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Schaum's Quantum Mechanics comes up with

$$\exp((-i/\hbar)\cdot \theta \cdot{\hat{L}} \cdot {\overrightarrow{n}})$$

as the formula of the rotation operator. Other sources I see don't have the negative sign. How did they get the negative sign in here?

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    $\begingroup$ The Wikipedia page, gives the sign alright. what sources are you talking about? $\endgroup$
    – ritvik1512
    Commented Jun 8, 2015 at 15:37
  • $\begingroup$ This is one but thanks for pointing out that the Wikipedia one does give the negative sign. I had overlooked that somehow. $\endgroup$
    – a00
    Commented Jun 10, 2015 at 11:53
  • $\begingroup$ Could you explain the Wikipedia derivation please? Specifically, I don't understand their step after "Taylor development gives". I understand Taylor series but the application is confusing here, especially with the parentheses notations. $\endgroup$
    – a00
    Commented Jun 10, 2015 at 13:30
  • $\begingroup$ Also, that Wikipedia article says ${p_x} = i \hbar (dT(0)/da)$, but shouldn't it be ${p_x} = -i \hbar (dT(0)/da)$? $\endgroup$
    – a00
    Commented Jun 10, 2015 at 13:39
  • $\begingroup$ Oops, above I meant to say this page is one source that does not have the negative sign. quantummechanics.ucsd.edu/ph130a/130_notes/node275.html $\endgroup$
    – a00
    Commented Jun 10, 2015 at 13:42

1 Answer 1

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Both negative sign and positive sign are correct. When you make an infinitesimal rotation with angle $d\phi$ about the z-axis, then both two following representations for transformed coordinates are true: $$ \left\{ \begin{array}{ll} x'=x-d\phi y \\ y'=y+d\phi x \end{array} \right. $$ and $$ \left\{ \begin{array}{ll} x'=x+d\phi y \\ y'=y-d\phi x \end{array} \right. $$ The former leads to the positive sign, while the later leads to the negative sign.

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  • $\begingroup$ I thought there was only one way to rotate about the z axis, by the right hand rule. Thank you for your statement though, which I'm sure is true. $\endgroup$
    – a00
    Commented Jun 10, 2015 at 13:31
  • $\begingroup$ This page is helping me understand why it's okay to take only a first order approximation here ($x - d\phi * y$ and y + $d\phi * x$). reedbeta.wordpress.com/2011/09/18/… $\endgroup$
    – a00
    Commented Jun 19, 2015 at 12:24
  • $\begingroup$ Another way I think about it is: You know that when you want to create an operator that would translate a wave function by $\Delta x$ in the positive x direction, you make a new wave function that returns the value of the old wave function, at every point $x_0$, the value $\psi(x_0 - \Delta x)$. i.e. if $\Delta x$ is positive, then by having the minus sign there, you actually end up shifting the wave function TO THE RIGHT (as desired, since $\Delta x$ is positive). What this ends up meaning is that when you write the translation operator it ends up $\endgroup$
    – a00
    Commented Jul 11, 2015 at 14:33
  • $\begingroup$ being $e^{-(i/\hbar) (\Delta x) (d/dx)}$ with the minus sign. It seems to be similar behavior for creating the Rotation operator. If you just have $e^{+(i/\hbar) (\Delta \phi)(d/d\phi)}$ naively, you end up rotating your wave function in the WRONG (opposite from desired) direction. $\endgroup$
    – a00
    Commented Jul 11, 2015 at 14:35
  • $\begingroup$ Thanks to march's comment on the other question "Negative sign in rotation operator again" that led to this interpretation. $\endgroup$
    – a00
    Commented Jul 12, 2015 at 13:23

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