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Ward identity

  1. Oct 31, 2014 #1

    kau

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    I was reading Schwartz's qft book. I saw the proof of ward identity taking pair annihilation as an example. he claimed he didn't assume that photon is massless in this derivation. but i have confusion with this statement. gauge invariance is a fact related to massless particles. now he has replaced $$\epsilon^{\mu}$$ with $$p^{\mu}$$ which is possible because of gauge invariance. it's like shifting $$\epsilon^{\mu}$$ to $$\epsilon^{\mu} + p^{\mu}$$ such a way so that it becomes equal to momentum of that particle. therefore in this derivation he has used gauge invariance and therefore assumed masslessness of photon. So why did he claim that it doesn't matter involved photon is physical or not??? wheredid i miss thepoint ??is there any other reason behind the above replacement??
     
  2. jcsd
  3. Nov 5, 2014 #2
    Thanks for the post! Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
  4. Nov 6, 2014 #3

    nrqed

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    What we do is not to change [itex] \epsilon^{\mu}[/itex] to [itex]\epsilon^{\mu} + p^{\mu} [/itex]. We clearly cannot do this since these two quantities have different units. What we do is to replace [itex] \epsilon^{\mu} [/itex] with [itex] p^{\mu} [/itex] as you initially said. When we do that in the amplitude, we obtain terms proportional to [itex] p^\mu p_\mu [/itex] which is zero for a massless particle.

    Hope this helps
     
  5. Nov 7, 2014 #4

    kau

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    Ok. then tell me how this replacement is justified on physical ground?
     
  6. Nov 7, 2014 #5

    nrqed

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    It is not a physical argument. It is a mathematical argument. Gauge invariance is a symmetry of the theory and when we impose it, it has consequences on the form of the lagrangian and this in turns leads to physical predictions. So we are not saying that physically [itex] \epsilon_\mu [/itex] is equal to [itex] p_\mu [/itex], we are saying that there is a symmetry of the theory such that if we replace the polarization by the momentum, the QED amplitudes must vanish. A consequence of this mathematical property is that the photon must be massless.
     
  7. Nov 7, 2014 #6

    kau

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    Ok let me rephrase my doubt. Gauge invariance is a symmetry of massless field. in principle gauge invariance is replacing $$A_{\mu} with A_{\mu}+\delta_{\mu}\alpha$$ ... What is the momentum space representation of ##A_{\mu}## .. it's simply ##\epsilon_{\mu}## .. and in momentum space $$\delta_{\mu}\alpha$$ where $$\alpha$$ is a function of x then this quantity is simply equivalent to ##p_{\mu}##... so in short gauge invariance is equivalent to replacing ##\epsilon_{\mu}## with ##\epsilon_{\mu}+p_{\mu}## .. that's why I said it equal in my first question... and the problem of units can be fixed by putting planck's constant and c... so I think this should justify the replacement. what do you think??? I might be wrong..but I don't see any discrepancy in the above argument. now the second thing is in Schwartz's qft book he said in derivation of ward's identity through example that he never assumed the fact that photon is massless... so the photons involved may be unphysical... but the moment he did above replacement he is using masslessness of field.. so what was his point?? my second doubt...
    Thanks..
     
  8. Nov 9, 2014 #7

    samalkhaiat

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    You ask me to participate in this thread! Since I have not seen the book, I cannot help you with issues related to the way the book explains the subject. However, since the subject is highly nontrivial, I will say few things about it.
    Let us consider a typical QED process such as the Compton scattering
    [tex]\gamma ( k ) + e ( p ) \to \gamma ( \bar{ k } ) + e ( \bar{ p } ) .[/tex]
    Its scattering amplitude is given by
    [tex]\langle f | ( S - 1 ) | i \rangle = \epsilon_{ \mu }^{ * } ( \bar{ k } ) \ M^{ \mu \nu } ( p , \bar{ p } ; k , \bar{ k } ) \ \epsilon_{ \nu } ( k ) , \ \ \ \ \ (1)[/tex]
    where
    [tex]M^{ \mu \nu } ( p , \bar{ p } ; k , \bar{ k } ) = ( - i e )^{ 2 } \int d^{ 4 } x d^{ 4 } y \ e^{ i ( \bar{ k } y - k x )} \langle \bar{ p } | T \left( J^{ \mu } ( y ) \ J^{ \nu } ( x ) \right) | p \rangle . \ \ \ (2)[/tex]
    Equation (2) is not actually true! The RHS usually contains non-covariant terms called “Seagulls”. Bellow, you will see the reason for not including these terms in (2).
    Usually, you are told the following: “because of gauge invariance” the amplitude does not change if, for any one or more photons, you make the substitution
    [tex]\epsilon_{ \mu } \to \epsilon_{ \mu } + c k_{ \mu } , \ \ \ \ (3)[/tex]
    where [itex]c[/itex] is a totally arbitrary (not necessarily constant) number. This means that the following relations must hold,
    [tex]\bar{ k }_{ \mu } M^{ \mu \nu } = M^{ \mu \nu } k_{ \nu } = 0 . \ \ \ \ (4)[/tex]
    Similar relations hold for any QED process with any number of external photons.
    This is highly non-trivial and often misunderstood subject. So, further explanation is required.
    In classical electromagnetic theory, the invariance under Eq(3) follows from the fact that the theory is invariant under the local gauge transformation
    [tex]A^{ \mu } \to A^{ \mu } + \frac{ 1 }{ e } \partial^{ \mu } \Lambda ( x ) . \ \ \ (5)[/tex]
    Indeed, Eq(3) is nothing but the Fourier transform of Eq(5) when [itex]A^{ \mu }[/itex] is a plane wave (*).
    However, in QED the above reasoning is incorrect because:
    i) the substitution (3) will only follow from (5) if [itex]\Lambda ( x )[/itex] is not a c-number function but is chosen to be a quantized scalar field linear in the photon creation and annihilation operators.
    ii) once a choice of gauge is made, the theory is no longer invariant under local gauge transformation.
    But, if we impose the gauge condition in such a way that global gauge invariance is respected, the invariance of the QED matrix elements under (3) follows from the following two conditions
    1) The conservation of the global [itex]U(1)[/itex] symmetry current,
    [tex]\partial_{ \mu } J^{ \mu } = 0 . \ \ \ \ \ (6)[/tex]
    2) The current [itex]J^{ \mu }[/itex] is itself gauge invariant, i.e.,
    [tex]\delta_{ \Lambda } J^{ \mu } \sim [ i J^{ 0 } ( x ) , J^{ \mu } ( 0 ) ] \delta ( x^{ 0 } ) = 0 . \ \ \ (7)[/tex]
    This also follows from the naïve use of the canonical anti-commutation relations. Indeed, for [itex]J^{ \mu } = \bar{ \psi } \gamma^{ \mu } \psi[/itex], one finds
    [tex][ J^{ \mu } ( x ) , J^{ \nu } ( 0 ) ]_{ x^{ 0 } = 0 } = \bar{ \psi } ( x ) ( \gamma^{ \mu } \gamma^{ 0 } \gamma^{ \nu } - \gamma^{ \nu } \gamma^{ 0 } \gamma^{ \mu } ) \psi ( x ) \delta^{ 3 } ( \vec{ x } ) .[/tex]
    Therefore,
    [tex][ J^{ 0 } ( x ) , J^{ \nu } ( 0 ) ]_{ x^{ 0 } = 0 } = 0 . \ \ \ \ \ \ (7a)[/tex]
    This is not actually true. Careful calculations, due to Schwinger, show that for [itex]\nu[/itex] spacelike, this commutator must contain derivatives of [itex]\delta[/itex]-functions. Such singular terms in the equal time commutator [itex][ J^{ 0 } ( x ) , J^{ i } ( y ) ][/itex] are called Schwinger terms. However, experience with many models has shown that the “Seagull” terms in Eq(2) “always” cancel the Schwinger terms. This is why we didn’t include the Seagulls in Eq(2).
    Okay, now we will use the above two conditions to prove the invariance of the S-matrix under the substitution (3). Contracting Eq(2) with [itex]\bar{ k }_{ \mu }[/itex] and using the relation
    [tex]i \bar{ k }_{ \mu } e^{ i \bar{ k } y } = \partial_{ \mu }^{ ( y ) } e^{ i \bar{ k } y } ,[/tex]
    then integrating by parts, we find
    [tex]\bar{ k }_{ \mu } M^{ \mu \nu } = i ( - i e )^{ 2 } \int d^{ 4 } x d^{ 4 } y \ e^{ i ( \bar{ k } y - k x ) } \ \langle \bar{ p } | \partial_{ \mu }^{ ( y ) } T \left( J^{ \mu } ( y ) \ J^{ \nu } ( x ) \right) | p \rangle . \ \ (8)[/tex]
    But, from the definition of the T-product, we have
    [tex]\partial_{ \mu }^{ ( y ) } T \left( J^{ \mu } ( y ) \ J^{ \nu } ( x ) \right) = T \left( \partial_{ \mu } J^{ \mu } \ J^{ \nu } ( x ) \right) + [ J^{ 0 } ( y ) , J^{ \nu } ( x ) ] \delta ( y^{ 0 } - x^{ 0 } ) . \ \ (9)[/tex]
    Now, using Eq(6) in the first term and Eq(7) in the second term of Eq(9), we find
    [tex]\partial_{ \mu }^{ ( y ) } T \left( J^{ \mu } ( y ) \ J^{ \nu } ( x ) \right) = 0 .[/tex]
    Thus the RHS of Eq(8) is zero and hence Eq(4) is proved.
    It is important to note that this result, i.e., the invariance of the amplitude under Eq(3), applies not to individual diagrams but only to the sums of all possible diagrams.
    A different and more fundamental approach for proving Eq(4), is to impose on the S-matrix the requirement of Lorentz invariance. This approach uses the fundamental nature of massless spin-1 vector fields in QED: Recall that in the radiation gauge, the polarization vector is given by
    [tex]\epsilon^{ 0 } ( k , \lambda ) = k_{ \mu } \epsilon^{ \mu } ( k , \lambda ) = 0 .[/tex]
    This clearly shows that the polarization “vector” (hence vector potential) is not a genuine Lorentz vector. Indeed, a Lorentz [itex]\Lambda[/itex] sends [itex]\epsilon^{ \mu }[/itex] into
    [tex]\bar{ \epsilon }^{ \mu } = \Lambda^{ \mu }{}_{ \nu } \epsilon^{ \nu } - \frac{ k^{ \mu } }{ k^{ 0 } } \Lambda^{ 0 }{}_{ \nu } \epsilon^{ \nu } .[/tex]
    Thus, in order for the S-matrix to remain Lorentz-invariant, it is necessary that the second term should give zero contribution, which leads to Eq(4), at least for [itex]k_{ \mu}[/itex] on the light cone.
    (*) See
    https://www.physicsforums.com/threa...t-spin-1-with-a-4-vector.772226/#post-4860248
     
    Last edited: Nov 9, 2014
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