral theorem. For general operators, you usually are confronted not just with discrete or continous spectra, but instead you have that + a bunch of other stuff, often called the residual spectrum. All bets are off when confronted with this, you can't just use naive physicist language of functional analysis in those cases.

mikeu

It does appear to be a proof by contradiction, at least for observables. If A is not Hermitean then [tex]\langle a|A=(A^\dagger|a\rangle)^\dagger\neq (A|a\rangle)^\dagger[/tex], so acting A to the left in the term [tex]\langle a|AB|a\rangle[/tex] doesn't yield [tex]a\langle a|B|a\rangle[/tex], as required to obtain the contradiction 0=1. Your conclusion is correct anyway, it's just not proven by this example (unless I missed something else, it is late...).

For 2D matrices, if A and B are completely arbitrary then

[tex]A=\left(\begin{array}{cc} a & b \\ c & d\end{array}\right),\
B=\left(\begin{array}{cc} w & x \\ y & z\end{array}\right),\
AB-BA=\left(\begin{array}{cc} bz-cy & b(w-x)+y(a-d) \\ c(x-w)+z(d-a) & -(bz-cy)\end{array}\right).[/tex]

Since the 1-1 entry is the negative of the 2-2 entry, this can never be proportional to the identity matrix.