Solving a Quantum Field Theory Recursion Relation Through Self Study

[Imanbk]

Hello Everyone!

I have a problem I am solving through a self study project from Lowell Brown's book entitled: Quantum Field Theory". It is a math question (basically) on recursion relations.

Homework Statement

The variational definition gives us the relation:
det[1-λK] = exp{tr ln[1-λK]}.

Where λ is a "small" number and K is the kernel.

The variational definition shows us that λ needs to be small in order for the power series of ln[1-λK] to converge. But on the other hand, one can show that the power series:

d(λ) = det[1-λK] = Ʃ(n=0 to inf) d_(n)λ^(n)

Always converges provided that K (the kernel) is sufficiently "well behaved".

Relavant question:

What we are are asked to do is to plug the above power series into the following differential eqaution:

d/dλ d(λ) = -d(λ)*Ʃ(m=0 to inf) [λ^(m)*trK^(m+1)]

to find a recursion relation relating d_(n+1), d_(n), and trK^(n+1).

The Attempt at a Solution

I showed why the differential equation has the form it has by integrating over λ and using boundary condition det1=1 (so that ln(det1)=0) to give us back the variational definition.

To solve for the recursion relation I tried several approaches. My closest approach was writing out the sum over n as:

Ʃ(n=0 to inf) [(n+1)*d_(n+1)*λ^(n)+d_(n)*λ^(n+m)*trK^(n+m)] = 0.

I tried to get the sum with all variables λ factored out so we get the sum in the following form:

Ʃ(...)*λ^(some power) = 0

so that I can use the theorem which states that: a polynomial is identically zero if and only if all of its coefficents are zero but I can't factor out λ^(m) from the second term.

I also supposed that for n ≠ m we have zero contributions to get:

Ʃ(n=0 to inf) [(n+1)*d_(n+1)+d_(n)*λ^(2)*trK^(n+m)]*λ^(n) = 0.

I have no reason to do this (yet) but I was just playing around to see if I can get the trK^(n+1) term which was asked for in the question.

I hope I gave enough info about this question. If not please let me know! I tried hard at this question with no solution. I was taught to solve recursion relations for terms in powers of λ which differ by constants such as λ^(n), λ^(n-1), λ^(n-2), etc. but not terms which differ by powers of λ of another summation variable (m).

Thanks a lot for help on this!
Imankb

 

[Imanbk]

Is there anyone who can help me with this question? I'm willing to think with whoever can shed some light on how to solve this recursion relation.
Thanks a bunch..

 

[Imanbk]

Hello Mentors:

I posted a question 5 days ago and have received no reply to date. I checked and re-checked the posting rules and I think I have followed all the posting rules. Can you please tell if there is anything wrong so I can better ask my question next time? Thank you, imanbk

 

[Imanbk]

Hello Everyone!

I have a problem I am solving through a self study project from Lowell Brown's book entitled: Quantum Field Theory". It is a math question (basically) on recursion relations.

1. Homework Statement
The variational definition gives us the relation:
det[1-λK] = exp{tr ln[1-λK]}.

Where λ is a "small" number and K is the kernel.

The variational definition shows us that λ needs to be small in order for the power series of ln[1-λK] to converge. But on the other hand, one can show that the power series:

d(λ) = det[1-λK] = \sum\limits_{n=0}^{\infy} d_(n)λ^(n)

Always converges provided that K (the kernel) is sufficiently "well behaved".

Relavant question:

What we are are asked to do is to plug the above power series into the following differential eqaution:

\frac{d}{dλ} d(λ) = -d(λ)*Ʃ(m=0 to inf) [λ^(m)*trK^(m+1)]

to find a recursion relation relating d_(n+1), d_(n), and trK^(n+1).

3. The Attempt at a Solution

I showed why the differential equation has the form it has by integrating over λ and using boundary condition det1=1 (so that ln(det1)=0) to give us back the variational definition.

To solve for the recursion relation I tried several approaches. My closest approach was writing out the sum over n as:

\sum\limits_{n=0}^{\infy} [(n+1)*d_(n+1)*λ^(n)+d_(n)*λ^(n+m)*trK^(n+m)] = 0.

I tried to get the sum with all variables λ factored out so we get the sum in the following form:

Ʃ(...)*\lambda^(some power) = 0

so that I can use the theorem which states that: a polynomial is identically zero if and only if all of its coefficents are zero but I can't factor out \lambda^{m} from the second term.

I also supposed that for n ≠ m we have zero contributions to get:

\sum\limits_{n=0}^{\infy} [(n+1)*d_(n+1)+d_(n)*\lambda^(2)*trK^(n+m)]\lambda^(n) = 0.

I have no reason to do this (yet) but I was just playing around to see if I can get the trK^(n+1) term which was asked for in the question.

I hope I gave enough info about this question. If not please let me know! I tried hard at this question with no solution. I was taught to solve recursion relations for terms in powers of λ which differ by constants such as \lambda^(n), \lambda^(n-1), \lambda^(n-2), etc. but not terms which differ by powers of λ of another summation variable (m).

Thanks a lot for help on this!

 

[voko]

What is "the kernel" (K)?

 

[Imanbk]

Hi Voko!

I'm looking at the text and at the end of the problem it says let K be an arbitrary kernel. At the end of the problem he starts his promised explanation on the "convergence of the power series in λ that rigorously defines the infinite determinant for an arbitrary kernel
K(x,x')". I can write out the explanation here if it is required.

Given that he puts his explanation at the end of the problem, I think this part of the problem can be solved without knowing what K is, and that knowing it's well-behaved is sufficient?

I looked up the equality:
det[1-λK] = exp{tr ln[1-λK]} and found it is also called the "standard matrix identity".
I also found this text on it (attached, please look starting at statement (m) on page 4).
Let me know if you have any other questions on the problem. Any help on solving this recursion relation would be greatly appreciated... thanks!

 

[Imanbk]

I found a document online which explains the problem on page 8 of the 52 available pages. I am now just working out the induction part of the proof.. not entirely sure how they expand the espression for a_(n+1)..

Link:
http://krein.unica.it/~cornelis/private/PDF/IEOT/ieot_26_136.pdf