Srednicki 2.8 / Inverse Lorentz Transformation

[tshafer]

Homework Statement

I need to demonstrate the relation $$[\varphi(x),M^{\mu\nu}]=\matchal{L}^{\mu\nu}\varphi(x)$$

where
$$\mathcal{L}^{\mu\nu}\equiv \frac{\hbar}{i}(x^\mu\partial^\nu-x^\nu\partial^\mu)$$.

Homework Equations

$$U(\Lambda)^{-1}\varphi(x)U(\Lambda) = \varphi(\Lambda^{-1}x)$$

$$\Lambda = 1+\delta\omega$$

$$\Lambda^\mu_\nu = \delta^\mu_\nu + \delta\omega^\mu_\nu$$

$$U(1+\delta\omega) = 1 + \frac{i}{2\hbar}\delta\omega_{\mu\nu}M^{\mu\nu}$$

The Attempt at a Solution

I can pretty easily evaluate the LHS of the expression to give

$$\varphi(x) + \frac{i}{2\hbar}\delta\omega_{\mu\nu} [\varphi(x),M^{\mu\nu}]$$,

but I am having a ton of trouble with deciphering what is meant by $$\varphi(\Lambda^{-1}x)$$, particularly how I am supposed to get derivatives of the type $$\partial^\mu$$ when I had thought $$x\equiv x^\mu$$ here, implying derivatives of the kind $$\partial_\mu$$.

I tried, for instance, defining an inverse transformation so

$$\bar{x}^\mu = {\Lambda_\nu}^\mu x^\nu = (\delta^\mu_\nu - \delta\omega^\mu_\nu)x^\nu$$ and finding a Taylor series about x, but I don't really know how to do this in terms of indices. I get

$$\varphi(\Lambda^{-1}x) = \varphi(x^\mu-\delta\omega^\mu_\nu x^\nu)$$ so that

$$\sim \varphi(x) - \delta\omega^\mu_\nu x^\nu\bar{\partial_\mu}\varphi(\Lambda^{-1}x)$$, which is sort of close, but not right.

If I magically change the inverse transformation to $$\Lambda_{\mu\nu}x^\nu = x_\mu - \delta\omega_{\mu\nu}x^\nu$$ it works, but I can't justify this.

I think the problem is with my inverse transformation and taking the derivative. Help would be greatly appreciated.

 

[mark726]

Hello, it's great to see that you are working on this problem. Let me try to help you out.

First of all, let's start with the definition of the inverse transformation. In this case, we have \bar{x}^\mu = \Lambda_\nu^\mu x^\nu = (\delta^\mu_\nu - \delta\omega^\mu_\nu)x^\nu. This means that we are transforming from the primed coordinates to the unprimed coordinates, which is the opposite of what we usually do. So, x^\mu represents the unprimed coordinates and \bar{x}^\mu represents the primed coordinates.

Now, let's consider the expression \varphi(\Lambda^{-1}x). What this means is that we are transforming the field \varphi(x) from the unprimed coordinates to the primed coordinates. In other words, we are transforming from x^\mu to \bar{x}^\mu. This is where the inverse transformation comes into play. Since we have defined \bar{x}^\mu in terms of x^\mu, we can use the inverse transformation to transform the field. The inverse transformation is given by \bar{x}^\mu = \Lambda_\nu^\mu x^\nu = (\delta^\mu_\nu - \delta\omega^\mu_\nu)x^\nu. So, we can write \varphi(\Lambda^{-1}x) = \varphi((\delta^\mu_\nu - \delta\omega^\mu_\nu)x^\nu).

Now, let's take a closer look at the Taylor series expansion that you have done. You have correctly identified that \varphi(\Lambda^{-1}x) = \varphi(x) - \delta\omega^\mu_\nu x^\nu \bar{\partial_\mu}\varphi(\Lambda^{-1}x). However, there is a small mistake in your expression. The derivative should be with respect to \bar{x}^\mu, not x^\mu. So, it should be \bar{\partial_\mu}\varphi(\Lambda^{-1}x). This is because we are transforming from x^\mu to \bar{x}^\mu, so the derivative should be taken with respect to the primed coordinates.

I hope this helps you understand the concept better. Let me know if you have any further questions. Good luck!