# How the normalization condition implies the following relation?

Using equation 2.35 from Peskin and Schroeder:

$$|\vec{p}\rangle=\sqrt{2 E_{\vec{p}}} a^{\dagger}_\vec{p} |0\rangle$$

$$U(\Lambda)|\vec{p}\rangle = |\Lambda \vec{p}\rangle,$$ where a Lorentz transformation $$\Lambda$$ was implemented as some unitary operator $$U(\Lambda)$$.

Two questions:

1. How do I deduce the second equation?

2. What is exactly happening in here? We are applying a boost $$\Lambda$$ to the particle with momentum $$\vec{p}$$, which is a unitary operator (that is, satisfies the relation $$U U^{\dagger} = U^{\dagger} U$$?

PS: I am well aware of this question, but I wouldn't like to work it backwards (at least, I was not able to understand like that).

2)

Say that a Lorentz boost on $$|p\rangle$$ is given by an operator $$A$$. If $$|p\rangle$$ was normalised to begin with, you would want your theory to maintain its normalisation regardless of it being boosted to $$|Ap\rangle$$.

So if $$\langle p| p\rangle = 1$$, then you would also want $$\langle A p| Ap\rangle =1$$, but $$\langle A p| Ap\rangle = \langle p A^\dagger|Ap\rangle = \langle p|A^\dagger A|p\rangle$$. Requiring this to be $$= 1 = \langle p| p\rangle$$ means that $$A^\dagger A=1$$, that is that $$A$$ is unitary.

So let's call it $$U(\Lambda)$$.

1)

In 2.35 you have defined the normalisation condition $$|\mathbf{p}\rangle = \sqrt{2 E_\mathbf{p}}a^\dagger_\mathbf{p}|0\rangle.$$

If you now boost this state to a momentum $$\mathbf{q}$$, you would want this state to also be normalised like:

$$|\mathbf{q}\rangle = \sqrt{2 E_\mathbf{q}}a^\dagger_\mathbf{q}|0\rangle.$$

From what learnt in the previous question, $$|\mathbf{q}\rangle = U(\Lambda)|\mathbf{p}\rangle.$$

So instead of $$\mathbf{q}$$, let's call it $$\Lambda\mathbf{p}$$. Why invent a new name and waste a new variable.

Hence: $$|\Lambda\mathbf{p}\rangle = U(\Lambda)|\mathbf{p}\rangle.$$