Induced representations of the wavefunction

1. Jun 12, 2010

LAHLH

Hi, hopefully this is the right board to ask this on.

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: $$\psi'(\vec{x})=\psi(R^{-1}\vec{x})$$(equation 1).

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

$$(R_{2}^{-1}\vec{x})_{x}=x\cos(\beta)-z\sin(\beta)=r(\cos(\beta)\sin(\theta)\cos(\phi)-\sin(\beta)\cos(\theta))$$
$$(R_{2}^{-1}\vec{x})_{y}=y=r\sin(\theta)\sin(\phi)$$
$$(R_{2}^{-1}\vec{x})_{z}=x\sin(\beta)+z\cos(\beta)=r(\cos(\beta)\cos(\theta)+\sin(\beta)\sin(\theta)\cos(\phi))$$

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

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

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

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

Last edited: Jun 12, 2010
2. Jun 12, 2010

LAHLH

For conveinience of any kind person who might answer this, the spherical harmonics used here are:
$$Y_{11}=-(3/8\pi)^{1/2}\sin(\theta)e^{i\phi}=-(3/8\pi)^{1/2}\sin(\theta)(\cos(\phi)+i\sin(\phi))$$
$$Y_{10}=(3/4\pi)^{1/2}\cos(\theta)$$
$$Y_{11}=+(3/8\pi)^{1/2}\sin(\theta)(\cos(\phi)-i\sin(\phi))$$

Therefore,

$$\cos(\theta)=(4\pi/3)^{1/2}Y_{10}$$
$$\sin(\theta)\cos(\phi)=\frac{1}{\sqrt{2}}(4\pi/3)^{1/2}(Y_{1,-1}-Y_{11})$$

and the wavefunctions obviously have various angular parts equal to these spherical harmonics, e.g. $$u_{211} ~ Y_{11}$$ etc

Last edited: Jun 12, 2010
3. Jun 12, 2010

kaksmet

Not sure I grasp what is the actual question, he equates the z part of the rotation and get to an answer which he expresses in spherical harmonics.. would like to help so perhaps you could clarify to me exactly what the question is.

4. Jun 12, 2010

LAHLH

Hi Kaksmet, thanks for taking the time to read my post. I know my question was kind of buried amongst a lot of text, and prob not very clearly put.

I guess what I'm asking is: Why does he equate the z component of the (pre)rotated position vector $$(R_{2}^{-1}\vec{x})_{z}$$? it seems arbitrary to me, why not equate $$(R_{2}^{-1}\vec{x})_{x}$$ or $$(R_{2}^{-1}\vec{x})_{y}$$ or some function of these things.

All equation (1) $$\psi'(\vec{x})=\psi(R^{-1}\vec{x})$$ says as far as I see, is that $$\psi$$ must be some function of $$R^{-1}\vec{x}$$ such that it has the same values of $$\psi'(\vec{x})$$. We know that $$\psi'$$ looks like the z component $$\cos(\theta)$$ of the position vector $$\vec{x}$$, but why does that actually imply that unprimed $$\psi$$ has to look like to z component of the (pre)rotated position vector $$R_{2}^{-1}\vec{x}$$?

I hope that makes sense, if not let me know and I'll have a rethink and try to rexpress what it is I'm failing to grasp here.

Thanks again

5. Jun 13, 2010

kaksmet

It is arbitrary, he could equate any one of these. However, if the original wavefunction is proportional to $$z$$, then equating the rotation of the $$z$$ will yield the new wavefunction.
What I mean is that since we start at $$Y_{10}$$, meaning that we start with $$kz$$, Then we want to rotate $$Y_{10}$$, but we could equally well rotate $$z$$. Therefore we can rotate $$z$$ and the answer will be our rotated wavefunction, up to a constant.

Yes, thats correct.

Maybe this is were things go wrong, From what you wrote in your first post, we start out from $$Y_{10}$$, i.e. we know that $$\psi$$ (without the prime) looks like the z component $$\cos(\theta)$$ of the position vector $$\vec{x}$$,

Does this make things any clearer?

6. Jun 13, 2010

kaksmet

An attempt to summarise what I said in formulas

$$\psi = X_{21}Y_{10} = X_{21} k z$$
where $$k$$ is a constant and X is just the radial component of the wavefunction, and hence
$$\psi' = X_{21}Y'_{10} = X_{21} k z' = X_{21} k (R_{2}^{-1}\vec{x})_{z}$$

7. Jun 13, 2010

LAHLH

Thanks so much, I think I totally get it now.

Yep that was were my thinking was getting muddled.

Thanks again

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