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| bio | website | |
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| age | ||
| visits | member for | 2 years, 5 months |
| seen | Aug 29 '11 at 13:32 | |
| stats | profile views | 133 |
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Dec 17 |
awarded | Yearling |
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Nov 19 |
awarded | Nice Question |
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Sep 2 |
awarded | Popular Question |
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Dec 17 |
awarded | Yearling |
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Jul 19 |
accepted | What's the definition of the time ordering operator for more than two particles? |
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Jul 15 |
asked | What's the definition of the time ordering operator for more than two particles? |
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Jun 5 |
comment |
Questions about the Dyson equation @marek The $\tau$ is imaginary time within the finite temperature/equilibrium framework. Which Fourier transformation do you mean to perform first? Can you in general not do partial summations to infinite order then? (Eg in the Hartree-Fock approximation, you're left with two diagrams in the self energy: is it usually only possible to calculate $G$ using the integral equation to as many orders as you can calculate? Or is there something I'm missing which means they can be summed to infinite order?) |
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Jun 4 |
comment |
Questions about the Dyson equation One more comment.. If the integral equation can't be written like my equation 1, then don't the iterations only give the usual perturbatively expansion and not an actual summation? I.e. Without the integrals, to first order perturbation, $G=G_0 + G_0 \Sigma G_0$ and to second order $G=G_0 + G_0 \Sigma G_0 + G_0 \Sigma G_0 \Sigma G_0$. |
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Jun 4 |
comment |
Questions about the Dyson equation I do mean in thermal equilibrium. For the last question, the sign of each diagram appears to be different in each book. In Mattuck you only need to multiply by $-1$ for loops, but in Negele, the sign is also related to the order of perturbation and the sign of permutation (though I'm not 100% sure what that means). |
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Jun 4 |
comment |
Questions about the Dyson equation @Marek thanks, from the book, it sounds like you can only get $(G_^{-1} - \Sigma)^{-1}$ under certain special circumstances, otherwise you get an integral equation rather than an algebraic one which can't be written like that.. Is that not the case? |
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Jun 3 |
revised |
Questions about the Dyson equation edited tags |
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Jun 2 |
asked | Questions about the Dyson equation |
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May 19 |
comment |
Can the work done between two non-equilibrium states be calculated? @genneth thanks, I'll look at that. |
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May 19 |
comment |
When can the source term of a partition function be put in? @Qmechanic I'm not sure what you mean ... Do you mean that I have to start at equation (1) then diagonalise the Hamiltonian plus source term? And that I can't start with the Hamiltonian in diagonal form? |
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May 19 |
accepted | Is there a relativistic (quantum) thermodynamics? |
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May 19 |
comment |
When can the source term of a partition function be put in? @wsc Thanks, but what if the transformations needed to make the Hamiltonian diagonal are more complicated than a simple Fourier transform? |
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May 18 |
asked | When can the source term of a partition function be put in? |
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May 6 |
revised |
Have the correlation functions of the XY spin chain model been calculated using a functional partition function with source terms? added 55 characters in body |
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May 5 |
asked | Have the correlation functions of the XY spin chain model been calculated using a functional partition function with source terms? |
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May 4 |
comment |
Do derivatives anticommute with Grassmann variables and complex numbers in a many-body path integral? @Lubos I've got another related question to this, but I'm not sure it's interesting/different enough to ask it as a new question: we have $\partial_\tau (\bar{\psi} \psi)=(\partial_\tau\bar{\psi})\psi+\bar{\psi}\partial_\tau \psi=0$. Can I write $(\partial_\tau \bar{\psi}) \psi=−\psi(\partial_\tau \bar{\psi})$ by moving the $\psi$ passed the other Grassmann number? Then $\psi(\partial_\tau \bar{\psi})=\bar{\psi}\partial_\tau \psi$. |
