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Lorentz transform of a scalar in QM

  1. May 28, 2012 #1
    If you Lorentz transform a scalar:

    [tex]U^{-1}(\Lambda)\phi(x)U(\Lambda)=\phi(\Lambda^{-1}x) [/tex]

    If you now perform another Lorentz transform, would it it look like this:

    [tex]U^{-1}(\Lambda')U^{-1}(\Lambda)\phi(x)U(\Lambda)U(\Lambda')=\phi(\Lambda'^{-1}\Lambda^{-1}x) [/tex] ?

    But isn't this wrong, because this expression is equal to:

    [tex]U^{-1}(\Lambda\Lambda')\phi(x)U(\Lambda\Lambda')=\phi([\Lambda\Lambda']^{-1}x) [/tex]

    and not:

    [tex]U^{-1}(\Lambda'\Lambda)\phi(x)U(\Lambda'\Lambda)=\phi([\Lambda'\Lambda]^{-1}x) [/tex]
     
  2. jcsd
  3. May 29, 2012 #2

    Bill_K

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    Science Advisor

    geoduck, There are two ways to describe a rotation: active and passive. In both methods the function gets rotated relative to the coordinates, the function goes one way and the coordinates go the other. The only distinction is which you call Λ and which you call Λ-1. What you've described is the active approach. The two methods are completely equivalent of course, but in some ways the passive approach looks more natural:

    Let x → x' = Λx. Then U(Λ) φ(x) U(Λ)-1 = φ(x') = φ(Λx).

    Two rotations in succession: U(Λ2)U(Λ1) φ(x) U(Λ1)-1U(Λ2)-1 = φ(Λ2Λ1x) = U(Λ2Λ1) φ(x) U(Λ2Λ1)-1
     
  4. May 29, 2012 #3
    Thanks. I think I see it now. As you said, function goes opposite of coordinates. So if you want them to visually go the same way, you have to inverse all the U's.

    I guess what it boils down to is that mathematically, we like to read from right to left, which would look like this:

    B-1A-1 φ(x) AB

    Starting from the right, that should read first B, then A. And this is true if there were a ket on the right end: that's the order you'd perform A and B

    But with regards to φ(x), that would read first A, then B.

    Now if by default we like U(A)U(B) to mean first B, then A when operating on something to the right, then in order to preserve that when dealing with a function φ(x) in the middle (rather than a ket at the edge), you'll have to put U(A)U(B) on the left hand side, and not the right, so:

    U(A)U(B)φ(x)U-1(B)U-1(A)=...

    but mathematically then you'd have to invert the stuff to the right of the equal sign, so you'd get:

    U(A)U(B)φ(x)U-1(B)U-1(A)=φ(ABx)
     
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