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Srednicki problem 3.3

  1. Dec 19, 2013 #1

    WannabeNewton

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    1. The problem statement, all variables and given/known data

    Use ##U(\Lambda)^{-1}\varphi(x)U(\Lambda) = \varphi(\Lambda^{-1}x)## to show that ##U(\Lambda)^{-1}a(\mathbf{k})U(\Lambda) = a(\Lambda^{-1}\mathbf{k})## and ##U(\Lambda)^{-1}a^{\dagger}(\mathbf{k})U(\Lambda) = a^{\dagger}(\Lambda^{-1}\mathbf{k})## and hence that ##U(\Lambda)|k_1...k_n\rangle = |\Lambda k_1...\Lambda k_n \rangle## where ##|k_1...k_n\rangle ## is a multi-particle state of ##n## particles with momenta ##k_1,...,k_n##.

    2. Relevant equations

    The real free scalar field: ##\varphi(x) = \int d\tilde{k} [a(\mathbf{k})e^{ikx} + a^{\dagger}(\mathbf{k})e^{-ikx}]## where ##d\tilde{k}## is the Lorentz-invariant measure and ##kx = k_{\mu}x^{\mu}##.

    3. The attempt at a solution

    I can't even start the problem because of what Srednicki writes for the transformation of the annihilation and creation operators under the unitary representation ##U(\Lambda)## of the Lorentz transformation ##\Lambda##. In particular, he writes ##a(\Lambda^{-1}\mathbf{k})## and ##a^{\dagger}(\Lambda^{-1}\mathbf{k})## but ##\mathbf{k}## is a 3-vector (in particular the 3-momentum) so what does ##\Lambda^{-1}\mathbf{k}## even mean? Matrix representations of Lorentz transformations have Lorentz indices and hence must be contracted with other Lorentz indices so as to result in Lorentz indices after the contraction e.g. ##k^{\mu'} = (\Lambda^{-1})^{\mu'}{}{}_{\nu}k^{\nu} = \Lambda_{\nu}{}{}^{\mu'}k^{\nu}##. But a 3-vector like ##\mathbf{k}## doesn't have Lorentz indices so I can't make any sense of ##\Lambda^{-1}\mathbf{k}##. Thanks in advance!
     
    Last edited: Dec 19, 2013
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  3. Dec 19, 2013 #2

    vanhees71

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    This is very sloppy notation. Of course, you must use the energy-momentum four vector to define the Lorentz transformation of momentum. For free particles, the four-momentum is onshell, i.e.,
    [tex]E(\vec{p})=\sqrt{m^2+\vec{p}^2} \; \Leftrightarrow \; p_{\mu} p^{\mu}=m^2.[/tex]
    Of course also the Lorentz transformed four vector,
    [tex]p'^{\mu} = {\Lambda^{\mu}}_{\nu} p^{\nu}[/tex]
    obeys the on-shell condition.

    Just use the "local" transformation property of the field operators and the mode decomposition of the field operator with respect to annihilation and creation operators with respect to the momentum eigenbasis, as well as the Lorentz invariance of the invariant integration measure over on-shell momenta,
    [tex]\frac{\mathrm{d}^3 \vec{p}}{2 E(\vec{p})}=\mathrm{d}^4 p \Theta(p^0) \delta(p^2-m^2).[/tex]
     
  4. Dec 19, 2013 #3

    WannabeNewton

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    Thanks vanhees! Should I Fourier transform ##\varphi(x)## to its momentum representation first? In chapter 3, Srednicki only gives the mode decomposition of ##\varphi(x)## in the coordinate representation: ##\varphi(x) = \int d\tilde{k} [a(\mathbf{k})e^{ikx} + a^{\dagger}(\mathbf{k})e^{-ikx}]##; will this problem be substantially easier if I work with ##\tilde{\varphi}(k) = \int d^4 x e^{-ikx}\varphi(x) ## and derive the mode decomposition of ##\tilde{\varphi}(k)## in terms of ##a(\mathbf{k})## and ##a^{\dagger}(\mathbf{k})## using the mode decomposition of ##\varphi(x)##?

    Also he defines the Lorentz invariant measure as ##d\tilde{k} = \frac{d^3 k}{(2\pi)^3 2\omega}## so I'm not sure how to go from this to the expression you wrote with the delta function, theta function, and ##d^4 k##.
     
  5. Dec 19, 2013 #4

    strangerep

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    Don't over-think this. I reckon this exercise can be done in just a few lines once you recognize some "cut-through" techniques.

    Hmm, now... how can I hint at those techniques without actually giving you the answer?

    Let's see...

    1) What else do you know about ##\Lambda##? I.e., what matrix properties does it satisfy?

    2) What can you say (if anything) about linear independence of ##e^{ikx}## and ##e^{-ikx}## ?

    3) Write ##kx## in more explicitly vectorial notation. E.g., similar to how we can write ##u \cdot v## as ##u^T v##. If we apply an inner-product preserving transformation to ##v##, how can we re-express the inner product with the transformation acting only ##u## ?

    That's a different issue, and (I think?) not essential for this exercise. Anyway, Srednicki mentions this in (3.16) and (3.17).

    BTW, keep in mind that the expressions for ##\phi## should perhaps be more explicitly written like this: $$
    \phi(x) ~=~ \int \Big[ \cdots \Big]_{\omega^2 - k^2 = m^2}
    $$ which emphasizes that ##\phi## has support only the mass hyperboloid in momentum space.
     
    Last edited: Dec 19, 2013
  6. Dec 19, 2013 #5

    WannabeNewton

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    Hi Strangerep thanks for replying. I was on a 6 hour bus ride from my university back to my home for Christmas break so I didn't see your post until now; however during my bus ride I had already started working on the problem using the method I mentioned earlier. I'll post what I have thus far and it would be awesome if you could help me figure out the last few steps because I'm stuck. In the meanwhile I'll use the prompts you gave in your post to see if I can do the problem using the faster/more efficient method that you had in mind. Thanks again!

    So the Fourier transform of the free scalar field will be given by ##\tilde{\varphi}(k) = \int d\tilde{k}'d^{4}x[a(\mathbf{k}')e^{i(k' - k)x} + a^{\dagger}(\mathbf{k}')e^{-i(k' + k)x}] \\= 2\pi \int dk'^{0}d^{3}k' \delta(k'^2 + m'^2)\theta(k'^0)[a(\mathbf{k}')\delta^{4}(k - k') +a^{\dagger}(\mathbf{k}')\delta^{4}(k + k') ] \\= 2\pi \int dk'^{0} \delta(k^2 + m^2) \theta(k'^0) [a(\mathbf{k}) \delta(k^0 - k'^0) +a^{\dagger}(\mathbf{-k}) \delta(k^0 + k'^0)] \\= 2\pi \delta(k^2 + m^2)[a(\mathbf{k})\theta(k^0) +a^{\dagger}(\mathbf{-k})\theta(-k^0)] ##

    where I have used (3.16).

    Furthermore, ##U^{-1}\tilde{\varphi}(k)U = \int d^{4}xe^{-ikx}U^{-1}\varphi(x)U = \int d^{4}xe^{-ikx}\varphi(\Lambda^{-1}x)##.

    Define ##x' = \Lambda^{-1}x## so that ##\int d^{4}xe^{-ikx}\varphi(\Lambda^{-1}x)=\int d^{4}x'e^{-ik(\Lambda x')}\varphi(x') = \int d^{4}x' e^{-i(\Lambda^{-1}k)x'}\varphi(x') = \tilde{\varphi}(\Lambda^{-1}k)##

    where I have used the fact that ##d^4 x = d^4 x'## under a Lorentz transformation since the space-time measure is Lorentz-invariant and ##\Lambda^{\mu}{}{}_{\nu}k_{\mu}x'^{\nu} = \Lambda_{\mu}{}{}^{\nu}k^{\mu}x'_{\nu} = (\Lambda^{-1})^{\nu}{}{}_{\mu}k^{\mu}x'_{\nu}##.

    So collecting everything together we have
    ##\tilde{\varphi}(k) = 2\pi \delta(k^2 + m^2)[a(\mathbf{k})\theta(k^0) +a^{\dagger}(\mathbf{-k})\theta(-k^0)] ## and ##U^{-1}\tilde{\varphi}(k)U = \tilde{\varphi}(\Lambda^{-1}k)##.

    This gives us ##\delta(k^2 + m^2)[\theta(k^0) U^{-1}a(\mathbf{k})U+\theta(-k^0)U^{-1}a^{\dagger}(\mathbf{-k})U] \\= \delta(k^2 + m^2)[\theta(\Lambda^{-1}k^0)a(\Lambda^{-1}\mathbf{k}) +\theta(-\Lambda^{-1}k^0)a^{\dagger}(-\Lambda^{-1}\mathbf{k})] ##

    This is where I'm stuck. I'm not sure how to go from the above equation to the desired result.
     
  7. Dec 20, 2013 #6

    strangerep

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    You could invoke something##^*## to restrict to orthochronous ##\Lambda##, and that would simplify your ##\theta## expressions. But that doesn't get you to the final result.

    TBH, I think that by Fourier transforming as your first step, you lost the ability to apply my hint (2).
    [Edit: Or maybe not. After fixing up your ##\theta## functions, you can note that for any given ##k_0##, only 1 of the terms is non-zero. That gets you a little closer, I think.]

    But to re-start with the very first eqn in the exercise, i.e.,
    $$
    U(\Lambda)^{-1} \phi(x) U(\Lambda) ~=~ \phi(\Lambda^{-1} x) ~.
    $$
    Note that the 2nd eqn in (3.34) is just the adjoint of the 1st, so you only need to prove the 1st. So try to get equations involving ##a## and ##a^\dagger## separately.

    [##^*##] [Edit: The "something" I mentioned above is on Srednicki p17. He excludes parity and time reversal from his treatment at this early stage -- to be revisited later.]
     
    Last edited: Dec 20, 2013
  8. Dec 20, 2013 #7

    vanhees71

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    Despite the fact that you use the east-coast instead of the west-coast metric (which I'm used to; what is Srednicky using?), you already used the relation for the invariant integration measure over momenta on the mass shell I gave. It's, however better to simply use the local transformation properties of the scalar field operator (it's constructed such that it fulfills that relation, because you aim at a local QFT from the very beginning!).

    Using also the east-cost convention for the metric this reads
    [tex]U^{-1}(\Lambda) \phi(x) U(\Lambda)=\phi(\Lambda^{-1} x).[/tex]
    This you translate into the mode decomposition for the uncharged scalar field
    [tex]\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi^3) 2 E(\vec{p})} \left [U^{-1}(\Lambda) a(\vec{p}) U(\Lambda) \exp(\mathrm{i} x \cdot p) + U^{-1}(\Lambda) a^{\dagger}(\vec{p}) U(\Lambda) \exp(-\mathrm{i} x \cdot p) \right]_{p^0=E(\vec{p})} \\ \stackrel{!}{=} \int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi^3) 2 E(\vec{p})} \left [a(\vec{p}) \exp(\mathrm{i} (\Lambda^{-1} x) \cdot p) + a^{\dagger}(\vec{p}) \exp(-\mathrm{i} (\Lambda^{-1}x) \cdot p) \right]_{p^0=E(\vec{p})}.[/tex]
    Now after some playing with the Lorentz-transformation properties of four-vectors on the right-hand side, doing an appropriate substitution in the momentum integral and then comparing the Fourier coefficients you should get the desired transformation properties of the annihilation and creation operators (note that [itex]U(\lambda)[/itex] is a unitary transformation!).
     
    Last edited by a moderator: Dec 20, 2013
  9. Dec 20, 2013 #8
    you have [itex]\tilde{\varphi}(k) = 2\pi \delta(k^2 + m^2)[a(\mathbf{k})\theta(k^0) +a^{\dagger}(\mathbf{-k})\theta(-k^0)][/itex],So for positive k0 you have
    [itex] 2\pi \delta(k^2 + m^2)a(\mathbf{k})=\tilde{\varphi}(k)[/itex].Now make the orthochronous Lorentz transformation on it and use it once again while doing to get the result.
     
  10. Dec 20, 2013 #9

    WannabeNewton

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    Thanks for the help guys! Strangerep thanks I'll redo the problem starting from your hint (2) right after I get one last thing cleared up based on what andrien said directly above: does an orthochronous Lorentz transformation necessarily preserve the positivity of ##k^0##? We have ##k^{0'} = \Lambda^{0'}_{0}k^{0} + \Lambda^{0'}_{i}k^{i}## and we know that ##\Lambda^{0'}_{0}k^{0} >0## but what about ##\Lambda^{0'}_{i}k^{i}##?

    Also, there is no loss of generality in restricting ourselves to positive ##k^0## right?

    Thanks again everyone.
     
    Last edited: Dec 20, 2013
  11. Dec 20, 2013 #10

    George Jones

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    Srednicky uses east-coast, which, as you note, is quite unusual for qft/high energy.

    Yes, hence the name "orthochronous". See theorem 1.3.3 of Naber.

    In the rest frame of the particle, what does this mean?
     
  12. Dec 20, 2013 #11

    WannabeNewton

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    Awesome thanks!

    Ok so in the rest frame of the particle, ##k^0 = m > 0## since rest mass is positive. Then using the theorem you referenced in Naber, restricting ourselves to orthochronous Lorentz transformations we have ##k^0 > 0## in all reference frames so we basically can only use ##k^0 > 0## right? It wouldn't even make sense to consider ##k^0 < 0##?

    Thanks George!
     
  13. Dec 20, 2013 #12

    strangerep

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    For this exercise you could also consider ##k^0 < 0##.
    Think about what the last line of your earlier post #5 simplifies to in this case...
     
  14. Dec 20, 2013 #13

    WannabeNewton

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    Right so if we happened to restrict ourselves to ##k^0 < 0## then we would get ##U^{-1} a^{\dagger}(-\mathbf{k})U = a(-\Lambda^{-1}\mathbf{k})## but if we restrict ourselves to ##k^0 < 0## will ##U^{-1} a^{\dagger}(-\mathbf{k})U = a(-\Lambda^{-1}\mathbf{k})## hold for all ##\mathbf{k}##? That's what we want right? We want it to hold for all possible 3-momenta ##\mathbf{k}##.

    Thanks!
     
  15. Dec 20, 2013 #14

    strangerep

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    I presume that's a typo on the rhs, and you meant: ## \dots = a^\dagger(-\Lambda^{-1}\mathbf{k})## ?

    Yes.
     
  16. Dec 20, 2013 #15

    WannabeNewton

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    Sorry, yes that was a typo. Also I think I phrased my question ambiguously. I don't get why we don't lose any generality by restricting ourselves to ##k^0 < 0## and subsequently deriving ##U^{-1}a^{\dagger}(-\mathbf{k})U = a^{\dagger}(-\Lambda^{-1} \mathbf{k})##. Why does this necessarily hold for all ##\mathbf{k}## even after restricting to ##k^0 < 0##? I only ask because ##\mathbf{k}## isn't independent of ##k^0##. Thanks!

    EDIT: also for the very last part of the problem i.e. to show that ##U|k_1...k_n \rangle = |\Lambda k_1...\Lambda k_n \rangle##, can I simply say that since ##U^{-1}a^{\dagger}(\mathbf{k})U = a^{\dagger}(\Lambda^{-1} \mathbf{k})## we have as a result ##U^{-1}a^{\dagger}(\Lambda \mathbf{k})U = a^{\dagger}(\mathbf{k})## hence
    ##U|k_1...k_n \rangle \\= U a^{\dagger}(\mathbf{k}_1)...a^{\dagger}(\mathbf{k}_n)|0\rangle \\= a^{\dagger}(\Lambda \mathbf{k}_1)Ua^{\dagger}(\mathbf{k}_2)...a^{\dagger}(\mathbf{k}_n)|0 \rangle \\= a^{\dagger}(\Lambda \mathbf{k}_1)...a^{\dagger}(\Lambda \mathbf{k}_n)|0\rangle \\= |\Lambda k_1...\Lambda k_n \rangle##
    ?
     
    Last edited: Dec 20, 2013
  17. Dec 21, 2013 #16

    strangerep

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    Note that I didn't say "restrict" to ##k^0 < 0##. I said "consider".

    But ok, let's go back to the full equation at the end of your post #5. This holds for all 4-vectors ##k## satisfying the mass-shell constraint ##k^2 = m^2##. We can rewrite that constraint as ##k^0 = \pm\sqrt{m^2 + |\mathbf{k}|^2}## . Therefore it is sufficient to consider your equation for these 2 cases separately. For each sign of ##k^0##, only one of ##\theta##'s is nonzero. So you get the transformation rule for ##a## and ##a^\dagger## separately. In each case, the rules apply to all 3-vector ##\mathbf{k}## -- only the sign of ##k^0## is different. And (just as well!), the rules turn out to be consistent with each other, as can be seen by taking adjoints.

    Yes, that's the basic idea, since the vacuum state is Lorentz-invariant.
     
  18. Dec 21, 2013 #17

    George Jones

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    Another way of looking at this, related to strangerep's comment

    is through

    What is ##\tilde{\varphi}(-k)##?

    These negative mass solutions are the infamous negative energy solutions, and are extremely important. Interpreted correctly, they give *positive* mass antiparticles. If you want, I can expand substantially on this.
     
  19. Dec 21, 2013 #18

    WannabeNewton

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    Ok so to make sure I understand, say we take an arbitrary ##\mathbf{k}## then there are two valid choices for ##k^0## that differ by sign. The positive choice gives us the equation for ##a## and the negative choice gives us the equation for ##a^{\dagger}## but regardless the equations will hold for all ##\mathbf{k}## since ##\mathbf{k}## was taken to be arbitrary and we can branch off into freedom to take both negative and positive ##k^0## in the on-shell relation.

    So, in other words, the set of all possible ##\mathbf{k}## along with the set of all associated positive ##k^0## gives us the equation for ##a## whereas the set of all possible ##\mathbf{k}## along with the set of all associated negative ##k^0## gives us the equation for ##a^{\dagger}##.

    Does that sound about right? And thanks a million for the help!

    Ah right since ##\tilde{\varphi}^{\dagger}(k) = \tilde{\varphi}(-k)##. Thanks!

    I would absolutely love it if you could! Srednicki's explanation of the issue was not at all clear to me (he does it very briefly in chapter 1). Thanks!
     
  20. Dec 21, 2013 #19

    strangerep

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    Yes.
     
  21. Dec 22, 2013 #20

    George Jones

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    So, in some sense, the negative mass hyperboloid doesn't give any information not already given by the positive mass hyperboloid. This is because this thread deals with a real scalar field, and, for real scalar fields, particles are their own antiparticles.

    This may take some time, but I (with the help of Folland and others) definitely will do this. It might be better if I put the material in this thread

    https://www.physicsforums.com/showthread.php?t=727571.
     
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