Weinberg QFT 1 Normalization one 1 particle states p. 66

I encounter a question regarding the derivation of the normalization of 1 Particle states in Weinbergs book (Formula 2.5.14).

Similar questions were asked in A question on page 65 of Weinberg's QFT volume 1 and Inner product of standard-momentum one-particle states in Weinberg but I didn't see my questioned answered.

To achieve the formula for the normalization of a general scalar product of 1 particle states with momenta $$p, p'$$ and polarizations $$\sigma,\sigma'$$: $$(\psi_{p',\sigma'},\psi_{p,\sigma})$$ proportional to $$\delta^3(p-p')$$ he expands it to $$N(p)(U^{-1}(L(p))\psi_{p',\sigma'},\psi_{p,\sigma})$$ by using $$\psi_{p,\sigma} = N(p) U(L(p)) \psi_{k,\sigma}$$ where $$U(L(p))$$ is the standard quantum lorentz transformation to transform the standard states with standard momentum k to arbitrary states with momentum which satisfy $$p^2 = k^2$$. Then he derives the delta function normalization coming from normalization of standard states. For this he uses that $$L^{-1}(p)p' = k'$$. This particular statement makes no sense to me.

If we say, both states describe the same particle then $$k' = k$$ but then $$L^{-1}(p)p' = k'$$ can not hold except when $$p = p'$$. Otherwise if both states describe different particles it is generally not true that $$L^{-1}(p)p' = k'$$ where $$k'$$ should be a standard momentum since we only have 6 standard momenta $$k$$ and the relation should hold in general for his argument to work. So if anyone can enlighten me I would be glad.

I had basically the same questions as you! In my case at least, it stemmed from not properly understanding the previous arguments, so I'll try to explain everything in an organized way. It took many hours of being confused, but I think I finally get it.

But first, let me be clear with this: $$k'$$ is not a standard momentum. The notation Weinberg chose is somewhat confusing here in my opinion, but the book is nearly perfect so we can forgive him :). I'll switch the notation a tiny bit. Heads up: the actual answer to your question comes after this first section, but I think this is very useful as well.

Introduction: how and why we use standard momentum states

The $$\sigma$$ indices in $$\Psi_{p,\sigma}$$ indicate degrees of freedom of a particle that are not included in its momentum, and we want to understand how these change under Lorentz transformations. To start, I'll use a $$\Phi$$ basis instead of the $$\Psi$$ basis to indicate eigenstates of the remaining non-momentum observables needed to cover the entire Hilbert space. In other words, we have some set of commuting observables $$\mathcal{O}$$ for which $$\mathcal{O}\Phi_{p,\alpha}=\alpha \Phi_{p,\alpha}$$.

The most general transformation that the state will undergo is $$U(\Lambda)\Phi_{p,\alpha} = \sum_{\alpha'} \sum_{p'} A(p',p,\alpha',\alpha,\Lambda) \Phi_{p',\alpha'},$$ where the sum over $$p'$$ is continuous over all possible momenta. But from previous arguments in the book, it is clear that $$U(\Lambda)\Phi_{p,\alpha}$$ has momentum $$\Lambda p$$, so we can write $$U(\Lambda) \Phi_{p,\alpha} = \sum_{\alpha'} C_{\alpha'\alpha}(\Lambda,p) \Phi_{\Lambda p, \alpha'}.$$ This is telling us that under a Lorentz transformation, it is possible that $$\alpha$$ indices will get mixed up. Note that what happens for a momentum index is very simple: $$p\to \Lambda p$$. Meanwhile for $$\alpha$$, we might get a complicated superposition of various states.

Let's simplify this: we choose a standard momentum $$p\equiv k$$ and standard transformation $$\Lambda\equiv L(k,p)$$ in the above equation, such that $$Lk=p$$. Having fixed $$k$$ and $$L$$, $$C_{\alpha'\alpha}$$ only depends on $$p$$ implicitly through $$L$$ (it no longer depends on a general transformation $$\Lambda$$). This is important. (But it will only become clear later.) Plugging in above, we get $$U(L(k,p)) \Phi_{k,\alpha} = \sum_{\alpha'} C_{\alpha'\alpha}(p) \Phi_{p,\alpha'}.$$ But now let's choose another discrete basis to label our states with: $$\Psi_{p,\sigma}$$ instead of $$\Phi_{p,\alpha}$$ (which would correspond to other observables, of course). We have $$\Psi_{p, \sigma} \equiv \sum_{\alpha} \tilde{B}_{\sigma \alpha}(p) \Phi_{p, \alpha}, \quad \Phi_{p, \alpha} \equiv \sum_{\sigma} B_{\alpha \sigma}(p) \Psi_{p, \sigma}.$$ Plugging in above, we get (assuming linear $$U$$ for simplicity) $$\sum_\sigma \left(B_{\alpha \sigma}(k) U(L) \Psi_{k,\sigma} - \sum_{\alpha'} C_{\alpha'\alpha}(p) B_{\alpha'\sigma}(p) \Psi_{p,\sigma}\right) = 0$$

Here's the trick: choose the $$\Psi_{p,\sigma}$$ basis such that $$B_{\alpha \sigma}(k) = \delta_{\alpha \sigma}$$, and for each other $$p$$, $$\sum_{\alpha'}C_{\alpha' \alpha}(p) B_{\alpha'\sigma}(p) = \frac{\delta_{\alpha \sigma}}{N(p)}.$$ Finally then with this new basis we have $$N(p)U(L(k,p)) \Psi_{k,\sigma} = \Psi_{p,\sigma}.$$ The new $$\Psi$$ basis is fantastic because under this Lorentz transformation, we have the simple relation above - no mixing of $$\sigma$$ indices. This is what Weinberg means when he says that we define $$\Psi_{p,\sigma}$$ in this way. But this isn't valid for any Lorentz transformation! Note that it was important to fix a momentum $$k$$ and transformation $$L$$ in the above arguments, so that the $$C_{\alpha'\alpha}$$ coefficients only depended on $$p$$. If they also depended on $$\Lambda$$, we would not be able to consistently choose a $$\Psi$$ basis that satisfies the above formula. In particular, we have that for some general $$\Lambda$$, $$U(\Lambda)\Psi_{k,\sigma}$$ will end up in a superposition of states with different $$\sigma$$ values.

We note that choosing a standard momentum $$k$$ and transformation $$L$$ doesn't allow us to reach all possible momenta $$p$$. We can only reach those momenta that have the same mass as $$k$$ and value of sgn($$k^0$$). Thus we require 6 classes of standard momenta and transformations. Within each class we might also have different particle species (e.g., particles of different positive mass), which require different standard momenta $$k$$ and transformations $$L$$.

The little group

The little group consists of transformations $$W$$ satisfying $$Wk=k$$. In other words, $$U(W)$$ acting on $$\Psi_{k,\sigma}$$ only mixes up $$\sigma$$ indices. A general Lorentz transformation has 6 independent parameters, so there are 6 generators. But the constraint $$Wk=k$$ imposes 3 independent conditions, resulting in $$W$$ having 3 parameters. We then expect the little group to have 3 generators. Indeed this is the case for all $$p\neq 0$$, where we have the groups SO(3) and ISO(2). The $$p=0$$ case imposes no restrictions on $$W$$ so we still have SO(3,1). The transformation rule will be of the form $$U(W)\Psi_{k,\sigma} = \sum_{\sigma'} D_{\sigma'\sigma}(W) \Psi_{k,\sigma'}$$ As a check to see if you're following: what should the value of $$D_{\sigma'\sigma}(W)$$ be when $$W$$ is a standard transformation?

Once we have the little group matrices, we're all set, since we can find how our general states transform! \begin{align} U(\Lambda) \Psi_{p,\sigma} &= N(p) U(\Lambda) U(L(k,p)) \Psi_{k,\sigma}\\ &=N(p) U(L(k,\Lambda p)) U(L^{-1}(k,\Lambda p)\Lambda L(k,p)) \Psi_{k,\sigma}\\ &=N(p) U(L(k,\Lambda p)) U(W(\Lambda,p)) \Psi_{k,\sigma} \\ &=N(p) \sum_{\sigma'} D_{\sigma'\sigma}(W(\Lambda,p)) U(L(k,\Lambda p)) \Psi_{k,\sigma'}\\ &= \frac{N(p)}{N(\Lambda p)} \sum_{\sigma'} D_{\sigma'\sigma}(W(\Lambda,p)) \Psi_{\Lambda p, \sigma'}, \end{align} where we identified the little group element $$W(\Lambda,p)=L^{-1}(k,\Lambda p) \Lambda L(k,p)$$.

Normalizing standard momentum states

I leave it to you to show the following: if we want the $$D$$ matrices to furnish a unitary representation of the little group, we require the states to be normalized as $$(\Psi_{k,\sigma},\Psi_{p,\sigma'}) = \delta^3(\vec{p}-\vec{k}) \delta_{\sigma\sigma'}.$$ Here $$p$$ is not a standard momentum. Why can this normalization be done? We know that both states are eigenstates of $$\vec{P}$$, so they should be orthogonal if they correspond to different eigenvalues. The delta function doesn't have a pre-factor dependent on $$k$$ because we can just absorb that into the definition of $$\Psi_{k,\sigma}$$ (and there's no $$p$$ dependent factor because for $$p\neq k$$, this is zero anyways). The deep part of this normalization, the part which really determines that the $$D$$ matrices furnish a unitary representation, is the $$\delta_{\sigma\sigma'}$$ factor.

Normalizing general 1-particle states

Now all we're missing is the case of $$(\Psi_{p,\sigma},\Psi_{p',\sigma'})$$. The product doesn't involve a standard momentum $$k$$, so the potential pre-factor dependent on $$p$$ that I mentioned above may pop up here, and it is not obvious a priori that the $$\sigma$$ labels will give a delta factor. Spoiler: a momentum dependent pre-factor does show up, but again we get rid of it by re-scaling $$\Psi_{p,\sigma}$$. This re-scaling is allowed because of the $$N(p)$$ factor we included in the definition of $$\Psi_{p,\sigma}$$ in the first section. But the delta factor for the $$\sigma$$ labels stay the same. Let's re-derive this here: \begin{align} (\Psi_{p',\sigma'},\Psi_{p,\sigma}) &= N(p)(\Psi_{p',\sigma'},U(L(k,p))\Psi_{k,\sigma}) \\ &=N(p) (U(L^{-1}(k,p))\Psi_{p',\sigma'},\Psi_{k,\sigma}) \\ &=\frac{N(p)N^*(p')}{N^*(L^{-1}(k,p)p')}\sum_\alpha D_{\alpha\sigma'}^*(W)(\Psi_{L^{-1}p',\alpha},\Psi_{k,\sigma})\\ &=\frac{N(p)N^*(p')}{N^*(q)}\sum_\alpha D_{\alpha\sigma'}^*(W)\delta^3(\vec{q}-\vec{k}) \delta_{\alpha\sigma}\\ &=\frac{N(p)N^*(p')}{N^*(q)} D_{\sigma\sigma'}^*(W) \delta^3(\vec{q}-\vec{k}). \end{align} Here we defined $$q=L^{-1}(k,p)p'$$. For the specific $$W(L^{-1},p')$$ here, you can check in a few steps that $$D_{\sigma\sigma'}(W)=\delta_{\sigma\sigma'}$$. Since $$(q-k)=L^{-1}(k,p)(p'-p)$$, the above quantity is nonzero only for $$p'=p$$, so we can write it as $$|N(p)|^2 \delta_{\sigma\sigma'} \delta^3(\vec{q}-\vec{k}).$$ (because when $$p=p'$$, $$q=L^{-1}(k,p)p'=L^{-1}(k,p)p=k$$ and $$N(k)=1$$). The final step in the normalization is relating the delta function above to $$\delta^3(\vec{p}-\vec{p}')$$.

I hope this helped! Let me know if anything above is unclear.

• Nice explanation! One question regarding the derivation of $\left(\Psi_{p^{\prime}, \sigma^{\prime}}, \Psi_{p, \sigma}\right)$, in the final step you have the factor $N^{*}(q)$ in the denominator (I also got the same factor) but it is absent in Weinberg's book. So what is the reason for this discrepancy? – Epsilon Oct 30 '20 at 17:22
• Glad you liked it :). The delta function enforces q=k, and N(k) = 1. – Physics Llama Oct 31 '20 at 21:05