Lorentz boosts as unitary transformation

Question

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).

Answer 1

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.

Answer 2

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.

Answer 3

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).

Answer 4

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.