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

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.

tiny-tim
Homework Helper
hi kof9595995! 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?

(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)​

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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?

tiny-tim
Homework Helper
(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),

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

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.

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.

Fredrik
Staff Emeritus
Gold Member
(2)A unitary transformation should preserve the orthonormality
(3)U does not preserve the orthonormality, ergo U is not unitary.
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.

tiny-tim
Homework Helper
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 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.

$\Psi_{p,\sigma}$ 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.

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

tiny-tim
Homework Helper
(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.

Fredrik
Staff Emeritus
Gold Member
$\Psi_{p,\sigma}$ is transformed from a set of orthonormal vectors $\Psi_{k,\sigma}$ via U, you can check (2.5.12).
(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.

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(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.

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.

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.

Fredrik
Staff Emeritus
Gold Member
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.

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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.
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".

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.
So the key point is that the "delta function normalization" by no means indicates the vectors are "orthonormal"(even in any generalized sense)?
They are orthogonal, and you get to pick the normalization.

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A. Neumaier
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 orthonormality condition is not inherited.

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.

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:
By the usual orthonormalization procedure of quantum mechanics, we may choose the states with standard momentum $k^{\mu}$ to be orthonomal, in the sense that $(\Psi_{k',\sigma'},\Psi_{k,\sigma})=\delta^{3}(\vec k'-\vec k)\delta_{\sigma'\sigma}$ (2.5.12)

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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.

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 $(\Psi_{i'}, \Psi_{j'})=\delta_{i'j'}$

strangerep
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 $(\Psi_{i'}, \Psi_{j'})=\delta_{i'j'}$

I don't think so...

Starting from Arnold's nonstandard inner product
$$\def\<{\langle} \def\>{\rangle} \<x|y\> ~=~ \sum_j d_j \bar x_j y_j$$
the rhs can be re-written in terms of transformed vectors as
$$\sum_{jkm} d_j \bar x'_k \bar U_{kj} U_{jm} y'_m$$
Because of the presence of $d_j$ in the summand, you can't turn the product of the U's into a Kronecker delta.

I don't think so...

Starting from Arnold's nonstandard inner product
$$\def\<{\langle} \def\>{\rangle} \<x|y\> ~=~ \sum_j d_j \bar x_j y_j$$
the rhs can be re-written in terms of transformed vectors as
$$\sum_{jkm} d_j \bar x'_k \bar U_{kj} U_{jm} y'_m$$
Because of the presence of $d_j$ in the summand, you can't turn the product of the U's into a Kronecker delta.
But then U would not be unitary.$(\Psi_i, \Psi_j)=\delta_{ij}$ means the inner product of two basis is either 0 or 1, depending on whether i=j. Because of the unitarity of U, $(\Psi_{i'}, \Psi_{j'})$ should also be either 0 or 1, depending on if they are transformed from the basis with same indices, so $(\Psi_{i'}, \Psi_{j'})=\delta_{i'j'}$

A. Neumaier
But then U would not be unitary.$(\Psi_i, \Psi_j)=\delta_{ij}$ means the inner product of two basis is either 0 or 1, depending on whether i=j. Because of the unitarity of U, $(\Psi_{i'}, \Psi_{j'})$ should also be either 0 or 1, depending on if they are transformed from the basis with same indices, so $(\Psi_{i'}, \Psi_{j'})=\delta_{i'j'}$

Unitarity is always defined relative to a specified inner product. What is unitary in one inner product is usually not unitary in any other inner product.

In particular, most unitary operators in the nonrelativistic inner product are not unitary in the covariant inner product, and conversely.

Unitarity is always defined relative to a specified inner product. What is unitary in one inner product is usually not unitary in any other inner product.

In particular, most unitary operators in the nonrelativistic inner product are not unitary in the covariant inner product, and conversely.

I don't understand and can't see what's wrong with my post 22, can you be more specific? My argument is that suppose U is a unitary operator induced by a Lorentz transform, then what I said in post 22 has to be the case, so that transition probability between two same state vectors(viewed in different frames) will be the same.