# Why is the Lorentz boost a non-unitary operator in the single electron theory?

My main reference on the subject is Sakurai's "Advanced Quantum Mechanics".

Consider a single electron described by a bispinor that obeys the Dirac equation. The operator corresponding to a Lorentz boost of rapidity $$\chi$$ in the direction of the $$k$$ axis is $$S_{Lor} = \cosh{\frac{\chi}{2}} - \alpha_k \sinh{\frac{\chi}{2}}$$ which is not unitary ($$S_{Lor}^{\dagger} \neq S_{Lor}^{-1}$$).

Correct me if I'm wrong, but this implies that two observers, connected by a Lorentz boost, will compute different expectation values for the same observables. How is this not a problem in a theory which is suppposed to be in agreement with special relativity?

$$S_{Lor}$$ should not be unitary if $$\overline\psi\psi$$ is to transform like the fourth component of a four-vector under Lorentz transformations.

I fail to understand what he means by this. Could someone elaborate on Sakurai's statement and explain to me why is it ok for a Lorentz boost to be a non-unitary operator?

Edit: the complete transformation rule for a Lorentz transformation $$\Lambda$$ (such that $$x' = \Lambda x$$) is $$\psi'(x') = S_{Lor}\psi(\Lambda^{-1}x')$$ that is: apart from the matrix $$S_{Lor}$$, which mixes the components of the Dirac spinor, you also have to transform the coordinates.

One of the comments mentions that the Hilbert space operator which says how state-vectors transform unders boost should be unitary. Who is this operator, how do I construct it? Surely it must be related to the above equation, as I believe that is the "complete rule" of how a Dirac spinor (the wavefunction of an electron) transforms under boosts.

• Why do you expect observables to be the same under Lorentz boosts? Energy is not a Lorentz invariant quantity, for example. Commented Apr 29, 2021 at 20:47
• But isn't the probability distribution something that all observers should agree? For example: an eigenstate of spin may be boosted into a state which is no longer an eigenstate. In an experiment to measure this component of spin, the first observer will obtain +1/2 100% of the time. The boosted observer will measure -1/2 some of the time. Doesn't this violate relativity somehow? Commented Apr 29, 2021 at 22:43
• I deleted one of my comments which I don't think was right. I think @ChiralAnomaly 's explanation is correct. Commented Apr 30, 2021 at 0:15

It's important to distinguish between two things:

1. The matrix that says how the components of a spinor get mixed under a Lorentz boost. For a boost in the $$k$$-direction with rapidity $$\theta$$, this matrix can constructed from Dirac matrices as $$M\equiv \exp\left(\gamma^0\gamma^k\frac{\theta}{2}\right) =\cosh\left(\frac{\theta}{2}\right) +\gamma^0\gamma^k\sinh\left(\frac{\theta}{2}\right).$$ This matrix is not unitary, and this is presumably what the author is referring to.

2. The operator on the Hilbert space that says how state-vectors transform under a Lorentz boost. This operator is unitary.

## Constructing the Hilbert space

The Dirac spinor field $$\psi(x)$$ is an operator on the Hilbert space, not a wavefunction. Before we had quantum field theory, people allegedly tried to interpret $$\psi(x)$$ as a wavefunction. That doesn't work. What does work is treating $$\psi(x)$$ as an operator on the Hilbert space, like this:

Take the field operators $$\psi(x)$$ to satisfy the canonical anticommutation relations. Write the field operator $$\psi(x)$$ (the general solution of the Dirac equation) as a sum of its positive- and negative-frequency parts $$\psi_\pm(x)$$. If $$|0\rangle$$ is the state annihilated by the positive-frequency parts $$\psi_+(x)$$ and $$(\psi_-(x))^\dagger$$, then any other state can be obtained from $$|0\rangle$$ by applying various combinations of the negative-frequency parts $$\psi_-(x)$$ and $$(\psi_+(x))^\dagger$$. In particular, $$\psi_-(x)|0\rangle$$ and $$(\psi_+(x))^\dagger|0\rangle$$ are single-particle states. We usually call one of them an antiparticle.

## Constructing the unitary boost operator

For any given Lorentz boost, there is a unitary operator $$U$$ on this Hilbert space that implements that transformation. The unitary transformation $$\psi(x)\to U^{-1}\psi(x)U$$ mixes the spinor components of $$\psi(x)$$ just like the matrix $$M$$ does. However, the unitary transformation does more than just mixing the spinor components of $$\psi(x)$$. It also transforms the $$x$$ in $$\psi(x)$$, as shown in the question.

Constructing the unitary operator $$U$$ explicitly is too involved for this brief post, so I'll just outline the idea: start with the stress-energy tensor $$T^{ab}$$, which you can get either from Noether's theorem or by varying with respect to the background spacetime metric. (The former is easier, but the latter is often more powerful.) Then the generator of rotations in the $$j$$-$$k$$ plane is $$J^{jk}\equiv \int d^3x\ \big(T^{0j}x^k-T^{0k}x^j\big),$$ and the generator of boosts in the $$0$$-$$j$$ plane is $$K^j\equiv \int d^3x\ \big(T^{00}x^j - T^{0j}x^0\big)$$ where $$x^0$$ is the time coordinate. The unitary operator that implements a boost in the $$j$$-direction with rapidity $$\theta$$ is $$U=\exp(iK^j\theta)$$.