Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Transverse Susceptibility, and doubts about retarded/advanced Green's Functions

  1. Oct 28, 2011 #1
    Hi,

    I have been working through lecture notes of a condensed matter field theory class, and unfortunately I haven't found references for some of the specialized calculations though the way they're done makes them seem common knowledge in the cond-mat community. I have started this thread because it turns out I have some very basic doubts about the interpretations of advanced and retarded Green's functions which haven't been answered by successive re-readings of the standard textbooks. I will be grateful if someone can answer some of these queries.

    1. To begin with, I want to write down the transverse susceptibility through a particle hole propagator in the ladder approximation. Suppose [itex]U[/itex] denotes the intra-site (or intra-orbital, depending on the context -- I am using a Hubbard model like system) energy term, then the transverse susceptibility [itex]\chi^{-+}[/itex] is given in the Fourier domain as

    [tex]\chi^{-+}(Q,\Omega) = \frac{\chi_{0}(Q,\Omega)}{1-U\chi_{0}(Q,\Omega)}[/tex]

    where

    [tex]\chi_{0}(Q,\Omega)=i\sum_{k_1}\int_{-\infty}^{\infty}\frac{d\omega_1}{2\pi}G_{\uparrow}^{0}(k_1,\omega_1)G_{\downarrow}^{0}(k_1-Q,\omega_1-\Omega)[/tex]

    is just i times the bare particle hole propagator (w/o self energy corrections) without any interaction lines (the first term in the ladder).

    2. Now, the exact forms of the Green's functions will depend on the band structure of the system being considered. In the most general case, we can write

    [tex]G_{\uparrow}^{0}(k_1,\omega_1) = \frac{\theta(k_1-k_F^{\uparrow})}{\omega_1-E_{k_1}^{\uparrow}+i\eta} + \frac{\theta(k_F^{\uparrow}-k_1)}{\omega_1-E_{k_1}^{\uparrow}-i\eta}[/tex]

    and

    [tex]G_{\downarrow}^{0}(k_1-Q,\omega_1-\Omega) = \frac{\theta(|k_1-Q|-k_F^{\downarrow})}{\omega_1-\Omega-E_{k_1-Q}^{\downarrow}+i\eta} + \frac{\theta(k_F^{\downarrow}-|k_1-Q|)}{\omega_1-\Omega-E_{k_1}^{\downarrow}-i\eta}[/tex]

    Here [itex]E_{k}^{\uparrow,\downarrow}[/itex] denote the band energies for up-spin and down-spin. I consider retarded and advanced contributions from both the up-spin and down-spin bands.

    But I have a problem with the interpretation of the individual terms, especially in the down-spin Green's function in the form that it has been written. The variables [itex]k_1, \omega_1[/itex] correspond to internal lines, and are integrated over. With respect to [itex]\omega_1[/itex], the first term in [itex]G_{\downarrow}^{0}(k_1-Q,\omega_1-\Omega)[/itex] is retarded, whereas with respect to [itex]\Omega[/itex] it is advanced.

    Would you call the first term advanced or retarded? What's the convention here?

    3. Substituting the full form of the Green's functions into the expression for [itex]\chi^0[/itex] we find that only terms (1x4 + 2x3) contribute as the contour integral (over [itex]\omega_1[/itex]) of a product of two purely retarded or purely advanced terms (wrt [itex]\omega_1[/itex]). So I am left with

    [tex]\chi_{0}(Q,\Omega) = \sum_{k_1}\left[\frac{\theta(k_1-k_F^{\uparrow})\theta(k_F^{\downarrow}-|k_1-Q|)}{E_{k_1}^{\uparrow}-E_{k_1-Q}^{\downarrow}-\omega-i\eta} + \frac{\theta(|k_1-Q|-k_{F}^{\downarrow})\theta(k_{F}^{\uparrow}-k_1)}{E_{k_1-Q}^{\downarrow}-E_{k_1}^{\uparrow}+\omega-i\eta}\right][/tex]

    3. Suppose I consider a saturated ferromagnet, i.e. one in which he up-spin band is partially below the Fermi level, but the down-spin band is completely above the Fermi level. At T = 0 K, all states below the Fermi level will be filled with particles, so, the down spin electron density is zero. Incidentally now, there is no real value of [itex]k_{F}^{\downarrow}[/itex] as the down spin band lies completely above the Fermi Level, and there is no intersection with it.

    For such a system, the down spin Green's function should only get a contribution from the hole states, as there can't be any particle (electron) states which are down spin. So, which term would you pick? There is some ambiguity with respect to which one is an electronic contribution and which one is a hole -- there is after all no Fermi wave vector for the down-spin band.


    In addition to answers and explanations for the questions I've posed, I would appreciate suggestions and also references for such calculations. I haven't found them in Atland & Simons or Fetter & Walecka.

    Eagerly awaiting a response! Thanks in advance.
     
    Last edited: Oct 28, 2011
  2. jcsd
  3. Oct 28, 2011 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    It's important to tell us which propagator you are looking it! In thermal QFT in the real-time formalism in general you need the Schwinger-Keldysh contour to get a correct description of the system. The time-ordered Green's function is *not* identical with the retarded one as in non-relativistic vacuum QFT.

    Here, you seem to consider the time-ordered propgator for a fermion system at 0 temperature. Then it's clear that the particle and hole parts of the time-ordered propagator is retarded and advanced, respectively.

    The only simplification compared to the finite-temperature case is that here, the Feynman propagator is identical with the time-ordered propagator and thus you directly obtain the retarded propagator from the time-ordered via a simple analytic continuation in different limits of the "analytic" propagator with energies defined in the complex plane. For details, have a look at the following manuscript, which however deals with relativistic particles,

    http://fias.uni-frankfurt.de/~hees/publ/green.pdf [Broken]
     
    Last edited by a moderator: May 5, 2017
  4. Oct 28, 2011 #3
    Thank you for your reply. I mentioned in my post that we are considering zero temperature (T = 0). Thanks for bringing the fact about thermal field theory to my attention. The transverse susceptibility I am using is defined as the Fourier transform of the time ordered Green's function

    [tex]\langle \phi_0|T[S^{-}(r,t)S^{+}(r',t')]|\phi_0\rangle[/tex]

    My confusion stems from a statement which was used to simplify the expression for the transverse susceptibility of a saturated ferromagnet. The statement in the lecture was that since the down-spin band is fully above the Fermi level, we should only consider the contribution of the advanced propagator as the down-spin has no associated electron population, and only the hole contribution is to be considered. In case of a saturated ferromagnet, there will be up-spin electrons, up-spin holes, down-spin holes but no down-spin electrons.

    But this makes sense only as long as I am able to distinguish between the hole and particle contributions.

    When I write [itex]G_{\downarrow}(Q,\Omega)[/itex] and consider the complex [itex]\Omega[/itex] plane, then of course I know what the hole contribution is, and what the particle contribution is.

    However, when I write [itex]G_{\downarrow}(q_1-Q,\omega_1-\Omega)[/itex] where (and this is important) [itex]q_1[/itex] and [itex]\omega_1[/itex] are integrated over, what is advanced with respect to [itex]\omega_1[/itex] will be retarded with respect to [itex]\Omega[/itex]. So what in this case will be the appropriate advanced to 'pick up'?

    This is presumably something simple, but I haven't been able to grasp it. I would appreciate if you could elaborate.
     
  5. Oct 28, 2011 #4

    DrDu

    User Avatar
    Science Advisor

    In real space, the Fourier transform of a retarded Greensfunction is an outgoing spherical wave. Hence it can't be split up into factors. The behaviour is determined by the sign of omega_1-Omega (relative to i epsilon) and not of either of the two components alone.
     
  6. Oct 28, 2011 #5
    For a saturated ferromagnet, in which the down spin band is completely above the Fermi level, the expression I have is

    [tex]\chi_{0}(q,\omega) = i\sum_{k'}\int_{-\infty}^{\infty}\frac{d\omega'}{2\pi}\left[\frac{1}{\omega'-(E_{k'}-\Delta) \pm i\eta}\right]\left[\frac{1}{\omega'-\omega-(E_{k'-q}+\Delta)+i\eta}\right][/tex]

    What is the argument for keeping only the [itex]+i\eta[/itex] term in the second square bracket?

    The trouble here is that [itex]\theta(|k|-k_{F}^{\downarrow})[/itex] does not make sense for the down-spin band since [itex]k_{F}^{\downarrow}[/itex] is complex: the down span band does not intersect the Fermi level at all, in case of a saturated ferromagnet.

    In short my question is: what will the expressions for the up-spin and down-spin Green's functions inside the susceptibility integral be, for a saturated ferromagnet:

    [tex]\chi_{0}(Q,\Omega) = i\sum_{q'}\int_{-\infty}^{\infty}\frac{d\omega'}{2\pi} G^{0}_{\uparrow}(q',\omega')G^{0}_{\downarrow}(q'-Q,\omega'-\Omega)[/tex]

    Note that the argument of the second Green's function involves [itex]\omega'-Q[/itex] and [itex]\omega'[/itex] is integrated over. Since the contour integral I'll look at will be in the [itex]\omega'[/itex] plane, whether the Green's function will be advanced or retarded here will be determined by the poles in the complex [itex]\omega'[/itex] plane. However, my physical frequency is [itex]\Omega[/itex], and mathematically I know that due to the sign difference between [itex]\omega'[/itex] and [itex]\Omega[/itex], what is advanced wrt [itex]\omega'[/itex] will be retarded with respect to [itex]\Omega[/itex]. So how do I make a statement like: "only the advanced propagator will contribute" or "only the retarded propagator will contribute"? What does this mean?
     
    Last edited: Oct 28, 2011
  7. Oct 29, 2011 #6
    Okay I have managed to figure this out. The critical idea that struck me was that the saturated ferromagnet can be looked upon as a system in which the down-spin band is being progressively shifted above the Fermi level. At some point, the minima of the band intersects the Fermi level, and any further shift results in the wavenumber becoming complex. I have used this idea to sketch a full derivation, which is attached here.

    Thanks everyone for your help!
     

    Attached Files:

  8. Oct 29, 2011 #7
    Only thing is...how do I know whether this is a diagram for [itex]\chi^{+-}[/itex] or [itex]\chi^{-+}[/itex]
     
  9. Oct 29, 2011 #8

    DrDu

    User Avatar
    Science Advisor

    If the Fermi energy falls into a band gap, addition of a single electron to the system will lift the Fermi energy to the lower bottom of the upper band. This is clearly unobservable in an infinite system. So I think it makes sense to set k_F to its value at the lower bottom of the band.
     
  10. Oct 29, 2011 #9

    DrDu

    User Avatar
    Science Advisor

    thinking about it, it would guess that the sign of eta changes, i.e eta=abs(eta) *sgn(omega'-Omega), so as to keep only the true retarded or advanced Greensfunction.
     
  11. Oct 30, 2011 #10
    Okay, good so this makes sense...thanks.

    The transverse susceptibility is defined as

    [tex]\chi^{+-}(rt,r't') = i\langle\psi_0|T[S^{+}(r,t)S^{-}(r',t')]|\psi_0\rangle[/tex]

    From this how is it obvious that the ladder diagrams for the time ordered propagator actually calculate [itex]\chi^{-+}(q,\omega)[/itex] and not [itex]\chi^{+-}(q,\omega)[/itex]?

    Also, how is it obvious that

    [tex]\chi^{-+}(q,\omega) = \chi^{+-}(-q,-\omega)[/tex]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Transverse Susceptibility, and doubts about retarded/advanced Green's Functions
  1. About susceptibility (Replies: 1)

Loading...