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 - The Fusion of Science and Community**

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

# Induced representations of the wavefunction

Loading...

Similar Threads - Induced representations wavefunction | Date |
---|---|

B Induced radiation | May 31, 2017 |

I Bases for SU(3) Adjoint representation | Aug 13, 2016 |

I Is nuclear fusion induced by gamma photons possible? | May 8, 2016 |

Gamma rays induced fission of silicon | Jul 24, 2014 |

Cosmic ray induced nuclear fusion | Sep 15, 2012 |

**Physics Forums - The Fusion of Science and Community**