# Lorentz boosts as unitary transformation

In the article "Identification of the velocity operator for an irreducible unitary representation of the Poincaré group", there is formula to implement a Lorentz transformation of the velocity as a unitary transformation.

https://aip.scitation.org/doi/10.1063/1.523342

It goes as follows: consider a Lorentz transformation upon the components of the velocity operator $$\mathbf{V}\rightarrow\mathbf{V}'$$. The velocity operator transform just as the velocity components in classical relativistic physics. The unitary transformation that gives the Lorentz transformation of $$\mathbf{V}$$ by a boost in the $$z$$ direction with a factor $$\beta$$ is given by

$$\mathbf{V}'=\exp\left(iK_{z}\tanh^{-1}\beta\right)\mathbf{V}\,\exp\left(-iK_{z}\tanh^{-1}\beta\right)$$

What I don't understand is why the velocity appears in a factor $$\tanh^{-1}\beta$$. It seem quite different from the other space-time transformation of the Lorentz group (or the Galilei group).

• Have you tried computing those matrix exponentials? (Use the Taylor series.) They should end up looking like familiar Lorentz boosts with nonzero matrix elements like $\gamma$ and $\gamma\beta$. This is a very instructive exercise to do. – G. Smith Oct 27 '20 at 21:20
• So far I've been able to show the equality of the first terms of the series expansion, but I yet have to find have to find how to prove the equality of the whole series. – AndresB Oct 28 '20 at 0:19
• The paper is behind a paywall. What does it have for the boost generator $K_z$ as an explicit matrix? I’m thinking of writing an answer to show you how the exponential is just an ordinary Lorentz transformation, and I want to use the same sign conventions as in the paper. – G. Smith Oct 28 '20 at 3:01
• – SRS Oct 28 '20 at 3:46
• It turns out that they're not very similar. I've posted two answers working out both of them. – G. Smith Oct 30 '20 at 20:55

## 4 Answers

The velocity transformation you presented,

$$\mathbf{V}'=\exp{(iK_{z}\tanh^{-1}\beta)}\,\mathbf{V}\,\exp{(-iK_{z}\tanh^{-1}\beta)}\tag1$$

is equivalent to the usual Lorentz transformation of velocity, which for a boost by $$\beta\hat z$$ is

$${V_x}'=\frac{\sqrt{1-\beta^2}\,V_x}{1-\beta V_z}\tag{2a}$$

$${V_y}'=\frac{\sqrt{1-\beta^2}\,V_y}{1-\beta V_z}\tag{2b}$$

$${V_z}'=\frac{V_z-\beta}{1-\beta V_z}\tag{2c},$$

but it's a messy calculation. I believe that the equivalence can only be shown term-by-term in an expansion in powers of $$\beta$$.

In comments, you explained that (1) is to be interpreted in terms of various operators which are possibly non-commuting, namely the relativistic velocity operator

$$\mathbf{V}\equiv\frac{\mathbf{P}}{H},\tag{3a}$$

the Lorentz boost generator

$$\mathbf{K}\equiv\frac12(H\mathbf{X}+\mathbf{X}H),\tag{3b}$$

and the relativistic energy operator

$$H\equiv(\mathbf{P}^2+m^2)^{1/2}.\tag{3c}$$

Here $$\mathbf X$$ is the position operator and $$\mathbf P$$ the relativistic momentum operator with the canonical commutation relations

$$[X_j,P_k]=i\,\delta_{jk}.\tag4$$

To show the equivalence of (1) and (2), use the identity

$$e^ABe^{-A}=B+[A,B]+\frac{1}{2!}[A,[A,B]]+\frac{1}{3!}[A,[A,[A,B]]]+\dots\tag5$$

which follows from the Taylor series for the function $$f(t)=e^{tA}Be^{-tA}$$. Equation (1) becomes

\begin{align}{V_i}'&=V_i+(i\tanh^{-1}\beta)[K_z,V_i]\\&+\frac{(i\tanh^{-1}\beta)^2}{2!}[K_z,[K_z,V_i]]\\&+\frac{(i\tanh^{-1}\beta)^3}{3!}[K_z,[K_z,[K_z,V_i]]]+\dots\,.\end{align}\tag6

So the next step is to calculate the commutator $$[K_z,V_i]$$ and further nestings. We have

$$[K_z,V_i]=\left[\frac12(Hz+zH),V_i\right]=\frac12\left(H[z,V_i]+[z,V_i]H\right)=i(\delta_{iz}-V_iV_z).\tag{7a}$$

To get this it was necessary to calculate

$$[z,V_i]=[z,H^{-1}P_i]=[z,H^{-1}]P_i+H^{-1}[z,P_i]=i\frac{\delta_{iz}-V_iV_z}{H}\tag{7b}$$

and this required calculating

$$[z,H^{-1}]=[z,(\mathbf{P}^2+m^2)^{-1/2}]=[z,P_z]\frac{\partial}{\partial P_z}(\mathbf{P}^2+m^2)^{-1/2}=-i\frac{P_z}{H^3}.\tag{7c}$$

That last commutator required the identity

$$[A,f(B)]=[A,B]\frac{\partial f}{\partial B}\tag8$$

which holds when $$[A,B]$$ commutes with both $$A$$ and $$B$$.

From (7a) we have

$$[K_z,V_x]=-iV_xV_z\tag{9a}$$

and

$$[K_z,V_z]=i(1-V_z^2).\tag{9b}$$

We're boosting along $$z$$, so the transverse $$y$$ direction works the same as the transverse $$x$$ direction. We'll just do $$x$$.

From (9a) and (9b) it is possible to grind out the nested commutators. I used Mathematica to do it up to ten nestings. I'll show four of them.

For $$V_x$$ the nestings are

\begin{align} [K_z,V_x]&=-iV_xV_z\tag{10a}\\ [K_z,[K_z,V_x]]&=V_x \left(1-2 V_z^2\right)\tag{10b}\\ [K_z,[K_z,[K_z,V_x]]]&=i V_x V_z \left(-5+6 V_z^2\right)\tag{10c}\\ [K_z,[K_z,[K_z,[K_z,V_x]]]]&=V_x \left(5-28 V_z^2+ 24 V_z^4\right)\tag{10d} \end{align}

and for $$V_z$$ the nestings are

\begin{align} [K_z,V_z]&=i(1-V_z^2)\tag{11a}\\ [K_z,[K_z,V_x]]&=2 V_z \left(1-V_z^2\right)\tag{11b}\\ [K_z,[K_z,[K_z,V_z]]]&=2 i \left(1-4 V_z^2+3 V_z^4\right)\tag{11c}\\ [K_z,[K_z,[K_z,[K_z,V_z]]]]&=8 V_z \left(2-5 V_z^2+3 V_z^4\right).\tag{11d} \end{align}

As you can see, there is no obvious pattern. It’s basically a mess.

You can then insert these nested commutators into (6) and expand $$\tanh^{-1}\beta$$ in powers of $$\beta$$. The result through order $$\beta^4$$ is

\begin{align} {V_x}'&=\beta V_x V_z +\beta ^2 \left(V_x V_z^2-\frac{V_x}{2}\right) +\beta ^3 \left(V_x V_z^3-\frac{V_x V_z}{2}\right)\\ &+\beta ^4 \left(V_x V_z^4-\frac{1}{2} V_x V_z^2-\frac{V_x}{8}\right) +O\left(\beta ^5\right)\tag{12a} \end{align}

\begin{align} {V_z}'&=V_z +\beta \left(V_z^2-1\right) +\beta ^2 \left(V_z^3-V_z\right) +\beta ^3 \left(V_z^4-V_z^2\right)\\ &+\beta ^4 \left(V_z^5-V_z^3\right) +O\left(\beta ^5\right).\tag{12b} \end{align}

(12a) is the Taylor expansion of (2a), and (12b) is the Taylor expansion of (2c). (I verified it through order $$\beta^{10}$$.) This shows the equivalence of the two ways, (1) and (2), of doing a relativistic velocity transformation.

I would be impressed if anyone can prove the equivalence without doing it order-by-order in powers of $$\beta$$.

This does not motivate why $$\tanh^{-1}\beta$$ appears in (1). But if it didn't appear in that way, (1) would not be equivalent to (2). I will write a separate answer explaining why you might expect $$\tanh^{-1}\beta$$ to appear as the factor multiplying the boost generator in the exponential.

• Amazing answer. I already had checked the expansion up to order $\beta^{2}$, but after that the terms become quite annoying. I guess the equality between terms of the same power could be worked out using Mathematical induction, but It does not seem so a simple task. In any case, I consider the question completely solved. – AndresB Oct 31 '20 at 0:33

If we write $$\beta\equiv v/c= \tanh s$$, then $$s$$ is the rapidity. If we combine two parallel boosts with rapidity $$s_1$$ and $$s_2$$ then the combined boost has rapidity $$s_1+s_2$$ so $$s=\tanh^{-1} \beta$$ is the natural variable to describe boosts.

To motivate why $$\tanh^{-1}\beta$$ appears in your formula as the factor multiplying the boost generator in the exponential, consider a much simpler transformation, namely the Lorentz transformation of coordinates. For a boost $$\beta\hat z$$, this is

$$\begin{pmatrix}t'\\x'\\y'\\z' \end{pmatrix}= \begin{pmatrix}\gamma & 0 & 0 & -\gamma\beta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ -\gamma\beta & 0 & 0 & \gamma \end{pmatrix} \begin{pmatrix}t\\x\\y\\z \end{pmatrix}\tag1$$

where $$\gamma\equiv(1-\beta^2)^{-1/2}$$.

It is easy to show that this boost matrix, which I'll call $$B(\beta,\hat z)$$, is just

$$B(\beta,\hat z)=e^{-K_z \tanh^{-1}\beta}\tag2$$

where

$$K_z=\begin{pmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 \end{pmatrix}\tag3$$

is a the generator of boosts in the $$z$$-direction.

(How do you get this generator? By observing that an infinitesimal boost by velocity $$\epsilon \hat z$$ is $$B(\epsilon,\hat z)\approx I-\epsilon K_z$$.)

The matrix exponential is defined by its power series,

$$e^{-\zeta K_z}\equiv I-\zeta K_z+\frac{\zeta^2}{2!}K_z^2-\frac{\zeta^3}{3!}K_z^3+\frac{\zeta^4}{4!}K_z^4+\dots,\tag4$$

where I'll write $$\zeta\equiv\tanh^{-1}\beta$$ for convenience.

Now $$K_z$$ has the nice property that

$$K_z^3=K_z\tag5$$

and thus all higher powers of $$K_z$$ reduce to either $$K_z$$ or $$K_z^2$$. Thus (4) becomes

\begin{align} e^{-\zeta K_z}&=I-\left(\zeta+\frac{1}{3!}\zeta^3+\frac{1}{5!}\zeta^5+\dots\right)K_z\\ &+\left(\frac{1}{2!}\zeta^2+\frac{1}{4!}\zeta^4+\frac{1}{6!}\zeta^6+\dots\right)K_z^2\tag{6a}\\ \\ &=I-(\sinh\zeta)K_z+(\cosh\zeta-1)K_z^2\tag{6b}\\ \\ &=\begin{pmatrix} \cosh\zeta & 0 & 0 & -\sinh\zeta\\ 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ -\sinh\zeta & 0 & 0 & \cosh\zeta \end{pmatrix}.\tag{6c} \end{align}

Finally, $$\tanh\zeta=\beta,$$ implies that $$\cosh\zeta=\gamma$$ and $$\sinh\zeta=\gamma\beta$$, so the boost in (2) is equivalent to that in (1).

• These are certainly Lorentz's transformations into hyperbolic form. :-) – Sebastiano Oct 30 '20 at 22:28
• Yes, and the cosh and sinh arise from the exponential series separating into terms of odd and even power because the boost generator satisfies $K_z^3=K_z$. – G. Smith Oct 30 '20 at 22:44

To expand on mike stone's answer, a Lorentz boost is simply a rotation, and the traditional $$dx/dt$$ velocity is a slope.

In 2D Euclidean space, you could parametrize origin-preserving rotations by the slope $$m=dy/dx$$ of the line to which they rotate the line $$y=0$$. If you did that, then the angle of the rotation would be related to the slope by $$θ=\tan^{-1} m$$, and the "sum" of two slopes $$m$$ and $$m'$$, meaning the slope corresponding to the composition of the rotations given by $$m$$ and $$m'$$, would be $$\frac{m+m'}{1-mm'}$$. In spacetime the situation is exactly the same, except that because of the change of sign in the metric, you end up needing $$\tanh$$ instead of $$\tan$$, and the sign in the denominator of the slope "addition" formula is flipped.

Writing a Lorentz boost as $$\exp i K_z α$$ is closely related to writing a rotation in the complex plane as $$\exp i θ$$. The generalization of this formula (the Euler formula) to other continuous symmetries is the theory of Lie algebras and Lie groups.