Unruh Effect

Introduction

The Unruh effect is one of the cleanest demonstrations that in quantum field theory, the notion of “particle” is not absolute. It depends on the observer.

An inertial observer can regard a field state as vacuum, meaning no particles are present. But a uniformly accelerating observer can describe the same state as thermal, as if immersed in a bath of particles at some nonzero temperature. That temperature is called the Unruh temperature:

$$T_U = \frac{\hbar a}{2\pi c k_B}.$$

That sounds bizarre the first time you hear it. The lecture’s goal is to show that it is not mystical. It comes from a very explicit chain of reasoning.

The lecture begins with the simplest possible field-theory setting: a massless scalar field in $1+1$ dimensions. It reviews the canonical quantization of that field in ordinary Minkowski coordinates. Then, before doing anything with accelerating observers, it reviews a separate quantum-mechanical fact about two-mode squeezed states: if two oscillator modes are entangled in the right way, and you only observe one of them, then the state of the observed mode is indistinguishable from a thermal state.

That is the first key ingredient.

The second ingredient is the geometry of accelerated motion in Minkowski spacetime. Uniformly accelerating observers do not naturally use the same coordinates as inertial observers. Instead they use accelerating coordinates, usually called Rindler coordinates, and these coordinates only cover part of Minkowski spacetime. The lecture makes this extremely concrete: an observer confined to one accelerating wedge can never observe the field in the opposite wedge, because that region lies beyond an event horizon.

That is the second key ingredient: inaccessible degrees of freedom.

The third ingredient is the field expansion itself. The same quantum field can be expanded in terms of inertial Minkowski modes or in terms of accelerating-frame modes in the two wedges. Relating the corresponding ladder operators produces combinations of annihilation and creation operators — a Bogoliubov-type mixing. The lecture shows that, when applied to the inertial vacuum, this mixing implies that the inertial vacuum is a tensor product of two-mode squeezed states of the left and right accelerating wedges.

Now the result from the first part of the lecture applies immediately: if the accelerating observer can only access one wedge, then after tracing over the inaccessible wedge, the state appears thermal. And the resulting temperature turns out to be

$$T = \frac{\hbar a}{2\pi c k_B}.$$

So the effect is not “acceleration creates particles out of nowhere” in some sloppy sense. The sharper statement is:

• different observers use different mode decompositions,
• the inertial vacuum becomes entangled between the two accelerating wedges,
• and restricting attention to the observable wedge turns that entangled pure state into a thermal mixed state.

That is the Unruh effect. :contentReference[oaicite:1]{index=1}

Learning Objectives

• Review the quantization of a massless scalar field in $1+1$ dimensions.
• Understand the construction and physical meaning of a two-mode squeezed state.
• Show why observing only one mode of a two-mode squeezed state produces a thermal density operator.
• Define accelerating (Rindler) coordinates in Minkowski spacetime.
• Understand why uniformly accelerating observers only have access to one wedge of spacetime.
• Rewrite the scalar field in accelerating coordinates in the two wedges.
• Understand how inertial and accelerating ladder operators are related.
• Show that the inertial vacuum satisfies squeezed-state annihilation conditions in the accelerating basis.
• Interpret the inertial vacuum as a tensor product of two-mode squeezed states of the two wedges.
• Derive the Unruh temperature

$$T_U = \frac{\hbar a}{2\pi c k_B}.$$

• Explain why the effect is generic for horizons and how it connects to Hawking radiation.

Prerequisite Knowledge

• Canonical quantization of free scalar fields
• Harmonic oscillator ladder operators
• Density matrices and reduced states
• Basic Lorentzian spacetime geometry
• Coordinate changes and Jacobians
• Tensor-product Hilbert spaces
• Thermal density operators
• Basic idea of event horizons

1. Review: massless scalar field in 1+1 dimensions

The lecture begins with the simplest field theory possible for the purpose at hand: a massless scalar field in one spatial dimension.

The Lagrangian density is

$$\mathcal L = \frac12 \left[ \left(\frac{\partial \phi}{\partial t}\right)^2 - c^2 \left(\frac{\partial \phi}{\partial x}\right)^2 \right].$$

The canonical momentum field is

$$\Pi = \frac{\partial \mathcal L}{\partial(\partial_t \phi)} = \frac{\partial \phi}{\partial t},$$

and the Hamiltonian density is

$$\mathcal H = \frac12 \left[ \Pi^2 + c^2\left(\frac{\partial \phi}{\partial x}\right)^2 \right].$$

After canonical quantization, the field operator is expanded in plane-wave modes:

$$\hat\phi(x,t) = \sqrt{\frac{\hbar}{2\pi}} \int_{-\infty}^{\infty} \frac{dk}{\sqrt{2c|k|}} \left( \hat a_k e^{i(kx-c|k|t)} + \hat a_k^\dagger e^{-i(kx-c|k|t)} \right),$$

and the Hamiltonian is

$$\hat H = \int_{-\infty}^{\infty} dk\, \hbar c|k|\, \hat a_k^\dagger \hat a_k + E_0.$$

So the lecture starts from a completely standard inertial quantization of a free scalar field. :contentReference[oaicite:2]{index=2}

This simplicity is important. The Unruh effect is not being blamed on interactions, masses, or gauge fields. It appears already in the simplest free relativistic field theory.

2. Review: two-mode squeezed states

Before discussing accelerated observers, the lecture reviews a quantum-mechanical construction involving two harmonic oscillators of the same frequency.

Their Hamiltonian is

$$\hat H = E\hat a_1^\dagger \hat a_1 + E\hat a_2^\dagger \hat a_2 = E(\hat n_1 + \hat n_2),$$

and the ground state is the usual product vacuum

$$|G\rangle = |0\rangle_1 |0\rangle_2,$$

annihilated by both $\hat a_1$ and $\hat a_2$. :contentReference[oaicite:3]{index=3}

The lecture then defines new operators

$$\hat c_1 = C_1 \hat a_1 + C_2 \hat a_2^\dagger, \qquad \hat c_2 = C_2 \hat a_1^\dagger + C_1 \hat a_2,$$

and notes that these obey canonical bosonic commutation relations provided

$$|C_1|^2 - |C_2|^2 = 1.$$

It then asks: which state satisfies

$$\hat c_1 |\Psi\rangle = \hat c_2 |\Psi\rangle = 0?$$

Expanding $|\Psi\rangle$ in the number basis and solving the recursion relations gives the answer:

$$|\Psi_A\rangle = \sqrt{1-|A|^2} \sum_{n=0}^\infty A^n |n\rangle_1 |n\rangle_2, \qquad A = -\frac{C_2}{C_1}, \qquad |A|<1.$$

This is the two-mode squeezed state. :contentReference[oaicite:4]{index=4}

That is the first technical pillar of the lecture.

3. Why one mode of a two-mode squeezed state looks thermal

The lecture then asks what happens if an observer can measure only oscillator 1, while oscillator 2 is inaccessible.

That means all observables have the form

$$\hat O = \hat O_1 \otimes \hat I_2.$$

Computing expectation values in $|\Psi_A\rangle$ yields

$$\langle \Psi_A | \hat O_1 | \Psi_A \rangle = (1-|A|^2)\sum_{n=0}^\infty (|A|^2)^n \langle n|\hat O_1|n\rangle.$$

So the reduced density operator for oscillator 1 is

$$\hat\rho = (1-|A|^2)\sum_{n=0}^\infty (|A|^2)^n |n\rangle\langle n|.$$

The lecture then identifies this with a thermal density operator by defining an effective temperature through

$$|A|^2 = e^{-E/(k_B T)}.$$

Then

$$\hat\rho = (1-e^{-E/k_BT}) \sum_{n=0}^\infty e^{-En/k_BT}|n\rangle\langle n|.$$

So the key conclusion is:

A two-mode squeezed state is indistinguishable, by measurements on only one mode, from a thermal state. :contentReference[oaicite:5]{index=5}

This is the crucial conceptual bridge. Once the inertial vacuum is shown to be a two-mode squeezed state in the accelerating basis, thermality will follow almost immediately.

4. Accelerating coordinates in Minkowski spacetime

The lecture now turns to spacetime geometry.

Start with ordinary Minkowski spacetime in $1+1$ dimensions,

$$ds^2 = dx^2 - c^2 dt^2.$$

Then introduce alternative coordinates $(\xi,\tau)$ for the right wedge $x>0$:

$$x = \frac{c^2}{a} e^{a\xi/c^2}\cosh\left(\frac{a\tau}{c}\right), \qquad t = \frac{c}{a} e^{a\xi/c^2}\sinh\left(\frac{a\tau}{c}\right).$$

Here $a$ is an arbitrary constant. The lecture notes that the curve $\xi=0$ is a trajectory of constant proper acceleration $a$, and that $\tau$ is the proper time of that accelerating observer. :contentReference[oaicite:6]{index=6}

To cover the left wedge $x<0$, one needs a second coordinate patch $(\xi',\tau')$:

$$x = -\frac{c^2}{a} e^{a\xi'/c^2}\cosh\left(\frac{a\tau'}{c}\right), \qquad t = \frac{c}{a} e^{a\xi'/c^2}\sinh\left(\frac{a\tau'}{c}\right).$$

So Minkowski space is naturally split into two accelerating wedges.

5. The metric in accelerating coordinates

The lecture differentiates the coordinate transformation and finds

$$ds^2 = e^{2a\xi/c^2}(d\xi^2 - c^2 d\tau^2) = e^{2a\xi'/c^2}(d\xi'^2 - c^2 d\tau'^2).$$

So the metric in each wedge looks like the Minkowski metric multiplied by a conformal factor. :contentReference[oaicite:7]{index=7}

In $1+1$ dimensions for a massless scalar field, this is especially convenient, because the conformal factor ends up dropping out of the action in the crucial way shown later in the lecture.

6. The event horizon for the accelerating observer

The lecture then emphasizes the physical meaning of the wedge decomposition.

An observer following a uniformly accelerated trajectory in the right wedge can never be affected by events in the left wedge. There is an event horizon. The slide states this very directly: nothing from the inaccessible wedge can ever affect, or be observed by, the accelerating observer. In the Schrödinger picture, the accelerating observer can only ever observe the field for $x>0$; no field operator at $x<0$ can ever be measured. :contentReference[oaicite:8]{index=8}

This is the second conceptual pillar of the lecture: the accelerating observer is forced to ignore one half of spacetime.

That inaccessible half will play the role of the hidden oscillator mode in the two-mode squeezed-state story.

7. Rewriting the action in accelerating coordinates

The lecture now rewrites the scalar field action in the accelerating coordinates. Starting from

$$S = \frac12 \int dx\,dt \left[ \left(\frac{\partial \phi}{\partial t}\right)^2 - c^2\left(\frac{\partial \phi}{\partial x}\right)^2 \right],$$

it computes the Jacobian

$$dx\,dt = e^{2a\xi/c^2} d\xi\,d\tau,$$

and the transformed derivatives using the chain rule. After substitution, a remarkable simplification occurs: the conformal factor cancels. The final result is

$$S = \frac12 \int d\xi\,d\tau \left[ \left(\frac{\partial \phi}{\partial \tau}\right)^2 - c^2\left(\frac{\partial \phi}{\partial \xi}\right)^2 \right] + \frac12 \int d\xi'\,d\tau' \left[ \left(\frac{\partial \phi}{\partial \tau'}\right)^2 - c^2\left(\frac{\partial \phi}{\partial \xi'}\right)^2 \right].$$

The lecture highlights the striking result: the action looks the same in the accelerating coordinates as in the inertial coordinates — except that now it is doubled into two independent wedge contributions. :contentReference[oaicite:9]{index=9}

This is the third big structural step.

8. Field quantization in the two accelerating wedges

Because the action splits into two identical copies, the field can be quantized independently in the two wedges.

The lecture writes the inertial field expansion

$$\hat\phi_I(x,t) = \sqrt{\frac{\hbar}{2\pi}} \int_{-\infty}^{\infty}\frac{dk}{\sqrt{2c|k|}} \left( \hat a_k e^{i(kx-c|k|t)} + \hat a_k^\dagger e^{-i(kx-c|k|t)} \right),$$

and the accelerating-frame field expansions

$$\hat\phi_{A+}(\xi,\tau) = \sqrt{\frac{\hbar}{2\pi}} \int_{-\infty}^{\infty}\frac{dk}{\sqrt{2c|k|}} \left( \hat b_k e^{i(k\xi-c|k|\tau)} + \hat b_k^\dagger e^{-i(k\xi-c|k|\tau)} \right),$$ $$\hat\phi_{A-}(\xi',\tau') = \sqrt{\frac{\hbar}{2\pi}} \int_{-\infty}^{\infty}\frac{dk}{\sqrt{2c|k|}} \left( \hat c_k e^{i(k\xi'-c|k|\tau')} + \hat c_k^\dagger e^{-i(k\xi'-c|k|\tau')} \right).$$

Correspondingly, the Hamiltonian becomes

$$\hat H = \int dk\, \hbar c|k|\, \hat a_k^\dagger \hat a_k = \int dk\, \hbar c|k|\, \hat b_k^\dagger \hat b_k + \int dk\, \hbar c|k|\, \hat c_k^\dagger \hat c_k.$$

So the same field is now expressed in terms of inertial modes $a_k$ or accelerating modes $b_k,c_k$ in the right and left wedges. :contentReference[oaicite:10]{index=10}

9. Same field, different coordinates, different mode operators

The lecture then makes the identification explicit:

$$\hat\phi_{A+}(\xi,\tau) = \hat\phi_I\!\left( \frac{c^2}{a}e^{a\xi/c^2}\cosh\frac{a\tau}{c}, \frac{c}{a}e^{a\xi/c^2}\sinh\frac{a\tau}{c} \right),$$

and similarly for the left wedge,

$$\hat\phi_{A-}(\xi',\tau') = \hat\phi_I\!\left( -\frac{c^2}{a}e^{a\xi'/c^2}\cosh\frac{a\tau'}{c}, \frac{c}{a}e^{a\xi'/c^2}\sinh\frac{a\tau'}{c} \right).$$

So these are not different fields. They are the same quantum field written in different coordinates on different patches of spacetime. :contentReference[oaicite:11]{index=11}

This is why there must be a nontrivial mapping among the corresponding ladder operators

$$\hat a,\hat a^\dagger,\hat b,\hat b^\dagger,\hat c,\hat c^\dagger.$$

10. Deriving the operator relation

The lecture then works out the transformation explicitly, starting for example from the inverse-Fourier formula for $\hat b(k)$ at $\tau=0$. Using the fact that $\tau=0$ also corresponds to $t=0$ and carefully applying the chain rule, it expresses $\hat b(k)$ in terms of inertial operators $\hat a(k')$ and $\hat a^\dagger(k')$. After some changes of variables and integral evaluation, it obtains a result of the schematic form

$$\hat b(k) = \text{(integral over }k') \Big[ e^{+\pi |k|c^2/(2a)}\, \hat a(\cdots) + e^{-\pi |k|c^2/(2a)}\, \hat a^\dagger(\cdots) \Big].$$

A parallel formula holds for the left-wedge operators $\hat c$, with the appropriate sign changes coming from the $x<0$ patch. :contentReference[oaicite:12]{index=12}

The exact integral kernel is not the conceptual point. The crucial point is that the accelerating annihilation operators are not pure annihilation operators in the inertial basis. They mix inertial annihilation and creation operators.

That is the Bogoliubov mixing responsible for the effect.

11. The inertial vacuum in the accelerating basis

Combining the right- and left-wedge formulas, the lecture arrives at the key relations

$$\big(\hat b(k) + e^{-\pi c^2|k|/a}\hat c^\dagger(-k)\big)|0_I\rangle = 0,$$ $$\big(\hat c(-k) + e^{-\pi c^2|k|/a}\hat b^\dagger(k)\big)|0_I\rangle = 0, \qquad \forall k,$$

where $|0_I\rangle$ is the inertial vacuum, defined by

$$\hat a(k)|0_I\rangle = 0 \quad \forall k.$$

The lecture then draws the decisive conclusion:

The inertial vacuum is a big tensor product of two-mode squeezed states in the accelerating frame. :contentReference[oaicite:13]{index=13}

This is exactly the pattern reviewed earlier for two oscillators.

12. Explicit squeezed-state form of the inertial vacuum

The lecture writes the inertial vacuum as

$$|0_I\rangle = \prod_k \sqrt{1-e^{-2\pi |k|c^2/a}} \sum_{n_k=0}^{\infty} e^{-n_k \pi |k|c^2/a} |n_k\rangle_{A+,k}\,|n_k\rangle_{A-,-k}.$$

So for each $k$, the inertial vacuum is a two-mode squeezed state entangling the right and left accelerating wedges. The full vacuum is the tensor product over all $k$. :contentReference[oaicite:14]{index=14}

This is the core structural result of the whole lecture.

13. Why the accelerating observer sees a thermal state

Now the two ingredients meet.

The accelerating observer in the right wedge can never measure the field in the left wedge $x<0$. Therefore they never observe the states $|n_k\rangle_{A-}$. For them, those degrees of freedom are inaccessible. So the pure inertial vacuum, when restricted to the accessible wedge, becomes indistinguishable from a thermal mixed state.

The lecture then reads off the temperature mode by mode. Comparing the weight

$$|A|^2 = e^{-2\pi |k| c^2/a}$$

with the thermal form

$$e^{-E/(k_B T)}, \qquad E(k)=\hbar c|k|,$$

gives

$$T(k) = \frac{E(k)}{k_B} \frac{1}{\ln(e^{2\pi |k|c^2/a})} = \frac{\hbar c|k|}{k_B}\frac{a}{2\pi |k|c^2} = \frac{\hbar a}{2\pi c k_B}.$$

The lecture emphasizes that this is actually independent of $k$, just as in a genuine thermal state. :contentReference[oaicite:15]{index=15}

So the accelerating observer sees the inertial vacuum as a thermal bath with temperature

$$T_U = \frac{\hbar a}{2\pi c k_B}.$$

That is the Unruh effect.

14. The Unruh temperature

The lecture then names this temperature explicitly:

$$T = \frac{\hbar a}{2\pi c k_B},$$

the Unruh temperature, after W. G. Unruh. It also states that the same result holds much more generally:

• for massive as well as massless fields,
• in $D+1$ dimensions as well as $1+1$,
• for spinor and vector fields as well as scalars,
• for fermions as well as bosons.

The lecture says that the form of the derivation remains the same in all cases, though the integrals become more complicated. The “miracle” that always reappears is tied to the exponential relation between inertial and accelerating coordinates near the horizon. This kind of temperature is therefore generic for event horizons. :contentReference[oaicite:16]{index=16}

This is an important conceptual upgrade: the Unruh effect is not a trick of the simplest scalar field. It is a generic horizon phenomenon.

15. Hawking temperature as the same kind of effect

The final slide makes the connection to black holes.

The lecture says that Hawking temperature is derived in an exactly similar way, by relating different expressions for the same quantum field in inertial and accelerating frames — though now in curved spacetime rather than in Minkowski space. Near the horizon of a black hole, the relation between freely falling and stationary coordinates has the same essential exponential structure. That leads again to two-mode squeezed states, this time entangling modes inside and outside the horizon.

So the vacuum for a freely falling observer appears thermal to an observer who stays outside the black hole. For a spherical black hole, the Hawking temperature is

$$T_H(M) = \frac{\hbar c^3}{8\pi k_B G M} = T_U(a_s),$$

where

$$a_s = \frac{GM}{r_S^2}, \qquad r_S = \frac{2GM}{c^2}$$

is the surface gravity at the Schwarzschild radius. :contentReference[oaicite:17]{index=17}

So Hawking radiation is not a separate miracle. It is the curved-spacetime cousin of the same horizon thermality.

Worked Examples

Example 1: Why one wedge looks thermal

For each mode $k$, the inertial vacuum has the form

$$|0_I\rangle_k = \sqrt{1-q_k} \sum_{n=0}^\infty q_k^{n/2}\, |n\rangle_{A+,k}|n\rangle_{A-,-k}, \qquad q_k = e^{-2\pi |k|c^2/a}.$$

This is exactly a two-mode squeezed state. If the accelerating observer only has access to the $A+$ wedge, then tracing over the $A-$ wedge gives

$$\hat\rho_k = (1-q_k)\sum_{n=0}^\infty q_k^n |n\rangle\langle n|.$$

That is the thermal density matrix of a harmonic oscillator mode.

Example 2: Reading off the temperature

For the scalar mode, the energy is

$$E(k)=\hbar c|k|.$$

The squeezed-state weight is

$$q_k=e^{-2\pi |k|c^2/a}.$$

Comparing this to the Boltzmann weight

$$e^{-E(k)/(k_B T)}$$

gives

$$\frac{\hbar c|k|}{k_B T} = \frac{2\pi |k|c^2}{a},$$

so

$$T=\frac{\hbar a}{2\pi c k_B}.$$

The $k$-dependence cancels, which is exactly what one needs for a genuine temperature.

Intuition

The lecture’s best intuition is this:

The inertial vacuum is not “empty” in an observer-independent way. It is a particular entangled state of field modes. For an inertial observer, the natural modes are global Minkowski modes, and that state is vacuum. For an accelerating observer, the natural modes live separately in the right and left wedges. In that basis, the same state is not a product vacuum. It is an entangled two-mode squeezed state linking inaccessible left-wedge modes to accessible right-wedge modes.

If you are forced by the horizon to ignore half the system, then a pure entangled state looks mixed. And in this case, the mixed state is exactly thermal.

So the Unruh effect is not about “motion heating up space” in some crude mechanical sense. It is about:

• observer-dependent mode decomposition,
• horizon-induced inaccessibility,
• and entanglement turning pure vacuum into thermal appearance.

Common Mistakes

• Thinking the Unruh effect means the inertial vacuum and accelerating vacuum are literally different physical universes. They are different mode descriptions of the same field.
• Confusing the field vacuum with the statement “nothing exists.”
• Forgetting that the accelerating observer only has access to one Rindler wedge.
• Missing that the thermality comes from tracing out inaccessible modes, not from adding random noise by hand.
• Thinking the result depends on the field being massless or scalar. The lecture explicitly says it is more general.
• Ignoring the role of two-mode squeezed states in the logic.
• Treating the horizon as a black-hole-only concept. Uniformly accelerated observers already have one in flat spacetime.
• Thinking the boost or coordinate transformation acts on Hilbert space the same way time evolution does. It does not.
• Missing that Hawking radiation is presented here as the same structural phenomenon near a black-hole horizon.

Short Summary

The lecture derives the Unruh effect for a massless scalar field in $1+1$ dimensions. It begins by reviewing canonical quantization of the field in inertial Minkowski coordinates and then reviews two-mode squeezed states, showing that if only one mode of such a state is observable, the reduced density operator is thermal. It next introduces accelerating coordinates for the right and left Rindler wedges of Minkowski spacetime and shows that a uniformly accelerating observer can only access one wedge because the other lies beyond an event horizon. Rewriting the scalar-field action in accelerating coordinates reveals that it splits into two identical wedge contributions. The same field can therefore be expanded either in inertial Minkowski modes with ladder operators $\hat a_k$, or in accelerating-frame modes in the right and left wedges with ladder operators $\hat b_k$ and $\hat c_k$. Explicitly relating these operators shows that the inertial vacuum satisfies

$$(\hat b(k)+e^{-\pi c^2|k|/a}\hat c^\dagger(-k))|0_I\rangle = (\hat c(-k)+e^{-\pi c^2|k|/a}\hat b^\dagger(k))|0_I\rangle =0.$$

This means that the inertial vacuum is a tensor product of two-mode squeezed states of the two accelerating wedges. Since the accelerating observer can never access the opposite wedge, the inertial vacuum is indistinguishable to that observer from a thermal state. Comparing the squeezed-state weights with Boltzmann weights yields the Unruh temperature

$$T_U = \frac{\hbar a}{2\pi c k_B}.$$

The lecture concludes by emphasizing that this mechanism is generic for event horizons and is the flat-spacetime analogue of Hawking radiation, whose temperature for a Schwarzschild black hole is

$$T_H(M)=\frac{\hbar c^3}{8\pi k_BGM}.$$

:contentReference[oaicite:18]{index=18}

Practice Problems

1. Why does the lecture begin with two-mode squeezed states before introducing accelerating observers?

2. Show conceptually why one mode of a two-mode squeezed state looks thermal when the other mode is inaccessible.

3. Why are two coordinate patches needed to describe uniformly accelerating observers in Minkowski spacetime?

4. What physical role does the event horizon play in the Unruh effect?

5. Why does the scalar-field action split into two wedge contributions in accelerating coordinates?

6. What is the meaning of the operators $\hat b_k$ and $\hat c_k$ in the lecture?

7. Why do the accelerating annihilation operators mix inertial annihilation and creation operators?

8. What does it mean to say the inertial vacuum is a tensor product of two-mode squeezed states in the accelerating frame?

9. Why is the reduced state seen by the accelerating observer thermal rather than just some arbitrary mixed state?

10. Derive the Unruh temperature from the squeezed-state weight

$$e^{-2\pi |k|c^2/a}.$$

11. Why does the final temperature not depend on $k$?

12. Explain how the lecture connects the Unruh effect to Hawking radiation.