Correct dimensions of a single particle state?

Question

Good day to everyone, this is my first time asking a question on the site so I hope the formalism isn't too sloppy.

During my studies in QFT it was often remarked that given a single particle covariant state such as $|1(p)>$ the correct renormalization condition is the following

$$<1(p)|1(p)>=(2\pi)^32w_p\delta ^3(0) \tag{1}$$

where $w_p$ is the energy of the state. Equation $(1)$ implies that the state $|1(p)>$ has mass dimensions $-1$. However recently, while studying the optical theorem i've found the following equation.

$Im(<Z|T|Z>)=M_Z\Gamma_Z \, \, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$

where $|Z>$ is a $Z$ boson state, $M_Z$ and $\Gamma_Z$ are the mass and deacy width of the $Z$ boson and $T$ is the non-trivial part of the $S-matrix$

From equation $(2)$ i gather that $|Z>$ must have mass dimension $1$ since the r.h.s. has mass dimension $2$ and $T$ is dimensionless, but this contrasts what I deduced from equation $(1)$. so which one is correct? Am i missing something?

Answer 1

The optical theorem, for a two particle forward scattering, is generally stated as something in the lines of

$$ {\operatorname{Im}}\ M(k_1,k_2\to k_1,k_2) = 2 E_{\operatorname{CM}}p_{\operatorname{CM}}\sigma_{\operatorname{tot}}(k_1,k_2\to\operatorname{anything}),\tag{$*$}$$

where we decompose the function $T$ as

$$T(k_{\operatorname{in}}\to k_{\operatorname{out}}) = (2\pi)^4\delta^{(4)}(k_{\operatorname{in}}- k_{\operatorname{out}}) M(k_{\operatorname{in}}\to k_{\operatorname{out}}). \tag{$**$}$$

The development of $(*)$ can be found in Peskin's and Schroeder's introduction to QFT. Now, the dimensionality of such $M$ and $T$, understood as functions as they are here, depends on the process under consideration. Indeed, we define such functions through the dimensionless operator $\mathbb{S} = 1 + i\mathbb{T}$ as

$$T(k_{\operatorname{in}}\to k_{\operatorname{out}})\stackrel{\operatorname{def}}{=}\langle k_{\operatorname{out}}|\mathbb{T}| k_{\operatorname{in}}\rangle$$

which is dimensionful. In the specific case of a $2\to 2$ forward scattering, and under the normalization you give in $(1)$ [that I believe too implies that a bra or a ket of $n$ different particles has dimension $-n$] the scattering matrix element would be

$$\langle k_1,k_2|\mathbb{S}| k_1,k_2\rangle = \langle k_1,k_2| k_1,k_2\rangle + i \langle k_1, k_2|\mathbb{T}| k_2,k_2\rangle$$

and consequently it should have the dimensionality of $\langle k_1,k_2| k_1,k_2\rangle$, namely equal to $-4$. The dimensionality of $M(k_1,k_2\to k_1,k_2)$, in turn, satisfies $$\operatorname{dim}\ M(k_1,k_2\to k_1,k_2) = 4 + \operatorname{dim}\ T(k_1,k_2\to k_1,k_2) = 0$$

due to the Dirac's delta in $(**)$. As a result, $(*)$ is dimensionally sound since its RHS is dimensionless.