# Bjorken Drell derivation - Lorentz transformation

1. Dec 21, 2014

### Maybe_Memorie

I'm trying to derive (14.25) in B&J QFT. This is

$U(\epsilon)A^\mu(x)U^{-1}(\epsilon) = A^\mu(x') - \epsilon^{\mu\nu}A_\nu(x') + \frac{\partial \lambda(x',\epsilon)}{\partial x'_\mu}$, where $\lambda(x',\epsilon)$ is an operator gauge function.

This is all being done in the radiation gauge, i.e. $A_0 = 0$ and $\partial_i A^i=0$, with $i \in {1,2,3}$.

$\epsilon$ is an infinitesimal parameter of a Lorentz transformation $\Lambda$.

Under this transformation, $A^\mu(x) \rightarrow A'^\mu(x')=U(\epsilon)A^\mu(x)U^{-1}(\epsilon)$.

The unitary operator $U$ which generates the infinitesimal Lorentz transformation

$x^{\mu} \rightarrow x'^{\mu} = x^{\mu} + \epsilon^{\mu}_{\nu}x^{\nu}$ is

$U(\epsilon)=1 - \frac{i}{2}\epsilon_{\mu\nu}M^{\mu\nu}$

where $M$ are the generators of Lorentz transformations. (I guess really I should have $M^{\mu\nu}=(M^{\rho\sigma})^{\mu\nu}$. M is a hermitian operator, so

$U^{-1}(\epsilon)=1 + \frac{i}{2}\epsilon_{\mu\nu}M^{\mu\nu}$

Now I tried writing out $U(\epsilon)A^\mu(x)U^{-1}(\epsilon)$ explicitly but it didn't really get me anywhere. The answer is supposed to have $x'$ as the argument of $A^\mu$ on the RHS but I only get $x$. I'm not sure how to Lorentz transform the function and the argument at the same time.

Underneath the formula in B&J it says the gauge term is necessary because $UA_0U^{-1}=0$ since $A_0=0$. I don't see why this warrants the need of a gauge term I'm guessing it's needed because otherwise there will be no conjugate momenta for the $A_0$. Okay I get that, but still don't understand where the initial formula comes from.

2. Dec 23, 2014

### samalkhaiat

If you know why the gauge function $\lambda$ appears in the Lorentz transformation law for the gauge fields, then the rest is rather easy. Let your coordinates transform (under some Lie group $G$) according to $$x \to \bar{ x } = g \ x , \ \ \ g \in G .$$Then, a multi-component field $\phi^{ a } ( x )$ transforms by some finite-dimensional matrix (representation of $G$) $D(g)$ according to $$\phi^{ a } ( x ) \to \bar{ \phi }^{ a } ( \bar{ x } ) = ( D ( g ) )^{ a }{}_{ b } \ \phi^{ b } ( x ) .$$ Write this as $$\bar{ \phi }^{ a } ( \bar{ x } ) = ( D ( g ) )^{ a }{}_{ b } \ \phi^{ b } ( g^{ - 1 } \bar{ x } ) .$$ Dropping the bars from the coordinates, you get $$\bar{ \phi }^{ a } ( x ) = ( D ( g ) )^{ a }{}_{ b } \ \phi^{ b } ( g^{ - 1 } x ) . \ \ \ (1)$$
As an operator-valued field, $\phi ( x )$ transforms by a unitary operator $U(g)$ (representing $G$) on Hilbert space: $$\bar{ \phi }^{ a } ( x ) = U^{ - 1 } ( g ) \ \phi^{ a } ( x ) \ U ( g ) . \ \ \ \ (2)$$ From (1) and (2) you get $$U^{ - 1 } ( g ) \ \phi^{ a } ( x ) \ U ( g ) = ( D ( g ) )^{ a }{}_{ b } \ \phi^{ b } ( g^{ - 1 } x ) . \ \ \ \ \ (3)$$ This is how an arbitrary field operator transforms (under an arbitrary Lie group $G$) in Hilbert space. In order to get the form given in your book, let $g^{ - 1 } = \Lambda$ and use the relations $$U ( \Lambda^{ - 1 } ) = U^{ - 1 } ( \Lambda ) , \ \ \ D ( \Lambda^{ - 1 } ) = D^{ - 1 } ( \Lambda )$$ to transform (3) into $$U ( \Lambda ) \ \phi^{ a } \ U^{ - 1 } ( \Lambda ) = ( D^{ - 1 } ( \Lambda ) )^{ a }{}_{ b } \ \phi^{ b } ( \Lambda x ) .$$ If the group of transformations is Lorentz’s $\bar{ x } = \Lambda x$, and the field is a genuine vector field $V^{ \mu } ( x )$, then $D ( \Lambda ) = \Lambda$ and $$U ( \Lambda ) \ V^{ \mu } ( x ) \ U^{ - 1 } ( \Lambda ) = ( \Lambda^{ - 1 } )^{ \mu }{}_{ \nu } \ V^{ \nu } ( \bar{ x } ) .$$ Now, using $( \Lambda^{ - 1 } )^{ \mu }{}_{ \nu } \approx \delta^{ \mu }_{ \nu } - \epsilon^{ \mu }{}_{ \nu }$, you arrive at the transformation law for a GENUINE Lorentz VECTOR $$U ( \epsilon ) \ V^{ \mu } ( x ) \ U^{ - 1 } ( \epsilon ) = V^{ \mu } ( \bar{ x } ) - \epsilon^{ \mu }{}_{ \nu } \ V^{ \nu } ( \bar{ x } ) . \ \ \ (4)$$ For the gauge field $A^{ \mu }$, we have the freedom to make a gauge transformation $A^{ \mu } \to A^{ \mu } + \partial^{ \mu } \lambda$. For example, we can choose the function $\lambda ( x )$ in such a way so that $A^{ 0 } ( x ) = 0$ in all Lorentz frames. This means that the gauge field can NOT be a genuine Lorentz vector. To account for this fact, the RHS of equation (4) has to be modified to read $$U ( \epsilon ) \ A^{ \mu } ( x ) \ U^{ - 1 } ( \epsilon ) = A^{ \mu } ( \bar{ x } ) - \epsilon^{ \mu }{}_{ \nu } \ A^{ \nu } ( \bar{ x } ) + \frac{ \partial \lambda ( \bar{ x } ; \epsilon ) }{ \partial \bar{ x }_{ \mu } } . \ \ \ (5)$$ Written in this form, I think eq(5) looks very ugly. A more convenient and nicer form of this equation is $$U^{ - 1 } ( \Lambda ) \ A^{ \mu } ( x ) \ U ( \Lambda ) = \Lambda^{ \mu }{}_{ \nu } \ A^{ \nu } ( \Lambda^{ - 1 } x ) + \partial^{ \mu } \lambda ( x ; \Lambda ) .$$ Of course, the precise form of the function $\lambda ( x ; \Lambda )$ depends on the choice of the gauge in the problem and, in general, it is not easy.
One final remark is the fact that Eq(5) DOES NOT follow from the equations the book says it follow from. This may explain why the proof was left as exercise.

Sam

3. Dec 24, 2014

### Maybe_Memorie

That's perfect, thanks a lot!

I don't actually have the book with me on hand and am trying to derive the form of $\lambda$ with the conditions $A^0=0$ and $\partial_iA^i=0$.

The $A^0=0$ condition means we must have $\epsilon^0_iA^i(x')=\partial'^0\lambda(x',\epsilon)$, where I'm writing $\partial'^\mu=\frac{\partial}{\partial x'_\mu}$.

The Lorentz transformation of $\partial_iA^i$ must be zero too, so we can say

$\partial'_i[A^i(x')-\epsilon^i_jA^j(x') + \partial'^i\lambda]=0$

The first term will be zero by the gauge condition we we're left with $\partial'_i\partial'^i\lambda =\epsilon^i_j\partial'_iA^j(x')$

I have a vague recollection of the answer having a time derivative in the book so I'm guessing I've made a mistake.

4. Dec 24, 2014

### samalkhaiat

You should try harder, it is only a trivial algebra. You have $$\bar{ A }^{ j } ( \bar{ x } ) = A^{ j } ( x ) + \epsilon^{ j }{}_{ \nu } \ A^{ \nu } ( x ) + \partial^{ j } \lambda ( x ) .$$ Now, $$\bar{ \partial }_{ j } \bar{ A }^{ j } ( \bar{ x } ) = \partial_{ \mu } ( A^{ j } + \epsilon^{ j }{}_{ \nu } \ A^{ \nu } + \partial^{ j } \lambda ) \frac{ \partial x^{ \mu } }{ \partial \bar{ x }^{ j } } .$$ Using $$\frac{ \partial x^{ \mu } }{ \partial \bar{ x }^{ j } } = \delta^{ \mu }{}_{ j } - \epsilon^{ \mu }{}_{ j } ,$$ and keeping only first order terms (i.e. $\epsilon^{ 2 } \approx \epsilon \lambda \approx 0$ ), you find $$\bar{ \partial }_{ j } \bar{ A }^{ j } ( \bar{ x } ) = \partial_{ j } ( A^{ j } + \epsilon^{ j }{}_{ k } \ A^{ k } + \partial^{ j } \lambda ) - \epsilon^{ \mu }{}_{ j } \ \partial_{ \mu } A^{ j } ,$$ or $$\bar{ \partial }_{ j } \bar{ A }^{ j } ( \bar{ x } ) - \partial_{ j } A^{ j } ( x ) = \epsilon^{ j }{}_{ k } \ \partial_{ j } A^{ k } - \nabla^{ 2 } \lambda ( x ; \epsilon ) - \epsilon^{ 0 }{}_{ j } \ \partial_{ 0 } A^{ j } - \epsilon^{ k }{}_{ j } \ \partial_{ k } A^{ j } .$$ Now the LHS vanishes by assumption and the first and the last terms on the RHS add up to zero (by changing the dummy indices), so you are left with $$\nabla^{ 2 } \lambda ( x ; \epsilon ) = - \epsilon^{ 0 }{}_{ j } \ \partial_{ 0 } A^{ j } ( x ) .$$ Thus $$\lambda ( x ; \epsilon ) = - \epsilon^{ 0 }{}_{ j } \ \nabla^{ - 2 } ( \partial_{ 0 } A^{ j } ) = - \epsilon^{ 0 }{}_{ j } \ \int d^{ 3 } y \frac{ 1 }{ 4 \pi \ | x - y | } \partial_{ t } A^{ j } ( t , y ) .$$ End of the story.
Sam