# I Schwinger effect verified by Unruh temperature?

Tags:
1. Dec 1, 2017

### jcap

According to https://arxiv.org/abs/1407.4569, equation (2.15), the Schwinger electron-positron pair production rate in Minkowski space, $N_S$, is given in natural units by
$$N_S=\exp(-\frac{m}{2T_U})$$
where the `Unruh temperature for the accelerating charge', $T_U$, is given by
$$T_U=\frac{1}{2\pi}\frac{qE}{m}$$
where $q$ is the electron charge, $m$ is the electron mass and $E$ is the applied electric field.

In principle, could the Schwinger effect be confirmed by measuring the temperature $T_U$ rather than trying to detect electron-positron pairs?

In SI Units:
$$T_U = \frac{1}{2\pi}\frac{\hbar}{c k_B}\frac{q E}{m}$$
If the static electric field $E=1$ MV/m then the Unruh temperature $T_U\sim 10^{-3}$K.

Last edited: Dec 1, 2017
2. Dec 1, 2017

### Demystifier

How would you measure that temperature?

3. Dec 1, 2017

### jcap

Could one use laser cooling to cool atoms down to the millikelvin range and then apply a large electric field of say $10^6$ V/m? The atoms should warm up due to the Unruh temperature, $T_U \sim 10^{-3}$K, induced by the static electric field. Perhaps this temperature rise would cause quantum decoherence that could then be detected?

(see https://en.wikipedia.org/wiki/Laser_cooling)

Last edited: Dec 1, 2017
4. Dec 1, 2017

### LeandroMdO

The relation between the Schwinger and Unruh effects is best thought of as a formal analogy. As first demonstrated by Nikishov, I believe, (see equation 11'), the probability that a pair is produced at a certain energy is given by what looks a lot like a thermal factor, so it's tempting to go ahead and define a temperature and say the produced particles are "at" that temperature. However, it's not a perfect analogy:

1. The energy comes squared in the formula, unlike a typical Boltzmann factor.
2. The pair distribution is degenerate with respect to the momentum component parallel to the electric field. This is because the Schwinger process is essentially a tunneling process, so that component is always zero at the "moment" of creation, and is subsequently accelerated by the field.
3. Unruh radiation is in a mixed state because Rindler space is divided in two wedges (so some information about the field is concealed from the observer, and in particular, those correlations that would permit them to identify the state as a vacuum state in disguise). This is at the heart of the thermal interpretation. In contrast, pairs produced by an electric background are indeed in a pure state.

So, in short, the analogy is cute but don't take it too seriously.

5. Dec 2, 2017

### jcap

Perhaps the background Unruh-like radiation is thermal even though the particle pairs themselves (if any are produced) are in a pure state.

6. Dec 2, 2017

### LeandroMdO

You can see clearly in the Heisenberg picture that it must be a pure state. You set up the theory in the vacuum state and what changes due to the applied electric field are the creation and annihilation operators. After some time you can turn off the electric field, and you find that the theory is no longer in the vacuum state with respect to the new creation and annihilation operators. But it's still very much the vacuum state with respect to the original ones.

There is some subtlety in the Unruh case because the Rindler space is divided into wedges. No such subtlety arises in the electric field calculation. There is no "background Unruh-like radiation". As I said, it's just a formal analogy.