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Problems in elementary QFT

  1. Jun 12, 2009 #1
    Hi all,
    I'm trying to teach myself the basics of QFT. I'm using Peskin and Schroeder, and having a few difficulties reproducing a couple of the calculations. I don't think I've made careless algebraic slips, so before I show my working explicitly and beg for proof-reading I'd like to ask a couple of general questions.

    Firstly, canonical quantisation.
    1)Does the field operator act multiplicatively, as the ladder operators act on the complex exponentials inside the integral?
    2)Is there an explicit relationship between the operators
    [tex]a_{p} , a_{-p}[/tex]
    and a corresponding one for their adjoints? (Possibly one valid only when integrating over all momentum?) When trying to compute the hamilitonian and momentum operators (starting from the expressions 2.27, 2.28), I'm getting expressions in terms of operators in p and in -p, and it's not clear to me that they're equivalent to those given.

    Secondly, are there anything wrong with the relation
    [tex]\varepsilon^{ijk}\varepsilon_{ipq}=\delta^{j}_{p}\delta^{k}_{q}-\delta^{j}_{q}\delta^{k}_{p}[/tex]
    as applied to the 3d levi-civita symbol living in the spatial components of minkowski space?
    Thanks in advance.

    EDIT:Thanks christo :smile:
     
    Last edited: Jun 12, 2009
  2. jcsd
  3. Jun 12, 2009 #2

    malawi_glenn

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    for [tex]
    a_{p} , a_{-p}
    [/tex] you just have to swhithc sign of p i the definition.

    for evaluating the hamiltonian, you need some tricks...

    [tex]\int d\vec{p}f(\vec{p}) = [/tex]
    change p -> -p

    find the outcome of this as an excerice ;-)
     
  4. Jun 12, 2009 #3

    Avodyne

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    Operators always act multiplicatively in quantum mechanics. Ladder operators act on the state; the complex exponentials are just numbers.
    There is no relation among these operators themselves. Inside an integral, it can be useful to change the dummy integration variable from p to -p in some terms, as already noted by malawi_glenn.
    Not with metric signature (-,+,+,+). Can't help you with (+,-,-,-).
     
  5. Jun 13, 2009 #4
    Thanks for the responses guys.

    Malawi_Glenn:
    The good news is, your post convinced me that I shouldn't pick up a minus sign when making the obvious substitution needed to evaluate the hamiltonian, which I'd come to believe but wasn't 100% sure about. I've also taken [tex]\omega_p = \omega_{-p}[/tex], which would make physical sense (and gives the right answer :tongue:) but I've not noticed explicitly stated anywhere. The thing I'm struggling more with is the momentum, when I have terms in both p and -p that I can't appear to separate.
    As for the swapping the signs of the momentum, if I change the sign of the momentum density (and take [tex]\omega_p = \omega_{-p}[/tex])then I get [tex]a_p=a_p^{\dagger}[/tex], which is obviously not true by the commutation relations. Is the right way to think about this that you take the integrands of the field and momentum density operators to be operators indexed by p, and that you just change the sign on the indices when you make this substitution?

    Avodyne:
    By "multiplicatively" I meant in the way the position operator is multiplicative, as opposed to the effect of applying e.g. a differential operator. Would the effect of the field operator be unchanged if you swapped the orders of the exponentials and the ladder operators?

    As for the epsilon business, I am working in signature (+ - - -). The reason I ask is the following calculation, trying to parametrise an infinitesmal rotation:
    [tex]R= I-\frac{i}{2}\omega_{\mu\nu}S^{\mu\nu}
    \[=I-\frac{i}{2}\omega_{ij}S^{ij}\][/tex]
    Computing the second term
    [tex]\[\omega_{ij}S^{ij}=\frac{1}{2}\varepsilon_{ijk}\varepsilon^{ijl}\theta^k\Sigma_l\]
    \[=\frac{1}{2}(\delta^{j}_{j}\delta^{k}_{l}-\delta^{j}_{l}\delta^{j}_{k})\theta^k\Sigma_l\]
    \[=0\][/tex]
    Here [tex]\Sigma_l[/tex] is the matrix
    [tex]
    \[ \left( \begin{array}{cc}
    \sigma_l & 0 \\
    0 & \sigma_l \end{array} \right)\][/tex]
    As an aside, I've just noticed that Peskin and Schroeder don't lower the index on the matrix when they contract it with the levi-civita tensor, if that's significant?
     
  6. Jun 13, 2009 #5

    malawi_glenn

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    A good thing to have when one studies peskin is the errata:

    http://www.slac.stanford.edu/~mpeskin/QFT.html

    Now the hamiltonian, I have it fully calculated somewhere in my old notes. And it is hard to help you if you don't show us your calculations.
     
  7. Jun 13, 2009 #6
    Yes, you can switch these without any harm. The ladder operators act on the Hilbert space, i.e. the states [tex]|\Psi\rangle[/tex].

    As for the Levi-Cevita tensor, let me give some info:
    One defines it in the usual sense that it's a completely anti-symmetric tensor and [tex]\epsilon_{123} = +1[/tex]. Furthermore, by definition, [tex]\epsilon^{\mu\nu\lambda}[/tex] is obtained through raising via the Minkowski metric. Since the metric is symmetric and [tex]\epsilon_{123} [/tex] is antisymmetric, it follows that the object [tex]\epsilon^{\mu\nu\lambda}[/tex] is fully antisymmetric. So we only need to compute one component, for instance:
    [tex]\epsilon^{123} = \eta^{11}\eta^{22}\eta^{33}\epsilon_{123} = \pm(\eta_{\mu\nu})\epsilon_{123}[/tex] where the sign depends on which conventions you use for the metric. You can, ofcourse, also start with [tex]\epsilon^{123} = 1 [/tex].

    So as you can see, the relation you wrote down still holds on the basis that both objects are completely antisymmetric. But the overall sign of the relation you wrote down will depend on the choice of the metric (i.e. the sign of the spacial part of the metric)
     
    Last edited: Jun 13, 2009
  8. Jun 13, 2009 #7
    Xepma:
    Thanks for your reply; switching the overall sign of the expression is the only effect of the metric that I could think of. Unfortunately, this means my calculation is wrong in some more fundamental respect, as I know the answer should be the matrix
    [tex]
    \[ \left( \begin{array}{cc}
    \frac{\sigma.\theta}{2} & 0 \\
    0 & -\frac{\sigma.\theta}{2} \end{array} \right)\]
    [/tex]
    where theta is a 3-vector specifying rotations around the x,y,z axes and sigma is the "vector of pauli matrices". (The overall thing is then obviously a 4x4 matrix given in block form).

    Malawi_glenn:
    If I'm allowed to set
    [tex]\omega_p = \omega_{-p}[/tex]
    and
    [tex]\int\frac{d^3p}{(2\pi)^3}a^{\dagger}_{-p}a_{-p}=\int^{+\infty}_{-\infty}\frac{d^3p}{(2\pi)^3}a^{\dagger}_{p}a_{p}[/tex]
    then I think my calculation for the Hamiltonian is correct. Here's my effort with the momentum:
    [tex]
    P= \[-\int d^3x\pi(x)\nabla\phi(x)\]
    \[ = -\int d^3x\int\frac{d^3pd^3p'}{(2\pi)^6}(-i)\sqrt{\frac{\omega_p}{2}}(a_p-a_{-p}^{\dagger})e^{ip.x}.\nabla\frac{1}{\sqrt{2\omega_{p'}}}(a_{p'}+a_{-p}^{\dagger})e^{ip'.x}\]
    \[=-\int d^3x\int\frac{d^3pd^3p'}{(2\pi)^6}ip'(-i)\sqrt{\frac{\omega_p}{2}}(a_p-a_{-p}^{\dagger}).\frac{1}{\sqrt{2\omega_{p'}}}(a_{p'}+a_{-p'}^{\dagger})e^{i(p+p').x}\]
    \[=-\int\frac{d^3pd^3p'}{(2\pi)^3}p'\delta(p'+p)\sqrt{\frac{\omega_p}{2}}(a_p-a_{-p}^{\dagger}).\frac{1}{\sqrt{2\omega_{p'}}}(a_{p'}+a_{-p'}^{\dagger})\]
    \[=\frac{1}{2}\int\frac{d^3p}{(2\pi)^3}p(a_pa_{-p}+a_pa_p^\dagger-a_{-p}^\dagger a_{-p}-a_{-p}^\dagger a^\dagger_p)\]
    [/tex]
    Here p,p',x are all 3-vectors.
    By the substiution p->-p above the middle two terms will yield the delta function which we will go on to neglect:
    [tex]\frac{1}{2}\int\frac{d^3p}{(2\pi)^3}p(a_pa_p^\dagger-a_{-p}^{\dagger}a_{-p} )=\frac{1}{2}\int\frac{d^3p}{(2\pi)^3}p(a_pa_p^\dagger-a_{p}^{\dagger}a_p)
    [/tex]
    This leaves me needing the statement
    [tex]
    \frac{1}{2}\int\frac{d^3p}{(2\pi)^3}p(a_pa_{-p}-a_{-p}^\dagger a_p^\dagger)= \int\frac{d^3p}{(2\pi)^3}pa_p^{\dagger}a_{p}
    [/tex]
    to be true, and I don't see why it is; equally I've checked the above working repeatedly, so if I've made a mistake there there's a good chance I've misunderstood something.
     
  9. Jun 13, 2009 #8

    malawi_glenn

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    ok i will have a serious look at it next week, stay tuned.

    Meanwhile, do you know of Srednicki's book? http://www.physics.ucsb.edu/~mark/qft.html

    he does a quite detailed calculation, you might be able to pick up many things from there
     
  10. Jun 13, 2009 #9
    I'm not so sure this is correct. Maybe the answer should not have a negative sign? In the Weyl representation, the bottom block should transform as the adjoint of the top block (the left and right chiral fields are related by adjoint operation). But the adjoint of the top block equals the top block, so the bottom block should also be the same as the top block.
     
  11. Jun 13, 2009 #10
    Sorry RedX, you're absolutely right. I was careless about writing that out (TeXing stuff is taking me ages as I'm really only just getting used to it!). It is, however, very definitely non-zero...
    Thanks for the recommendation malawi-glenn, Srednicki's book is out on loan at my uni library at the moment but I'll try and have a look at some point.

    Thanks in general for all this help guys, I've just finished my 3rd year at uni so some of this stuff is ... stretching :biggrin:
     
  12. Jun 13, 2009 #11

    malawi_glenn

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    but you can obtain it at his website.... free
     
  13. Jun 13, 2009 #12
    Wow. I'll have a look at that later tonight. Thanks!
     
  14. Jun 13, 2009 #13

    malawi_glenn

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    I know I got some hints from his calculations when I did this the first time
     
  15. Jun 13, 2009 #14
    The negative sign you have for the spin matrix on the bottom right corner should be positive, and not negative, so you did the calculation right. However, my reasoning for why it should have been positive is sketchy so don't listen to that.

    For boosts however, the negative sign is correct.
     
  16. Jun 13, 2009 #15

    Avodyne

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    No. In the third term, you get an extra minus sign from p->-p, because you have a coefficient p. So you do not get a commutator here. In fact the sum of the 2nd and 3rd terms add up to the total you want. And [itex]p a_p a_{-p}[/itex] vanishes when integrated over p, because it is odd in p. Same for its hermitian conjugate, so the 1st and 4th terms are zero.
     
  17. Jun 13, 2009 #16
    Brilliant. Thanks Avodyne!
     
  18. Jun 13, 2009 #17
    I can get the corresponding calculation for boosts out fine, which makes me wonder if it's something in the way I'm parametrising the infinitesmal rotation. But I can't work out for the life of me what it is!
     
  19. Jun 14, 2009 #18

    malawi_glenn

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    So you don't need me anymore? ;-)
     
  20. Jun 14, 2009 #19
    You're not escaping that easily! :tongue:
     
  21. Jun 16, 2009 #20
    The dirac delta function is in 3d, so summing over the repeated index j gives the correct answer. Sorted :smile:
    Thanks for all your help guys!
     
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