Skip to main content
Physics

Return to Question

Became Hot Network Question
added 3 characters in body
Source Link
Proto
Proto
  • 71
  • 2

I often read something like "the Feynman propagator is the Green's function of the Klein-Gordon equation", so I try to write it as a sum over eigenfunctions, as should be possible for any Green's function. The eigenvalue problem is $$(\square + m^2)\psi(x) = -\lambda \psi(x),$$ which can be solved easily by writing $$\psi(x) = e^{-ip\cdot x},$$ with $p = (p_0, \vec p)$, implying an eigenvalue of $\lambda = p^2 - m^2$. Then I recall that, for a discrete spectrum $\lambda_n$, the Green's function looks like $$G(x, y) = \sum_{n} \frac{\psi_n(x)\psi_n(y)^*}{\lambda_n}$$

and hope something similar also holds for a continuous spectrum, writing $$G(x, y) = \int d\lambda\frac{\psi(x)\psi(y)^*}{\lambda}.$$$$``G(x, y) = \int d\lambda\frac{\psi(x)\psi(y)^*}{\lambda}."$$ Cearly this wrong but close -- the $\lambda$ in the denominator becomes the $p^2 - m^2$ I expect. Besides anything about my approach which may be wrong, my main question is: how do I get the correct measure to show up?

I often read something like "the Feynman propagator is the Green's function of the Klein-Gordon equation", so I try to write it as a sum over eigenfunctions, as should be possible for any Green's function. The eigenvalue problem is $$(\square + m^2)\psi(x) = -\lambda \psi(x),$$ which can be solved easily by writing $$\psi(x) = e^{-ip\cdot x},$$ with $p = (p_0, \vec p)$, implying an eigenvalue of $\lambda = p^2 - m^2$. Then I recall that, for a discrete spectrum $\lambda_n$, the Green's function looks like $$G(x, y) = \sum_{n} \frac{\psi_n(x)\psi_n(y)^*}{\lambda_n}$$

and hope something similar also holds for a continuous spectrum, writing $$G(x, y) = \int d\lambda\frac{\psi(x)\psi(y)^*}{\lambda}.$$ Cearly this wrong but close -- the $\lambda$ in the denominator becomes the $p^2 - m^2$ I expect. Besides anything about my approach which may be wrong, my main question is: how do I get the correct measure to show up?

I often read something like "the Feynman propagator is the Green's function of the Klein-Gordon equation", so I try to write it as a sum over eigenfunctions, as should be possible for any Green's function. The eigenvalue problem is $$(\square + m^2)\psi(x) = -\lambda \psi(x),$$ which can be solved easily by writing $$\psi(x) = e^{-ip\cdot x},$$ with $p = (p_0, \vec p)$, implying an eigenvalue of $\lambda = p^2 - m^2$. Then I recall that, for a discrete spectrum $\lambda_n$, the Green's function looks like $$G(x, y) = \sum_{n} \frac{\psi_n(x)\psi_n(y)^*}{\lambda_n}$$

and hope something similar also holds for a continuous spectrum, writing $$``G(x, y) = \int d\lambda\frac{\psi(x)\psi(y)^*}{\lambda}."$$ Cearly this wrong but close -- the $\lambda$ in the denominator becomes the $p^2 - m^2$ I expect. Besides anything about my approach which may be wrong, my main question is: how do I get the correct measure to show up?

added 776 characters in body; edited tags
Source Link
Proto
Proto
  • 71
  • 2

Let's say I wanted to compute theoften read something like "the Feynman propagator is the Green's function of the Klein-Gordon equation", so I try to write it as a sum over eigenfunctions, as should be possible for any Green's function. The eigenvalue problem is $G(x, y) = \int\frac{d^4p}{(2\pi)^4\frac{e^{-ip\cdot(x - y)}{p^2 - m^2}$$$(\square + m^2)\psi(x) = -\lambda \psi(x),$$ which can be solved easily by using eigenfunction methodswriting $$\psi(x) = e^{-ip\cdot x},$$ with $p = (p_0, \vec p)$, implying an eigenvalue of $\lambda = p^2 - m^2$. Then I recall that, for a discrete spectrum $\lambda_n$, the Green's function looks like $$G(x, y) = \sum_{n} \frac{\psi_n(x)\psi_n(y)^*}{\lambda_n}$$

and hope something similar also holds for a continuous spectrum, writing $$G(x, y) = \int d\lambda\frac{\psi(x)\psi(y)^*}{\lambda}.$$ Cearly this wrong but close -- the $\lambda$ in the denominator becomes the $p^2 - m^2$ I expect. Besides anything about my approach which may be wrong, my main question is: how do I get the correct measure to show up?

Let's say I wanted to compute the Feynman propagator $G(x, y) = \int\frac{d^4p}{(2\pi)^4\frac{e^{-ip\cdot(x - y)}{p^2 - m^2}$ by using eigenfunction methods.

I often read something like "the Feynman propagator is the Green's function of the Klein-Gordon equation", so I try to write it as a sum over eigenfunctions, as should be possible for any Green's function. The eigenvalue problem is $$(\square + m^2)\psi(x) = -\lambda \psi(x),$$ which can be solved easily by writing $$\psi(x) = e^{-ip\cdot x},$$ with $p = (p_0, \vec p)$, implying an eigenvalue of $\lambda = p^2 - m^2$. Then I recall that, for a discrete spectrum $\lambda_n$, the Green's function looks like $$G(x, y) = \sum_{n} \frac{\psi_n(x)\psi_n(y)^*}{\lambda_n}$$

and hope something similar also holds for a continuous spectrum, writing $$G(x, y) = \int d\lambda\frac{\psi(x)\psi(y)^*}{\lambda}.$$ Cearly this wrong but close -- the $\lambda$ in the denominator becomes the $p^2 - m^2$ I expect. Besides anything about my approach which may be wrong, my main question is: how do I get the correct measure to show up?

Post Undeleted by Proto
Post Deleted by Proto
Source Link
Proto
Proto
  • 71
  • 2

Feynman propagator as a sum over eigenfunctions

Let's say I wanted to compute the Feynman propagator $G(x, y) = \int\frac{d^4p}{(2\pi)^4\frac{e^{-ip\cdot(x - y)}{p^2 - m^2}$ by using eigenfunction methods.