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About Wick's Theorem, Time Order Operator, Normal Ordering and Green's Function

  1. Oct 30, 2014 #1
    So if I understood well, Normal ordering just comes due to the conmutation relation of a and a⁺? right? Is just a simple and clever simplification.

    Wick Theorem is analogue to normal ordering because it is related to the a and a⁺ again (so related to normal ordering, indeed).

    However I do not know why and exactly where the Time order operator comes. Is this like onion layers? Normal ordering -> time order operator -> wick's theorem ?

    And can someone explain me this equation? Please.

    $$ <0|T\big\{\phi^{(0)}(x)\phi^{(0)}(y)\big\}|0>=G^{(0)}_F(x-y)$$

    This really vanishes just because of a(p)|0>=0?

    $$<0|:\phi(x)\phi(y):|0>=0$$
     
  2. jcsd
  3. Oct 31, 2014 #2
    to my understanding, wick's theorem gives a way to represent the time ordered combination of field operators. it turns out, via wick's theorem that you can think of the time ordered product as a sum of normal ordered products and contractions.

    since we are interested in the vacuum expectation value to calculate amplitudes, wick's theorem is nice because we can effectively ignore the normal ordered terms (i.e. the vacuum expectation value of a normal ordered product is zero), so the time ordering for the vacuum expectation value is simply a sum over contractions. The contraction for two fields is equal to the feynman propagator, which is the first equation you have written. Only the normal ordered part vanishes, the contractions remain and are generally nonzero.
     
    Last edited: Oct 31, 2014
  4. Oct 31, 2014 #3
    why? in mathematical terms.
     
  5. Nov 1, 2014 #4
    Well, there are a few possibilities for the normal ordered product. To convey the point it is enough to think of the normal ordered product of two creation/annihilation operators.

    for the first example we have
    apap
    where ap corresponds to the creation of a particle with momentum p.

    now take the vacuum expectation value of the normal ordered product.

    <0|N(apap)|0> = <0|apap|0> = <0|2p> because the state with no particles (i.e. <0|) is orthogonal to the state with two particles of momentum p (i.e. |2p>)

    now let's take the example apap

    <0|N(apap)|0> = <0|apap|0> = 0 because ap is operating on the vacuum state first. ap sends the vacuum state to 0 by definition of what we mean by the vacuum state.

    the last example is trivial

    <0|N(apap)|0> = <0|apap|0> = 0

    We can also think of the operators as operating on the bra vector instead of the ket vector, in which case


    <0|apap|0> = <2p|0> = 0

    The same reasoning follows through for products of an arbitrary number of creation/annihilation operators, showing that the vacuum expectation value of a normal ordered product of an arbitrary set of creation/annihilation operators is always zero.
     
  6. Nov 2, 2014 #5
    Ok, so it is just that easy to say the exponentials times |0> are equal to the exponentials in the ground state, however the normal ordering makes the annhilation operators produce a zero factor which makes the whole expression vanishes.

    Right?
     
  7. Nov 2, 2014 #6
    I was operating the first equation in the official post and I reached this relation:

    $$ \langle 0 | G_F^{(0)}(x-y)|0 \rangle = G_F^{(0)}(x-y) \langle 0|0 \rangle = G_F^{(0)}(x-y) $$

    Is this correct?
     
  8. Nov 3, 2014 #7
    More generally, the normal ordering produces a vanishing expectation value because either it annihilates the vacuum state sending it to zero, or it changes the bra or ket vector into a state that is orthogonal to the vacuum state (like in the first example) and so this also gives zero expectation value.

    Unless we're talking about exponentials of operators that rotate the quantum state in Hilbert space, the exponentials don't do anything except act like a number.

    The equation you have there is certainly correct.
     
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