- 80

- 1

In quantum mechanics we have a time displacement operator and a space displacement operator, which are respectively:

[itex]

\hat{T}(t) = e^{-i\hat{H}t}

[/itex]

[itex]

\hat{D}(\underline{x}) = e^{-i\hat{\underline{p}}\cdot\underline{x}}

[/itex]

In quantum field theory there is a spacetime displacement operator:

[itex]

\hat{U}(a_{\mu}) = e^{-i\hat{P}^{\mu}a_{\mu}}

[/itex]

So as I understand this can be written out as:

[itex]

\hat{U}(a_{\mu}) = e^{-i\hat{P}^{0}a_{0}-i\hat{P}^{j}a_{j}}

[/itex]

Or, depending on the metric convention:

[itex]

\hat{U}(a^{\mu}) = e^{-i\hat{P}^{0}a^{0}+i\hat{P}^{j}a^{j}}

[/itex] , or:

[itex]

\hat{U}(a^{\mu}) = e^{+i\hat{P}^{0}a^{0}-i\hat{P}^{j}a^{j}}

[/itex]

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?