Hi, hopefully this is the right board to ask this on.(adsbygoogle = window.adsbygoogle || []).push({});

I'm currently reading Groups, representations and Physics by Jones, and trying to get my head around induced transformations of the wavefunction. The problem is I seem to understand nearly all of what he's saying except the crucial part I guess. So firstly I understand that after a rotation R, the value of the new wavefunction at a rotated point is the same as that of the old wavefunction at the original point, leading to: [tex] \psi'(\vec{x})=\psi(R^{-1}\vec{x})[/tex](equation 1).

I also understand that if we work in spherical polars and look at a rotation [tex] R_{2}(\beta)[/tex] (about the y-axis), then the components of the position vector (when rotated back as needed for equation (1) look like:

[tex] (R_{2}^{-1}\vec{x})_{x}=x\cos(\beta)-z\sin(\beta)=r(\cos(\beta)\sin(\theta)\cos(\phi)-\sin(\beta)\cos(\theta))[/tex]

[tex] (R_{2}^{-1}\vec{x})_{y}=y=r\sin(\theta)\sin(\phi)[/tex]

[tex] (R_{2}^{-1}\vec{x})_{z}=x\sin(\beta)+z\cos(\beta)=r(\cos(\beta)\cos(\theta)+\sin(\beta)\sin(\theta)\cos(\phi))[/tex]

Now finally, assuming we start with some 2p electron (n=2,l=1) and with wavefunction proportional to [tex]\cos(\theta)[/tex] (so also m=0). One could also write this as [tex] Y_{10}=(3/4\pi)^{1/2}\cos(\theta)[/tex]. So calling this, [tex] u_{210}^{'}(\vec{x}) [/tex] (this corresponds to our post rotation wavefunction of equation (1). It's also obvious that [tex] u_{210}^{'}(\vec{x}) [/tex] is just proportional to z since [tex] z=r\cos(\theta)[/tex].

Now what I don't understand is that Jones, then just considers [tex] (R_{2}^{-1}\vec{x})_{z}=r(\cos(\beta)\cos(\theta)+\sin(\beta)\sin(\theta)\cos(\phi))=r(4\pi/3)^{1/2}\left(\cos(\beta)Y_{10}+\frac{1}{\sqrt{2}}\sin(\beta)(Y_{1,-1}-Y_{1,1})\right)[/tex], which is fine but....

How can he just equate this, [tex](R_{2}^{-1}\vec{x})_{z}[/tex] with the original wavefunction? Why not say, [tex](R_{2}^{-1}\vec{x})_{x}[/tex] etc. I mean equation (1), [tex] \psi'(\vec{x})=\psi(R^{-1}\vec{x})[/tex], simply states [tex] \psi [/tex] is a function of [tex] R^{-1}\vec{x} [/tex], not any particular componant of that pre-rotated position vector. Although of course [tex] \psi'[/tex] was a function of [tex]z[/tex] only, there's no reason I can see that [tex]\psi[/tex] should only be a function of zundera rotation about the y-axis.

Many thanks for any help resolving this, I suspect I am just missing something for small to unlock this.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Induced representations of the wavefunction

**Physics Forums | Science Articles, Homework Help, Discussion**