# Prove that Slater determinants form a complete basis

I want to prove that the basis vectors of the antisymmetrized, $N$-particle vector space generated by the Slater determinants form a complete basis.

Attempt: I started from the definition of a Slater determinant as $$\lvert\psi\rangle = \frac{1}{\sqrt{N!}} \textrm{det} \left[ \,\,\lvert u_1\rangle\, |u_2\rangle\, ... |u_N\rangle \,\,\right],$$ and want to prove that $$\sum_{u_1\, u_2\, ... ,u_N} \frac{1}{N!}|\psi\rangle \langle \psi| = I.$$ Ok, first I think that $N!$ shouldn't be there in the equation immediately above because each Slater determinant already contains it. Am I right? Secondly about solving this problem itself, being expanded each determinant will have $N!$ terms. So that each $|\psi\rangle \langle\psi|$ contains $(N!)^2$ terms, $N!$ of which are "self" terms having the form $$\sum_{u_1\, u_2\, ... ,u_N} |u_1\rangle\, |u_2\rangle\, ... |u_N\rangle \langle u_1|\, \langle u_2|\, ... \langle u_N| = I$$ Since there are $N!$ terms with such form, they will add up to $N!I$. Now I am concerned with the "cross" terms in the product of the two determinant. For if they are all zero, then it's straightforward to see that we will get something like $N!I/N! = I$ hence proved. But are the cross terms indeed zero?

EDIT: To illustrate what I have been working on, I will take a special case of N=2. $$\psi_{kl}(x_1,x_2) = \frac{1}{\sqrt{2}}(u_k(x_1)u_l(x_2)-u_k(x_2)u_l(x_1))$$ Then $$\sum_{k,l} \psi_{kl}(x_1,x_2) \psi^*_{kl}(x_1,x_2) = 1$$ (note that here I don't use extra 2!). Inserting the expression for $\psi_{kl}(x_1,x_2)$ we will get four terms. The first two are $$\frac{1}{2}\left(\sum_{k,l} u_k(x_1)u_l(x_2)u^*_k(x_1)u^*_l(x_2) + \sum_{k,l} u_l(x_1)u_k(x_2) u^*_l(x_1)u^*_k(x_2) \right)= 1$$ because $\sum_{k,l} u_k(x_1)u_l(x_2)u^*_k(x_1)u^*_l(x_2) = \langle x_1|\sum_{k} |u_k\rangle \langle u_k| |x_1\rangle \langle x_2|\sum_{l} |u_l\rangle \langle u_l| |x_2\rangle = \langle x_1|x_1\rangle \langle x_2|x_2\rangle = 1$. The other two terms are the cross terms $$\frac{1}{2}\left(\sum_{k,l} u_k(x_1)u^*_k(x_2)u^*_l(x_1)u_l(x_2) + \sum_{k,l} u_l(x_1)u^*_l(x_2) u^*_k(x_1)u_k(x_2) \right)= 0$$ Because $\sum_{k,l} u_k(x_1)u^*_k(x_2)u^*_l(x_1)u_l(x_2) = \langle x_1|x_2\rangle \langle x_2|x_1\rangle = 0$. In this last step I used the fact that $x_1$ must be different from $x_2$ otherwise all eigenfunctions vanish. Summing all those four terms, I get 1 as required.

• did you write those equations yourself or were they in the text of the exercise? I think it is important to consider that the "completeness" you want to prove is in the antisymmetric tensor product space, which has dimension $\binom{M}{N}$ if $M$ is the dimension of the single-particle Hilbert spaces (that is, the number of states in which each $u_i$ can be). This is much smaller than the whole tensor product Hilbert space that you would have for example for distinguishable particles, with dimension $M^N$.
– glS
Nov 2 '16 at 15:06
• Yes it's antisymmetrized, N-particle space as I indicated above. Nov 3 '16 at 0:08
• The equation you wrote in your edit for the "completeness of $\psi$" is wrong. How did you obtain it?
– glS
Nov 3 '16 at 8:38
• Which equation are you referring to? There are more than one completeness relation involved there. Nov 3 '16 at 9:28
• I see. Still, the equation for the wavefunction $\sum_{kl}\lvert\psi(x_1,x_2)\rvert^2 = 1$ is not equivalent to the completeness relation you wanted to prove. You can see it in analogy to the 1-particle case: you wrote the equivalent of $\sum_k \psi_k(x_1)\psi_k^*(x_1)=1$, which is very different from $\sum_k \psi_k(x_1)\psi_k^*(x_2)=\delta(x_1-x_2)$
– glS
Nov 3 '16 at 11:52

One has to be very careful with those sums: Those factors $1/N!$ are necessary in those sums involving the one operator because there is only one linear independent state for one set of one particle states $\{|u_1\rangle,|u_2\rangle,\dots,|u_N\rangle\}$. So the following is correct: \begin{align} \mathbf{1}&=\frac{1}{N!}\sum_{u_1,u_2,\dots,u_N}|u_1,u_2,\dots,u_N\rangle_a {}_a\langle u_1,u_2,\dots,u_N|\\ &=\sum_{u_1<u_2<\dots<u_N}|u_1,u_2,\dots,u_N\rangle_a {}_a\langle u_1,u_2,\dots,u_N|. \end{align} The the sum over "ordered" one particle states prevents multi counting of identical N-particle basis states and the factor $1/N!$ corrects the multi counting.

Where $|u_1,u_2,\dots,u_N\rangle_a$ is the antisymmetrized N-particle state (slater determinant if you want to think in position space): $$|u_1,u_2,\dots,u_N\rangle_a=\frac{1}{\sqrt{N!}}\sum_\pi \mathrm{sgn}(\pi)\mathbf{P}_\pi|u_1,u_2,\dots,u_N\rangle.$$

As mentioned in the comments by glS $|u_1,u_2,\dots,u_N\rangle_a$ is only the complete basis of the antisymmetric part of the N-particle Hilbert space. I am not really sure how to properly show that $|u_1,u_2,\dots,u_N\rangle_a$ actually is a complete basis of this antisymmetric N-particle Hilbert space. But no matter what one has to be really care full to not count over depended states.

• I think he didn't say anything about Slater determinant being the only complete basis in the space we are considering. Nov 3 '16 at 0:09
• Can you please see again my post. I have added my work for N=2. I don't use extra 2! in the denominator, yet why could I prove it? Nov 3 '16 at 0:58
• I agree with this, but I'd say that there is an additional complication: that "$\mathbf1$" is the identity in the antisymmetric Hilbert space, represented by the identity in that space, but whose matrix representation is very different from the identity in the regular tensor product space. It is likely much easier to prove completeness by showing that the set of antisymmetric $N$-particle states is orthonormal and has the correct number of elements
– glS
Nov 3 '16 at 11:05
• Ok I think proving that it's orthonormal is not too difficult. But what does it mean when you say correct number of elements because certainly there are infinite number of such antisymmetric vectors for a given N. Also, if that identity is not that ordinary identity matrix. Then if you express the closure relation in position space which is scalar, what will the right hand side be? Is it not unity? Nov 3 '16 at 11:36

The nontrivial point to understand here is that the antisymmetrized $N$-particle states do not form a basis for the complete space of $N$-particle states, but only for its antisymmetric part. Consequently, they will not satisfy a completeness relation in the former, but only in the latter.

Consider the simpler example of 2 fermions over two modes. The only possible state is in this case: $$\lvert12\rangle_{\mathcal A} = \frac{1}{\sqrt2} (\lvert 12\rangle - \lvert 21\rangle ).$$ The matrix representation (in the total tensor product Hilbert space) of this state is: $$\lvert12\rangle_{\mathcal A}\langle12\rvert = \begin{pmatrix}0&0&0&0\\0&1&-1&0\\0&-1&1&0\\0&0&0&0\end{pmatrix},$$ which is quite clearly not the identity one could naively expect.

Let us also work out the slightly more complex case of 2 fermions over 3 modes. There are now $\binom{3}{2}=3$ basis states: $$\lvert12\rangle_{\mathcal{A}} = \frac{1}{\sqrt2}(\lvert12\rangle-\lvert21\rangle),\\ \lvert13\rangle_{\mathcal{A}} = \frac{1}{\sqrt2}(\lvert13\rangle-\lvert31\rangle), \\ \lvert23\rangle_{\mathcal{A}} = \frac{1}{\sqrt2}(\lvert23\rangle-\lvert32\rangle).$$

The projector corresponding to the first one has now matrix representation:

$$\lvert12\rangle_{\mathcal A}\langle12\rvert = \begin{pmatrix}0&0&0&0&0&0&0&0&0\\0&1&0&-1&0&0&0&0&0\\0&0&0&0&0&0&0&0&0\\0&-1&0&1&0&0&0&0&0\\0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0\\0&0&0&0&0&0&0&0&0\end{pmatrix},$$ and similar matrices for the other two states, that again you can easily seen do not sum up to anything similar to an identity matrix.

What you can do is showing that these antisymmetric states nonetheless for a complete basis for the antisymmetric component of the Hilbert space, $\mathcal{H}_{\mathcal A}$. Let us call $N$ the number of femions and $M$ the number of modes in which each fermion can be. It follows that $\mathcal{H}_{\mathcal A}$ has dimension $\binom{M}{N}$, which is also the number of possible antisymmetric states (you can see this remembering that the binomial factor $\binom{M}{N}$ counts the number of ways in which you can choose groups of $N$ objects among a collection of $M$).

The normalization follows from the definition of antisymmetric state: $${}_{\mathcal A}\langle u_1\cdots u_N \rvert u_1\cdots u_N \rangle_{\mathcal A} = \frac{1}{N!} \sum_{\sigma,\tau} (-1)^{\sigma+\tau} \prod_{k=1}^N \langle u_{\sigma(k)}\rvert u_{\tau(k)}\rangle\\ = \frac{1}{N!} \sum_{\sigma,\tau} (-1)^{\sigma+\tau} \prod_{k=1}^N \delta_{\sigma(k),\tau(k)} = \frac{1}{N!} \sum_{\sigma} 1 = 1.$$ To finally prove the orthogonality we consider two states, $\lvert u_1\cdots u_N\rangle_{\mathcal A}$ and $\lvert v_1\cdots v_N\rangle_{\mathcal A}$ where there is at least one element of the latter, say $v_1$, which is different from all the elements of the former: $\forall i \in \{1,...,N\}, \,\, v_1 \neq u_i.$ Note that this is equivalent to say that the two states are different. If this is true, than we have:

$${}_{\mathcal A}\langle u_1\cdots u_N \rvert v_1\cdots v_N \rangle_{\mathcal A} = \frac{1}{N!} \sum_{\sigma,\tau} (-1)^{\sigma+\tau} \prod_{k=1}^N \langle u_{\sigma(k)}\rvert v_{\tau(k)}\rangle = 0,$$ where in all terms of the above sum, the product is always trivially zero, because it will contain a braket of the form $\langle u_j \rvert v_1\rangle$, which always vanishes because $v_1$ is different from all the $u_j$.

• Thank you for taking time to write this answer. Actually I am interested in $M \to \infty$ such as that in multielectron atoms. Nov 3 '16 at 13:08
• @nougako the same reasoning is valid for any $M$
– glS
Nov 3 '16 at 13:09