# Elementary doubts i i got confronted with while reading qft

1. Sep 6, 2014

### kau

why should Proca eqn be like ∂γ Fγμ + m2 Aμ = 0 but not ∂γ Fμγ + m2 Aμ = 0 ???? another doubt is (λ-1 ω λ)μγ = λ-1 ρμ ωρσ λσγ ??? why in λ-1 transformation got upper index in the second place but not in the first place????
if someone clear my doubts...I would be thanful...
regards..
Kau

2. Sep 6, 2014

### Freddieknets

Your first problem is easily solved:

First note that the gauge field tensor is antisymmetric, i.e. $F_{\gamma\mu}=-F_{\mu\gamma}$.

If $\partial^\gamma F_{\gamma\mu} + m^2 A_\mu =0$, then $\partial^\gamma F_{\gamma\mu} = -m^2 A_\mu$.

Filling this in the second equation, we have $\partial^\gamma F_{\mu\gamma} +m^2 A_\mu=-\partial^\gamma F_{\gamma\mu} +m^2 A_\mu =m^2 A_\mu+m^2 A_\mu\neq 0$.

The second equation needs a minus sign, i.e.

$$\partial^\gamma F_{\gamma\mu} + m^2 A_\mu =0$$
$$\partial^\gamma F_{\mu\gamma} - m^2 A_\mu =0$$

are both correct.

For your second question, i.e. $(\lambda^{-1}\omega\lambda)_{\mu\gamma}$ I don't understand what the problem is.

Last edited: Sep 6, 2014
3. Sep 6, 2014

Just use $$at the beginning and [ /tex] (without the space) at the end. Or $for inline equations. The is not rendered by MathJax. 4. Sep 6, 2014 ### Freddieknets Jup, actually I knew that, switched too fast from thesis writing to physicsforums.. :-) 5. Sep 6, 2014 ### kau I said in the second one suppose λ is lorentz transformation , λ-1 is inverse lorentz transformation and δωμγ is some parameter boost or rotation. now if you write its components explicitly. then it would be λ-1 μρ δωμγ λγ σ ... ok.. look the difference in λ-1 μρ μ is written upstair in 2nd position and ρ is written downstair ad 1st position which says that λ-1 μ ρ = λμ ρ ... but in λγ σ here y is in upstair 1st postion and σ is in downstair second position .. but this I understand it represents direct transformation. so if you write γ in the second position and σ in the 1st position that would imply and inverse lorentz transformation. my ques when we wrote that inverse transformation part why we wrote λ-1 μρ but not λ-1μ ρ ? I am sorry if you still didn't get it.. I am using this first time.. so don't know proper way to write eqns. so my eqns may look misleading... but if you get my point please answer me. thanks. Last edited: Sep 6, 2014 6. Sep 6, 2014 ### ChrisVer Another way to see the Proca equation, is by taking the Lagrangian of a spin-1 massive vector field without sources: [itex] L= -aF_{\mu \nu} F^{\mu \nu} + \frac{1}{2} m^{2} A_{\mu} A^{\mu}$ The first term contains the gauge invariant kinetic terms and the second is the mass term. $F_{\mu \nu} \equiv \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$ Now if you try to find the EOM: $\partial_{\rho}\frac{\partial L}{\partial (\partial_{\rho} A_{\sigma})}- \frac{ \partial L}{\partial A_{\sigma}}=0$ You have for the 2nd term: $\frac{ \partial L}{\partial A_{\sigma}}= m^{2} A^{\sigma}$ and for the 1st term: $-a \partial_{\rho}\frac{\partial (F_{\mu \nu} F^{\mu \nu})}{\partial (\partial_{\rho} A_{\sigma})}=-2a \partial_{\rho} F^{\mu \nu} \frac{\partial (\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})}{\partial (\partial_{\rho} A_{\sigma})} = -2a \partial_{\rho} F^{\mu \nu} (\delta^{\rho}_{\mu} \delta^{\sigma}_{\nu}-\delta^{\rho}_{\nu} \delta^{\sigma}_{\mu})=-2a \partial_{\rho} (F^{\rho \sigma} - F^{\sigma \rho})= -4 a \partial_{\rho} F^{\rho \sigma}$ So that's a reason why someone can choose $a= \frac{1}{4}$ and have: $-\partial_{\rho} F^{\rho \sigma} - m^{2} A^{\sigma} =0$ And you get what you ask for. Now what would your "2nd" choice mean? it would have to mean that $a= - \frac{1}{4} <0$ and so the 1st term in the Lagrangian would have a positive sign. That's bad I guess because then the vacuum would have to have negative energy ... Also in general you can reach this lagrangian (without the mass) from the "classical" electromagnetic lagrangian with the correct signs which leads to the term: $-\frac{1}{4}F^{2}$ As for your second question... The positioning of the indices can be very tough to follow, and sometimes people in notes or books can mix them up without a problem, because afterall what they want to say becomes clear... The only time someone has to be careful with the indices is when they denote the representation and not the rows-columns. It's difficult for you to try explaining in words what you are asking, so it's better to show us what you mean. To me, as you write them, there is no 1st or 2nd position. In general what you wrote as $(\Lambda ^{-1} \omega \Lambda)_{\mu \nu}$ means the mu nu component of the matrix $(\Lambda ^{-1} \omega \Lambda)$...So take the $\Lambda$ its inverse and mix them with the $\omega_{\rho \sigma}$ as: $[\Lambda^{-1}] [\omega] [\Lambda]$ where by brackets I mean matrices. Last edited: Sep 6, 2014 7. Sep 7, 2014 ### samalkhaiat The second equation leads to wrong sign for the mass term. Remember, you always need to satisfy the dispersion relation $E^{2} = p^{2} + m^{2}$. The “first place” is already occupied by the index $\mu$. [tex]\left( \lambda^{ - 1 } \omega \lambda \right)_{ \mu \nu } = ( \lambda^{ - 1 } )_{ \mu }{}^{ \rho } \omega_{ \rho \sigma } \lambda^{ \sigma }{}_{ \nu } .$$
For Lorentz transformation, we have
$$( \lambda^{ - 1 } )_{ \mu }{}^{ \rho } = \lambda^{ \rho }{}_{ \mu } .$$
So, the $\mu \nu$ matrix element is
$$\left( \lambda^{ - 1 } \omega \lambda \right)_{ \mu \nu } = \omega_{ \rho \sigma } \lambda^{ \rho }{}_{ \mu } \lambda^{ \sigma }{}_{ \nu } .$$
In the Euclidean space, i.e. when we don’t distinguish between upstairs from down stairs indices, the $\mu \nu$ matrix element of $(ABC)$ is found by usual multiplication of matrices
$$( A B C )_{ \mu \nu } = A_{ \mu \rho } B_{ \rho \sigma } C_{ \sigma \nu } .$$
In Minkowski space, this matrix element has the following equivalent forms
$$A_{ \mu }{}^{ \rho } B_{ \rho \sigma } C^{ \sigma }{}_{ \nu } = A_{ \mu \rho } B^{ \rho \sigma } C_{ \sigma \nu } = A_{ \mu \rho } B^{ \rho }{}_{ \sigma } C^{ \sigma }{}_{ \nu } = A_{ \mu }{}^{ \rho } B_{ \rho }{}^{ \sigma } C_{ \sigma \nu } .$$

Sam

8. Sep 7, 2014

### kau

tensor notation contraction and ordering of elements

Last edited: Sep 7, 2014
9. Sep 7, 2014

### ChrisVer

You can write it as such, you only have to be careful about what you actually did... In your case you renamed mu into nu and vice versa, and also exchanged the indices of the omega [which would have to give you a minus which I don't see].
As for the other question: $\lambda^{a}_{b} C_{ac} = C_{bc}$. Do you want more explanation on that?

As for the position of the elements of the tensor, yes in most cases you can do that. Except for if your elements are not commuting. No such case comes in my mind now.

Last edited: Sep 7, 2014
10. Sep 7, 2014

### kau

here you have assume the metric to be of form (-,+,+,+).
if you work with (+,-,-,-) probably you could have written it like

$-\partial_{\rho} F^{\sigma \rho} - m^{2} A^{\sigma} =0$ ~ to get the energy relation correct. so in any case checking energy relation would give the right choice. isn't it??

11. Sep 7, 2014

### ChrisVer

I don't see where you used the metric in what you did [flipping the indices of the F without changing the sign is not a part of metric choice, but the fact that F is an antisymmetric tensor]. In fact the metric choice doesn't really play any difference - the ones you said are opposite to each other, and in the Lagrangian the metrics appear twice in the kinetic term: $F_{\mu \nu} F^{\mu \nu} = \eta^{\mu \rho} \eta^{\nu \sigma} F_{\mu \nu} F_{\rho \sigma}$
Also energy results don't depend on the choice of the metric.

Last edited: Sep 7, 2014
12. Sep 7, 2014

### kau

yeah if you change ${ \mu}$ and ${ \nu}$ in w and x and ∂ in all places then you will get $-L^{ \nu \mu}$ which is equivalent to $L^{ \mu \nu}$ .. but my question is in the first place why I can't write $δω_{ \mu \nu} x^{ \nu} ∂^{ \mu}$??? why writing in this way leads to an expression which is not quite right??? [/QUOTE]
if you want to explain it,please do. I understand this result. but really i do not have clue why that quantity should not be$C_{cb}$??

Last edited: Sep 7, 2014
13. Sep 7, 2014

### kau

so, I have following eqn of motion
$-∂_{\rho}F^{\rho \sigma }-m^{2}A^{\sigma}=0$
therefore, $-∂^{2} A^{/sigma}-m^{2}A^{\sigma}=0$ using Lorentz gauge.
now if $∂^{2}=-E^{2}+p^{2}$ then I will have correct form $E^{2}-p^{2}=m^{2}$ for other choice of metric I will get an extra negative sign.

14. Sep 7, 2014

### ChrisVer

Because $\lambda$ tensor will act on the 1st components of the $C$... Or in other words you will have the tensor product which will give you [in components] $S^{a}_{dac} = S_{dc}$. At least that's how I understand it.
As for your last, then my metric choice is (+ - - - ), because the final result was $(\partial F)^{\sigma} + m^{2} A^{\sigma} =0$
Now if you suddenly change the sign of the metric, there should appear an extra change in the sign of the mass term... because in the EOM you had to take the derivative of $A^{2} = \eta_{\mu \nu} A^{\mu} A^{\nu}$ wrt $A^{\sigma}$. The result always remains the wanted one, that's why I said the energy result doesn't depend on the metric choice... Otherwise you wouldn't be able to compare energetically results of different metric choices.
Ah and I forgot the other question... Because then you are not getting a well defined matrix multiplication... you have to take the transpose of omega ... or else you are writting that the rows will multiply rows or collumns will multiply collumns...
In matrices when you have $A^a C_{ba} B^{b}= A^{T} C^{T} B \ne A^{T}CB$

Last edited: Sep 7, 2014
15. Sep 7, 2014

### kau

ok .. I got you. but for the last question I still find any reason which stops me from writing
$∂ω_{\mu \nu} x^{\nu} ∂^{\mu}$ .... say$∂ω = C_{ba}$ and $x =A^a$ and $∂{\mu}=B^{b}$ then it's all fine. I can understand that what you are saying is true but I am not getting what is wrong in my choice?? would you little elaborate this part??

16. Sep 7, 2014

### ChrisVer

Nevertheless, in Schrednicki's you have 34.1, where you insert 34.5:

$(I - \frac{i}{2} \delta \omega_{\mu \nu} M^{\mu \nu}) \psi_{a}(x) (I + \frac{i}{2} \delta \omega_{\mu \nu} M^{\mu \nu}) = (\delta^{~b}_{a} + \frac{i}{2} \delta \omega_{\mu \nu} (S^{\mu \nu})^{~b}_{a}) \psi_{b}$

$\psi_{a}\frac{i}{2} \delta \omega_{\mu \nu} M^{\mu \nu}-\frac{i}{2} \delta \omega_{\mu \nu} M^{\mu \nu} \psi_{a} = -\delta \omega_{\nu \mu} x^{\nu} \partial^{\mu} \psi(x)_{a}+ \frac{i}{2} \delta \omega_{\mu \nu} (S^{\mu \nu})^{~b}_{a}\psi_{b}$

So you are asking why you get it like that?
In general what he does is a Taylor expansion of $\psi ( (I+\delta \omega^{T}) x)$ from that you get the indices...

Last edited: Sep 7, 2014
17. Sep 7, 2014

### samalkhaiat

No, you can’t. $\omega_{ \mu \nu } x^{ \nu } \partial^{ \mu } = - \omega_{ \mu \nu } x^{ \mu } \partial^{ \nu }$
Use the fact that contracted (dummy) indices can be relabelled freely. So, you can write
$$\omega_{ \mu \nu } x^{ \mu } \ \partial^{ \nu } \phi = \omega_{ \rho \sigma } \ x^{ \rho } \ \partial^{ \sigma } \phi . \ \ \ (1)$$
Using the identity, $A = ( A + A ) / 2$, we can rewrite (1) as
$$\omega_{ \mu \nu } \ x^{ \mu } \ \partial^{ \nu } \phi = \frac{ 1 }{ 2 } \left( \omega_{ \mu \nu } \ x^{ \mu } \ \partial^{ \nu } \phi + \omega_{ \rho \sigma } \ x^{ \rho } \ \partial^{ \sigma } \phi \right) .$$
Now, letting $\rho = \nu$ and $\sigma = \mu$, we get
$$\omega_{ \mu \nu } \ x^{ \mu } \ \partial^{ \nu } \phi = \frac{ 1 }{ 2 } \left( \omega_{ \mu \nu } \ x^{ \mu } \ \partial^{ \nu } \phi + \omega_{ \nu \mu } \ x^{ \nu } \ \partial^{ \mu } \phi \right) .$$
But $\omega_{ \nu \mu } = - \omega_{ \mu \nu }$. Thus
$$\omega_{ \mu \nu } \ x^{ \mu } \ \partial^{ \nu } \phi = \frac{ 1 }{ 2 } \omega_{ \mu \nu } \left( x^{ \mu } \partial^{ \nu } - x^{ \nu } \partial^{ \mu } \right) \phi . \ \ \ \ \ \ (2)$$

You don’t need to mention any textbook for me. Instead, I think it is better to show you a step by step derivation of this equation. Under the Lorentz group, finite-component fields on space-time transform by finite-dimensional (matrix) representations
$$\bar{ \phi }_{ a } ( \bar{ x } ) = D_{ a }{}^{ b } \phi_{ b } ( x ) , \ \ \ \ \ \ \ (3)$$
where $D$ is a representation matrix
$$D_{ a }{}^{ c }( \Lambda_{ 1 } ) \ D_{ c }{}^{ b }( \Lambda_{ 2 } ) = D_{ a }{}^{ b }( \Lambda_{ 1 } \Lambda_{ 2 } ) .$$
Infinitesimally, we may write
$$D_{ a }{}^{ b } = \delta_{ a }^{ b } - \frac{ i }{ 2 } \omega_{ \mu \nu } ( S^{ \mu \nu } )_{ a }{}^{ b } , \ \ \ \ \ \ \ (4)$$
where $S^{ \mu \nu }$ are the appropriate spin matrices for the field $\phi$. They satisfy the Lorentz algebra.
However, since $\phi_{ a }( x )$ (for all a’s) is an operator-valued field, it also transforms by (infinite-dimensional) unitary representation,$U( \Lambda )$, of the Lorentz group.
$$\bar{ \phi }_{ a } ( \bar{ x } ) = U^{ - 1 } \ \phi_{ a } ( \bar{ x } ) \ U . \ \ \ \ \ (5)$$
In terms of the abstract Lorentz generators, $M^{ \mu \nu }$, and the infinitesimal parameters $\omega_{ \mu \nu }$, we may write the unitary operator as
$$U( \Lambda ) = 1 - \frac{ i }{ 2 } \omega_{ \mu \nu } \ M^{ \mu \nu } . \ \ \ \ \ (6)$$
From (3) and (5), we find the finite transformation rule for the field operator
$$\bar{ \phi }_{ a } ( \bar{ x } ) = U^{ - 1 } \ \phi_{ a } ( \bar{ x } ) \ U = D_{ a }{}^{ b } \ \phi_{ b } ( x ) . \ \ \ (7)$$
Using
$$x = \Lambda^{ - 1 } \ \bar{ x } ,$$
we rewrite (7) as
$$\bar{ \phi }_{ a } ( \bar{ x } ) = U^{ - 1 } \ \phi_{ a } ( \bar{ x } ) \ U = D_{ a }{}^{ b } \phi_{ b } ( \Lambda^{ - 1 } \ \bar{ x } ) .$$
Dropping the bars from the coordinates, we get
$$\bar{ \phi }_{ a } ( x ) = U^{ - 1 } \ \phi_{ a } ( x ) \ U = D_{ a }{}^{ b } \ \phi_{ b } ( \Lambda^{ - 1 } x ) . \ \ \ \ (8)$$
Now, we will try to find the infinitesimal version of (8). Using
$$( \Lambda^{ - 1 } )^{ \mu \nu } x_{ \nu } = x^{ \mu } - \omega^{ \mu \nu } \ x_{ \nu } ,$$
we expand $\phi ( \Lambda^{ - 1 } x )$ to first order as
$$\phi_{ b } ( \Lambda^{ - 1 } x ) = \phi_{ b } ( x ) - \omega^{ \mu \nu } \ x_{ \nu } \ \partial_{ \mu } \phi_{ b } ( x ) . \ \ \ \ \ (9)$$
Substituting the equations (4), (6) and (9) in equation (8) and keeping only first order terms, we find
$$\bar{ \phi }_{ a } ( x ) - \phi_{ a } ( x ) = \frac{ i }{ 2 } \omega_{ \mu \nu } \ [ M^{ \mu \nu } , \phi_{ a } ( x ) ] = - \omega^{ \mu \nu } \ x_{ \nu } \ \partial_{ \mu } \phi_{ a } - \frac{ i }{ 2 } \omega_{ \mu \nu } ( S^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) . \ (10)$$
Using $\omega^{ \mu \nu } = - \omega^{ \nu \mu }$ and
$$\omega^{ \nu \mu } \ x_{ \nu } \ \partial_{ \mu } = \omega_{ \nu \mu } \ x^{ \nu } \ \partial^{ \mu } = \omega_{ \mu \nu } \ x^{ \mu } \ \partial^{ \nu } ,$$
equation (10) becomes
$$\delta \phi_{ a } ( x ) = \frac{ i }{ 2 } \omega_{ \mu \nu } \ [ M^{ \mu \nu } , \phi_{ a } ( x ) ] = \omega_{ \mu \nu } \ x^{ \mu } \ \partial^{ \nu } \phi_{ a } - \frac{ i }{ 2 } \omega_{ \mu \nu } ( S^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) .$$
Now, if we use (2) in the first term on the right-hand-side, we find
$$\delta \phi_{ a } ( x ) = \frac{ i }{ 2 } \omega_{ \mu \nu } \ [ M^{ \mu \nu } , \phi_{ a } ( x ) ] = \frac{ 1 }{ 2 } \omega_{ \mu \nu } \ ( x^{ \mu } \partial^{ \nu } - x^{ \nu } \partial^{ \mu } ) \phi_{ a } ( x ) - \frac{ i }{ 2 } \omega_{ \mu \nu } \ ( S^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) .$$
So, the equation you are after follows from
$$[ i M^{ \mu \nu } , \phi_{ a } ( x ) ] = ( x^{ \mu } \partial^{ \nu } - x^{ \nu } \partial^{ \mu } ) \phi_{ a } ( x ) - i ( S^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } ( x ) .$$

Neither! The question does not make sense. $\lambda^{ n }{}_{ a } C_{ n u } = C_{ a u }$ if and only if $\lambda^{ n }{}_{ a }$ is equal to the Kronecker delta $\delta^{ n }_{ a }$.

Yes, for numerical tensors and matrix elements the order does not matter.

Sam

18. Sep 7, 2014

### ChrisVer

Samal I think the problem is that you derived:
$[\phi_{a}, M^{\mu \nu}] = L^{\nu \mu} \phi_{a} + i (S^{\mu \nu})^{~~b}_{a} \phi_{b}$

while in Schrednicki [the one kau mentions to have problem with] is:
$[\phi_{a}, M^{\mu \nu}] = L^{\mu \nu} \phi_{a} + i (S^{\mu \nu})^{~~b}_{a} \phi_{b}$

However I also reached your result.

19. Sep 7, 2014

### samalkhaiat

With the D matrix and the U operator are given by
$$D = 1 - \frac{ i }{ 2 } \omega_{ \mu \nu } \ S^{ \mu \nu } \ , \ \ U = 1 - \frac{ i }{ 2 } \omega_{ \mu \nu } \ M^{ \mu \nu } ,$$
and in any metric convention, the correct transformation is given by
$$[i M^{ \mu \nu } , \phi_{ a } ] = ( x^{ \mu } \ \partial^{ \nu } - x^{ \nu } \ \partial^{ \mu } ) \phi_{ a } - i ( S^{ \mu \nu } )_{ a }{}^{ b } \phi_{ b } .$$
Notice that I did not write $L^{ \mu \nu }$ because its definition depends on the metric one uses.
Any way, who is Steven ?

Sam

20. Sep 12, 2014

### kau

let me tell you what I actually mean in this part.
$\lambda^{\mu \nu} X^{\alpha}{}_{mu}= Y^{\nu \alpha} or y^{\alpha \nu}$
now I guess my question makes sense..
and you have written that numerical and matrix component tensor cases order does not matter. ok fine.. but for just to get one clarification i am asking you this.. so the point is we should only worry about tensor or matrix component ordering when they are non commutating.???
thanks.

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