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I am trying to find the commutator of $K$, which is just the infinitesimal generator of Lorenz boosts, and $\alpha_p^\dagger$, the creation operator for a particle of momentum $p$ defined by: $$\alpha_p^\dagger|k_1\cdots k_n\rangle=|pk_1\cdots k_n\rangle$$

I am using the relation that: $$U(\Lambda)\alpha^\dagger_pU(\Lambda)^\dagger=\alpha^\dagger_{\Lambda\cdot p}$$ where $(\Lambda\cdot p)^i=\Lambda_\mu^iP^\mu$ is a three vector. I started with defining a path given by:

$$\Lambda(t)=\exp(it(\beta\cdot K))$$

So that I can rewrite the aforementioned relation as:

$$\exp(it(\beta\cdot K))\alpha^\dagger_p\exp(it(\beta\cdot K))=\alpha^\dagger_{\Lambda(t)\cdot p}$$ Then I took a derivative with respect to $t$ at $t=0$, which I believe gives me something like: $$i(\beta\cdot K)\alpha^\dagger_p-i\alpha_p^\dagger(\beta\cdot K)=\frac{d}{dt}\Big|_ {t=0}\alpha^\dagger_{\Lambda(t)\cdot p}$$ Via the chain rule, I rewrite the RHS as:

$$\frac{d}{dt}\Big|_{t=0}\alpha^\dagger_{\Lambda(t)\cdot p}=\lim_{t\rightarrow 0} \left(\frac{d\left(\Lambda(t)\cdot p\right)^i}{dt}\frac{\partial}{\partial ( \Lambda(t)\cdot p)^i}\alpha^\dagger_{\Lambda(t)\cdot p}\right)$$

The time derivative can be calculated by:

$$ \frac{d}{dt}\left(\Lambda(t)\cdot p\right)^i=\frac{d}{dt} \exp(it(\beta\cdot K))^i_\mu P^\mu$$ $$=(i(\beta\cdot K)\exp(it(\beta\cdot K)))^i_\mu P^\mu$$ Taking the limit as $t\rightarrow 0$, and inserting this expression into the RHS gives:

$$ \frac{d}{dt}\Big|_{t=0}\alpha^\dagger_{\Lambda(t)\cdot p}=i(\beta\cdot K)^i_\mu P^\mu\frac{\partial}{\partial p^i}\alpha^\dagger_p$$ $$=i\left((\beta\cdot K)P\right)^i\left(\nabla_p\alpha^\dagger_p\right)_i$$

Selecting values of $\beta$ so that I can pick out the components $K_1,K_2,K_3$ of $K$ I obtain that $$[K_1,\alpha^\dagger_p]=-iE_p\frac{\partial}{\partial p_1} \alpha^\dagger_p$$ $$[K_2,\alpha^\dagger_p]=-i\left(p_2\frac{\partial}{\partial p_1}+ p_1\frac{\partial}{\partial p_2}\right)\alpha^\dagger_p$$ $$[K_3,\alpha^\dagger_p]=-i\left(p_3\frac{\partial}{\partial p_2} +p_2\frac{\partial}{\partial p_3}\right)\alpha^\dagger_p$$

Which I think is incorrect, since my professor said that $K$ should be a dimensional operator. If someone could point out where I went wrong, or provide some instruction in how to go about this problem that would be incredibly helpful. My apologies if I have missed something trivial.

Edit:

So I chose the wrong basis vectors of $\mathfrak{sl}(2,\mathbb{C})$ when looking at each $K_i$. With the correct ones we obtain: $$[K_i,\alpha_p^\dagger]=-iE_p\frac{\partial}{\partial p_i}\alpha^\dagger_p$$ hence:

$$[K,\alpha_p^\dagger]=-iE_p\nabla_p\alpha^\dagger_p$$

which looks much better.

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  • $\begingroup$ Under rotations, the $K_i$ form a 3D vector. The right sides of your equations don’t seem to have any kind of relationship. $\endgroup$
    – Ghoster
    Commented Sep 19, 2022 at 22:55
  • $\begingroup$ @Ghoster Unsure what you mean by this. To me the $K_i$ are 3 of the six basis vectors for $\mathfrak{sl}(2,c)$ $\endgroup$
    – Chris
    Commented Sep 19, 2022 at 23:08
  • $\begingroup$ Look at their commutators with the $J_i$. They commute like other 3-vectors. This is because if you rotate the axes, a boost in one direction just becomes a boost in a different direction. $\endgroup$
    – Ghoster
    Commented Sep 19, 2022 at 23:17
  • $\begingroup$ @Ghoster I see what you are saying now, can you point out where I went wrong then? $\endgroup$
    – Chris
    Commented Sep 19, 2022 at 23:26
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    $\begingroup$ Check your $K_i$ matrices. $K_2$ should deal with $t$ and $y$ components in the same way that $K_1$ deals with $t$ and $x$ components. $\endgroup$
    – Ghoster
    Commented Sep 20, 2022 at 3:04

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