Special Euclidean group

Problem: Show that the Lorentz transformation $W$ that keeps $k^{\mu}\equiv[1, 0,  0,  1]^T$ unchanged forms a two-dimensional Euclidean group SE(2).







Solution:

As shown in this post, the general solution of the Lorentz transformation $W$ that satisfies $W^{\mu}_{\,\,\,\,\nu}\,k^{\nu}=k^{\mu}$ can be parametrized in three parameters $\alpha,\beta,\theta$ as \begin{equation}W(\alpha, \beta, \theta)=\left[\begin{matrix} 1+(\alpha^2+\beta^2)/2  & -\alpha & -\beta & -(\alpha^2+\beta^2)/2  \\  -\alpha & 1 & 0 & \alpha \\ -\beta & 0 &  1 & \beta \\ (\alpha^2+\beta^2)/2 & -\alpha & -\beta & 1-(\alpha^2+\beta^2)/2 \end{matrix}\right]\left[\begin{matrix} 1 &  &  & \\  & \cos\theta & -\sin\theta & \\ & \sin\theta & \cos\theta  &   \\ & &  & 1\end{matrix}\right]\,.\tag{1}\end{equation}

To work out the Lie algebra, we expand $W(\alpha, \beta, \theta)$ in the first order of $\alpha, \beta, \theta$ as \begin{eqnarray}W(\alpha, \beta, \theta)&\approx&\left[\begin{matrix} 1  & -\alpha & -\beta & 0 \\  -\alpha & 1 & 0 & \alpha \\ -\beta & 0 &  1 & \beta \\  0 & -\alpha & -\beta & 1 \end{matrix}\right]\left[\begin{matrix} 1 &  &  & \\  & 1 & -\theta & \\ & \theta & 1  &   \\ & &  & 1\end{matrix}\right]\approx \left[\begin{matrix} 1  & -\alpha & -\beta & 0 \\  -\alpha & 1 & -\theta & \alpha \\ -\beta & \theta &  1 & \beta \\  0 & -\alpha & -\beta & 1 \end{matrix}\right]\\  \\&\equiv& I - i \alpha A -i\beta B - i\theta J\,,\end{eqnarray} where the generators are \begin{eqnarray}A\equiv\left[\begin{matrix} & -i & & \\ -i & & & i \\ & & & \\ & -i & & \end{matrix}\right]\,,\quad B\equiv\left[\begin{matrix} & & -i & \\ & & & \\ -i & & & i \\ & & -i & \end{matrix}\right]\,,\quad J\equiv\left[\begin{matrix} & & & \\ & & -i & \\ & i & & \\ & & & \end{matrix}\right] \,.\tag{2}\end{eqnarray} The generator $J$ represents a rotation of the x-y plane, which is obvious that keeps $[1, 0, 0, 1]^T$ unchanged. In contrast, the generators $A$ and $B$ are less obvious: $A$ represents a boost along x-direction plus a rotation of the x-z plane; $B$ represents a boost along y-direction plus a rotation of the y-z plane.

With the explicit matrix forms (2), one can verify that \begin{equation}[A,\,B] = 0\,,\quad [J,\,A]=i\,B\,,\quad [J,\,B]=-i\,A\,.\tag{3} \end{equation} As we will see below, the commutation relations (3) are exactly the Lie algebra of two-dimensional special Euclidean group SE(2).

SE(2) group

By definition, the two-dimensional special Euclidean group SE(2) consists of two-dimensional rotations plus translations. In terms of the homogenous coordinate $[x, y, 1]$, the transformation of SE(2) can be represented as \begin{equation}\left[\begin{matrix}x' \\ y' \\ 1\end{matrix}\right]=\left[\begin{matrix}\cos\theta & -\sin\theta & \alpha \\  \sin\theta & \cos\theta & \beta \\ 0 & 0 & 1\end{matrix}\right]\left[\begin{matrix}x \\ y \\ 1\end{matrix}\right]\,,\end{equation} where $\theta$ is the rotation angle along the z-axis and $[\alpha,\beta]^T$ is the translation vector within the x-y plane. 

Expand the above transformation matrix to the first order of $\alpha, \beta, \theta$: \begin{equation}\left[\begin{matrix}\cos\theta & -\sin\theta & \alpha \\  \sin\theta & \cos\theta & \beta \\ 0 & 0 & 1\end{matrix}\right]\approx\left[\begin{matrix}1 & -\theta & \alpha \\  \theta & 1 & \beta \\ 0 & 0 & 1\end{matrix}\right]\equiv I - i \alpha A - i\beta B - i\theta J\,,\end{equation} where the generators are \begin{equation}A=\left[\begin{matrix}0 & 0 & i \\  0 & 0 & 0 \\ 0 & 0 & 0\end{matrix}\right]\,,\quad B=\left[\begin{matrix}0 & 0 & 0 \\  0 & 0 & i \\ 0 & 0 & 0\end{matrix}\right]\,,\quad J=\left[\begin{matrix}0 & -i & 0 \\  i & 0 & 0 \\ 0 & 0 & 0\end{matrix}\right]\,.\tag{4} \end{equation} The generator $J$ represents a rotation of the x-y plane and $A\,(B)$ represents a translation along the x(y)-direction.

With the explicit matrix forms (4), one can work out the Lie algebra of SE(2): \begin{equation}[A,\,B] = 0\,,\quad [J,\,A]=i\,B\,,\quad [J,\,B]=-i\,A\,,\tag{5} \end{equation} which are the same as (3). 

Therefore, we can conclude that

  • The matrices (4) are the 3-dimensional representation of SE(2) group, and the parameters $\alpha,\beta$ are related a 2-dimensional translation.
  • The matrices (2) are the 4-dimensional representation of SE(2) group, and the parameters $\alpha,\beta$ are related to the Lorentz boosts followed by rotations.

Irreducible representation of SE(2)

To find out the irreducible representation of SE(2), we define a set of new operators \begin{equation}T^2\equiv A^2+B^2\,,\quad T_{\pm}\equiv A \pm i\,B\,,\end{equation} then the Lie algebra (5) becomes \begin{equation} [T^2, \,T_{\pm}] = 0\,,\quad [T_+,\,T_-]=0\,,\quad [T^2,\,J] = 0\,,\quad [J,\,T_{\pm}]=\pm T_{\pm}\,. \end{equation}

Since $T^2$ and $J$ commute, we can introduce a state $|t,j\rangle$ such that \begin{equation}T^2\,|t,j\rangle=t^2\,|t,j\rangle\,,\quad  J\,|t,j\rangle=j\,|t,j\rangle\,.\tag{6}\end{equation} The state $|t,j\rangle$ in (6) is the basis of the SE(2) irreducible representation. What about the state $T_{\pm}|t,j\rangle$? Consider \begin{eqnarray}T^2\,T_{\pm}|t,j\rangle&=&T_{\pm}\,T^2|t,j\rangle = T_{\pm}\,t^2|t,j\rangle = t^2\, T_{\pm}|t,j\rangle\,,\\ J\,T_{\pm}|t,j\rangle &=& T_{\pm}\,J|t,j\rangle \pm T_{\pm}|t,j\rangle=T_{\pm}\,j|t,j\rangle \pm T_{\pm}|t,j\rangle = \left(j\pm 1\right)\,T_{\pm}|t,j\rangle\,, \end{eqnarray} we can conclude that $T_{\pm}|t,j\rangle$ is the state $|t,j\pm 1\rangle$ up to a constant. The constant can be obtained by considering the inner product \begin{equation}\langle t,j|T^{\dagger}_{\pm}\,T_{\pm}|t,j\rangle= \langle t, j|\,T^2\,|t,j\rangle = t^2\langle t, j|t,j\rangle = t^2\,.\end{equation} As a result, we have \begin{equation}T_{\pm}\,|t,j\rangle=\pm\frac{t}{i}\,|t,j\pm 1\rangle\,,\end{equation} where the phase factor $\pm 1/i$ is chosen for other convenience.

The simplest case is the "degenerate" representation when $t=0$. In this case, we have \begin{equation}J\,|0,j\rangle = j\,|0, j\rangle\,,\quad T^2|0,j\rangle = T_{\pm}|0,j\rangle=A|0,j\rangle=B|0,j\rangle = 0\,.\end{equation} The degenerate representation of SE(2) is \begin{equation}\langle 0, j'| \exp\left[-i\alpha A - i\beta B - i\theta J\right]|0, j\rangle=e^{-i\,j\,\theta}\delta_{j',j}\,.\end{equation}

Location: 美国加利福尼亚帕罗奥图

Comments

Post a Comment

Popular posts from this blog

529 Plan

This post is for knowledge sharing only. It is not intended to be investment or tax advice. 529 Plans are tax-advantaged accounts specific for education saving: Contribution are made with after-tax dollars. Growth within the account is tax-free. Withdraw for qualified education expenses is also tax free. Roth IRA First 529 Plans are very similar to Roth IRA . But Roth IRA withdraw under  the 5-year rule  is tax free for any purpose. Therefore, unless for the state tax deduction in some states, it is always better to max out Roth IRA first (through backdoor and mega backdoor  for high income earners) before funding a 529 plan. Note: there is NO California state tax deduction for any 529 plan. 529 or Not? Here are some considerations in reality before funding a 529 plan: What if no qualified education expenses In the worst case that there will be no qualified education expenses, you can still take money out for non-qualified expenses, but you have to pay: federal income tax + 10% federa
Read more

How to offset W2 tax

This post is for knowledge sharing only. It is not intended to be tax or investment advice. 1. Contribute to pre-tax 401K and HSA As well known, pre-tax 401K and health saving account (HSA) are the only a few tax shelters for W2 workers.  2. Tax-loss harvesting If your total capital losses exceed your capital gains in a tax year, you can deduct up to \$3,000 (married filing jointly) from your ordinary income. If your losses are more than $3,000, you can carry over the remaining amount to future years, continuing to offset your income until all the losses are used up. A popular method to generate these deductible losses is  tax-loss harvesting . This involves selling a security at a loss to offset gains or reduce ordinary income, and optionally purchasing a similar, but not "substantially identical", security to avoid missing out on the market. However, be aware of the  wash-sale rule , which disallows claiming a loss if you buy the same or a substantially identical security
Read more

Retirement Accounts

Image
This post is for knowledge sharing only. It is not intended to be investment or tax advice. 0. Account Overview  Table 1 Common retirement  accounts in U.S. I summarizes the common retirement accounts in Table 1:  The second row in Table 1 lists the company-sponsored accounts that are offered through employers. The third row in Table 1 lists the individual retirement accounts (IRA) that is unrelated to employments. The columns in Table 1 are about different tax advantages as described below. WARNING: because of the tax advantages, there are also limitations and specific rules imposed by IRS for early withdrawal before the retirement age. For accounts listed in the second column,  Contributions are made with before-tax dollars . The money taken out will be taxed as regular income in the year of withdrawal. For accounts listed in the third column,  Contributions are made with after-tax dollars . The money taken out is  tax free . For after-tax 401K in the fourth column, Contributions are
Read more