Questions about the Unruh effect derivation in Wald's QFT in curved spacetime

Question

So I'm currently reading chapter 5 of Wald's book on QFT in curved spacetime and I'm terribly confused with the notation in the last steps of his Unruh effect derivation.

Context:

In eq. 5.1.26, he expresses a two particle state as (using index notation): $$ \varepsilon^{ab} = \prod_i\exp(-\pi\omega_i/a)2(\psi_{iI})^{(a}(\psi_{iII})^{b)}, $$ where $\{\psi_{iI}\}$ and $\{\psi_{iII}\}$ are orthonormal basis for the one-particle-Hilbert spaces,$\mathcal{H}_I$ and $\mathcal{H}_{II}$, obtained from the quantization in the right and left Rindler wedges, respectively. This two particle state comes from a discussion in section 4.4 in which he computes the action of a unitary transformation $U:\mathcal{F}(\mathcal{H}_1)\rightarrow\mathcal{F}(\mathcal{H}_2)$ (more precisely, a Bogoliubov transformation) on the vacuum state of $\mathcal{F}(\mathcal{H}_1)$ (the symmetric Fock space of $\mathcal{H}_1$). Assuming $|0\rangle_1$ is the vacuum state of $\mathcal{F}(\mathcal{H}_1$), he gets (eq 4.4.23) $$ U|0\rangle_1=(1,0,\varepsilon^{ab}/\sqrt{2},0,\varepsilon^{(ab}\varepsilon^{cd)}\sqrt{3/8},...), $$ In his Unruh effect derivation,$\mathcal{H}_2=\mathcal{H}_I\otimes\mathcal{H}_{II}$ and $\mathcal{H}_1$ is the one-particle-Hilbert space obtained from quantization in Minkowski space.

Short question: Why $\varepsilon^{ab}$ has that form written above? How would it look like in Dirac notation?

Elaboration: Since $\varepsilon^{ab}$ can be seen as a map from $\bar{\mathcal{H_2}}\times\bar{\mathcal{H_2}}$ to $\mathbb{C}$, it is an element of $\mathcal{H_2} \otimes \mathcal{H_2}$ and I thought that it should be written as a linear combination of those bases mentioned above (like with a Schauder basis). Instead, there is a product symbol of which I don't know from where it came. I'm not even sure if the product symbol implies tensor product or something else.

I really appreciate if anyone can explain these things and hope my question is clearly put.

Answer 1

Product Symbol

I believe that is just a typo. In my copy of the book, Eq. (5.1.26) reads $$\epsilon^{ab} = \sum_i \exp(- n \omega_i/a) 2 (\psi_{i \text{I}})^{(a}(\psi_{i \text{II}})^{b)}, \tag{1}$$ with a sum rather than a product, which makes much more sense in this context. No clue what a product could mean here, but the sum is just a superposition of elements of $\mathcal{H}_{2} \otimes \mathcal{H}_{2}$.

Meaning of $\epsilon^{ab}$

As for the actual expression for $\epsilon^{ab}$ and how to write it in Dirac notation, let us begin by recalling that $\mathcal{E} = \bar{D} \bar{C}^{-1}$ and $\epsilon^{ab}$ is the associated two-particle state of $\mathcal{E}$. From Wald's Eqs. (5.1.24)–(5.1.25) we know that $$DC^{-1} \psi_{i\text{I}} = e^{-\frac{\pi\omega_i}{a}} \bar{\psi}_{i\text{II}} \quad \text{and} \quad DC^{-1} \psi_{i\text{II}} = e^{-\frac{\pi\omega_i}{a}} \bar{\psi}_{i\text{I}}.$$

These expressions characterize the map $DC^{-1}\colon \mathcal{H}_2 \to \mathcal{H}_2$. If we take the conjugate of this map (check Wald's App. A), we get $\bar{D}\bar{C}^{-1} = \mathcal{E}$, which acts according to $$\mathcal{E} \bar{\psi}_{i\text{I}} = e^{-\frac{\pi\omega_i}{a}} \psi_{i\text{II}} \quad \text{and} \quad \mathcal{E} \bar{\psi}_{i\text{II}} = e^{-\frac{\pi\omega_i}{a}} \psi_{i\text{I}}.$$

In index notation, this can be written as $$\epsilon^{ab} (\bar{\psi}_{i\text{I}})_b = e^{-\frac{\pi\omega_i}{a}} (\psi_{i\text{II}})^a \quad \text{and} \quad \epsilon^{ab} (\bar{\psi}_{i\text{II}})_b = e^{-\frac{\pi\omega_i}{a}} (\psi_{i\text{I}})^a.$$

I find it easier to justify Eq. (1) by seeing it works than by deriving it. From the expression in Eq. (1), we can see that, indeed, \begin{align} \epsilon^{ab} (\bar{\psi}_{i\text{I}})_b &= \sum_j \exp(- n \omega_j/a) 2 (\psi_{j \text{I}})^{(a}(\psi_{j \text{II}})^{b)} (\bar{\psi}_{i\text{I}})_b, \\ &= \sum_j \exp(- n \omega_j/a) (\psi_{j \text{I}})^{a}(\psi_{j \text{II}})^{b} (\bar{\psi}_{i\text{I}})_b + \sum_j \exp(- n \omega_j/a) (\psi_{j \text{I}})^{b}(\psi_{j \text{II}})^{a} (\bar{\psi}_{i\text{I}})_b, \\ &= 0 + \sum_j \exp(- n \omega_j/a) \delta_{ij} (\psi_{j \text{II}})^{a}, \\ &= \exp(- n \omega_i/a) (\psi_{i \text{II}})^{a}. \end{align} Similarly for when one applies it to (\bar{\psi}_{i\text{II}})_b. I used the fact that $\bar{\phi}_a\psi^a = \langle\phi\vert\psi\rangle$ (Eq. (A.3.2)) and the fact that the $\lbrace\psi_{i,\text{I}},\psi_{i,\text{II}}\rbrace$ provide an orthonormal basis.

Dirac Notation

As for an expression in Dirac notation, it is hinted at from the expression $\bar{\phi}_a\psi^a = \langle\phi\vert\psi\rangle$. It is simply $$\epsilon = \sum_i \exp(- n \omega_i/a) (\vert\psi_{i \text{I}}\rangle\otimes\vert\psi_{i \text{II}}\rangle + \vert\psi_{i \text{II}}\rangle\otimes\vert\psi_{i \text{I}}\rangle), $$ where the symmetrization and the tensor product are written explicitly and we used that $\psi^a \equiv \vert\psi\rangle$. I kept $\epsilon$ outside of a ket because it lives in $\mathcal{H}_2 \otimes \mathcal{H}_2$, while the kets I wrote live in $\mathcal{H}_2$.