Query about commutation of Scalar rest mass with Position

[Jay R. Yablon]

I am trying to decide between two view of a problem, and would like
input.

Consider the canonical commutation relationship

x^v p^u = p^u x^v + i hbar kronecker^vu (1)

consider also an on-shell mass:

m^2 = p^u p_u (2)

Now, let us suppose we have a term

x^v m^2 = x^v p^u p_u . (3)

Can m^2, since it is a scalar, be commuted to the left of x^v. That is,
does

x^v m^2 = m^2 x^v ? (4)

Or, as I suspect, starting from (3), does one have to use:

x^v m^2 = x^v p^u p_u
= [p^u x^v + i hbar kronecker^vu] p_u
= p^u x^v p_u + i hbar kronecker^vu p_u
= p^u [p_u x^v + i hbar kronecker^v_u] + i hbar kronecker^vu p_u
= p^u p_u x^v + i hbar p^u kronecker^v_u + i hbar kronecker^vu p_u
= m^2 x^v + 2i hbar p^v ? (5)

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 May 27, 10:06 pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
> I am trying to decide between two view of a problem, and would like
> input.
>
> Consider the canonical commutation relationship
>
> x^v p^u = p^u x^v + i hbar kronecker^vu (1)

At this point you already depart from quantum mechanics, relativistic
or not. There is no x^v operator when v is the time index. But, say
that you take this relation as an axiom (*).

> consider also an on-shell mass:
>
> m^2 = p^u p_u (2)
>
> Now, let us suppose we have a term
>
> x^v m^2 = x^v p^u p_u . (3)
>
> Can m^2, since it is a scalar, be commuted to the left of x^v.

OK, here you take on another axiom (**), [m^2,x^v] = 0.

> That is, does
>
> x^v m^2 = m^2 x^v ? (4)
>
> Or, as I suspect, starting from (3), does one have to use:
>
> x^v m^2 = x^v p^u p_u
> = [p^u x^v + i hbar kronecker^vu] p_u
> = p^u x^v p_u + i hbar kronecker^vu p_u
> = p^u [p_u x^v + i hbar kronecker^v_u] + i hbar kronecker^vu p_u
> = p^u p_u x^v + i hbar p^u kronecker^v_u + i hbar kronecker^vu p_u
> = m^2 x^v + 2i hbar p^v ? (5)

Logically, both (4) and (5) are correct starting from axioms (*) and
(**). Congratulations, you've just proven that 0 = 1 (subtract (4)
from (5) and contract with p_v).

Such inconsistencies are not present when you stick with usual QM
axioms. Basically, this little calculation shows how much care must be
taken when arbitrarily departing from them.

Igor

[torre@cc.usu.edu]

On May 27, 8:06 pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
> I am trying to decide between two view of a problem, and would like
> input.
>
> Consider the canonical commutation relationship
>
> x^v p^u = p^u x^v  + i hbar kronecker^vu  (1)
>
> consider also an on-shell mass:
>
> m^2 = p^u p_u   (2)
>
> Now, let us suppose we have a term
>
> x^v m^2 = x^v p^u p_u .   (3)
>
> Can m^2, since it is a scalar, be commuted to the left of x^v.  

Etc.

Normally in the quantum mechanics (or the classical
mechanics) of a relativistic particle the relation between
rest mass and 4-momentum is treated as a *constraint*,
not as an identity. Then these sorts of issues and associated
paradoxical computations do not arise.

For a thorough treatment of constrained Hamiltonian systems
and their quantum formulation, see, "Quantization of Gauge
Systems", by Henneaux and Teitelboim.

charlie torre

[Jay R. Yablon]

"Igor Khavkine" <igor.kh@gmail.com> wrote in message
news:63654c4d-3c1e-4a0b-8a86-1646bf060109@l64g2000hse.googlegroups.com...
> On May 27, 10:06 pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
>> I am trying to decide between two view of a problem, and would like
>> input.
>>
>> Consider the canonical commutation relationship
>>
>> x^v p^u = p^u x^v + i hbar kronecker^vu (1)
>
> At this point you already depart from quantum mechanics, relativistic
> or not. There is no x^v operator when v is the time index. But, say
> that you take this relation as an axiom (*).
>
>> consider also an on-shell mass:
>>
>> m^2 = p^u p_u (2)
>>
>> Now, let us suppose we have a term
>>
>> x^v m^2 = x^v p^u p_u . (3)
>>
>> Can m^2, since it is a scalar, be commuted to the left of x^v.
>
> OK, here you take on another axiom (**), [m^2,x^v] = 0.
>
>> That is, does
>>
>> x^v m^2 = m^2 x^v ? (4)
>>
>> Or, as I suspect, starting from (3), does one have to use:
>>
>> x^v m^2 = x^v p^u p_u
>> = [p^u x^v + i hbar kronecker^vu] p_u
>> = p^u x^v p_u + i hbar kronecker^vu p_u
>> = p^u [p_u x^v + i hbar kronecker^v_u] + i hbar kronecker^vu p_u
>> = p^u p_u x^v + i hbar p^u kronecker^v_u + i hbar kronecker^vu p_u
>> = m^2 x^v + 2i hbar p^v ? (5)
>
> Logically, both (4) and (5) are correct starting from axioms (*) and
> (**). Congratulations, you've just proven that 0 = 1 (subtract (4)
> from (5) and contract with p_v).
>
> Such inconsistencies are not present when you stick with usual QM
> axioms. Basically, this little calculation shows how much care must be
> taken when arbitrarily departing from them.
>
> Igor
>
So, where is the arbitrary departure that causes 1=0? Is it the x^0 not
being an operator that does this? Anything else?

And, are you saying that the canonical commutation relationship is NOT
generally covariant?

If so, what is the proper covariant formulation? How does time commute
with energy? Does energy commute with any space coordinates? Does
momentum commute with the time coordinate?

Jay.

[Igor Khavkine]

On May 29, 7:54 am, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:

> So, where is the arbitrary departure that causes 1=0? Is it the x^0 not
> being an operator that does this? Anything else?

I refer you to my previous reply for the answers.

> And, are you saying that the canonical commutation relationship is NOT
> generally covariant?

Are you assuming canonical commutation relations ARE covariant? In
canonical quantization, the transformation properties of the
commutators should be compared the corresponding Poisson brackets (and
usually are the same, at least up to higher order corrections in
hbar). Try to answer the same question in the context of Poisson
brackets.

> If so, what is the proper covariant formulation? How does time commute
> with energy? Does energy commute with any space coordinates? Does
> momentum commute with the time coordinate?

If you want a covariant formulation, read about constrained mechanical
systems. Rovelli's book on Quantum Gravity, Dirac's Lectures on
Quantum Mechanics, and Henneaux and Teitelbiom's Quantization of Gauge
Systems are good references in order of increasing sophistication.

Much of the time the constrained system approach is not necessary.
Think carefully about whether you need it.

Igor

[Jay R. Yablon]

<torre@cc.usu.edu> wrote in message
news:9c9da895-b361-48eb-9a59-daabec79e497@e39g2000hsf.googlegroups.com...
On May 27, 8:06 pm, "Jay R. Yablon" <jyab...@nycap.rr.com> wrote:
> I am trying to decide between two view of a problem, and would like
> input.
>
> Consider the canonical commutation relationship
>
> x^v p^u = p^u x^v + i hbar kronecker^vu (1)
>
> consider also an on-shell mass:
>
> m^2 = p^u p_u (2)
>
> Now, let us suppose we have a term
>
> x^v m^2 = x^v p^u p_u . (3)
>
> Can m^2, since it is a scalar, be commuted to the left of x^v.

Etc.

Normally in the quantum mechanics (or the classical
mechanics) of a relativistic particle the relation between
rest mass and 4-momentum is treated as a *constraint*,
not as an identity. Then these sorts of issues and associated
paradoxical computations do not arise.

For a thorough treatment of constrained Hamiltonian systems
and their quantum formulation, see, "Quantization of Gauge
Systems", by Henneaux and Teitelboim.

charlie torre

Charlie,

I found the book on Amazon. Can you please point out some specific
points where the book references the relationship:

m^2 = p^u p_u

and this is discussed in the context of being a constraint among
canonical variables, in this case, presumably, among m and

p^u, u=0,1,2,3

and

g_uv, u,v=0,1,2,3

Thanks,

Jay.

[pirillo]

Jay,

One thing to note is that in relativistic QM of single partikl m^2 is
the identity operator
times the number m^2 which commutes with all operators on that
sector. When you
promote M^2 to many particle theory, its no longer the identity, so it
may not commute
with x's. Will kommute with the p's though!

[pirillo]

So,..Mr Yablon proved that

[x_m, p_n] = eta_{mn} (theres ihbar but that dont
matter for us here)

[M^2, x_n] = 0

and p^2 = M^2

with m,n ranging 0,1,2,3

is a logically inconsistent set of statements.
(what makes this go wrong compared to the usual picture,
is that in the usual picture m,n range from 123 in [x_m, p_n] =
eta_{mn} ,
while the indices range from 0123
in p^2 = M^2 )

Now if what he wants to know is some kind of fix to this, that is
consistent with the usual picture,

just "assume the usual picture"

[x_m, p_n] = eta_{mn} for m,n=123 --------leave m,n=4 and
combination of 4 with 123 undetermined

[M^2, x_n] = 0 for n=123

p^2 = sum_k p_k^2 eta_{kk} = M^2 where the k=0123

then we ask M^2 x_n = Sum_{j \= n, 0} p_j^2 x_n + p_0^2 x_n + p_n^2
x_n

(Here make the added assumption that [M^2, x_0] = 0)

Now the above becomes x_n M^2 = Sum_{j \= n, 0} x_n p_j^2 +
eta_{00} p_0^2 x_n + x_n p_n^2 + 2p_n

Then, whatever freedom is left are basically your choices for the
operator x_0