# Do wavefunctions shrink under Lorentz boosts?

Suppose I have a relativistic wavefunction for some massive particle, defined by Dirac equation. Suppose the particle is free from interactions.

Let us suppose that the probability density takes the shape of a spatially localized wavepacket, let's say a spherical Gaussian distribution (or any localized distribution) with a size dispersion $$\sigma$$ (standard deviation for Gaussian). I know already that the wavefunction and the equations are invariant but would $$\sigma$$ shrink under a Lorentz boosts?

The total wavefunction does not have to be localized, but I would like to know if the envelope of dense points of the probability distribution are invariant under Lorentz boosts.

I guess I will have to find a $$\psi$$ as a sum of plane waves solutions to Dirac equation, such that $$\psi^\dagger\psi\propto \exp\left(-\frac{|x|^2}{2\sigma^2}\right)$$ (if that is even possible), and then see how this distribution changes under Lorents boosts. But this seems complicated already as it is not as simple as changing $$x\to x'$$ as the Gaussian is a sum of many plane waves with different frequencies. Maybe another distribution would be easier to handle? Any feedback?

Some are suggesting that the single particle picture is lost, but would this happen for free particles?

• The tricky thing about the Lorentz boosts is that their action on the wave function is not unitary. That means, for example, that the integral over all space of $\psi^{\dagger}\psi$ will change under the transformation, and you lose the simple probabilistic interpretation of the wave function.
– Buzz
Oct 12, 2021 at 23:10
• @Buzz but shouldn’t the number of particles be conserved in free space ? If the particle is a single electron, the integral of the probability should be invariant to conserve charge. Oct 13, 2021 at 8:26
• Oct 13, 2021 at 9:41
• @Mauricio: Quantum field theory is a many particle theory and not a single particle theory. This means particles are annihilated as well as being created. Oct 14, 2021 at 20:16

Wave functions could be very complicated things. I don't intend to address all possibilities. But, for the case of the wave function of a single particle, let's say a single photon, one can consider the situation as follows. We start by defining the single photon state by $$|\psi\rangle = \hat{a}_F^{\dagger} |\text{vac}\rangle$$ where the creation operator is defined in a Lorentz covarinat manner by $$\hat{a}_F^{\dagger} = \int F(\mathbf{k}) \hat{a}^{\dagger}(\mathbf{k}) \frac{d^2k}{(2\pi)^3\omega} .$$ One can define the ladder operator with a Lorentz covariant commutation relation as $$[ \hat{a}(\mathbf{k}_1),\hat{a}^{\dagger}(\mathbf{k}_2)] = (2\pi)^3\omega\ \delta(\mathbf{k}_1-\mathbf{k}_2) .$$ These ladder operators produce a single photon basis $$|\mathbf{k}\rangle = \hat{a}^{\dagger}(\mathbf{k}) |\text{vac}\rangle .$$ The Fourier domain wave function of the single photon state is then given by $$F(\mathbf{k}) = \langle\mathbf{k}|\psi\rangle .$$ The coordinate space wave function is obtained from the Fourier domain wave function $$\psi(\mathbf{x},t) = \int F(\mathbf{k}) \exp(i\omega t - i\mathbf{k}\cdot \mathbf{x}) \frac{d^2k}{(2\pi)^3\omega} .$$

As you can see, everything is defined in a Lorentz covariant manner. So, when a Lorentz transformation is applied to this wave function, an equivalent Lorentz transformation is applied to the Fourier domain variables, so that the expressions remain covariant. Under these circumstances, the wave function that one would obtain as the result of a Lorentz boost would be contracted in the same way as one would find for a classical field.

These arguments can also be extended to cases where all the particles in the state have the same spectral function $$F(\mathbf{k})$$. So that the coherent state that describes laser light, for example, would also behave is this way as we would expect.

• Just to be clear that just means that in my equation above, I just need to replace in my exponential, $|x|^2=x^2+y^2+z^2$ by $\gamma (x-vt)^2+y^2+z^2$? (for Lorentz boost in x-axis) Oct 15, 2021 at 15:52
• Yes, that would be valid for a boost in the $x$-direction. There would be a modification of $t$ as well. Oct 16, 2021 at 3:12
• I think your $\delta$ argument needs a $1$ instead of $2$ by one of the $\mathbf{\mathrm k}$s. Oct 20, 2021 at 12:39
• Yes, thanks @Ruslan. Fixed it. Oct 21, 2021 at 10:52

If the wavefunction is physically present in spacetime then it too should be affected by Lorentz transformations. However, it's not, it supervenes upon it. Nevertheless, it will still be affected by a Lorentz boost. For example, if we took the double slit experiment in the lab frame and boosted this in the direction parallel to the screen, we see that for everything to remain physically consistent, that is, due to the principle of relativity, the wave function expressed in terms of space will contract since the distance betweens the peaks on the image on the screen will also contract. Moreover, since bodies contract in the direction of motion and bodies are made up of atoms we see that the electron orbits must also contract in the direction of motion, and as these orbits are given by a quantum wave function, we again see that the wave function suffers Lorentz contraction.

More generally, the issue about how quantum fields are affected by curved spacetime is an open question, it's a question of quantum gravity. What we can do is semi-classical approximations on static curved backgrounds. It's through this that Hawking derived his entropy formula for black holes.

Moreover, the Davies-Unruh effect shows that an accelerating particle will see a thermal bath and is conceptually equicalent to Hawkings derivation. Essentially, the structure of the vacuum changes.

• Why are you discussing quantum gravity, blackholes and curved spacetime? I was just considering special relativity. Oct 14, 2021 at 19:27
• @Mauricio: I was discussing situations where the phenomena that you are asking about might occur. I am not discussing quantum gravity per se nor curved spacetime. You'll notice that the first sentence discusses the situation with regards to special relativity. Oct 14, 2021 at 20:13
• @Mauricio: I've added a few extra lines to show the connection. The point is, that the wave function does change but not in the naive way that you were expecting. Oct 14, 2021 at 20:25
• You seem to be discussing accelerated frames, I was just referring to simple Lorentz boosts Oct 15, 2021 at 7:05
• @Mauricio: I've added a few sentences to show how the wave function will be affected - satisfied? Oct 15, 2021 at 11:21