for Leg Spinners 17/01/11

Lecture Notes 7: The Unruh Effect

Lecturer: Prakash Panangaden Scribe: Shane Mansfield

1 Defining the

Recall from the last lecture that choosing a complex structure defines the positive and nega- tive frequency solutions of your theory. This choice is not unique and one needs some input from the physics to pick out the “right” complex structure. We consider what happens when two observers have different notions of positive and negative frequencies. Suppose we have two complete, orthonormal sets of of complex solutions to the Klein-Gordon equa- tion, {fi} and {qj}. By orthonormal we mean that the union of the fi and their complex conjugates satisfy the following equations

∗ ∗ hfi, fji = − fi , fj = δij.

So the Klein-Gordon “inner product” only behaves like an inner product when restricted to positive frequency solutions. By complete we mean that they form a basis in the complex vector of all solutions. If φˆ(x) is our field operator, then the creation and annihilation operators with respect to the f-basis are defined by

ˆ X  † ∗ φ(x) = aˆifi +a ˆi fi , i and the vacuum |0i is defined as the unique state that is killed by all the annihilation operators:

aˆi|0i = 0.

Since we are dealing with complete sets, we can write the fi and gj in terms of each other: P ∗ ∗ P  ∗ ∗ ∗  gj = i (αijfi + βijfi ) gj = i αijfi + βijfi P  ∗ ∗ P  ∗ ∗ ∗  fi = j αjigj + βjigj fi = j αjigj + βjigj .

So, for example, αij = hgj, fii.

7 - 1 We could equally well expand the field operator in the g-basis. Then we have

ˆ X ˆ ˆ† ∗ φ(x) = bjgj + bjgj , j ˆ ˆ† for some other creation and annihilators bj and bj, and we also have another vacuum defined 0 by ˆbj|0i = 0. The question we are interested in answering is: what does a one vacuum look like in the other basis?

2 Bogolioubov Transformations

It is a routine calculation to find that P h ˆ ˆ†i ˆ P h ˆ ˆ†i aˆi = j αjibj + βjibj bj = i αijbi + βijbi † P h ∗ ˆ† ∗ ˆ i ˆ† P h ∗ ˆ† ∗ ˆ i aˆi = j αjibj + βjibj bj = i αijbi + βijbi . These are called the Bogoliubov transformations. Now we have two number operators:

ˆ † ˆ 0 ˆ†ˆ Ni =a ˆi aˆi Nj = bjbj.

So we can take one vacuum and with it calculate the expectation value of the other number operator:

0 0 X 2 h0| Nˆi|0i = |βij| . j We see that under the Bogolioubov transformations the annihilation operators pick up a creation part, and the coefficients give rise to the non-zero right-hand side here. It is this mixing of the creation and annhilation operators that is responsible for particle creation. The Bobolioubov transformation formalism was first used in order to describe particle by Leonard Parker in the 1960s.

3 Rindler

Recall that a vector field is a Killing field if the Lie derivative of the metric along that vector field vanishes. In Minkowski spacetime we have the boost Killing field: ∂ ∂ t + z . ∂z ∂t The integral curves of this Killing field are timelike in the right and left Rindler wedges (RRW and LRW, see figure 1). So in these regions of the spacetime we could use these

7 - 2 Figure 1: Worldlines of uniformly accelerated observers. curves to define a time coordinate. They curves are curves of constant acceleration, so for a uniformly accelerating observer these are the “natural” time coodinates. An accelerating observer will see the diagonals of the figure as horizons. So for such an observer, the Rindler wedge is his “natural” home. Note that each of the left and right wedges is a globally hyperbolic static spacetime in its own right. The coordinates of a uniformly accelerated observer are the (ρ, η, x, y). These are related to the Minkowski coordinates (t, z, x, y) by the transformations

t = ρ sinh η, z = ρ cosh η.

The of the Rindler spacetime is

ds2 = ρ2dη − dρ2 − dx2 − dy2.

Lines of constant acceleration are lines of constant ρ, and the value of the acceleration is ρ−1. In terms of the scaled coordinates (τ, η, x, y), where the acceleration a is factored out:

ρ = a−1eaξ, η = aτ.

ξ = 0 corresponds to acceleration a. We will use τ to define positive and negative frequency, and thus the vacuum for an accelerating observer. We will consider the difference between the vacuum of Minkowski coordinates, and that of Rindler coordinates. It turns out that, to the accelerated observer, the Minkowski background looks like a thermal bath: this is what’s known as the Unruh effect or the

7 - 3 Fulling-Davies-Unruh effect. We derive this for a two-dimensional spacetime with massless fields. This is a toy demonstration of the Unruh effect. It holds true in the four dimensional massive case, but the analysis is quite a bit longer.

4 Toy Demonstration of the Unruh Effect

We write the momentum operator in terms of right-going and left-going parts; each with its own creation and annihilation operators. Z ∞ ˆ dk ˆ −ik(t−z) ˆ −ik(t+z) ˆ† −ik(t−z) ˆ† −ik(t+z) Φ(t, z) = √ b−ke + bke + b−ke + bke . 0 4πk When we change coordinates from (t, z) to (τ, ξ) we get  ∂ ∂   ∂ ∂  − φ = 0 → − φ = 0. ∂t2 ∂z2 ∂τ 2 ∂ξ2 This is due to conformal invariance. It does not happen for four dimensional massive fields, and for that reason they are a bit more difficult to deal with. If we use u = τ + ξ U = t + z v = τ − ξ V = τ + ξ then we can write the momentum in the RRW in terms of the other coordinates as Z ∞ ˆ dk  R −ikV R ikv  Φω(v) = √ αωke b? + βωke b? , 0 4πk R R and similarly for the LRW. To calculate αωk and βωk is just an integration now. For example,

πω ω ie 2a a−i  iω  αR = √ a Γ 1 − . ωk 2π ωk k a The thing to note here is the exponential. It is exactly the form you get in the Boltzmann distribution for blackbody . ˆ From b|0iM = 0 we have

 πω  R − a R† aˆω − e aˆω |0iM = 0; so if Z ω Z ω R 0 R 0 0 0 aˆf = f(ω )ˆaω0 dω , where f(ω ) dω = 1, 0 0 then Z ∞ dω|f(ω)|2 h0|aˆR†aˆR|0i = , M f f M 2πω/a 0 e − 1 which is the spectrum of a black body with a.

7 - 4 Figure 2: The Unruh-Wald detector. The two-state quantum mechanical system is coupled to a Klein-Gordon field, and has a transition probability due to this field. In this picture, a particle is the effect of a detector interacting with a field.

5 The Rindler Observer

An accelerating observer would use Rindler coordinates. Unruh and Wald considered what would happen if an accelerating observer had a particle detector. By a detector they meant a simple two-state quantum mechanical system (states | ↑i, | ↓i), coupled to a Klein-Gordon field (see figure 2). Absorbing a particle from the field, the quantum system could change state | ↓, ni → | ↑, n − 1i.

They showed that such a state, if accelerated in , will detect a particle with probability given by the black body spectrum. This raises another issue: how a Minkowski observer interprets the absorption of a Rindler particle. Unruh and Wald showed that the Minkowski observer interprets this as the emission of a Minkowski particle. The problem now is: who thinks that the energy of the field has increased or decreased? In fact, both observers will agree that the energy in the field has increased. It seems counter-intuitive in that the Rindler observer has absorbed a particle from the field, and yet the field energy goes up. In general, however, when a particle is absorbed from a heat bath, the energy goes up. The following toy calculation is a simple demonstration of the fact.

7 - 5 Example

Consider the state 1 |0i + √ |ni n where n >> 1 is the number of quanta, each having energy E. The expectation value of energy is hEi ' E. If, however, you detect a particle, then

hEnergyi ' (n − 1)E >> E.

6 Rindler Vacuum?

An interesting question to consider is whether a Rindler observer can even prepare a vacuum. Of course, if not, then the observer can’t prepare a pure state of any kind. This is a problem because in theory, all of our protocols begin with preparing pure states. If the Rindler vacuum state can be prepared, then which has lower energy? Surely observers should agree on the ordering of energy levels, if not on the scaling. It has been calculated that the vacuum energy of the Rindler spacetime diverges as the horizons are approached. This may be taken as an indication that the Rindler vacuum is unphysical. However, much remains to be understood.

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