# Commutator of Boost Generator with Creation operator

## Homework Statement

Given that U upon acting on the creation operator gives a creation operator for the transformed momentum $$U(\Lambda) a_p^\dagger U(\Lambda)^\dagger = a_{\boldsymbol{\Lambda} \mathbf{p}}^\dagger$$
and ##\Lambda ## is a pure boost, that is ## U(\Lambda) = e^{i \boldsymbol{\phi} \cdot \mathbf{K}}##, find
$$\left[\mathbf{K}, \, a_p^\dagger\right]$$

## The Attempt at a Solution

The first idea we had was to expand the left side giving
$$a_p^\dagger + i\left[\boldsymbol{\phi} \cdot \mathbf{K}, \, a_p^\dagger\right] + \dots = a_{\boldsymbol{\Lambda} \mathbf{p}}^\dagger$$

However we are not able to meaningfully expand the right hand side, so we have hit a dead end. We don't know how to assign meanings to derivatives of ## a_{\mathbf{p}}^\dagger##. Is this the wrong approach?

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vanhees71
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2019 Award
I'd say, you have to use the local transformation properties of the free-field operators and then use the appropriate mode decomposition in terms of creation and annihilation operators.

I guess supposedly it's okay to leave expressions in terms of derivatives of the ladder operators. So I have used something like this.
$$U\left( \Lambda\right) a^\dagger_{\vec{p}}\,SU\left( \Lambda\right)^\dagger = a^\dagger_{\vec{p}} +i \beta_i \left[K^i, a^\dagger_{\vec{p}}\right] + \dots = a^\dagger_{\vec{p}} + \frac{\partial a^\dagger_{\vec{p}}}{p^\mu}\delta p^\mu + \dots$$

However, now the question remains, what is ##\vec{K} ##? Someone had the idea to use the fact that the vacuum Lorentz transforms into itself, but I wasn't able to follow the reasoning all the way to obtaining an expression for ##\vec{K} ##.

strangerep
I guess supposedly it's okay to leave expressions in terms of derivatives of the ladder operators. So I have used something like this.
$$U\left( \Lambda\right) a^\dagger_{\vec{p}}\,SU\left( \Lambda\right)^\dagger = a^\dagger_{\vec{p}} +i \beta_i \left[K^i, a^\dagger_{\vec{p}}\right] + \dots = a^\dagger_{\vec{p}} + \frac{\partial a^\dagger_{\vec{p}}}{p^\mu}\delta p^\mu + \dots$$

However, now the question remains, what is ##\vec{K} ##? Someone had the idea to use the fact that the vacuum Lorentz transforms into itself, but I wasn't able to follow the reasoning all the way to obtaining an expression for ##\vec{K} ##.
Here's my \$0.02 ....

First, it would help to know what textbook or lecture notes you're working from. :grumpy:

Second, start simple by considering only a boost in a particular direction, say the z direction. And recognize that ##\Lambda_z## is a function of ##\phi##. Then,
$$U\Big(\Lambda_z(\phi)\Big) ~=~ e^{i \phi K_z} ~.$$Also, write your creation operator as a function, i.e., ##a^\dagger_p \equiv a^\dagger(p)##.

The 1st equation in your OP then becomes
$$e^{i \phi K_z} \, a^\dagger(p) \, e^{-i \phi K_z} ~=~ a^\dagger\Big( \Lambda_z(\phi) \, p \Big)$$Now, differentiate both sides by ##\phi##, and then set ##\phi=0##. Then you'll need (or must look up) an explicit expression for the boost ##\Lambda_z(\phi)## as a matrix operating on ##p##. I.e., when acting in ordinary Minkowski space.

If you reach the tearing-out-hair stage, you can always consult Eugene Stefanovich's notes on Relativistic Quantum Dynamics (findable on the arXiv or via Google Scholar). Somewhere in there, he derives an explicit expression for the boost operators in terms of a/c operators.