# Weinberg Lectures on QM (2013 ed.), Equation 3.6.18

1. Feb 18, 2016

### jouvelot

Hi,

I don't get how one goes from 3.6.17 to 3.6.18 on Page 80 (Galilean invariance) regarding the zeroing of boost generator commutators. I do get that this is a special case of the Lorentz invariance (which I understand), but this particular step eludes me.

2. Feb 18, 2016

### vanhees71

This is about Galileo invariance. In the case of Lorentz invariance two boosts, if not in the same direction, do NOT commute!

Here you have
$$\hat{U}(\vec{v}) \hat{U}(\vec{v}')=\hat{U}(\vec{v}) \hat{U}(\vec{v}')=\hat{U}(\vec{v}+\vec{v}')$$
for all $\vec{v} \in \mathbb{R}^3$. Expanding
$$\hat{U}(\vec{v})=\exp(-\mathrm{i} \vec{v} \cdot \hat{\vec{K}})$$
of both transformations up to 1st order in $\vec{v}$ you get
$$[\hat{K}_i,\hat{K}_j]=0.$$

3. Feb 18, 2016

### jouvelot

Hi,

Thanks for your fast relay :)

Indeed, I get this, but if you actually try to use infinitesimal definitions of $$U(v)$$ as $$1 -ivK + O(v^2),$$ as suggested in Weinberg's book, and do the very simple math in $$U(v)U(v') = U(v+v'),$$I don't see how this forces the commutator to become 0.

Thanks.

4. Feb 18, 2016

### JorisL

What you use is the following

$$U(\vec{v})U(\vec{v}^\prime) =U(\vec{v}+\vec{v}^\prime) = U(\vec{v}^\prime+\vec{v}) = U(\vec{v}^\prime)U(\vec{v})$$

In other words, Galilei boosts commute.

You can write this as $\left[ U(\vec{v}), U(\vec{v}^\prime)\right] = 0$.
By expanding $U(\vec{v}) = \mathbf{1}-i\vec{v}\cdot \vec{K} + O(v^2)$ and inserting that into the commutator you should get there.

5. Feb 18, 2016

### jouvelot

Hi,

I see. Usually, for instance with Equation 3.6.19, you can actually get the value of the commutator as a consequence of replacing the infinitesimal unitary operator in the invariance constraint (Equation 3.6.16).

In 3.6.18, when one actually does the expansion of the multiplication of the two infinitesimal boosts in the constraint $U(v)U(v') = U(v+v')$, the last factor is in fact $O(v^2)$, and can thus be neglected, yielding 0 and validating the constraint. The value of the commutator is not a direct consequence, though.

Thanks.

6. Feb 18, 2016

### JorisL

I don't quite get what you're telling. But let me show what I did.

\begin{align*} 0 &= [U(\vec{v}),U(\vec{v}^\prime)]\\ &= \left[ \mathbf{1}-i\left( \vec{v}+\vec{v}^\prime\right) \cdot\vec{K} - \left(\vec{v}\cdot \vec{K}\right) \left(\vec{v}^\prime\cdot \vec{K}\right) + \ldots \right]\\ &- \left[ \mathbf{1}-i\left( \vec{v}+\vec{v}^\prime\right) \cdot\vec{K} - \left(\vec{v}^\prime\cdot \vec{K}\right) \left(\vec{v}\cdot \vec{K}\right) + \ldots \right]\\ &= \sum_{i,j}v_iv_j^\prime [K_i,K_j] \end{align*}

Since this holds for all vectors $\vec{v}$ and $\vec{v}^\prime$ it follows that the commutators should vanish.

7. Feb 18, 2016

### jouvelot

Yes, because you know that the commutator of two general U(v) translations is 0 and subsequently proved that this implies the generators have to commute.

What I was doing was simply replacing $U(v)$ and $U(v')$ in the constraint $$U(v)U(v') = U(v+v')$$ with their infinitesimal definitions, hoping to get that this constraint is true only if a particular condition is verified, which I hoped would be the fact that $[K,K'] = 0$. But this doesn't work that way here, it seems, contrarily to the other cases that appear before in the book.

Thanks a lot for for explanation and the time taken :)