# Does the Unruh effect need the accelerating observer's Hilbert space to be smaller than the inertial one?

I've encounter two different approaches to the Unruh effect and I feel like they are not consistent with one another.

Bogoliubov Transformation

In this approach the basic statement is that the vacuum state for a Minkowski observer $$|0\rangle_M$$ is written as a superposition of excited states in the basis of the Rindler observer

$$\begin{equation} |0\rangle_M=\sum_{n_1=0}^\infty \dots \sum_{n_r=0}^\infty f(n_1,\dots,n_r) |n_1,\dots,n_r\rangle_R \end{equation}$$

Schematically, $$n_i$$ represents the number of excitations in the $$i$$ state. $$r$$ represents the last state but it could be infinite too. In this approach, it is straightforward to calculate the expectation value of the number of particles in the Rindler state $$k$$ $$\hat{n}_k^{\text{(Rindler)}}$$ in the Minkowski vacuum

$$\begin{equation} \langle0|_M \hat{n}_k^{\text{(Rindler)}}|0\rangle_M=\sum_{n_1=0}^\infty \dots \sum_{n_k=0}^\infty \dots \sum_{n_r=0}^\infty f^2 (n_1,\dots,n_k,\dots,n_r) n_k \end{equation}$$

given the right function $$f$$ it is possible for this expectation value to give a thermal number of particles, that is:

$$\begin{equation} \langle0|_M \hat{n}_k^{\text{(Rindler)}}|0\rangle_M=\frac{1}{e^{\frac{k}{2\pi a}}-1} \end{equation}$$

where $$a$$ is the acceleration of the Rindler obserber. Note that I don't really care about the function $$f$$. I'm just checking that it is possible to have a thermal expectation value for the number operator of one basis if we calculate it respect to a state from a different basis. Also, note that the fact that I'm writting the Minkowski vacuum as a sum of particle states in the Rindler basis means (I think) that the Rindler states form a complete basis or in other words, that the two Hilbert spaces have the same size.

Tracing the left wedge in Rindler space

This method is a bit more abstract and much more general. One starts with the Minkowski vacuum and traces over everything not covered by the Rindler coordinates, leaving us with the reduced density matrix on that region of spacetime, which is

$$\begin{equation} \rho_{\text{Rindler}}=e^{-2\pi K} \end{equation}$$

where $$K$$ is the charge associated with boosts. Now, if I remember quantum mechanics correctly, if a state is pure, it is pure for every choice of basis so the statement $$|0\rangle_M$$ becomes a thermal (i.e. mixed) state in the Rindler basis can't be true. It can only be true if we are also tracing over the patch of spacetime not covered by Rindler coordinates.

So the question is: How are these two methods giving the same thermal result? where is the "tracing out" in the first one? Is it implicitely using that the Rindler basis is smaller than the Minkowski basis?

Note: Even if the Rindler basis is indeed smaller than the Minkowski basis, it doesn't look like it is necessary in the derivation I provided. The question remains: If we calculate the expectation value of the number of particles operator of one basis respect to a pure state on a different basis for the same Hilbert space, can we obtain a thermal distribution?

• In the first approach you actually need modes in the two Rindler wedges to expand the inertial observer's plane waves. Then you need to trace out the left Rindler wedge as well. For a full explanation of the first approach see the chapter on quantum field theory in curved spacetimes in Carroll's GR textbook.
– Gold
Sep 10, 2021 at 22:24
• Hi! Thanks for your reply. I was actually following Carroll's calculation for the first method and I do understand that you need two wedges of Rindler to fully describe Minkowski. However, I fail to see at which point he is tracing over one wedge. Specifically, I'm looking at equation 9.163 where he calculates the expectation value. Sep 10, 2021 at 22:32
• Also, my question points at the fact that I didn't have to specify that the Rindler basis is smaller, in my first approach. And I still got a thermal distribution. So that also confuses me. Sep 10, 2021 at 22:33
• The trace is, in fact, implied. The number operator that Carroll is computing the mean value of is that associated to one of the Rindler wedges. In that sense such an operator acts as the identity in the complementary wedge. Whenever you have a composite Hilbert space ${\cal H} = {\cal H}_A\otimes{\cal H}_B$ and an observable of the form $O_A\otimes 1_B$ its mean value with respect to a state in ${\cal H}$ is the mean value of $O_A$ with respect to the state obtained by tracing out ${\cal H}_B$.
– Gold
Sep 10, 2021 at 23:29

First let us have a clear description of the situation. First we quantize a free scalar field by expanding it into plane waves, which are positive frequency solutions of the wave equation with respect to the inertial observer. This is the standard quantization encountered in QFT and it gives us a Fock space $${\cal H}$$ which, in particular, contains the vacuum $$|0_M\rangle$$.

You can alternatively quantize the field expanding it into modes which are positive frequency with respect to the Rindler observers. But now you have the two wedges and certainly each of them alone is not enough to reproduce the Minkowski quantization since plane waves need modes supported in both Rindler wedges to be expanded. In fact one often supposes that the most natural thing happens: $${\cal H}={\cal H}_L\otimes{\cal H}_R$$ where $${\cal H}_L$$ is the Fock space constructed from the modes in the left Rindler wedge and likwewise for $${\cal H}_R$$. There are issues with such a decomposition and it looks like it does not really exist rigorously, but let us go on with it, since it is the simplest and most natural way to proceed.

Now $$|0\rangle_M$$ is a state in $${\cal H}$$. If the system is in that state, observers restricted to each Rindler wedge will actually perceive the reduced states $$\rho_L = \operatorname{Tr}_R|0_M\rangle\langle 0_M|$$ and $$\rho_R = \operatorname{Tr}_L|0_M\rangle\langle 0_M|$$.

At this point we ask: what is thermality? A state is thermal when it is of the form $$\rho = \frac{1}{Z}e^{-\beta H}$$. If $$\rho_R$$ is thermal and you pick a number operator $$n_i^R$$ associated to any of the modes in the right Rindler wedge, such an operator will have the kind of mean value you point out. Still having such a mean value does not alone imply a thermal state. In fact, you could consider things like $$n_i^R n_j^R$$ which now inform on correlation between modes. Thermality also constrains these.

So the difference between the two approaches is basically that (1) is, in fact, testing if the mean number of excitations matches that of a thermal state, but it has not proven thermality, while approach (2) is verifying thermality explicitly.

Now, about how (1) is working in practice and why the trace doesn't seem to be there, it is all about the fact that $$n_i^R$$ is an operator defined only in $${\cal H}_R$$ and it lifts to the composite Hilbert space $${\cal H}$$ trivially on the other sector, i.e., $$1_L\otimes n_i^R$$. One often does not bother writing this explicitly and just abuses notation denoting $$n_i^R$$ both the operator in $${\cal H}_R$$ and its extension to $${\cal H}$$ acting trivially on $${\cal H}_L$$. Then we really have $$\langle 0_M| n_i^R|0_M\rangle = \operatorname{tr}_R(\rho_R n_i^R)$$

So in summary there are two main points: approach one just tests the mean values while approach two proves thermality explicitly, and approach one may be employed without explicitly writing the trace because it is in fact implicit in the fact that $$n_i^R$$ act trivially on the other sector.

• Hi! Thanks for your great answer. So, as a last question: It seems like we can start with a vacuum pure state in basis A of space H, calculate the expectation number of particles in basis B, respect to the vacuum of A and get a thermal number of particles (which doesn't imply thermality). Even if basis A and B have the same size, we can get a thermal number of particles but only if B is smaller than A can we actually get a thermal state (i.e. change the pure state into a mixed state). Is this correct? Sep 11, 2021 at 15:52
• Hi, you're welcome ! Well, I must confess I'm confused by this idea of "bases of same size" or "a basis smaller than the other". All the basis in this discussion are actually infinite. The Hilbert spaces in question are infinite-dimensional, so I can't see how these ideas would apply. So in a sense the basis of the Rindler spaces ${\cal H}_L$ or ${\cal H}_R$ and that of ${\cal H}$ associated to inertial observers are, in fact, "of the same size". I would say that thermality is not connected to this idea of one basis being smaller than another.
– Gold
Sep 11, 2021 at 16:30
• I see. What I mean is that you can get one basis after tracing over a subspace of the other. Which intuitively for me means that one was “bigger” but you are absolutely right that concept makes little sense for infinite size spaces. Sep 11, 2021 at 16:39
• When you have a Hilbert space ${\cal H}$ and a mixed state $\rho$ you can always ask whether there is a another space $\tilde{\cal H}\simeq {\cal H}^C\otimes {\cal H}$ and a pure state $|\psi\rangle$ such that tracing over the complementary ${\cal H}^C$ yields back $\rho$. This is precisely what purification is about, and mathematically the answer is yes. See this question of mine on the subject physics.stackexchange.com/questions/424007/…
– Gold
Sep 11, 2021 at 16:44
• Also see these notes by Tom Hartman regarding the thermofield double hartmanhep.net/topics2015/17-eternalblackholes.pdf which is relevant for the particular topic you are studying
– Gold
Sep 11, 2021 at 16:46