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

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.

• Wald uses his abstract index notation for Hilbert spaces as well by what I remember, and it can indeed become confusing. Still, these are just tensor products. The quantity $\varepsilon^{ab}$ has two indices because it is an element of a tensor product space. The objects $\psi_{iI}^a$ and $\psi_{iII}^b$ are vectors on the individual factor spaces and $\psi_{iI}^{(a}\psi_{iII}^{b)}$ is (I believe) their symmetrized tensor product.
– Gold
Mar 18, 2022 at 17:25
• Yes, that I kind of understand. What I don't understand is that product symbol appearing in the definition of $\varepsilon^{ab}$. Mar 18, 2022 at 17:40
• Are you sure of that expression? In my copy, (5.1.26) reads $\epsilon^{ab} = \sum_i \exp(- n \omega_i/a) 2 (\psi_{i I})^{(a}(\psi_{i II})^{b)}$, with a sum, not a product Mar 18, 2022 at 19:51

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 $$\mathcal{H}_2$$.

• So I was going crazy over a typo... Thank you for your answer! Mar 18, 2022 at 23:07