# A Entanglement and density matrix in QFTs

1. Dec 2, 2016

### ShayanJ

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

2. Dec 2, 2016

### MathematicalPhysicist

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?

3. Dec 2, 2016

### ShayanJ

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.

4. Dec 3, 2016

### MathematicalPhysicist

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.

5. Dec 3, 2016

### ShayanJ

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.

6. Dec 4, 2016

### MathematicalPhysicist

Well, I am not sure; but in QFT wouldn't we have fields instead of the states $\phi_n$?

7. Dec 16, 2016

### samalkhaiat

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 $\beta$ is given by $$\rho (x,y) \propto \langle x | e^{-\beta H} | y \rangle .$$ The proportionality constant which ensures $$\mbox{Tr}( \rho ) = \int dx \ \rho (x,x) = 1 ,$$ is given by $$\frac{1}{Z} = \frac{1}{\mbox{Tr}( e^{-\beta H})} .$$ One can (if one wants to) express $\rho$ as path integral over the variables $x(\tau)$ which take “boundary” values $y$ at $\tau = 0$, and $x$ at $\tau = \beta$. So, you should have no problem with path integral expression $$\rho (x , y) = \frac{\langle x | e^{-\beta H} | y \rangle }{Z} = (1/Z) \int [dx(\tau)] \ \delta (x(0) - y) \delta (x(\beta) - x) \ e^{- \int^{\beta}_{0} \ L d\tau} . \ \ \ (1)$$ Taking the trace, i.e., setting $x = y$ and integrating, has, therefore, the effect of “gluing the edges” along $\tau = 0$ and $\tau = \beta$ and forming a "cylinder" of circumference $\beta$.
Okay, let us formally generalize the above to a bosonic field theory in (1+1)-dimension: the operator $\hat{x}$ will be “replaced” by a complete set of local observables $\{ \hat{\varphi}(x) \}$, and $\hat{x}|x\rangle = x |x\rangle$ with the eigenvalue equations $$\{ \hat{\varphi}(x) \} \left( \otimes_{x}| \{ \varphi (x) \} \rangle \right) = \{ \varphi (x) \} \left( \otimes_{x}| \{ \varphi (x) \} \rangle \right) .$$ So, instead of (1), you will have
\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*}
Again, the delta functions are there to enforce the boundary values $\varphi^{'}(x)$ at $\tau = 0$, and $\varphi^{''}(x)$ at $\tau = \beta$. And the trace is found by setting $\varphi^{''}(x) = \varphi^{'}(x)$ and integrating over these variables except, in this case, you really have edges to glue along $\tau = 0$ and $\tau = \beta$.
So, if you want to calculate $\mbox{Tr}(\rho^{n})$, 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
8. Dec 17, 2016

### vanhees71

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

9. Dec 17, 2016

### ShayanJ

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
10. Dec 17, 2016

### ShayanJ

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!

11. Dec 17, 2016

### vanhees71

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.

12. Dec 17, 2016

### ShayanJ

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