# Confusion (1) from Weinberg's QFT.(unitary representation)

1. Jul 15, 2011

### kof9595995

Just started my way ploughing through Weinberg's book, for which my progress can be measured in lines since I got so many confusions. I suppose there'll be more confusions awaiting me, so I wish all my confusions or questions from his book will make a series, and I'll label them as "Confusion(1),(2).......from Weinberg's QFT". Hopefully this series can also benefit others, but the unfortunate thing is these posts are probably only accessible to those who have the book on hand, because his notations are quite different from others, and sometimes issues can be dispersed in the book so I can't quote everything. Anyway here goes the first question.
Page 64, eqn (2.5.5)
$\Psi _{p,\sigma } \equiv N(p)U(L(p))\Psi _{k,\sigma }$
where N(p) is a normalization factor and U is a unitary operator. My question is if U is really unitary ,why do we need the normalization factor?

Edit: the title should be .... Weinberg's QFT.(unitary operator), I was having some issues with representations earlier.

2. Jul 15, 2011

### tiny-tim

hi kof9595995!

(Weinberg's book is accessible online, at http://books.google.com/books?id=h9...nother book on quantum field theory"&f=false" )

ah, if you'd ploughed your way on to page 67 , you'd have seen …

The normalisation factor N(p) is sometimes chosen to be N(p) = 1, but then we would need to keep track of the factor p0/k0 in scalar products. Instead, I will adopt the more usual convention that N(p) = √(k0/p0)​

Last edited by a moderator: Apr 26, 2017
3. Jul 15, 2011

### kof9595995

err..I hadn't got there yesterday. But now I still have questions, if N(p) is chosen to be 1, then
$(\Psi _{p',\sigma '} ,\Psi _{p,\sigma } ) = \frac{{p^0 }}{{k^0 }}\delta _{\sigma '\sigma } \delta ^3 (\vec p' - \vec p)$
Doesn't this mean U transforms a set of normalized vectors to a set of unnormalized ones? Then in what sense is U a unitary operator?

4. Jul 16, 2011

### tiny-tim

(just got up :zzz: …)

ah but that normalisation factor doesn't change U, it only changes the states U acts on

for example, in ordinary 2D polar coordinates …

suppose U(r,θ) = (r,θ+20°),

and ψθ = (tan29θ,θ),

(instead of (1,θ))

then (U(tan29θ11),U(tan29θ22)) is still equal to ((tan29θ11),(tan29θ22)), isn't it?

5. Jul 16, 2011

### meopemuk

The problem here is that we are dealing with continuous spectrum of momentum and that vectors $\Psi _{p,\sigma }$ are not really normalized. Their norm is actually infinite, and we need to be careful when selecting the normalization constant. If you assume that N(p)=1, then you'll get a contradiction. For example, you may want to define the identity operator as (I omit spin indices and write $\Psi _{\mathbf{p}} = |\mathbf{p} \rangle$ for simplicity)

$$I = \int d\mathbf{p} |\mathbf{p} \rangle \langle \mathbf{p}|$$

You also want this operator to be invariant with respect to boosts

$I' = U(\Lambda) I U^{-1} (\Lambda) = I = \int d\mathbf{p} |\mathbf{p} \rangle \langle \mathbf{p}|$ .............(1)

However, if you use your assumption N(p)=1 you'll get (by changing integration variables $\mathbf{q} = \Lambda \mathbf{p}$)

$I' = U(\Lambda) \int d\mathbf{p} |\mathbf{p} \rangle \langle \mathbf{p}| U^{-1} (\Lambda) = \int d\mathbf{p} | \Lambda \mathbf{p} \rangle \langle \Lambda \mathbf{p}| = \int d\mathbf{q} \det \left| \frac{d \Lambda^{-1} \mathbf{q}}{d \mathbf{q}} \right| | \mathbf{q} \rangle \langle \mathbf{q}|$ ....................(2)

where

$$\det \left| \frac{d \Lambda^{-1} \mathbf{q}}{d \mathbf{q}} \right| = \frac{\omega_{\Lambda^{-1}q}}{\omega_q}$$

is the Jacobian of the variable change $\mathbf{p} \to \mathbf{q}$. So, in order to bring (2) to the desired form (1) you need to cancel this Jacobian in the integrand by assuming that $N(p) = \omega_p^{-1/2}$.

Eugene.

6. Jul 16, 2011

### kof9595995

I know setting N(p) always equal to 1 will lead to contradiction, that's why I'm not convinced that U is unitary, but Weinberg said U is unitary, and this is what is confusing me.
To make my confusion unambiguous, here's my train of thoughts:
(1)orthonormal condition is $(\Psi _{p',\sigma '} ,\Psi _{p,\sigma } ) = \delta _{\sigma '\sigma } \delta ^3 (\vec p' - \vec p)$
(2)A unitary transformation should preserve the orthonormality
(3)U does not preserve the orthonormality, ergo U is not unitary.

7. Jul 16, 2011

### Fredrik

Staff Emeritus
It must preserve the orthonormality of every orthonormal set in the Hilbert space. Do you have a reason to think that it doesn't? The $\Psi_{\mathbf p,\sigma}$ aren't members of the Hilbert space.

8. Jul 16, 2011

### tiny-tim

hi kof9595995!

ah, you're looking at $\Psi _{p,\sigma }$ (2.5.5) and $\Psi _{k,\sigma }$ in (2.5.12)

unfortunately, Weinberg is defining them differently, but using the same notation

the second lot are defined to be orthonormal, the first lot aren't

9. Jul 16, 2011

### kof9595995

$\Psi_{p,\sigma}$ is transformed from a set of orthonormal vectors $\Psi_{k,\sigma}$ via U, you can check (2.5.12). However, only after putting a normalization factor N(p) can he arrived another set of orthonormal vectors (2.5.19), which makes me question why on earth U is unitary.

10. Jul 16, 2011

### kof9595995

I don't see it, can you answer post #9?

11. Jul 16, 2011

### tiny-tim

(2.5.12) gives you (2.5.19).

(2.5.5) doesn't.

(2.5.5) and (2.5.12) are defined differently, even though Weinberg has used the same notation.

12. Jul 16, 2011

### Fredrik

Staff Emeritus
(2.5.12) implies that the "norm" of each $\Psi_{k,\sigma}$ is infinite. That means that we're not talking about the norm on a Hilbert space. I don't think the set of all $\Psi_{k,\sigma}$ can be described as an orthonormal set in any Hilbert space. It certainly isn't an orthonormal subset of the Hilbert space on which U is supposed to be unitary. In fact, none of the $\Psi_{k,\sigma}$ is a member of that Hilbert space.

Last edited: Jul 16, 2011
13. Jul 16, 2011

### meopemuk

Fredrik is right. The momentum "basis vectors" $\Psi_{k} \equiv | \mathbf{p} \rangle$ have infinite norm, so they do not belong to the Hilbert space. True Hilbert space normalized vectors have the form

$$|\Psi \rangle = \int d \mathbf{p} \psi (\mathbf{p}) | \mathbf{p} \rangle$$

where "wave function" $\psi (\mathbf{p})$ satisfies condition

$$\int d \mathbf{p} |\psi (\mathbf{p}) |^2 = 1$$

For such vectors the unitarity of boost transformations can be proven easily. First one can show that Weinberg's selection of N(p) implies the following "wave function" transformation law

$$U(\Lambda) \psi (\mathbf{p}) = \sqrt{\frac{\omega_{\Lambda^{-1}p}}{\omega_p}} \psi (\Lambda^{-1}\mathbf{p})$$

Then the inner product of two normalized Hilbert space vectors is invariant with respect to boosts:

$$\langle \Psi' | \Phi' \rangle = \int d \mathbf{p} \psi^*(\Lambda^{-1}\mathbf{p})\phi(\Lambda^{-1}\mathbf{p}) \frac{\omega_{\Lambda^{-1}p}}{\omega_p} = \int d \mathbf{q} \psi^*(\mathbf{q})\phi(\mathbf{q}) = \langle \Psi | \Phi \rangle$$

Eugene.

14. Jul 17, 2011

### kof9595995

So the key point is that the "delta function normalization" by no means indicates the vectors are "orthonormal"(even in any generalized sense)? Ok, maybe I can accept that. It's just strange to me because a Kronecker delta should indicate orthonormality, and we always say Dirac delta is a continuous version of Kronecker, yet the orthonrmality condition is not inherited.

15. Jul 17, 2011

### Fredrik

Staff Emeritus
The key point is that you can't apply a fact about Hilbert spaces ("unitary operators preserve the inner product") to structures that aren't Hilbert spaces. Obviously, it makes sense to describe the $\Psi_{k,\sigma}$ as "orthonormal in a generalized sense", but that just defines what we mean by those words.

Last edited: Jul 17, 2011
16. Jul 17, 2011

### kof9595995

But U does preserve inner product because $\delta^3(\vec k'-\vec k)=\frac{p^0}{k^0}\delta^3(\vec p'-\vec p)$, but the problem is this preserved inner product does not look like a orthonormal condition in the new coordinates,because of $\frac{p^0}{k^0}$(I suppose a reasonable "orthonormal condition", if exists, should only contain delta functions, with no prefactor).
"The key point is that you can't apply a fact about Hilbert spaces to structures that aren't Hilbert spaces.Obviously, it makes sense to describe the Ψk,σ as "orthonormal in a generalized sense", but that just defines what we mean by those words." This could be the solution, but to make the analogy to Hilbert space more exact, I would have to think an "orthonormal condition" for momentum eigenstates doesn't exist, the delta functions only define a "orthogonal condition", so you see the orthogonality is preserved and we don't need to worry about "normality".

17. Jul 18, 2011

### Avodyne

Life is much easier if you just take N(p)=1, and use the normalization
$$\langle k|k'\rangle=(2\pi)^3 2k^0 \delta^3(\vec k-\vec k')$$
where |k> is the state of a single particle of momentum k. Then the right-hand side is Lorentz invariant.
They are orthogonal, and you get to pick the normalization.

Last edited: Jul 18, 2011
18. Jul 20, 2011

### A. Neumaier

The Dirac delta describes (formal) orhonormality in the nonrelativistic inner product, but one needs an additional factor before the delta to have the same for the covariant inner product. This is like using in C^n the inner product $$<x|y>= sum_j d_j x_j^* y_j$$; then the Kronecker delta no longer describes orhonormality, but needs an additional factor.

19. Jul 20, 2011

### kof9595995

I guess now the issue is what's appropriate definition for "orthonormal", I tend to think orthonormality should have something to do with unity, clearly in this sense Dirac delta is more appropriate than the covariant one. Besides, Weinberg hims seems to refer Dirac delta as the orthonormal condition, as he wrote:

Last edited: Jul 20, 2011
20. Jul 20, 2011

### kof9595995

Does it? I think Kronecker delta is already covariant, if $U(\Lambda)\Psi_i=\Psi_{i'}$ and $(\Psi_i, \Psi_j)=\delta_{ij}$, then naturally $(\Psi_{i'}, \Psi_{j'})=\delta_{i'j'}$

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