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A Entanglement and density matrix in QFTs

  1. Dec 2, 2016 #1

    ShayanJ

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    I'm reading this paper. But I haven't read anything on how to calculate the density operator in a QFT or how to calculate its trace. Now I can't follow this part of the paper. Can anyone clarify?
    Thanks
     
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  3. Dec 2, 2016 #2

    MathematicalPhysicist

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    Well to answer your question, it seems one first needs to learn CFT, I have the book of Di Franchesko but I still didn't read it, so I guess I cannot help here, what is your knowledge of CFT?
     
  4. Dec 2, 2016 #3

    ShayanJ

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    I know some CFT but actually its not necessary for this question. My question is about calculating the density matrix of any QFT. Maybe @samalkhaiat can help.
     
  5. Dec 3, 2016 #4

    MathematicalPhysicist

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    I don't know if it helps or not, but I searched google for density functional in QFT, and got this first hit:
    https://arxiv.org/abs/1503.02925

    Seems like a good reference, don't know if specifically here it helps.
     
  6. Dec 3, 2016 #5

    ShayanJ

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    Density functional is different from density matrix(operator). In QM, density operator is calculated by the formula ## \rho=\sum_n p_n |\phi_n\rangle \langle \phi_n |## where ## \{|\phi_n\rangle\} ## is a set of not necessarily orthonormal states. I want to know how to calculate the corresponding quantity in QFT.
     
  7. Dec 4, 2016 #6

    MathematicalPhysicist

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    Well, I am not sure; but in QFT wouldn't we have fields instead of the states ##\phi_n##?
     
  8. Dec 16, 2016 #7

    samalkhaiat

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    Tadashi Takayanagi is good young physicist, I was impressed by his lectures at CERN Winter School on Supergravity 2012. He presented the subject very well, I suggest (if you have not done so) you look at them. As you might know, to explain this properly, one needs to draw pictures. Unfortunately I cannot do this, but this does not prevent us from making formal analogy with QM. In statistical physics, the density matrix in a thermal state at (inverse) temperature [itex]\beta[/itex] is given by [tex]\rho (x,y) \propto \langle x | e^{-\beta H} | y \rangle .[/tex] The proportionality constant which ensures [tex]\mbox{Tr}( \rho ) = \int dx \ \rho (x,x) = 1 ,[/tex] is given by [tex]\frac{1}{Z} = \frac{1}{\mbox{Tr}( e^{-\beta H})} .[/tex] One can (if one wants to) express [itex]\rho[/itex] as path integral over the variables [itex]x(\tau)[/itex] which take “boundary” values [itex]y[/itex] at [itex]\tau = 0[/itex], and [itex]x[/itex] at [itex]\tau = \beta[/itex]. So, you should have no problem with path integral expression [tex]\rho (x , y) = \frac{\langle x | e^{-\beta H} | y \rangle }{Z} = (1/Z) \int [dx(\tau)] \ \delta (x(0) - y) \delta (x(\beta) - x) \ e^{- \int^{\beta}_{0} \ L d\tau} . \ \ \ (1)[/tex] Taking the trace, i.e., setting [itex]x = y[/itex] and integrating, has, therefore, the effect of “gluing the edges” along [itex]\tau = 0[/itex] and [itex]\tau = \beta[/itex] and forming a "cylinder" of circumference [itex]\beta[/itex].
    Okay, let us formally generalize the above to a bosonic field theory in (1+1)-dimension: the operator [itex]\hat{x}[/itex] will be “replaced” by a complete set of local observables [itex]\{ \hat{\varphi}(x) \}[/itex], and [itex]\hat{x}|x\rangle = x |x\rangle[/itex] with the eigenvalue equations [tex]\{ \hat{\varphi}(x) \} \left( \otimes_{x}| \{ \varphi (x) \} \rangle \right) = \{ \varphi (x) \} \left( \otimes_{x}| \{ \varphi (x) \} \rangle \right) .[/tex] So, instead of (1), you will have
    [tex]
    \begin{align*}
    \rho \left( \varphi^{''}(x^{''}) , \varphi^{'}(x^{'}) \right) &= \frac{ \langle \varphi^{''}(x^{''}) | e^{-\beta H} | \varphi^{'}(x^{'}) \rangle}{Z} \\
    &= \frac{1}{Z} \int [d\varphi (x , \tau)] \prod_{x}\left[ \delta \left( \varphi (x,0) - \varphi^{'}(x^{'}) \right) \delta \left( \varphi (x ,\beta) - \varphi^{''}(x^{''}) \right) \right] e^{-\int_{0}^{\beta} L_{E} d\tau} \ \ \ \ (2)
    \end{align*}
    [/tex]
    Again, the delta functions are there to enforce the boundary values [itex]\varphi^{'}(x)[/itex] at [itex]\tau = 0[/itex], and [itex]\varphi^{''}(x)[/itex] at [itex]\tau = \beta[/itex]. And the trace is found by setting [itex]\varphi^{''}(x) = \varphi^{'}(x)[/itex] and integrating over these variables except, in this case, you really have edges to glue along [itex]\tau = 0[/itex] and [itex]\tau = \beta[/itex].
    So, if you want to calculate [itex]\mbox{Tr}(\rho^{n})[/itex], you will need to take the products of n path integral of the form (2) which will be equivalent to path integral over n-sheeted Reimann surface.
     
    Last edited: Dec 16, 2016
  9. Dec 17, 2016 #8

    vanhees71

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    For an intro to relativistic quantum many-body theory (QFT), I can recommend three textbooks

    J. I. Kapusta and C. Gale, Finite-Temperature Field Theory; Principles and Applications, Cambridge University Press, 2 ed., 2006.
    M. LeBellac, Thermal Field Theory, Cambridge University Press, Cambridge, New York, Melbourne, 1996.

    M. Laine and A. Vuorinen, Basics of Thermal Field Theory, vol. 925 of Lecture Notes in Physics, 2016.
    http://dx.doi.org/10.1007/978-3-319-31933-9

    or my lecture notes

    http://th.physik.uni-frankfurt.de/~hees/publ/off-eq-qft.pdf
     
  10. Dec 17, 2016 #9

    ShayanJ

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    It seems to me you're talking about something else!

    1) The calculations in sec.3.1 of the paper are done for a zero temperature CFT, i.e. ##-\infty<t_E<\infty##.

    2) The density matrix for the field in all points is ## [\rho]_{\phi_0\phi_0'}=\Psi(\phi_0)\overline \Psi(\phi'_0) ##. Since ## \displaystyle \Psi(\phi_0)\propto \int_{t_E=-\infty}^{\phi(t_E=0,x)=\phi_0(x)} D \phi e^{-S[\phi]}## and ## \overline\Psi(\phi'_0)\propto \displaystyle \int^{t_E=\infty}_{\phi(t_E=0,x)=\phi'_0(x)} D \phi e^{-S[\phi]} ##, the density matrix should be ## \displaystyle [\rho]_{\phi_0\phi'_0}\propto \int_{t_E=-\infty}^{\phi(t_E=0,x)=\phi_0(x)} \int^{t_E=\infty}_{\phi'(t_E=0,x)=\phi'_0(x)} D \phi D \phi' e^{-S[\phi]} e^{-S[\phi']}##.

    3) The path integral with those dirac deltas, is actually the reduced density matrix which is calculated by integrating out all the degrees of freedom outside of region A and the density matrix for all points of space should have no dirac delta! This means that in the density matrix above, we should put ## \phi(x)=\phi'(x) ## and ## \phi_0(x)=\phi'_0(x) ## except for ## x\in A ##. Taking a full trace gives ## \displaystyle Tr \rho=\int_{t_E=-\infty}^{\phi(t_E=0,x)=\phi_0(x)} \int^{t_E=\infty}_{\phi'(t_E=0,x)=\phi'_0(x)} D \phi D \phi' e^{-S[\phi]} e^{-S[\phi']} \prod_{x} \delta( \phi(t_E,x)-\phi'(t_E,x) )=\int_{t_E=-\infty}^{t_E=\infty}D \phi e^{-2S[\phi]}##. But because we don't want the full trace and just want to integrate out degrees of freedom outside of A and have a reduced density matrix for A, we should avoid integrating over all fields for ## x\in A ##. We also need two boundary conditions at zero(## \phi_{\pm}(x) ##). So we put those dirac deltas in the formula for the full trace and get:
    ## \displaystyle [\rho_A]_{\phi_+\phi_-}\propto \int_{t_E=-\infty}^{t_E=\infty}D \phi e^{-2S[\phi]} \prod_{x\in A} \delta(\phi(+0,x)-\phi_+(x))\delta(\phi(-0,x)-\phi_-(x)) ##.

    But as you can see, instead of ## e^{-S[\phi]} ##, I have ## e^{-2S[\phi]} ##. Also I'm not sure about all these because its just my guts filling the blanks!!!
    I also don't quite understand how ## \displaystyle \int_{t_E=-\infty}^{\phi(t_E=0,x)=\phi_0(x)} \int^{t_E=\infty}_{\phi'(t_E=0,x)=\phi'_0(x)} \to \int_{t_E=-\infty}^{t_E=\infty} ## because the second integral should simply give 1 and shouldn't extend the integration interval of the other integral. But it seems that should be done to get the desired result!
     
    Last edited: Dec 17, 2016
  11. Dec 17, 2016 #10

    ShayanJ

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    My previous post seems wrong to me now. What actually happens, is that first we write the density matrix as follows:

    ## \displaystyle [\rho]_{\phi_0 \phi_0'}\propto \int_{t_E=-\infty}^{\phi(t_E=0,x)=\phi_0(x)} \int_{\phi'(t_E=0,x)=\phi_0'(x)}^{t_E=\infty} D\phi D\phi' e^{-S[\phi]}e^{-S[\phi']}##.

    To take a full trace, we set ## \phi_0(x)=\phi_0'(x) ## and integrate over ## \phi_0(x) ##, so we have:

    ## \displaystyle Tr[\rho] \propto \int D\phi_0 \int_{t_E=-\infty}^{\phi(t_E=0,x)=\phi_0(x)} \int_{\phi'(t_E=0,x)=\phi_0(x)}^{t_E=\infty} D\phi D\phi' e^{-S[\phi]}e^{-S[\phi']} ##.

    This somehow should reduce to ## \displaystyle Tr[\rho]\propto \int D\phi e^{-S[\phi]} ##, but I have no idea how!
     
  12. Dec 17, 2016 #11

    vanhees71

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    You integrate "over all paths", where in QFT the paths are in field-configuration space. Your first expression describes the integration over all paths first involving only the part where ##t<0## and at ##t=0## the field is fixed as ##\phi_0(t=0,\vec{x})=\phi_0(\vec{x})## and then a part where ##t>0## and ##\phi_0(t=0,\vec{x})=\phi_0'(\vec{x})##, leading to a functional of ##\phi_0## and ##\phi_0'##.

    Now if you want the trace (i.e., the partition sum) you set ##\phi_0=\phi_0'##, i.e., you integrate first for ##t<0## and ##t>0## with fixed field configuration ##\phi_0## for both parts. Finally you integrate also over ##\phi_0##. So altogether you integrate in this final step over all field configurations without restrictions, and that's the meaning of the trace.
     
  13. Dec 17, 2016 #12

    ShayanJ

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    Thanks guys. I now understand it clearly.
    But I just realized there is another problem. In the paper in the OP, the ground state wave functional is given by ## \displaystyle \Psi(\phi_0(x))=\int_{t_E=-\infty}^{\phi(t_E=0,x)=\phi_0(x)} D\phi e^{-S[\phi]} ##. As you can see the argument of the wave functional appears as the ## t_E=0 ## boundary condition of the path integral. But the calculations mentioned in this thread, give a wave functional whose argument appears as the ## \pm \infty ## boundary condition of the path integral. Where does this discrepancy come from?

    EDIT:
    Is ## t=it_E ## only a convention for Wick rotation or is it the only right way? Because if I'm allowed to do the Wick rotation with ## t=-it_E ##, the above problem would be solved!
     
    Last edited: Dec 17, 2016
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