In the file linked below, I have written out a particular calculation of
a Gaussian integral, and would like to know if (6) in this file is is
correct. (If link does not work, right click to download and then
open.)

http://jayryablon.files.wordpress.co...al-correct.pdf

This is not rocket science nor is it particularly earthshaking. But, I
am having some conceptual trouble thinking about this integral (6) in
which the plane wave coefficient B=0, and would appreciate if someone
can help me straighten out my thinking about this.

Best regards,

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

First I would be careful with your eq. 5. I believe that your last
step is illegal, because

exp(A+B)=exp(A)exp(B) is true only if A and B commute. [A,B]=0

As for your other problem, you can not treat B as a parameter and
differentiate w.r.t. to it and at the same time set it equal to a
fixed value and still differentiate w.r.t. to it.

This is like letting

g(x)=(d/dx)f(x) (1)

and

g(x)=(d\dx)f(x=1)=0 (2)

Eq. 1 and 2 are not the same thing.

kp

"Jay R. Yablon" wrote in message ...

Could you *please* consider crossposting as opposed to multiposting?
http://www.blakjak.demon.co.uk/mul_crss.htm

Thanks.

Dirk Vdm

Dear Dirk Van de moortel:

On May 15, 12:21*pm, "Dirk Van de moortel" dirkvandemoor...@ThankS-NO-
SperM.hotmail.com wrote:
"Jay R. Yablon" wrote in message ...

Could you *please* consider crossposting as opposed
to multiposting?
* *http://www.blakjak.demon.co.uk/mul_crss.htm

I had only seen him posting here and in sci.physics.foundations. spf
does not allow posts to multiple newsgroups, flags them as spam.

Certainly, if he posts more places than these two, cross-posting
allows responses to be funnelled to a common thread... for "quality
control purposes".

David A. Smith

"kp" wrote in message
...
First I would be careful with your eq. 5. I believe that your last
step is illegal, because

exp(A+B)=exp(A)exp(B) is true only if A and B commute. [A,B]=0

Good point, I agree. exp(A+B)=exp(A)exp(B) is a habit. That does not
change the base calculation, which is correct aside from the last step,
because in this case, exp(A) is operating on exp(B).

As for your other problem, you can not treat B as a parameter and
differentiate w.r.t. to it and at the same time set it equal to a
fixed value and still differentiate w.r.t. to it.

This is like letting

g(x)=(d/dx)f(x) (1)

and

g(x)=(d\dx)f(x=1)=0 (2)

Eq. 1 and 2 are not the same thing.

kp

Yes, I agree. That calculation has been nagging at me for about ten
days.

I tried a correction to the calculation, at the link below:

http://jayryablon.files.wordpress.co...-correct-2.pdf

Please advise if you think that works for the the fixed parameter B=0.
I think the question here is whether it is OK to take the square roots
of the operator in the way I am doing. That is, if we have two
operators O_1 and O_2 operating on x, and if

O_1 x = O_2 x

can we also say (or at least in this circumstance at the above link can
we say) that:

sqrt(O_1) x = sqrt (O_2) x

Thanks for the help!

Jay.

"kp" wrote in message
...
First I would be careful with your eq. 5. I believe that your last
step is illegal, because

exp(A+B)=exp(A)exp(B) is true only if A and B commute. [A,B]=0


Question. Would the following be true?:

Suppose I start with exp(A+B). Suppose [A,B]0.

I can still write:

exp(A+B)=exp(A)exp(B) (1)

However, this means that

exp(A)exp(B) exp(B)exp(A) = exp(B+A) (2)

which one can see by taking the series expansion for exp(x).

So, what this really means is that I cannot commute A and B in the
exponent, *even under addition*. I have to start with a particular
order for A and B, and then be careful thereafter to maintain that
order, or, if I do not, I have to know what the commuting relationship
actually is, and apply it. But, I can still go back and forth using
(1), so long as I am careful with the term order.

Yes?

Thanks,

Jay.

On May 16, 5:53*am, "Jay R. Yablon" wrote:
"kp" wrote in message

...

First I would be careful with your eq. 5. I believe that your last
step is illegal, because

exp(A+B)=exp(A)exp(B) is true only if A and B commute. [A,B]=0

Question. *Would the following be true?:

Suppose I start with exp(A+B). *Suppose [A,B]0.

I can still write:

exp(A+B)=exp(A)exp(B) * (1)

However, this means that

exp(A)exp(B) exp(B)exp(A) = exp(B+A) * *(2)

which one can see by taking the series expansion for exp(x).

So, what this really means is that I cannot commute A and B in the
exponent, *even under addition*. *I have to start with a particular
order for A and B, and then be careful thereafter to maintain that
order, or, if I do not, I have to know what the commuting relationship
actually is, and apply it. *But, I can still go back and forth using
(1), so long as I am careful with the term order.

Yes?

No. See

http://en.wikipedia.org/wiki/Baker-C...sdorff_formula


kp

On May 15, 11:45*pm, "Jay R. Yablon" wrote:
"kp" wrote in message

...

First I would be careful with your eq. 5. I believe that your last
step is illegal, because

exp(A+B)=exp(A)exp(B) is true only if A and B commute. [A,B]=0

Good point, I agree. *exp(A+B)=exp(A)exp(B) is a habit. *That does not
change the base calculation, which is correct aside from the last step,
because in this case, exp(A) is operating on exp(B).



As for your other problem, you can not treat B as a parameter and
differentiate w.r.t. to it and at the same time set it equal to a
fixed value and still *differentiate w.r.t. to it.

This is like letting

g(x)=(d/dx)f(x) (1)

and

g(x)=(d\dx)f(x=1)=0 (2)

Eq. 1 and 2 are not the same thing.

kp

Yes, I agree. *That calculation has been nagging at me for about ten
days.

I tried a correction to the calculation, at the link below:

http://jayryablon.files.wordpress.co...alculation-of-...


No I wouldn't say this is correct. If you want to calculate your
integral with a B=0, you can still use Zee's formula you just have to
take all the derivatives w.r.t B first then set B=0.

kp

On May 15, 8:53 pm, "Jay R. Yablon" wrote:
"kp" wrote in message

...

First I would be careful with your eq. 5. I believe that your last
step is illegal, because

exp(A+B)=exp(A)exp(B) is true only if A and B commute. [A,B]=0

Question. Would the following be true?:

Suppose I start with exp(A+B). Suppose [A,B]0.

I can still write:

exp(A+B)=exp(A)exp(B) (1)

However, this means that

exp(A)exp(B) exp(B)exp(A) = exp(B+A) (2)

which one can see by taking the series expansion for exp(x).

So, what this really means is that I cannot commute A and B in the
exponent, *even under addition*. I have to start with a particular
order for A and B, and then be careful thereafter to maintain that
order, or, if I do not, I have to know what the commuting relationship
actually is, and apply it. But, I can still go back and forth using
(1), so long as I am careful with the term order.
Yes?

Personally, I find it easier to specify
A(u) B(v) =/= A(v) B(u).
That keeps the calculations easy and
highlights the asymmetry that evolved.

I think Jay you have a good developement.

As I understand, only scalars are permissible
as exponents, hence action - being invariant -
is the usual "unit" in quantum exponents.

Call "h" action, then physically, +/- h
corresponds to a system absorbing or emitting
a photon with it's energy, spin, charge etc.

Emission and absorption are physically
opposite, and so antisymmetrical, but your
equations are likely doing both at the same
time.
Regards
Ken S. Tucker