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I am having trouble deriving the general Lorentz boost for a spinor with rapidity $\rho $ using the hyperbolic functions. I know that the matrix: $$\exp [-\rho/2 \mathbf{n}\cdot \mathbf{\sigma }]=\cosh (\rho /2) I -\sinh (\rho /2) \mathbf n \cdot \mathbf \sigma. $$

What I have so far is that the above equality holds by reordering the Taylor series. I have also checked the action on the individual components $(x,y,z)$, for example $\Lambda =\exp (-\rho /2 \sigma _z)$. I have $$X=\left[\begin{matrix}t+z & x-iy\\ x+iy & t-z\end{matrix}\right]=tI+x\sigma _x+y\sigma _y+z\sigma _z$$ is the representation for a 4-vector $\mathbf x=(t,x,y,z)$. Then, the action of multiplication by $X'=\Lambda .X.\Lambda ^\dagger $ is the same as performing that hyperbolic rotation along the $z$-axis.

I have a general rotation-matrix by just multiplying the three coordinate rotations, but I cannot figure out how prove that this product equivalent to action $\Lambda .X.\Lambda ^\dagger $, where $\Lambda =\exp [-\rho/2 \mathbf{n}\cdot \mathbf{\sigma }]$

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  • $\begingroup$ So, are you asking how to convert your calculations for $\exp(-\rho\,\sigma_z/2)$ to corresponding calculations for a boost in a general direction? It's not altogether clear what you're asking, because you seem to understand the operations involved (i.e. that the operators act on the spinors by the Adjoint representation of $\mathrm{SL}(2,\,\mathbb{C})$) $\endgroup$ Commented May 31, 2017 at 3:57
  • $\begingroup$ I wish to derive the formula explicitly, for $/mathbf n$, right now I have only successfully derived it for the individual axes x,y,z. $\endgroup$ Commented May 31, 2017 at 4:06

2 Answers 2

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Any vector in $\mathbb{R}^4$ can be represented by a $2\times2$ hermitian matrix and vice versa. So, there exists a bijection (one-to-one and onto correspondence) between $\mathbb{R}^4$ and the space of $2\times2$ hermitian matrices, let it be $\mathbb{H}$ : \begin{equation} \mathbf{x}=(x_0,x_1,x_2,x_3)=(x_0,\boldsymbol{x})\in \mathbb{R}^4\;\boldsymbol{\longleftrightarrow} \; \mathrm X= \begin{bmatrix} x_0\boldsymbol{+}x_3 & x_1 \boldsymbol{-}ix_2 \\ x_1 \boldsymbol{+}ix_2 & x_0\boldsymbol{-}x_3 \end{bmatrix} \in \mathbb{H} \tag{01}\label{01} \end{equation} From the usual basis of $\mathbb{R}^4$ \begin{equation} \mathbf{e}_{0}=\left(1,0,0,0\right),\quad \mathbf{e}_{1}=\left(0,1,0,0\right),\quad \mathbf{e}_{2}=\left(0,0,1,0\right),\quad \mathbf{e}_{3}=\left(0,0,0,1\right) \tag{02}\label{02} \end{equation} we construct a basis for $\mathbb{H}$ \begin{align} \mathbf{e}_0 &= \left(1,0,0,0\right)\qquad \boldsymbol{\longleftrightarrow} \qquad \sigma_0= \begin{bmatrix} 1 & \!\!\hphantom{\boldsymbol{-}}0 \vphantom{\tfrac{a}{b}}\\ 0 & \!\!\hphantom{\boldsymbol{-}}1\vphantom{\tfrac{a}{b}} \end{bmatrix}=I \tag{03.0}\label{03.0}\\ \mathbf{e}_1 &=\left(0,1,0,0\right)\qquad \boldsymbol{\longleftrightarrow} \qquad \sigma_1= \begin{bmatrix} 0 & \!\!\hphantom{\boldsymbol{-}}1 \vphantom{\tfrac{a}{b}}\\ 1 & \!\!\hphantom{\boldsymbol{-}}0\vphantom{\tfrac{a}{b}} \end{bmatrix} \tag{03.1}\label{03.1}\\ \mathbf{e}_2 &=\left(0,0,1,0\right)\qquad \boldsymbol{\longleftrightarrow} \qquad \sigma_2= \begin{bmatrix} 0 & \!\!\boldsymbol{-} i \vphantom{\tfrac{a}{b}}\\ i & \!\!\hphantom{\boldsymbol{-}} 0\vphantom{\tfrac{a}{b}} \end{bmatrix} \tag{03.2}\label{03.2}\\ \mathbf{e}_3 &= \left(0,0,0,1\right)\qquad \boldsymbol{\longleftrightarrow} \qquad \sigma_3= \begin{bmatrix} 1 & \!\!\hphantom{\boldsymbol{-}} 0 \vphantom{\frac{a}{b}}\\ 0 & \!\!\boldsymbol{-} 1\vphantom{\frac{a}{b}} \end{bmatrix} \tag{03.3}\label{03.3} \end{align} where $\:\boldsymbol{\sigma}\equiv(\sigma_{1},\sigma_{2},\sigma_{3})\:$ the Pauli matrices.

Suppose now that the vector $\:\mathbf{x}=(x_0,x_1,x_2,x_3)=(x_0,\boldsymbol{x})\:$ is Lorentz transformed along an axis with unit vector $\:\boldsymbol{n}=(n_1,n_2,n_3)$ with rapidity $\zeta$ \begin{align} x'_0 & \boldsymbol{=}x_0\cosh\!\zeta \boldsymbol{-}\sinh\!\zeta(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{x}) \tag{04.1}\label{04.1}\\ \boldsymbol{x}'&\boldsymbol{=}\boldsymbol{-}x_0\sinh\!\zeta\;\boldsymbol{n}\boldsymbol{+} \boldsymbol{x}\boldsymbol{+}(\cosh\!\zeta\boldsymbol{-}1)\;(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{x})\;\boldsymbol{n} \tag{04.2}\label{04.2} \end{align}

Now, let to the vectors $\:\mathbf{x},\mathbf{x}^{\prime}\:$ correspond the matrices \begin{align} \mathrm X & \boldsymbol{\equiv} \sum\limits_{k=0}^{k=3}x_k\sigma_k \boldsymbol{=} x_0 I\boldsymbol{+} \boldsymbol{x}\boldsymbol{\cdot} \boldsymbol{\sigma} \boldsymbol{=} \begin{bmatrix} x_0\boldsymbol{+}x_3 & x_1 \boldsymbol{-}ix_2 \\ x_1 \boldsymbol{+}ix_2 & x_0\boldsymbol{-}x_3 \end{bmatrix} \tag{05.1}\label{05.1}\\ \mathrm X' & \boldsymbol{\equiv} \sum\limits_{k=0}^{k=3}x'_k\sigma_k \boldsymbol{=} x'_0 I\boldsymbol{+} \boldsymbol{x}'\boldsymbol{\cdot} \boldsymbol{\sigma} \boldsymbol{=} \begin{bmatrix} x'_0\boldsymbol{+}x'_3 & x'_1 \boldsymbol{-}ix'_2 \\ x'_1 \boldsymbol{+}ix'_2 & x'_0\boldsymbol{-}x'_3 \end{bmatrix} \tag{05.2}\label{05.2} \end{align} From \eqref{04.1} and \eqref{04.2} we have \begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathrm X' \boldsymbol{\!=\!}I x_0\cosh\!\zeta \boldsymbol{\!-\!}I(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{x})\sinh\!\zeta\boldsymbol{\!-\!} x_0\sinh\!\zeta(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\boldsymbol{+} (\boldsymbol{x}\boldsymbol{\cdot}\boldsymbol{\sigma})\boldsymbol{+}(\cosh\!\zeta\boldsymbol{-}1)(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{x})(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{\sigma}) \tag{06}\label{06} \end{equation}
For convenience we enumerate the items of \eqref{06} \begin{equation} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathrm X' \boldsymbol{\!=\!} \underbrace{I x_0\cosh\!\zeta}_{\boxed{_1}} \boldsymbol{\!-\!}\underbrace{I(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{x})\sinh\!\zeta}_{\boxed{_2}} \boldsymbol{\!-\!} \underbrace{x_0\sinh\!\zeta(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{\sigma})}_{\boxed{_3}} \boldsymbol{\!+\!} \underbrace{(\boldsymbol{x}\boldsymbol{\cdot}\boldsymbol{\sigma})}_{\boxed{_4}} \boldsymbol{\!+\!}\underbrace{(\cosh\!\zeta\boldsymbol{-}1)(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{x})(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{\sigma})}_{\boxed{_5}} \tag{06}\label{repeat06} \end{equation}
Using properties \begin{align} \cosh\!\zeta & \boldsymbol{=} \cosh^2\!\dfrac{\zeta}{2}+ \sinh^2\!\dfrac{\zeta}{2}\boldsymbol{=} 1\boldsymbol{+}2\sinh^2\!\dfrac{\zeta}{2} \tag{07.1}\label{07.1}\\ 1 & \boldsymbol{=} \cosh^2\!\dfrac{\zeta}{2}-\sinh^2\!\dfrac{\zeta}{2} \tag{07.2}\label{07.2}\\ \sinh\!\zeta & \boldsymbol{=} 2\sinh\!\dfrac{\zeta}{2}\cosh\!\dfrac{\zeta}{2} \tag{07.3}\label{07.3} \end{align} \begin{align} (\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{x})\,I &=\frac{(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{\sigma})(\boldsymbol{x}\boldsymbol{\cdot}\boldsymbol{\sigma})\boldsymbol{+}(\boldsymbol{x}\boldsymbol{\cdot}\boldsymbol{\sigma})(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{\sigma})}{2} \nonumber\\ &=\frac{\mathrm N(\mathrm X-x_0 I)\boldsymbol{+}(\mathrm X-x_0 I)\mathrm N}{2}=\frac{\mathrm N \mathrm X\boldsymbol{+}\mathrm X \mathrm N}{2}-x_0\mathrm N \tag{08}\label{08} \end{align} where \begin{equation} \mathrm N \boldsymbol{\equiv} \boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{\sigma}\boldsymbol{=} \begin{bmatrix} n_3 & n_1\boldsymbol{-}i\,n_2\\ n_1\boldsymbol{+}i\,n_2&\boldsymbol{-} n_3 \end{bmatrix}\,,\qquad \mathrm N^2\boldsymbol{=} I \tag{09}\label{09} \end{equation} we have \begin{align} \boxed{1} & \boldsymbol{=} I x_0\cosh\!\zeta \boldsymbol{=}I x_0\left(\cosh^2\!\dfrac{\zeta}{2}+ \sinh^2\!\dfrac{\zeta}{2}\right) \tag{10.1}\label{10.1}\\ \boxed{2} & \boldsymbol{=} I(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{x})\sinh\!\zeta\boldsymbol{=}\left(\mathrm N\mathrm X\boldsymbol{+}\mathrm X\mathrm N-2x_0\rm N\right)\sinh\!\dfrac{\zeta}{2}\cosh\!\dfrac{\zeta}{2} \tag{10.2}\label{10.2}\\ \boxed{3} & \boldsymbol{=} x_0\sinh\!\zeta(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\boldsymbol{=}2x_0 \rm N\sinh\!\dfrac{\zeta}{2}\cosh\!\dfrac{\zeta}{2} \tag{10.3}\label{10.3}\\ \boxed{4} & \boldsymbol{=}(\boldsymbol{x}\boldsymbol{\cdot}\boldsymbol{\sigma})\boldsymbol{=}(\boldsymbol{x}\boldsymbol{\cdot}\boldsymbol{\sigma})\left(\cosh^2\!\dfrac{\zeta}{2}- \sinh^2\!\dfrac{\zeta}{2}\right) \tag{10.4}\label{10.4}\\ \boxed{5} & \boldsymbol{=} (\cosh\!\zeta\boldsymbol{-}1)(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{x})(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\boldsymbol{=}\left(\mathrm{NXN} \boldsymbol{+}\mathrm X-2x_0 I\right)\sinh^2\!\dfrac{\zeta}{2} \tag{10.5}\label{10.5} \end{align} so \begin{align} \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\mathrm X'& \!\!\!\!\!\!\!\boldsymbol{=} \boxed{1}\boldsymbol{-}\boxed{2}\boldsymbol{-}\boxed{3}\boldsymbol{+}\boxed{4}\boldsymbol{+}\boxed{5}\boldsymbol{=}\bigl[Ix_0\boldsymbol{+}(\boldsymbol{x}\boldsymbol{\cdot}\boldsymbol{\sigma})\bigr]\cosh^2\!\dfrac{\zeta}{2}\boldsymbol{+}\bigl[Ix_0\boldsymbol{-}(\boldsymbol{x}\boldsymbol{\cdot}\boldsymbol{\sigma})\bigr]\sinh^2\!\dfrac{\zeta}{2} \nonumber\\ & \boldsymbol{-} \left(\mathrm N\mathrm X\boldsymbol{+}\mathrm X\mathrm N-2x_0\rm N\right)\sinh\!\dfrac{\zeta}{2}\cosh\!\dfrac{\zeta}{2}\boldsymbol{-}2x_0 \rm N\sinh\!\dfrac{\zeta}{2}\cosh\!\dfrac{\zeta}{2} \nonumber\\ & \boldsymbol{+}\left(\mathrm{NXN} \boldsymbol{+}\mathrm X-2x_0 I\right)\sinh^2\!\dfrac{\zeta}{2}\nonumber\\ & \boldsymbol{=}\overbrace{\bigl[Ix_0\boldsymbol{+}(\boldsymbol{x}\boldsymbol{\cdot}\boldsymbol{\sigma})\bigr]}^{\mathrm X}\cosh^2\!\dfrac{\zeta}{2}\boldsymbol{+}\overbrace{\bigl[\mathrm X\boldsymbol{-} Ix_0\boldsymbol{-}(\boldsymbol{x}\boldsymbol{\cdot}\boldsymbol{\sigma})\bigr]}^{0}\sinh^2\!\dfrac{\zeta}{2} \nonumber\\ & \boldsymbol{-}\left(\mathrm N\mathrm X\boldsymbol{+}\mathrm X\mathrm N\right)\sinh\!\dfrac{\zeta}{2}\cosh\!\dfrac{\zeta}{2}\boldsymbol{+}\mathrm{NXN}\sinh^2\!\dfrac{\zeta}{2} \tag{11}\label{11} \end{align} or \begin{align} \mathrm X'&\boldsymbol{=}\mathrm X \cosh^2\!\dfrac{\zeta}{2}\boldsymbol{-}\left(\mathrm N\mathrm X\boldsymbol{+}\mathrm X\mathrm N\right)\sinh\!\dfrac{\zeta}{2}\cosh\!\dfrac{\zeta}{2}\boldsymbol{+}\mathrm{NXN}\sinh^2\!\dfrac{\zeta}{2}\nonumber\\ & \boldsymbol{=} \left(I\cosh\frac{\zeta}{2}\boldsymbol{-}\mathrm N\sinh\frac{\zeta}{2} \right)\mathrm X\cosh\!\dfrac{\zeta}{2}\boldsymbol{-}\left(I\cosh\frac{\zeta}{2}\boldsymbol{-}\mathrm N\sinh\frac{\zeta}{2}\right)\mathrm{XN}\sinh\frac{\zeta}{2} \tag{12}\label{12} \end{align} that is \begin{equation} \mathrm X'\boldsymbol{=}\left(I\cosh\frac{\zeta}{2}\boldsymbol{-}\mathrm N\sinh\frac{\zeta}{2} \right)\mathrm X\left(I\cosh\frac{\zeta}{2}\boldsymbol{-}\mathrm N\sinh\frac{\zeta}{2} \right) \tag{13}\label{13} \end{equation} Equation \eqref{13} is written in compact form \begin{equation} \mathrm X'=W\;\mathrm X\;W \tag{14}\label{14} \end{equation}
with \begin{equation} W\equiv I\cosh\frac{\zeta}{2}\boldsymbol{-}\mathrm N\sinh\frac{\zeta}{2}=I\cosh\frac{\zeta}{2}\boldsymbol{-}(\boldsymbol{n}\boldsymbol{\cdot}\boldsymbol{\sigma})\sinh\frac{\zeta}{2} \tag{15}\label{15} \end{equation}

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  • $\begingroup$ In my opinion, your answe would be improved if you boxed your answer, and wrote the question preceding your answer. For example, "In summary, a general Lorentz boost for a spinor in Hyperbolic form with Pauli matrices is given by....." $\endgroup$ Commented May 28 at 4:01
  • $\begingroup$ It would be very helpful to me if you were to let me know which of these equations represents your actual answer to the question. Thank you. $\endgroup$ Commented Jun 3 at 15:38
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To convert your calculations for $\exp(-\rho\,\sigma_z/2)$ to corresponding calculations for a boost in a general direction, first recall that in $\mathrm{SO}(1,\,3)$ a boost in a general direction has the matrix:

$$\Lambda = R\,\Lambda\,R^{-1}\tag{1}$$

where $R$ is a rotation that rotates a frame whose $z$-axis corresponds with the boost direction to the frame you wish to compute in. Since the Adjoint representation, which maps the double cover $\mathrm{SL}(2,\,\mathbb{C})$ to $\mathrm{SO}(1,\,3)$, is a homomorphism, you can use the same equation as (1) to find the preimage (or one of the two preimages) in $\mathrm{SL}(2,\,\mathbb{C})$ that maps to your general boost in $\mathrm{SO}(1,\,3)$.

So now you need to know the what a preimage of a rotation operator looks in $\mathrm{SU}(2)\subset\mathrm{SL}(2,\,\mathbb{C})$; unsurprisingly, a rotation in $\mathrm{SU}(2)$ is given by:

$$R(\theta, \hat{n}) = \exp\left(-i\,\frac{\theta}{2}\,\hat{n}\cdot\sigma\right) = \cos\left(\frac{\theta}{2}\right)-\sin\left(\frac{\theta}{2}\right)\,\hat{n}\cdot\sigma\tag{2}$$

So that your calculation yields:

$$\Lambda = \left(\cos\left(\frac{\theta}{2}\right)-\sin\left(\frac{\theta}{2}\right)\,\hat{n}\cdot\sigma\right)\,\left(\cosh\left(\frac{\rho}{2}\right)-\sinh\left(\frac{\rho}{2}\right)\,\hat{n}\cdot\sigma_z\right)\,\left(\cos\left(\frac{\theta}{2}\right)+\sin\left(\frac{\theta}{2}\right)\,\hat{n}\cdot\sigma\right)\tag{3}$$

which I'll leave you to multiply out.

But take heed that the calculation's result naturally wipes out the rotational degree of freedom one subtly introduces - any frame that is a rotation about the $z$ axis of any other frame wherein the boost is along the $z$ axis is itself a frame wherein the boost is along the $z$-axis. This is because if we replace the rotation in (2) by any rotation of the form $R(\theta, \hat{n})\,\exp(-\phi\,\sigma_z/2)$ for $\phi\in\mathbb{R}$, the $\exp(-\phi\,\sigma_z/2)$ and $\exp(+\phi\,\sigma_z/2)$ that are introduced into (3) both commute with the central $z$-direction boost in (3) and the calculation's result is unchanged. Thus you can apply these ideas with any rotation that aligns the $z$ axis as you need, and the calculation's result will be independent of this choice.

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  • $\begingroup$ It makes sense to rotate the coordinate axes so that $\textbf n$ points along $\hat z$, then apply the Lorentz transformation $\cosh \rho /2 - \sinh \rho /2 \sigma _z$, then rotate the axes back to their original orientation. I still need to work out the details as to why this $\Lambda $ is equal to $\cosh \rho /2-\sinh \rho /2 \textbf n . \sigma$ $\endgroup$ Commented May 31, 2017 at 5:22
  • $\begingroup$ @AndrewDynneson Apply the two outer rotation matrices in (3) to the two terms of the center boost separately. So you end up with $\cosh (\rho /2) I -\sinh (\rho /2) \mathbf{\tilde{n}} \cdot \mathbf{\sigma}$ where $\mathbf{\tilde{n}}\cdot\mathbf{\sigma}$ s the image of $\hat{n}\cdot\sigma$ under conjugation by the rotation in (3) $\endgroup$ Commented May 31, 2017 at 13:23

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