Rigorous Feynman pathintegral derivation

In summary, this person is having trouble doing an exercise from a physics textbook. They are not able to derive the path integral for a free hamiltonian with a Hamiltonian of the form 1/2 m p^2 + V(q). They are able to write the exponential for an operator but it does not equal the taylor exp. They are able to find an answer using the Trotter expansion and Stone's theorem.
  • #1
nahsihorst
4
0
Hey,

I'm trying to do exercise I.2.1. from Zee's QFT in a nutshell but I ran into a problem. The exercise is to derive the QM path integral with a Hamiltonian of the form 1/2 m p^2 + V(q). In the textbook he shows the proof for a free hamiltonian. He gets to a point where he has (I left out the integral for |p><p|)
[itex]e^{-i \delta t (\hat p^2 /2m)} |q> = e^{-i \delta t (\hat p^2 /2m)} |p><p|q> = e^{-i \delta t (p^2 /2m)} |p><p|q>[/itex] ([itex] \hat p [/itex]is an operator) which is obviously true. But in my case I have
[itex]e^{-i \delta t (\hat p^2 /2m + V(\hat q))} |q> \neq e^{-i \delta t( \hat p^2 /2m + V(q))} |q>[/itex]
as the commutator of [itex]\hat p[/itex] and [itex]\hat q[/itex] does not vanish. Thus I have no idea of how to prove this in general. In some QFT lecture notes I found they expand the exponential to first order, substitute[itex] \hat q = q [/itex]and [itex]\hat p = p [/itex]and write it again as an exponential. But I don't like this last step and want to do it more rigorous. Any hints?

Thanks :)
 
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  • #2
Well that actually sounds reasonable like a reasonable step. Since δt is infinitesimal, writing them as exponentials is somewhat of a lie in the first place.
 
  • #3
True, writing them as exponentials is reasonable. But I don't like to reverse this in the end, writing
[itex]1+\delta x = e^x[/itex]

Could you explain why writing them as exponentials is not correct?
 
  • #4
When a physics book uses the term "infinitesimal", it means nothing more than that the expression that follows contains only a finite number of terms from a Taylor expansion around 0. For example, "for infinitesimal x, we have exp x=1+x" means that [tex]e^x=1+x+\mathcal O(x^2).[/tex] If it's OK to replace exp x with 1+x, then it's also OK to replace 1+x with exp x. The idea is that the neglected terms go to zero in the limit x→0. (You are working with an expression that follows a "[itex]\lim_{x\rightarrow 0}[/itex]" in the actual calculation, right?)

(I didn't look at the details of this specific problem. I'm just making a comment about what appears to be the main issue).
 
  • #5
But I don't like to reverse this in the end, writing 1+δx=ex
At that point he's got an infinite product, and to get back to an exponential I guess he's using the identity lim N-> ∞ (1 + a/N)N = ea
 
  • #6
Bill_K said:
At that point he's got an infinite product, and to get back to an exponential I guess he's using the identity lim N-> ∞ (1 + a/N)N = ea

Thanks a lot. I totally forgot about that relation. Just one last question:
Usually one defines the exponential of an operator with the help of the taylor series. Can one also use this definition and is it equal to the taylor exp.?
 
  • #7
nahsihorst said:
Can one also use this definition and is it equal to the taylor exp.?
I'm pretty sure it would work for bounded operators, but I haven't tried to prove it. I'm guessing that it doesn't work in general for unbounded operators. Actually, I think that applies to the power series definition of the exponential too; it works for bounded operators but not in general for unbounded operators.

It seems to work for unbounded operators in some special cases. For example, define D by Df(x)=f'(x) for all smooth f. Then the power series definition combined with Taylor's formula tells us that f(x)=(exp(D)f)(0) for all x.

In those cases when the power series definition doesn't work, the exponential is defined by Stone's theorem.
 
  • #8
I recommend you look up the "Trotter Expansion" of a matrix exponential. The path integral can be understood as a Trotter expansion in which resolutions of the identity (alternating between Q and P) are inserted. The Trotter expansion can be derived from the Baker-Campbell-Hausdorff formula, where operators 'X' and 'Y' will refer to the kinetic and potential energy in a small time slices.

http://en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula
 
  • #9
Thanks at all. I think the Suzuki-Trotter expansion finally solved the problem for me.
 

1. What is the Rigorous Feynman pathintegral derivation?

The Rigorous Feynman pathintegral derivation is a mathematical approach used in theoretical physics to calculate the probability amplitudes for quantum systems. It is based on Feynman's path integral formulation of quantum mechanics, which states that the probability of a particle moving from one point to another is equal to the sum of all possible paths the particle could take.

2. How is the Rigorous Feynman pathintegral derivation different from the standard Feynman path integral?

The standard Feynman path integral is a formal mathematical expression that is not well-defined in some cases, leading to potential inaccuracies in calculations. The Rigorous Feynman pathintegral derivation provides a more rigorous and mathematically consistent approach to these calculations, ensuring more accurate results.

3. What are the advantages of using the Rigorous Feynman pathintegral derivation?

One of the main advantages of using the Rigorous Feynman pathintegral derivation is its ability to handle complex quantum systems with more accuracy. It also allows for the inclusion of important physical effects, such as temperature and interactions between particles, which are not easily accounted for in the standard Feynman path integral.

4. Are there any limitations or drawbacks to using the Rigorous Feynman pathintegral derivation?

One limitation of the Rigorous Feynman pathintegral derivation is that it can be quite mathematically involved and challenging to apply in practice. It also requires a good understanding of quantum mechanics and mathematical techniques, making it less accessible to non-experts.

5. In what fields of science is the Rigorous Feynman pathintegral derivation commonly used?

The Rigorous Feynman pathintegral derivation is commonly used in theoretical physics, particularly in the study of quantum systems and their behavior. It is also used in fields such as condensed matter physics, particle physics, and quantum field theory.

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