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Spacetime displacement operator in QFT

  1. Jun 11, 2013 #1
    I'm trying to fit together my understanding of quantum mechanics, quantum field theory, given my lacking maths education.

    In quantum mechanics we have a time displacement operator and a space displacement operator, which are respectively:
    \hat{T}(t) = e^{-i\hat{H}t}
    \hat{D}(\underline{x}) = e^{-i\hat{\underline{p}}\cdot\underline{x}}

    In quantum field theory there is a spacetime displacement operator:
    \hat{U}(a_{\mu}) = e^{-i\hat{P}^{\mu}a_{\mu}}

    So as I understand this can be written out as:
    \hat{U}(a_{\mu}) = e^{-i\hat{P}^{0}a_{0}-i\hat{P}^{j}a_{j}}
    Or, depending on the metric convention:
    \hat{U}(a^{\mu}) = e^{-i\hat{P}^{0}a^{0}+i\hat{P}^{j}a^{j}}
    [/itex] , or:
    \hat{U}(a^{\mu}) = e^{+i\hat{P}^{0}a^{0}-i\hat{P}^{j}a^{j}}

    Now while the 1st expression is in agreement with what I know from quantum mechanics, the latter two have sign differences, and also what's surprising is that there are sign ambiguities depending on the convention of the metric - as I understand it - depending on it, either time is displaced forwards and space backwards; or time backwards and space forwards.

    I'm guessing that for some reason we should only take the first expression, but I don't understand why, and what is the significance of the latter two? Why are t and x in the quantum mechanics formulae for D and T necessarily covariant? Is it that when we multiply vectors like in these formulae - one quantity must be covariant, and the other contravariant?
  2. jcsd
  3. Jun 11, 2013 #2


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    When you contract a covariant vector with a contravariant one, there's never a minus sign.

    AμBμ = A1B1 + A2B2 + A3B3 + A0B0.

    In order to contract two covariant vectors, or two contravariant vectors, you will need to use the metric, and therefore a minus sign may appear:

    ημνAμBν = η11A1B1 + η22A2B2 + η33A3B3 + η00A0B0 = A1B1 + A2B2 + A3B3 - A0B0
  4. Jun 11, 2013 #3
    I believe the versions with sign differences are correct, but there's some confusing notational issues here that make it confusing.

    There are two objects here, and it seems like you might be jumping back and forth between them. One is a vector, [itex] a^\mu [/itex], and the other is called a "dual vector" [itex] a_\mu [/itex]. A dual vector is a linear map that takes a vector, and gives you a number. So the combination of a dual vector and a vector gives you a number, which you are familiar with as [itex] a^\mu a_\mu [/itex]. So far, there is nothing tricky about this: you're just taking the components of the vector, the components of the dual vector, multiplying the corresponding elements and summing them.

    But it starts to get complicated once you add in the metric, which is a map from a vector to a dual vector (or a dual vector to a vector). Once you have this metric, you implicitly define a dual vector for every vector you define, because you just use the metric to map your vector to it's corresponding dual vector. What you may be seeing is that authors are rewriting the same expressions in terms of either the vector components, or the dual vector components, taking advantage of this implicit definition. However, the components are not the same (which is evident by the changing signs in each of your expressions), so it pays to be diligent, explicit, and consistent.
  5. Jun 11, 2013 #4
    Thanks a lot - it did refresh some of the maths that I've forgotten.

    In fact I'm using Weinberg's book. And he only wrote the equation with μ's, while I was trying to make a connection with quantum mechanics.

    So I think my question now is - in the quantum mechanics expressions - which ones are dual vectors and which ones are just vectors? From this description it seems to me that H and p are dual vectors, while t and x are just vectors?
  6. Jun 11, 2013 #5
    Actually, it doesn't matter (in this context) which you call vectors and which you call dual vectors. It turns out vectors are also linear maps from dual vectors into numbers. The important point is once you write down a vector, keep track of it. If an index is raised or lowered, then there has been an application of the metric to it, and it's components will change.

    (To be more precise: dual vectors also form a vector space. So you could imagine the "dual dual" vectors, which are linear maps from the dual vectors to numbers. But this space is isomorphic, ie the same, as the original vector space, so you can call whichever one you want a vector and the other a dual vector.)
  7. Jun 11, 2013 #6
    A good book for the stuff I'm talking about is Sean Carroll's general relativity book. Probably a better QFT book for your level would be Srednicki, not Weinberg.
  8. Jun 11, 2013 #7
    Great! So if I understood properly it's either that p is dual and x "straight", or p "straight" and x dual - but it can't be that both are "straight" or dual.

    So from that it would follow that in fact the expression with both "-" signs would be valid in terms of correspondence to quantum mechanics - as it has both covariant (dual) and contravariant ("straight") components, and not all contravariant.

    Is that true?

    Thanks for the recommendation. It's just that I really like Weinberg - no matter how challenging he is, he's 100% my style! But it would be great to read some proper differential geometry book to clear up the doubts.
  9. Jun 11, 2013 #8
    It sounds like you have the right idea. Personally, I usually think of both as vectors first to figure out their components, then apply the metric to one of them to get the dual vector, then apply the dual vector to the other vector. If you follow this, you'll find one relative minus sign, because the metric is diag(-1,+1,+1,+1).

    For example, suppose I wanted to take the inner product of the 4 momentum, [itex] p^\mu p_\mu [/itex]. I would start by saying [itex] p^\mu = (p_0, p_1, p_2, p_3) [/itex] then I'd compute the dual vector: [itex] p_\mu = \eta_{\mu\nu} p^\nu = (-p_0, p_1, p_2, p_3) [/itex]. Now I sum the components: [itex]p^\mu p_\mu = -p_0^2 + p_1^2 + p_2^2 + p_3^2 [/itex].

    Notice I was careful to stick close to my definitions, so I know the correct minus sign appeared.
  10. Jun 11, 2013 #9
    Hm, ok to be honest I'm still not sure. So let's now move on to [itex] \hat{P}^{\mu}a_{\mu}[/itex], and use the convention (-1,1,1,1)

    We have: [itex] \hat{P}^{\mu} = ( \hat{P}^0 , \hat{P}^1 , \hat{P}^2 , \hat{P}^3 ) = ( \hat{H} , \hat{p}_x , \hat{p}_y , \hat{p}_z) [/itex] , where in the latter x,y,z are only labels.

    And then: [itex] a^{\mu} = (t , x, y, z) [/itex] - that's the spacetime distance over which we displace. But then: [itex] a_{\mu} = (-t , x, y, x) [/itex] and we get the sign differences, which don't occur in ordinary quantum mechanics:

    [itex] \hat{P}^{\mu}a_{\mu} = -\hat{H}t + \hat{p}_x x + \hat{p}_y y + \hat{p}_z z [/itex]

    Alternatively we could have given the vector over which we displace in the covariant form right away: [itex] a_{\mu} = (t , x, y, x) [/itex] and that would be consisent with QM. But why should we give it one way and not the other?
  11. Jun 11, 2013 #10


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    If you use ##x^{\mu}## then you will have to use ##\hat{P_{\mu}}## and ##\hat{P_{\mu}} = \eta_{\mu\nu}\hat{P^{\nu}}## meaning ##\hat{P_0} = -\hat{P^0}##, ##\hat{P_i} = \hat{P^i}## so you will still get ##\hat{P_\mu}x^{\mu} = -\hat{P_{0}}x^{0} + \hat{P_{i}}x^{i}##. Note that ##\hat{P_{\mu}}x^{\mu} = \eta_{\mu\nu}\hat{P^{\nu}}x^{\mu} = \hat{P^{\mu}}x_{\mu}## so you will get the same result regardless.
  12. Jun 11, 2013 #11
    Thanks WannabeNewton, that clarifies everything! You meant [itex]\hat{P_\mu}x^{\mu} = [/itex]+[itex]\hat{P_{0}}x^{0} + \hat{P_{i}}x^{i}[/itex] though?
    Last edited: Jun 11, 2013
  13. Jun 11, 2013 #12


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    Ah sorry yes. I meant to write ##-\hat{P^{0}}##.
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