I have read a few proofs of the Idempotent propertry of the antisymetrisation operator used in Hartree Fock, I feel I am either missing something or the proofs are a bit hand wavy.

$$ \hat{A}\hat{A} = \frac{1}{(N!)^{1/2}}\sum_{P} (-1)^p\hat{P}\frac{1}{(N!)^{1/2}}\sum_{P'}(-1)^{p'}\hat{P'} = \frac{1}{N!} \sum_{P}\sum_{P'}(-1)^{p+p'}\hat{P}\hat{P'} = \frac{1}{N!}\sum_P\sum_R(-1)^r\hat{R},$$

$$ \hat{R} = \hat{P}\hat{P'}.$$

I understand the progression to this point, and I also understand that the symmetrisation operator is idempotent ,i.e. the following operator

$$\hat{S} = \frac{1}{N!}\sum_{P} \hat{P}\frac{1}{N!}\sum_{P'}\hat{P'}. $$

What I fail to understand is the following-

$$ \hat{A}\hat{A} = \frac{1}{N!}\sum_P\sum_R(-1)^r\hat{R} = \frac{1}{N!}\sum_P(N!)^{1/2}\hat{A}=(N!)^{1/2}\hat{A}.$$

I dont understand how you go from the first to the second expression. I am having trouble with the (-1) term, but I am comfortable with the symmetrising operator which doesn't have (-1) multiplying the even terms.

Also am I thinking about the $P$ operators the right way, for example is the following use correct?

$$ P_{12}P_{23}[1,2,3,4] = P_{12}[1,3,2,4] = [3,1,2,4] .$$


I am having trouble understanding how $$\sum_R(-1)^r\hat{R} = \sum_P(-1)^p\hat{P} $$

I understand that both the expressions have the same permutations, but I do not understand how the signs will be consistent. As an example consider the following -

The above expression is true irrespective of the value of P', the following equation should hold

$$\sqrt{N!}\hat{A}=\sum_P(-1)^p\hat{P}=\sum_R(-1)^r\hat{R}=\sum_P(-1)^{p+p'}\hat{P'}\hat{P} = (-1)^{p'}\hat{P'}\sum_P(-1)^{p}\hat{P}=\sqrt{N!}(-1)^{p'}\hat{P'}\hat{A}$$

this is true for any P', taking P' to be $$P_{12}$$ we have

$$ (-1)\hat{P_{12}}\sum_P(-1)^{p}\hat{P} = -\sqrt{N!}\hat{P_{12}}\hat{A} = \sqrt{N!}\hat{A}$$

Taking [1,2,3,4] as an example we get

$$\hat{A}[1,2,3,4] = [1,2,3,4] - ([2,1,3,4] + [1,3,2,4] + [1,2,4,3] + [4,2,3,1] + [3,2,1,4] + [1,4,3,2]) + ([2,3,1,4]+[1,3,4,2] + [3,2,4,1]+[2,4,3,1]+[3,1,2,4] + [1,4,2,3] + [4,2,1,3] + [4,1,3,2])-([2,3,4,1]+[3,4,1,2]+[4,1,2,3] + [4,3,2,1] + [4,3,1,2] + [2,4,1,3]+[2,1,4,3]+[3,4,2,1]+[3,1,4,2])$$

$$ -\hat{P'_{12}}\hat{A}[1,2,3,4] = -[2,1,3,4] + ([1,2,3,4]+[3,1,2,4]+[2,1,4,3]+[2,4,3,1]+[2,3,1,4] + [4,1,3,2]) - ([3,2,1,4]+[3,1,4,2]+[2,3,4,1]+[4,2,3,1]+[1,3,2,4]+[4,1,2,3]+[2,4,1,3]+[1,4,3,2])+([3,2,4,1]+[4,3,1,2]+[1,4,2,3]+[3,4,2,1]+[3,4,1,2]+[4,2,1,3]+[1,2,4,3]+[4,3,2,1]+[1,3,4,2]) $$

though these expressions have the same permutations but the terms do not have the same sign. For example [2,1,4,3] is negative in the first expression but positive in the second.

  • $\begingroup$ The sum over $P$ in the definition of $A$ goes over all permutations. Since $R=PP'$ goes over all permutations when $P'$ does, the sum over $R$ in the first expression of you next-to-last equation is the very definition of $A$. $\endgroup$ – user154997 Aug 11 '17 at 9:35

First let me state some things I think you already correctly understood, just to make sure we are on the same page.

The step $\sum_P \sum_{P'} (-1)^{p+p'} \hat{P} \hat{P}' = \sum_P \sum_R (-1)^r \hat{R}$ requires recognizing that summing over the product all possible combinations of permutations $P$ and $P'$ is the same as summing over all permutations only once (because the permutation operators form a group). But, all permutations then turn up multiple times, hence the sum over $P$ remains, but the argument of the sum no longer depends on $P$. So the sum over $R$ obviously means a sum over all permutations again.

Now, in the step you asked about, you just need to identify $$ \sum_R(-1)^r \hat{R} = \sum_P(-1)^p \hat{P} = \sqrt{N!} \hat{A}. $$ The last step uses that the number of permutations (so the number of summands in the sum over $P$) is exactly $N!$.

Your use of $\hat{P}$ in this context is partly correct. In the sense it is used in the proof, it would of course have more indices ($N$), because $\sum_P \hat{P}$ is supposed to produce all permutations.


The equations up to your example are correct. After all, this is exactly what an antisymmetrizer does. It shuffles things so that when you transpose 2 elements afterwards, the whole thing changes sign.

Your example, however, is not correct. In the last bracket are some wrong signs. $[3,4,1,2]$ is an even permutation. Transpose 3 with 1 and 2 with 4 and you have $[1,2,3,4]$. Also even are $[4,3,2,1]$ and $[2,1,4,3]$. This explains your discrepancy. If you do it with all the right signs, you will find that indeed, your calculations are correct.

The important thing here is that every permutation has a unique parity. It does not matter whether you call it $p$ or $r$, if $R$ produces the same permutation as $P$ does, then $p$ must have the same parity as $r$.

  • $\begingroup$ Thanks for the answer, I have edited my original question as i could not add a comment that exceeded 600 characters. I am having trouble understanding how $$\sum_R(-1)^r\hat{R} = \sum_P(-1)^p\hat{P} $$ $\endgroup$ – ace_101 Aug 11 '17 at 22:38
  • $\begingroup$ I updated my answer, hope this clears things up. $\endgroup$ – noah Aug 11 '17 at 23:26
  • $\begingroup$ Thanks that clears things up for me. I was not placing the parities correctly, i was placing them according to derangement of numbers rather than permutations. $\endgroup$ – ace_101 Aug 14 '17 at 17:11

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