Why is the ROHF spin density of the propenyl radical not symmetric? I was preparing some material for my students and in order to illustrate the concept of spin density and how the ROHF method gives a purer description of this property than the UHF method. I prepared a simple calculation for the $\ce{CH3}$ and $\ce{CH2CHCH2}$ radicals.

The calculations for both radicals performed properly. Given the symmetry of the molecules, I expected the distributions to be symmetric. That expectation was borne by the results of the calculations for the methyl radical and the UHF method of the allyl radical. However, in the case of the allyl radical the spin density surface is asymmetric as shown in the following picture obtained with GaussView 5.



The canonical MO have the proper symmetry (see below)



Can anybody point to an explanation of these behaviour or a mistake in the calculation?

P.D. Find enclose a copy of the Gaussian input field I used in the calculation.

%chk=c3h5_rohf.chk
# opt=tight rohf/6-31++g(d,p) guess=mix 

ROHF

0 2
 C              
 H                  1    1.06999999
 C                  1    rCC    2  ACCH
 H                  3    rCH    1  AHCC    2    0.0    
 H                  3    rCH    1  AHCC    2  180.0    
 C                  1    rCC    2  ACCH    3  180.0   
 H                  6    rCH    1  AHCC    3  180.0   
 H                  6    rCH    1  AHCC    3    0.0    

 rCC 1.40140000
 rCH 1.0700000
 ACCH 120.0
 AHCC 120.0
I believe your wave function is not the one you would expect in ROHF. The reason for this is guess=mix which will try to construct a broken symmetry wavefunction from the HOMO and LUMO. Why the SCF converges to that solution, and why this is stable (-116.460412224 ha), I don't know, and I have currently not the time to investigate (I will try to come back to that), but remove the keyword and you should obtain the symmetric correct solution (-116.457980797 ha). Inspect your output file, and you will notice, that the calculation is run in Cs symmetry, not C2v. (Also HF might not be the best method to show delocalisation.)