3
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Gold
    Mar 18, 2022 at 17:25
  • $\begingroup$ Yes, that I kind of understand. What I don't understand is that product symbol appearing in the definition of $\varepsilon^{ab}$. $\endgroup$ Mar 18, 2022 at 17:40
  • $\begingroup$ 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 $\endgroup$ Mar 18, 2022 at 19:51

1 Answer 1

1
$\begingroup$

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 $\mathcal{H}_2$.

$\endgroup$
1
  • $\begingroup$ So I was going crazy over a typo... Thank you for your answer! $\endgroup$ Mar 18, 2022 at 23:07

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.