Integrating over all 4-momentum space
arkobose
To answer the question, one needs to know the physical context within
which this integral occurs. Without further information the integral is
meaningless. Most probably it's also divergent withouth the on-shell
constraint. Also it's not clear what this factor (-5/2) means. I guess
it's the power (5/2), i.e., [...]**(5/2) is meant.
==================================================================
Hello,
I am having a little confusion due to the mass-shell condition in
evaluating a definite integral. Let us say we need to evaluate the
following integral, over the entire 4-momentum space (with c = 1):
\int dp0*dp1*dp2*dp3[1 + a^2(p1^2 + p2^2 + p3^2 - p0^2)](-5/2).
There are two things one can do. One can evaluate the above integral
w.r.t each variable, between the limits -infinity and +infinity. I have
not done this yet, so don't know if this leads the integral to diverge.
On the other hand, if one uses the mass-shell condition, the above
integral becomes:
\int dp0*dp1*dp2*dp3[1 - a^2*m^2](-5/2).
It is obvious that this integral diverges. Which is correct?
Thank you.
arkobose
I apologize for the typo. The factor of (-5/2) is indeed an exponent.
As for the definite integral, I have checked. Both methods yield a
divergent result.
arkobose
Since the moderator correctly reminded me that the context was
important, it helped me ask the right question.
The position eigenkets of a free particle, in momentum space, are
plane waves. We see that these eigenkets are not normalizable, since
the norm diverges when we integrate over the entire momentum space.
However, the momentum of a free particle is constant. Since the
momentum has a unique value for a free particle, how are we justified
in integrating over momentum between minus infinity and plus infinity?
Oh No
>The position eigenkets of a free particle, in momentum space, are plane
>waves. We see that these eigenkets are not normalizable, since the norm
>diverges when we integrate over the entire momentum space. However, the
>momentum of a free particle is constant. Since the momentum has a
>unique value for a free particle, how are we justified in integrating
>over momentum between minus infinity and plus infinity?
>
not have a unique value for a free particle. In fact, as you have
already pointed out, in a position eigenstate we a plane wave in
momentum space, so that all values are represented.
Physically the answer to your question is that we never have a perfect
eigenstate of position; we have a wave function localised to a small
region. Similarly we never have a perfect eigenstate of momentum.
Mathematically we can deal with idealised position eigenkets and
momentum eigenkets using rigged Hilbert space, which is based on the
theory of distributions.
Informally we can pretend everything works just as we would like it to,
because in practice it mostly does.
Regards
--
Charles Francis
moderator sci.physics.foundations.
charles (dot) e (dot) h (dot) francis (at) googlemail.com (remove spaces and
braces)
arkobose
> Thus spake arkobose <arkob...@gmail.com>>The position eigenkets of a free particle, in momentum space, are plane
Dear Francis,
Thanks for clarifying. However, I still have questions.
Let us consider a free particle of mass m in three dimensions. The
momentum 4-vector for this particle will obey the mass-shell
condition. Now, when we write the normalization integral for the
position eigenket of this free particle in momentum space, should we
first integrate over p0 between -Sqrt(px^2 + py^2 + pz^2 + m^2) and
Sqrt(px^2 + py^2 + pz^2 + m^2), keeping px, py and pz constant, and
then integrate over px, py and pz between -Infinity and +Infinity
(which essentially means that p0 is also varying between -Infinity and
+Infinity)? Or should we go ahead and integrate over each momentum
component between -Infinity and +Infinity?
Though the eventual result may be the same in many cases, I want to
know which is procedurally correct. This is because one can find
measures on which the two methods will give different answers. I am
inclined to think that we should follow the first route, since
otherwise, the information that the momentum 4-vector obeys the mass-
shell condition will be lost.
Best,
Oh No
>Let us consider a free particle of mass m in three dimensions. The
>momentum 4-vector for this particle will obey the mass-shell condition.
>Now, when we write the normalization integral for the position eigenket
>of this free particle in momentum space, should we first integrate over
>p0 between -Sqrt(px^2 + py^2 + pz^2 + m^2) and Sqrt(px^2 + py^2 + pz^2
>+ m^2), keeping px, py and pz constant, and then integrate over px, py
>and pz between -Infinity and +Infinity (which essentially means that p0
>is also varying between -Infinity and +Infinity)? Or should we go ahead
>and integrate over each momentum component between -Infinity and
>+Infinity?
>
It depends a little bit how you set the theory up. Really the inner
product is an integral over 3-space at constant time. In a relativistic
treatment it is usual to use the covariant integral
Integral d^3p / (2p^0) = Integral d^4p delta(p^2 - m^2)
which is more like your first route, only actually we start on the lhs,
not the rhs which we only introduce to show covariance.
Personally I prefer a different normalisation, described at
http://rqgravity.net/TheDiracEquation#CovarianceOfTheDiracEquation
because it simplifies formulae through the calculation of experimental
results. Ultimately the formulae of quantum theory are related to
reality through the calculations of probabilities, and probability only
makes a great deal of sense at given time, so I don't see the need to
choose a covariant normalisation.