Questions for Igor about canonical commutation

[Jay R. Yablon]

Dear Igor (or anyone else who wants to give it a try):

I have a few questions with reference to a thread that you participated
in some time ago, at:

http://www.physicsforums.com/archive/index.php/t-142092.html or
http://groups.google.com/group/sci.physics.research/browse_frm/thread/d78cbfecf703ff6a#.

1) In post #2, you say "The x_i and p_i operators, for i=1,2,3, *even
in the relativistic case* are still canonically conjugate pairs so the
still have the commutation relations [x_i,p_j] = i*hbar*d_ij." (emphasis
added)

Steve Carlip, in a post at
http://groups.google.com/group/sci.physics.relativity/browse_frm/thread/a2db013e54cfd129#,
says "in relativistic quantum field theory, x^u are not operators, and
therefore don't have nontrivial commutators with anything."

Unless you are talking about single-particle theory and Carlip about
multiple-particle theory, these two thoughts seem to perhaps be in
conflict. Might you be able to clarify?

2) You also say "neither t nor H are independent observables. The
former is just an independent scalar variable . . ." I want to be
clear: is t a 1x1 dimensional scalar, a scalar multiplied by an
infinite-dimensional identity matrix I_oo (along the lines of the x
operator shown at
http://scienceworld.wolfram.com/physics/PositionOperator.html, in which
case it is still commuting with any other infinite-dimensional matrix),
or something else? I.e., is t really

t = t I_oo ? (1)

3) You allude to, and Eugene Stefanovich in post 6 points out, that the
way to ensure [x_i, m^2] = 0, is for the commutators with p_0 to be:

[x^i, P_0] = P^i/P_0 = -i*hbar V^i (velocity) (2)

(Eugene has the wrong sign, BTW, I added the - to correct this). Has
anyone, so far as you are aware of, done a calculation for the
commutation with *linear* rest mass, i.e.,

[x_i, m] = 0 (3)

which, I believe, would have to make use of Dirac's equation written as:

m psi = gamma^u p_u psi ? (4)

If so, what result comes from that? I have tried it, and the result
seems to be:

gamma^0 [x^i,P_0] psi = -i gamma^i psi (5)

or, in terms of alpha^i=gamma^0 gamma^i

[x^i,P_0] psi = -i alpha^i psi, (6)

contrast (6) with (2) and note that in (6) the commutator only makes
sense as an eigenvalue of alpha^i in relation to eigenvectors defined
from psi. If you wish to see the calculation, see (2.6) and (2.7) in
http://jayryablon.files.wordpress.com/2008/06/cannonical-commutators.pdf
which was already posted in the recent SPR thread "Canonical
Commutators, Rest Mass Commutation, and Dirac's Equation."

How would you interpret this indication that (at least some) canonical
commutation relationships may need to be specified in relationship to
associated Eigenvectors and not merely in the abstract?

Thanks,

Jay.
____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.nycap.rr.com/jry/FermionMass.htm

[Igor Khavkine]

On Jun 15, 3:01 pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
> Dear Igor (or anyone else who wants to give it a try):
>
> I have a few questions with reference to a thread that you participated
> in some time ago, at:
>
> http://www.physicsforums.com/archive/index.php/t-142092.htmlorhttp://groups.google.com/group/sci.physics.research/browse_frm/thread....
>
> 1) In post #2, you say "The x_i and p_i operators, for i=1,2,3, *even
> in the relativistic case* are still canonically conjugate pairs so the
> still have the commutation relations [x_i,p_j] = i*hbar*d_ij." (emphasis
> added)
>
> Steve Carlip, in a post athttp://groups.google.com/group/sci.physics.relativity/browse_frm/thre...,
> says "in relativistic quantum field theory, x^u are not operators, and
> therefore don't have nontrivial commutators with anything."
>
> Unless you are talking about single-particle theory and Carlip about
> multiple-particle theory, these two thoughts seem to perhaps be in
> conflict. Might you be able to clarify?

There is no conflict. I was talking about a single-particle system (as
I believe you were, if not you should have been more specific) and
Steve Carlip was clearly talking about "quantum field theory", which
is equivalent to a quantum multi-particle system.

> 2) You also say "neither t nor H are independent observables. The
> former is just an independent scalar variable . . ." I want to be
> clear: is t a 1x1 dimensional scalar, a scalar multiplied by an
> infinite-dimensional identity matrix I_oo (along the lines of the x
> operator shown athttp://scienceworld.wolfram.com/physics/PositionOperator.html, in which
> case it is still commuting with any other infinite-dimensional matrix),
> or something else? I.e., is t really
>
> t = t I_oo ? (1)

This is just the standard way to represent a number (real or complex)
as an operator, in a way consistent with the usual vector space axioms
involving scalars.

> 3) You allude to, and Eugene Stefanovich in post 6 points out, that the
> way to ensure [x_i, m^2] = 0, is for the commutators with p_0 to be:
>
> [x^i, P_0] = P^i/P_0 = -i*hbar V^i (velocity) (2)
>
> (Eugene has the wrong sign, BTW, I added the - to correct this). Has
> anyone, so far as you are aware of, done a calculation for the
> commutation with *linear* rest mass, i.e.,
>
> [x_i, m] = 0 (3)

The operator m is a scalar (of the form (real number)*I_oo, as you've
put it above). That, by itself, is plenty sufficient to show that
[O,m^k]=0, for *any* operator O and any power k of m.

> [...snip introduction of Dirac equation...]

Let me repeat this: m is a scalar, [O,m] = 0 for any operator O.
Introducing anything else into the picture is merely muddying the
waters.

Igor

[Jay R. Yablon]

"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:d1c24826-4c36-4b56-aa43-69d619211de7@m44g2000hsc.googlegroups.com...
> On Jun 15, 3:01 pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
.. . .
> The operator m is a scalar (of the form (real number)*I_oo, as you've
> put it above). That, by itself, is plenty sufficient to show that
> [O,m^k]=0, for *any* operator O and any power k of m.
>
>> [...snip introduction of Dirac equation...]
>
> Let me repeat this: m is a scalar, [O,m] = 0 for any operator O.
> Introducing anything else into the picture is merely muddying the
> waters.
>
> Igor
>

I agree completely. So, I have linked a 2.5 page calculation at:

http://jayryablon.files.wordpress.com/2008/06/linear-mass-commutator-calculation.pdf

It is simple and to the point. Start with [O,m] = 0 for any operator O
including O=x_u. Take Dirac's equation to be true. Take the canonical
commutation relationship:

[x_i,p_j]=i hbar eta_ij i,j=1,2,3 (1)

to be true, where diag (eta_ij) = (1,1,1) are the space components of
the Minkowski metric tensor. From this, one may directly deduce that:

[x_k,p_0] psi = -i alpha_k psi (2)

where alpha_k = gamma^0 gamma_k, and the gamma^u are the Dirac matrices.

Equation (2) suggests that the [x_k,p_0] commutators times i=sqrt(-1)
must be equal to the eigenvalues of the Dirac alpha_k, which alpha_k, by
the way, also sit along the 0k components of the

-2i sigma_uv = [gamma_u, gamma_v]

and that these [x_k,p_0] commutators can therefore only be considered in
relation to their associated eigenstate vectors.

Just as Dirac's equation reveals some features that cannot be seen
strictly from the Klein Gordon equation, the calculation here seems to
reveal some features about the canonical commutators that the usual
calculation based on [O,m^2]=0 and m=p^u.p_u cannot, by itself, reveal.

I would like some input on whether this is on the right track or whether
I am overlooking something.

Jay.

[Igor Khavkine]

On Jun 21, 5:59 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
> "Igor Khavkine" <igor...@gmail.com> wrote in message
> news:d1c24826-4c36-4b56-aa43-69d619211de7@m44g2000hsc.googlegroups.com...

> > Let me repeat this: m is a scalar, [O,m] = 0 for any operator O.
> > Introducing anything else into the picture is merely muddying the
> > waters.

> I agree completely. [...Dirac equation comes up again...]

Let me respectfully disagree with your agreement, since it leads you
to once again introduce the Dirac equation. Or maybe I'm just getting
confused here...

> Take Dirac's equation to be true. Take the canonical
> commutation relationship:
>
> [x_i,p_j]=i hbar eta_ij i,j=1,2,3 (1)
>
> to be true, where diag (eta_ij) = (1,1,1) are the space components of
> the Minkowski metric tensor. From this, one may directly deduce that:
>
> [x_k,p_0] psi = -i alpha_k psi (2)
>
> where alpha_k = gamma^0 gamma_k, and the gamma^u are the Dirac matrices.

Jay, unfortunately, you've missed the subtle point that the x and p of
(2) are not the same as the x and p of (1). To get the right
relationship between the x and p operators satisfying the correct
commutation relations with the Lorentz group generators and the
operators "multiply by x" and "differentiate by x" for solutions of
the Dirac equation, you need to study the Foldy-Wouthuysen
transformation. It is described in detail in several books on
relativistic quantum mechanics as well as in some papers on the arXiv.
Arnold Neumaier also has several great relevant references in his
Theoretical Physics FAQ.

Igor

[Jay R. Yablon]

"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:3a1e0293-2aa3-47dd-8a85-24309c3a78ab@2g2000hsn.googlegroups.com...
> On Jun 21, 5:59 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>> "Igor Khavkine" <igor...@gmail.com> wrote in message
>> news:d1c24826-4c36-4b56-aa43-69d619211de7@m44g2000hsc.googlegroups.com...
>
>> > Let me repeat this: m is a scalar, [O,m] = 0 for any operator O.
>> > Introducing anything else into the picture is merely muddying the
>> > waters.
>
>> I agree completely. [...Dirac equation comes up again...]
>
> Let me respectfully disagree with your agreement, since it leads you
> to once again introduce the Dirac equation. Or maybe I'm just getting
> confused here...
>
>> Take Dirac's equation to be true. Take the canonical
>> commutation relationship:
>>
>> [x_i,p_j]=i hbar eta_ij i,j=1,2,3 (1)
>>
>> to be true, where diag (eta_ij) = (1,1,1) are the space components of
>> the Minkowski metric tensor. From this, one may directly deduce
>> that:
>>
>> [x_k,p_0] psi = -i alpha_k psi (2)
>>
>> where alpha_k = gamma^0 gamma_k, and the gamma^u are the Dirac
>> matrices.
>
> Jay, unfortunately, you've missed the subtle point that the x and p of
> (2) are not the same as the x and p of (1). To get the right
> relationship between the x and p operators satisfying the correct
> commutation relations with the Lorentz group generators and the
> operators "multiply by x" and "differentiate by x" for solutions of
> the Dirac equation, you need to study the Foldy-Wouthuysen
> transformation. It is described in detail in several books on
> relativistic quantum mechanics as well as in some papers on the arXiv.
> Arnold Neumaier also has several great relevant references in his
> Theoretical Physics FAQ.
>
> Igor
>
Thank you Igor, I will study this.

Let me please ask one more question to try to pinpoint the place where I
have missed the "subtle point that the x and p of (2) are not the same
as the x and p of (1)."

I look at (1) and of course see p_j; j=1,2,3 amidst the commutator:

[x_i,p_j]=i hbar eta_ij; i,j=1,2,3 (1)

Then, I look at Dirac's equation:

m psi = gamma^u p_u psi (3)

and also see a p_u; u=0,1,2,3 and presume that the subset p_j; j=1,2,3
from Dirac's equation are the same as the p_j; j=1,2,3 of (1).

Then, I apply:

[m,x^u]=0 (4)

and calculate from there.

Not even getting to the x^u, you are telling me that I am equating
apples to oranges here and p_j in (1) and (3) are not the same. Yes?

Thanks,

Jay.

[Jay R. Yablon]

"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:3a1e0293-2aa3-47dd-8a85-24309c3a78ab@2g2000hsn.googlegroups.com...
> On Jun 21, 5:59 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>> "Igor Khavkine" <igor...@gmail.com> wrote in message
>> news:d1c24826-4c36-4b56-aa43-69d619211de7@m44g2000hsc.googlegroups.com...
>
>> > Let me repeat this: m is a scalar, [O,m] = 0 for any operator O.
>> > Introducing anything else into the picture is merely muddying the
>> > waters.
>
>> I agree completely. [...Dirac equation comes up again...]
>
> Let me respectfully disagree with your agreement, since it leads you
> to once again introduce the Dirac equation. Or maybe I'm just getting
> confused here...
>
>> Take Dirac's equation to be true. Take the canonical
>> commutation relationship:
>>
>> [x_i,p_j]=i hbar eta_ij i,j=1,2,3 (1)
>>
>> to be true, where diag (eta_ij) = (1,1,1) are the space components of
>> the Minkowski metric tensor. From this, one may directly deduce
>> that:
>>
>> [x_k,p_0] psi = -i alpha_k psi (2)
>>
>> where alpha_k = gamma^0 gamma_k, and the gamma^u are the Dirac
>> matrices.
>
> Jay, unfortunately, you've missed the subtle point that the x and p of
> (2) are not the same as the x and p of (1). To get the right
> relationship between the x and p operators satisfying the correct
> commutation relations with the Lorentz group generators and the
> operators "multiply by x" and "differentiate by x" for solutions of
> the Dirac equation, you need to study the Foldy-Wouthuysen
> transformation. It is described in detail in several books on
> relativistic quantum mechanics as well as in some papers on the arXiv.
> Arnold Neumaier also has several great relevant references in his
> Theoretical Physics FAQ.
>
> Igor
>
Igor,

I found two good links for Newton-Wigner and Foldy-Wouthuysen. The
first shows the calculation itself of this transformation:

http://www.physics.ucdavis.edu/~cheng/230A/RQM7.pdf

The second, an excellent and lucid exposition of the physics (why this
is of interest), is to be found at:

http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.27.3209&rep=rep1&type=pdf

Those who have followed these discussions may be interested in these,
especially the latter.

Thanks,

Jay.
____________________________
Jay R. Yablon
Email: jyablon@nycap.rr.com
co-moderator: sci.physics.foundations
Weblog: http://jayryablon.wordpress.com/
Web Site: http://home.nycap.rr.com/jry/FermionMass.htm