2
$\begingroup$

In arXiv:1409.1231 in section 3.5 above equation (3.34) Daniel Harlow says

More precisely if the left and right wedges are completely uncorrelated, as in the state 3.33, then the typical difference between neighboring fields on either side is of order the typical field fluctuation, which is given by $\frac{1}{\epsilon}$ where $\epsilon$ is a UV length cutoff, so we have $$\partial_x\phi|_{x=0}\propto \frac{1}{\epsilon^2} $$

Since we assumed there is no entanglement between Right and Left wedges there should not be any 2 point correlation between points near the Rindler horizon but on opposite sides. But why should $\partial_x\phi|_{x=0}$ diverge?

$\endgroup$

2 Answers 2

2
$\begingroup$

Good question. There are a few things to make clear ( I am not an expert in Algebraic QFT but I will try my best, hoping someone will correct me eventually).

  1. Note that Harlow says "left and right wedges are completely uncorrelated" which to me is more stringent than your assumption of "no entanglement". Taking for just a moment Harlow's statement as correct, then there cannot be any divergence of any Wightman function evaluated on points which are at opposite wedges, since those functions are by hypothesis factorized in left and right Wightman functions.

  2. However, his statement is incorrect. In QFT in general you cannot think $\rho_L \otimes \rho_R$ as the product of reduced density matrices. Each of these states is not a density matrix, but a state (a positive linear functional) on the localized algebra of observables $A_L$ or $A_R$. A less confusing notation would be $\omega_L \otimes \omega_R$. In general these algebras are von Neumann algebras which do not admit states represented as trace-class density matrices (they are type III factors). See Haag's book "Local quantum physics".

  3. The product state $\omega_L \otimes \omega_R$ is still well-defined(*) in the algebra generated by the algebras at each wedge $A_L$ and $A_R$ (this means the smallest von Neumann algebra containing both algebras $A_L$ and $A_R$, and this is equivalent to the double commutant of the union of the algebras $(A_L \cup A_R)''$). Let me be more precise: we have that for $x\in A_L$ and $y \in A_R$, $(\omega_L \otimes \omega_R)(xy)=\omega_L(x)\omega_R(y)$, and $\omega_L \otimes \omega_R$ extends to a normal state in $(A_L \cup A_R)''$ (see the work "Product states by local algebras" by Buchholz, Commun.Math.Phys. 36 (1974) 287-304).

  4. The mentioned state is not uncorrelated despite being a product state. It would be uncorrelated if we could write the Hilbert space as a tensor product $\mathcal{H}_L \otimes \mathcal{H}_R $ and each state on the wedges was a density matrix. But again, in QFT this is not the case, and any state $\omega$ in physics should be a Hadamard state, meaning that in the coincident limit the two-point function diverges in a particular way. This is what I think is behind the divergence in Harlow's notes (despite his previous misleading statements).

I recommend also Witten's notes

(*) EDIT: actually, it is well-defined if we are talking about two causal diamonds. In the case of two wedges it has been proven in Bochholz paper that there is no product state (see counter-example b there), with the exception of the 1+1 case.

$\endgroup$
1
  • $\begingroup$ Thanks, I will check Witten's notes and Haag's book. $\endgroup$ Aug 9, 2021 at 7:26
2
+50
$\begingroup$

Harlow's equation (3.34) is a schematic way of expressing the idea that the gradient term in the Hamiltonian contributes $\sim 1/\epsilon^4$ to the expectation value of the energy density at the L/R interface when the state is a product state. I'll explain this in detail using lattice QFT, which implements an explicit version of the "UV length cutoff" that Harlow calls $\epsilon$.

Definition of the model

The Hamiltonian for the free scalar field with mass $m$ is $$ \newcommand{\bfu}{\mathbf{u}} \newcommand{\bfx}{\mathbf{x}} \newcommand{\half}{\frac{1}{2}} H = \epsilon^3\sum_\bfx H_\bfx + \text{const} $$ where $H_\bfx$ is the energy density operator \begin{align} H_\bfx &= K_\bfx + V_\bfx \\ K_\bfx &= -\half \left(\frac{1}{\epsilon^{3}}\frac{\partial}{\partial\phi_\bfx}\right)^2 +\half m^2\phi_\bfx^2 \\ V_\bfx &= \half\sum_{\bfu} \frac{(\phi_{\bfx+\bfu}-\phi_\bfx)^2}{\epsilon^2} \tag{1} \end{align} where the index $\bfx$ runs over lattice sites, $\bfu$ runs over a set of three basis vectors for the lattice, and $\epsilon$ is the distance between neighboring sites. The operator $\sim -(i/\epsilon^3)\partial/\partial\phi$ is the lattice version of $\dot\phi$, the canonical conjugate of $\phi$, and the quantity $V_\bfx$ is the lattice version of the spatial gradient term $\sim(\nabla\phi)^2$.

A state is represented by a normalizable (square-integrable) complex-valued function of the field variables: $\Psi[\phi]$. The inner product of two states is defined by integrating over all of the field variables: $$ \langle\Psi_1|\Psi_2\rangle \equiv \int \prod_\bfx (\epsilon\,d\phi_\bfx)\ \Psi_1^*[\phi]\Psi_2[\phi]. \tag{2} $$ A factor of $\epsilon$ is included with each factor of $d\phi_\bfx$ in the measure because the field varaibles $\phi_\bfx$ in (1) have the same units as $1/\epsilon$, but we want the inner product to be dimensionless. We can get rid of those factors of $\epsilon$ in the inner product by switching to a normalization in which the field variables $\phi_\bfx$ are dimensionless. Then equations (1) are replaced by \begin{align} H_\bfx &= K_\bfx + V_\bfx \\ K_\bfx &= -\half \left(\frac{1}{\epsilon^{2}}\frac{\partial}{\partial\phi_\bfx}\right)^2 +\half \frac{m^2}{\epsilon^2}\phi_\bfx^2 \\ V_\bfx &= \half\sum_{\bfu} \frac{(\phi_{\bfx+\bfu}-\phi_\bfx)^2}{\epsilon^4} \tag{3} \end{align} and the inner product (2) is replaced by $$ \langle\Psi_1|\Psi_2\rangle \equiv \int \prod_\bfx d\phi_\bfx\ \Psi_1^*[\phi]\Psi_2[\phi]. \tag{4} $$

The vacuum state

The vacuum state $\Psi_0[\phi]$ is the state that minimizes the expectation value of the Hamiltonain $H$. The vacuum state can be determined explicitly in this model, as shown in the body of this question, which however uses continuous space and $m=0$. Here, instead of writing down the explicit expression, I'll use some intuition to deduce its key properties.

Temporarily suppose $V_\bfx=0$. Then the Hamiltonian $H$ would describe a set of uncoupled harmonic oscillators, one per lattice site $\bfx$. The ground state would be the product of the individual harmonic oscillator ground states, each a function of $\phi_\bfx$ for just one lattice site.

The term $V_\bfx$ describes interactions among those harmonic oscillators. To find the vacuum state of the full Hamiltonian, including the $V_\bfx$ term, we need to balance the expectation value of the $V_\bfx$ term with the other terms. To do this, we need to replace the product state with a state that drops rapidly to zero whenever the $\phi_\bfx$s at neighboring sites differ much from each other, so that the numerator $(\phi_{\bfx+\bfu}-\phi_\bfx)^2$ of the gradient term is always small. (In other words, we need all of the oscillators to be entangled with each other.) The state that achieves this balance — the state that minimizes the total energy when the interaction term $V_\bfx$ is included — is the vacuum state.

From now on, suppose that the constant term in (1) has been chosen so that the energy of the vacuum state is zero.

Energy density in a product state

Now, following Harlow, divide space into two halves L/R and consider a product state $$ \Psi[\phi]=\Psi_L[\phi]_L \Psi_R[\phi]_R \tag{5} $$ where $[\phi]_L$ means all of the $\phi_\bfx$s with $x\in L$, and similarly for $[\phi]_R$. The vacuum state cannot be written in this form, because a state of this form cannot drop rapidly to zero whenever the $\phi_\bfx$s at neighboring sites differ much from each other. For a product state (5), in the integrand of the expectation value, nothing constrains the numerator $(\phi_{\bfx+\bfu}-\phi_\bfx)^2$ of the gradient term to be small at the L/R interface. Equation (3) therefore says that the expectation value of that term is therefore of order $1/\epsilon^4$ at the interface when the state has the form (5), exactly as Harlow claimed. This shows that a product state is nothing like the vacuum state, at least not near the interface. The product state necessarily has a firewall.

The meaning of Harlow's equation

Now we can see what Harlow really meant by $$ \partial_x\phi\big|_{x=0}\propto\frac{1}{\epsilon^2}. \tag{3.34} $$ This "equation" is a schematic way of conveying the idea that the quantity $\phi_{\bfx+\bfu}-\phi_\bfx$ in the numerator of the gradient term is not small at the L/R interface when the state $\Psi$ is a product state, as I explained above. Harlow is using the normalization that I used in equation (1), so that $\partial\phi$ has dimensions $1/\epsilon^2$, and (3.34) is understood to be a statement about the typical numeric value of this quantity in the integrand of the expectation value, where the $\phi$s are integration variables.

$\endgroup$
5
  • $\begingroup$ I somewhat intuitively understood your explanation. I haven't studied lattice quantum field theory, can you suggest where to read it and what portion of it to read to understand your explanation clearly? $\endgroup$ Oct 4, 2021 at 5:27
  • $\begingroup$ @KasiReddySreemanReddy The keywords are "Hamiltonian lattice QFT," but the sources that came up when I searched either focus on other kinds of fields (gauge fields, fermions) or don't use the Hamiltonian formulation. But it's easy: start with the lagrangian $L=\dot\phi^2-(\nabla\phi)^2-m^2\phi^2$, discretize space (replace the gradient with its finite-difference version) so that the spatial coordinates become discrete indices, and then canonically quantize as usual. $\endgroup$ Oct 4, 2021 at 13:47
  • $\begingroup$ @KasiReddySreemanReddy Just like in nonrelativistic QM, the commutator $[\phi_x,\dot\phi_y]=i\epsilon^{-3}\delta_{xy}$ says that the Hilbert space can be represented by functions of the real variables $\phi_x$, with $\dot\phi_x\sim -i\epsilon^{-3}\partial/\partial\phi_x$. The only new things compared to nonrelativistic QM are (1) now the variables are field-amplitudes instead of particle-coordinates, (2) there are lots of them, and (3) we need to keep track of factors of $\epsilon$ so that the lagrangian and commutation relations have the correct continuum limit ($\epsilon\to 0$). $\endgroup$ Oct 4, 2021 at 13:47
  • $\begingroup$ @KasiReddySreemanReddy I'm sure I've seen this in the literature somewhere before, but it's one of those things that I mainly learned just by working though it myself. I wish I knew where to find a good introduction. $\endgroup$ Oct 4, 2021 at 13:51
  • $\begingroup$ OK, thanks Chiral Anomaly. $\endgroup$ Oct 4, 2021 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.