It appears very intuitive to me that, just as a tangent space can be constructed on every point on the manifold, a Hilbert space should also be implementable as it is also a vector space. This would mean that each "tangent function" would have its own transformation properties, and a connection associated so that we can parallel transport them. Does this make sense? How can this be done?

My attempt at going around this has been to inspire myself on the tangent bundle structure. I mean that just as the tangent bundle is $$TM \:\colon = \bigcup\limits_{x\in \mathcal{M}} \{x\} \times T_xM,$$ my idea is to have $\mathcal{H}$ as our fiber, the manifold $\mathcal{M}$ as the base space, and $\mathcal{M} \times \mathcal{H}$ as the total space, so that, in analogy, we can have a $\mathcal{H}_x$ at each point in the space.

To me this is important because then we can implement bra-kets on the spacetime, introduce creation and annihilation operators, and decompose the QFT functions into modes. If there is a way to do this without considering my question I would greatly appreciate if you could explain this.

  • 2
    $\begingroup$ If you want the fancy geometric process for this you can look up what is called geometric quantization $\endgroup$
    – Slereah
    Commented Dec 9, 2021 at 10:14
  • 2
    $\begingroup$ As mentioned in the comments above, geometric quantization is definitely something you might be interested in. Furthermore, let me just mention that the tangent space of a manifold is always a finite-dimensional vector space. However, in quantum mechanics we usually deal with infinite-dimensional Hilbert spaces. Hence, there is not such a direct application to quantum mechanics or QFT as you suggest. Mathematically, there is also the notion of "Hilbert manifolds", which are spaces locally homeomorphic to Hilbert spaces (see for example "infinite-dimensional manifolds" on wikipedia) $\endgroup$ Commented Dec 14, 2021 at 12:57
  • $\begingroup$ @Slereah, does geometric quantization also quantize quantum field theory wavefunctions? Up to know, I've only seen examples of the quantization of observables in the regular quantum mechanical theory. $\endgroup$ Commented Dec 15, 2021 at 12:31
  • $\begingroup$ Related: Hilbert Space and distribution theory in QFT $\endgroup$ Commented Feb 26, 2022 at 19:29

2 Answers 2


There are other approaches to doing Quantum Field Theory in Curved Spacetime and I've never seen anyone trying to define a "Hilbert bundle" (so to speak) in order to achieve it. My guess, but this is just a guess, is that the state of the entire field would need to be on $\bigotimes_{x\in\mathcal{M}} \mathcal{H}_x$ and I don't think this is well-defined, not to mention the fact that fields are operator-valued distributions, and hence it would be somewhat complicated to rigorously deal with them point by point. Furthermore, it should be pointed out that states are defined globally, not locally.

There are quite a few references on QFTCS (and there is the tag ). Some of them can be found on the questions Suggested reading for quantum field theory in curved spacetime and Modern treatment of effective QFT in curved spacetime. This answer will be a brief outline of the approach taken by R. Wald's Quantum Field Theory in Curved Spacetime and Black Hole Thermodynamics on how to define a QFT in curved spacetime and a few consequences. The section on stationary spacetimes also draws some inspiration and interpretation from a seminal paper by Asthekar & Magnon.

Wald's book only deals with free fields for simplicity (and perhaps because the theory of interacting fields was better established only a few years after the book was written), and hence so will I. In all cases, I'll assume spacetime to be globally hyperbolic, which essentially means the classical equations of motion for the quantum fields of interesting admit a well-posed initial value problem throughout the entire spacetime. The reason behind this assumption is that we'll use the space of solutions to the classical equations of motion to define the Hilbert space.

Stationary Spacetimes

Let us first assume we are dealing with a stationary spacetime, whose Killing field we'll denote by $\xi^\mu$. In this situation, the spacetime possesses a time-translation symmetry, and as a consequence we can define some notion of energy. In particular, the classical energy of some field at time $t$ will be $$E = \int_{t} T_{\mu\nu} \xi^\mu n^\nu \sqrt{h} ~ \text{d}^3 x,$$ where $h_{\mu\nu}$ is the induced metric on the spacelike hypersurface of constant $t$, $h$ is its determinant, $n^\mu$ is the vector normal to said hypersurface, and $T_{\mu\nu}$ is the stress-energy-momentum tensor of the field.

We need to construct a Hilbert space for the theory. Since we'd like to have creation and annihilation operators, we are actually looking for a Fock space. Since Fock spaces have the form $$\mathcal{F} = \mathbb{C} \oplus \mathcal{H} \oplus (\mathcal{H} \otimes \mathcal{H}) \oplus \cdots,$$ we pretty much just need to build $\mathcal{H}$.

There is a cool trick to do it: we must first get some adequate vector space. It turns out that, with some adjustments due to technical issues, the vector space formed by the solutions to the classical equations of motion, $\mathcal{S}$, almost does the job (see Chap. 2 of Wald's book to see this in the context of ordinary QM, Chap. 3 for the modifications to QFT, Chap. 4 for the modifications to QFTCS). However, we still need to equip it with an inner-product and figure out a way of complexifying it (i.e., choosing a complex structure), since $\mathcal{S}$ is initially a real vector space and QM uses complex vector spaces.

The trick to deal with these issues is to exploit the Poisson brackets defined on $\mathcal{S}$ to restrict the possible choices of inner-product and complex structure in a way that choosing the complex structure will be sufficient to fully determine the inner-product.

In short, here's the situation we are in: we have a vector space, but we still need to turn it into a complex Hilbert space. We managed to employ the classical structure of the theory to restrict the possible inner-products and complex structures. From a more traditional field-theoretic point of view, one could say that so far we've done canonical quantization (we imposed commutation relations and so on), but we still don't know the states because we were not able to write the fields in terms of creation and annihilation operators.

To fix the inner-product (or to determine the creation and annihilation operators, in this stage of the construction both things are essentially equivalent), we can exploit the stationary symmetry. Since we know the classical energy of the field, we can demand that the inner-product we'll choose will give to that particular state of the field (recall that we are building $\mathcal{H}$ out of $\mathcal{S}$) the very same energy as an expectation value. This turns out to be sufficient to single out an inner-product for $\mathcal{H}$, hence leading us to the desired Fock space. From the creation and annihilation point of view, the presence of the stationary Killing field lets us define which modes are of positive-frequency and which modes are of negative-frequency, hence leading to the possibility of splitting the field in the form $\phi \sim e^{- i \omega t} a + e^{+ i \omega t} a^\dagger$.

The Unruh Effect

Notice that this construction heavily depends on the stationary Killing field. One might then ask: what if we had more than one stationary Killing field?

This happens in Minkowski spacetime: in the region $x > |t|$ (Cartesian coordinates), known as the right Rindler wedge, the Killing field associated with boosts in the $x$ direction is timelike, and hence can also be chosen as a stationary Killing field (at least in the $x > |t|$ region). Let me denote this Killing field by $\beta^\mu$. If we use $\beta^\mu$ instead of the Poincaré time translation Killing field (I'll write $\xi^\mu$), we'll get to a different Hilbert space. These Hilbert spaces are not isomorphic, meaning they do lead to different theories. Furthermore, one can show that the vacuum state of the Fock space constructed out of $\xi^\mu$ is actually a thermal state of the Fock space constructed out of $\beta^\mu$: while an observer static with respect to $\xi^\mu$ is freezing on a universe with no particles at all, an observer static with respect to $\beta^\mu$ (which is actually an observer accelerating at a constant acceleration) is being burned alive with particles. This is known as the Unruh effect, and it is a consequence of the fact that Quantum Field Theory is a theory of fields, not of particles. It also reflects the fact that some of the features you asked about ("introduce creation and annihilation operators and decompose the QFT functions into modes") are highly dependent on symmetries, which will not be available on generic curved spacetimes.

The thermal nature of this state is quite related to the fact that states are global constructions: the $\xi^\mu$ vacuum is defined on the entire Minkowski spacetime, but to consider is as a state on the Fock space of $\beta^\mu$ we must restrict its action to the region $x > |t|$. This is done, intuitively, by tracing out the modes associated with the rest of the spacetime. When one does this tracing out, the resulting state is (formally) a thermal density matrix at the Unruh temperature $$T_U = \frac{a}{2\pi},$$ where units with $c = G = \hbar = k_B = 1$ are being employed.

Let me also mention the existence of the tag .

The Hawking Effect

I'll oversimplify another similar situation: suppose you have a stationary spacetime containing a star, which eventually undergoes gravitational collapse and, after some time, reaches a stationary state as a black hole. The non-stationarity of spacetime during the collapse makes the initial and final stationary Fock spaces be different, leading to another astonishing prediction: if the field starts in a vacuum before the collapse, it becomes thermal after the collapse. This was originally predicted by Hawking in the 1970s and, as a consequence, is now known as the Hawking effect. The temperature measured by static observers infinitely far away from the black hole is given by the Hawking temperature, $$T_H = \frac{1}{8\pi M}.$$

As with the Unruh effect, this prediction employs the global nature of states of quantum fields. This time, however, one is tracing out the modes that fell down the black hole.

Notice the existence of the tag .

Quantum Fields in Schwarzschild Spacetime

As an extra example, let us consider how to define a quantum field theory on a Schwarzschild spacetime and mention an interesting consequence (namely, the extension of the Unruh effect to the Schwarzschild spacetime).

The Schwarzschild spacetime has the metric $$\text{d}s^2 = - \left(1 - \frac{2M}{r}\right)\text {d}t^2 + \left(1 - \frac{2M}{r}\right)^{-1}\text {d}r^2 + r^2 \text{d}\Omega^2,$$ and from the absence of terms containing $t$ on the components we can tell it is a stationary metric. As a consequence, one can define a QFT on Schwarzschild spacetime by simply following the procedure outlined on the previous paragraphs: the time-translation invariance of the spacetime allows one to pick a preferred inner-product for the one-particle Hilbert space $\mathcal{H}$, allowing one to build a Fock space and even get a notion of particles.

It can be shown (see pp. 126–129 of Wald's QFTCS book and references therein) that there is a single state defined on the entire spacetime that is both compatible with the stationary symmetry and non-singular, known in this context as the Hartle–Hawking state. "Compatible with the stationary symmetry" can be understood by thinking that just as we often ask for a Poincaré invariant vacuum on flat spacetime, we can now ask for a Killing invariant state. Non-singular means that the state allows for the stress-energy tensor of the quantum field to be renormalized when the field is in this state (such states are said to satisfy the "Hadamard condition", which is rather technical and I'm avoiding diving too deep in here, but Wald's book provides more detail on Sec. 4.6).

One can also show, by making a computation fully analogous to the derivation of the Unruh effect on flat spacetime, that the restriction of the Hartle–Hawking state to the outside of the event horizon is a thermal state at the Hawking temperature, $$T_H = \frac{1}{8\pi M}.$$ This is discussed in considerable detail on Wald's book, Sec. 5.3.

It is worth noticing that this is not the temperature measured by observers. A static observer would measure the temperature $$T = \frac{1}{8\pi M \sqrt{-\xi^\mu \xi_\mu}},$$ which includes a correction due to the gravitational redshift. This also means that, very close to the black hole, the formula resembles what one has in the Unruh effect on flat spacetimes, since $\frac{1}{4 M \sqrt{-\xi^\mu \xi_\mu}}$ will be the acceleration necessary for the observer to maintain their static orbit.

Different Vacua in Schwarzschild

Notice that this is not the same thing as the Hawking effect: the Hawking effect corresponds to what one gets for a black hole arising from gravitational collapse, while this time we are dealing with an eternal black hole. Not only are the derivations intrinsically different, but also the states of the quantum field are different.

The absence of Poincaré symmetry means we no longer have a way to pinpoint a single possible vacuum state. On the Unruh effect on a Schwarzschild spacetime, we chose our vacuum by requiring it to be Killing invariant and everywhere non-singular, and arrived at the Hartle–Hawking vacuum. However, on different physical situations, different vacua are of interest.

For a collapsing star, the spacetime does not possess a "white hole region". This means the Hartle–Hawking vacuum is no longer that interesting because it includes correlations between the outside of the even horizon with the "white hole region". Hence, it is not really physical. Then what could be choose?

Previously, we chose the Hartle–Hawking vacuum based on its Killing invariance and non-singularity. However, it should be pointed out there are other Killing invariant states, which are singular on some regions of the spacetime. The Unruh vacuum, for example, is Killing-invariant, but singular on the white hole horizon. Of course, this is not an issue for a collapsing star, since the white hole region is not physical in this case. Hence, this is the vacuum we are working with for the Hawking effect.

(Yes, ironically, the Hartle–Hawking vacuum is the vacuum for the Unruh effect, while the Unruh vacuum is the vacuum for the Hawking effect).

These different physical states lead to different physical consequences. The Hartle–Hawking vacuum is state with respect to all field modes. In other words, an observer feels to be surrounded by a heat bath coming from all around them, not only from the black hole. The Unruh vacuum, on the other hand, is thermal only with respect to modes "arising from the black hole", meaning an observer will feel the black hole as being hot, but the infinite space won't be hot.

For completeness, another state of interest in the Schwarzschild spacetime is the Boulware vacuum, which corresponds to the absence of particles on the exterior of the event horizon. It is Killing-invariant, but it is singular on both the white hole horizon and on the black hole horizon.

Path Integral Methods

While this answer is mostly focused on some rather technical aspects, it is worth pointing out that there are also derivations of some of these results employing path integral techniques, for example. If I'm not mistaken, a derivation of the Unruh effect for the Schwarzschild spacetime is available at a 1976 paper by Hartle and Hawking, DOI: 10.1103/PhysRevD.13.2188. I haven't read it yet (it is currently waiting on my to-do list haha), but I believe it is worth mentioning it anyway, so this answer also makes some more contact with more usual techniques in Quantum Field Theory.

Non-Stationary Spacetimes

One might then ask about how to deal with QFTCS when the spacetime is not stationary. In this case, one must abandon the particle and Fock space interpretations (unless you are using them as an approximation) and Wald defines the theory in terms of operator algebras of observables, which is a more rigorous approach to Quantum Field Theory.

In this approach, one can recover a Hilbert space by choosing some state to work as the vacuum and performing the so-called Gelfand–Naimark–Segal (GNS) construction. For well-behaved choices of vacuum (namely, for states that are pure and Gaussian (also known as quasifree)), this Hilbert space will be a Fock space, but there is no natural choice of Fock space in principle.

Your notion of local states can now be recovered in some sense: one of the features of this approach is that given the algebra of observables on the entire manifold, $\mathcal{A}(\mathcal{M})$, and some "subspacetime" $\mathcal{N}$, there is a subalgebra $\mathcal{A}(\mathcal{N})$ of $\mathcal{A}(\mathcal{M}$) corresponding to the observables on the region $\mathcal{N}$. By performing the GNS construction on $\mathcal{A}(\mathcal{N})$, you can obtain a Hilbert space $\mathcal{H}_{\mathcal{N}}$ associated with the region $\mathcal{N}$.

All of this is wildly different from a "Hilbert bundle", but given the final phrase of your question ("If there is a way to do this without considering my question I would greatly appreciate it if you could explain this."), I hope it can be helpful.

  • $\begingroup$ Just what I was looking for, thank you. The discussion is valid for a schwarzschild ST right? $\endgroup$ Commented Dec 30, 2021 at 8:50
  • $\begingroup$ @tonetillo4 You are welcome! Yes, a Schwarzschild spacetime is stationary, and hence one can define a quantum field theory on it by using the generic prescription for stationary spacetimes. If you are particularly interested on the case for a Schwarzschild spacetime, I don't mind editing the answer to add an extra section mentioning an example of a consequence obtained in this case (namely, one can get a generalization of the Unruh effect) $\endgroup$ Commented Dec 30, 2021 at 9:08
  • $\begingroup$ there aren't enough points to give you in the forum if you could write specifically about the schwarzschild case hahaha. I would appreciate that very much. $\endgroup$ Commented Dec 30, 2021 at 11:42
  • 1
    $\begingroup$ @tonetillo4 I've added a new section on it! =D $\endgroup$ Commented Dec 31, 2021 at 0:37
  • 1
    $\begingroup$ @RyderRude A KG wavefunction doesn't uniquely determine the state of the system (you also need its derivative), so you are correct that it fails as a wavefunction. This is one of the failures of Relativistic QM that led to the development of field theory. However, as you said, we are making a correspondence between the classical solutions and the one-particle Hilbert space to get to a Fock space in which evolution is indeed unitary $\endgroup$ Commented Jan 24, 2023 at 18:31
  1. Start off with a vector bundle over spacetime $M$. This is defined as a locally trivial bundle where every fibre is a topological vector bundle and this in turn means that the scalar multiplication and vector addition is continuous. We also need to assume that these operations are globally continuous. Examples of these are the tangent bundle $TM$ or the exterior spaces over this, $\wedge^k TM$ amongst others.

  2. Next we define a Banach bundle. This is a vector bundle over spacetime where each fibre is a Banach space. This means each fibre is equipped with a norm for which that vector space is complete. We also need to assume that these norms are globally continuous. This means for any sequence of based vectors $v_i$ such that $|v_i| \rightarrow 0$ and $p(v_i) \rightarrow b$ then $v_i \rightarrow 0_b$ where $0_b$ is the zero of the fibre over $b$. Here $p$ is the projection of the total space of the bundle onto its base.

  3. Finally, we define a Hilbert bundle by taking a Banach bundle and specifying in addition that each fibre is a separable Hilbert space. Equivalently, we simply need to stipulate that the norm in each fibre satisfies the parallelogram law since in that case we can obtain the inner product by polarising the norm.


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.