Let $\mathcal{H}_N$ be the $N$ particle Hilbert space. So a quantum state $\left| \Psi \right>$ may be representated by $$\left| \Psi \right> = \left| k_1 \right>^{(1)}\left| k_2 \right>^{(2)}...\left| k_N \right>^{(N)}$$ , where the $\left| k_i \right>^{(n)}$ each form a basis of the single particle Hilbert space. In order to obtain the fully symmetric subspace $\mathcal{H}_N^S$ the permutation Opertaror $$\hat P \left[ \left| k_1 \right>^{(1)}\left| k_2 \right>^{(2)}...\left| k_N \right>^{(N)} \right] = \left| k_{P_1} \right>^{(1)}\left| k_{P_2} \right>^{(2)}...\left| k_{P_N} \right>^{(N)}$$ and the symmetrisation Operator $$ \hat S = \frac{1}{N!} \sum_P \hat P$$ are introduced.

I am asked to prove that the symmetrisation Operator is hermitian. My first idea was that the permutation operator is hermitian and therefore the symmetrisation Operator, as a sum of hermitian Operators, is too. However I strugle to prove it and by now am unsure if $\hat P$ is hermitian after all. Any help is most welcome.


With some inspiration from the answers I think I got a prove. I think it is different from what the answers are aiming at, but I don't see how to do it in another way. So here my shot: Every permutation Operator can be decomposed into a number of Exchange Operators $\hat E_{ij}$ that exchanges two particles: $$\hat P = \prod_k \hat E_{i_k j_k}$$, with $$\hat E_{ij} \left| k_1 \right>^{(1)}...\left| k_i \right>^{(i)}...\left| k_j \right>^{(j)}...\left| k_N \right>^{(N)} = \left| k_1 \right>^{(1)}...\left| k_{j} \right>^{(i)}...\left| k_{i} \right>^{(j)}...\left| k_N \right>^{(N)}$$ To prove that $\hat P$ (and therefore $\hat S$) is hermitian it is sufficient to prove that $\hat E_{ij}$ is. This finally can be done explicetly via $$\left< k_1' \right|^{(1)}...\left< k_i' \right|^{(i)}...\left< k_j' \right|^{(j)}...\left< k_N' \right|^{(N)} \hat E_{ij} \left| k_1 \right>^{(1)}...\left| k_i \right>^{(i)}...\left| k_j \right>^{(j)}...\left| k_N \right>^{(N)}\\ =\delta(k_1',k_1)... \delta(k_i', k_j)... \delta(k_j', k_i)...\delta(k_N', k_N) $$ ,which is the same as $$\left< k_1 \right|^{(1)}...\left< k_i \right|^{(i)}...\left< k_j \right|^{(j)}...\left< k_N \right|^{(N)} \hat E_{ij} \left| k_1' \right>^{(1)}...\left| k_i' \right>^{(i)}...\left| k_j' \right>^{(j)}...\left| k_N' \right>^{(N)}\\ =\delta(k_1,k_1')... \delta(k_i, k_j')... \delta(k_j, k_i')...\delta(k_N, k_N') $$ Since both terms are reals, we may freely take the complex conjugate and arrive at the hermitian condition for the matrix elemets. I think this should prove it for $\hat E_{ij} \Rightarrow \hat P \Rightarrow \hat S$.

Edit 2

Darn, the prove doesn't work since the $\hat E_{ij}$ don't commute.


4 Answers 4


You may be getting dragged down by your notation. Why not consider the simplest possible case, with only two particles? Essentially, then, you have the two-particle basis $\newcommand{\ket}[1]{|#1\rangle}\{\ket{k_1}\ket{k_2}\}$, and a single permutation operator that acts as $$ \hat{P}\ket{k_1}\ket{k_2}=\ket{k_2}\ket{k_1}. $$ Can you use this to compute the matrix elements $\langle k_1'k_2'|\hat{P}|k_1k_2\rangle$ and $\langle k_1k_2|\hat{P}|k_1'k_2'\rangle$? How should they be related for $\hat{P}$ to be hermitian?

The general case is relatively similar. The permutation operator $\hat{P}$ corresponding to some permutation $P$ acts on basis states as $$ \hat{P}\ket{k_1\cdots k_N}=\ket{k_{P(1)}\cdots k_{P(N)}}. $$ This does not reduce to the two-electron case but an exactly analogous argument holds.

  • $\begingroup$ Yea, I didn't make up this notation, but unfortunaly I am stuck to because my Prof uses it. For your example it is ok: $\langle k_1'k_2'|\hat{P}|k_1k_2\rangle=\delta(k_1',k_2) \delta(k_2',k_1)$, which is the same as $\langle k_1k_2|\hat{P}|k_1'k_2'\rangle=\delta(k_1,k_2') \delta(k_2,k_1')$. Therefore since both are real, $\hat P^*_{k_1'k_2', k_1k_2}=\hat P_{k_1k_2,k_1'k_2'}$ which means precisly that $\hat P$ is hermitian. However this is only the case $N=2$ and only for one specific permutation $P$. In the case $N=2$ the only other Permutation is the trivial one, but what about $N>2$ $\endgroup$
    – Haatschii
    Mar 21, 2013 at 11:42
  • $\begingroup$ This is all you really need. The notation is more complicated for $N>2$ but the essentials remain. Each $\hat{P}$ will act as above for two electrons and as the identity on the rest. Since the tensor product of hermitian operators is hermitian, you're essentially done. $\endgroup$ Mar 21, 2013 at 12:11
  • $\begingroup$ So you mean it has to be brocken down to exchange operators (see my edit)? $\endgroup$
    – Haatschii
    Mar 21, 2013 at 12:18
  • $\begingroup$ @EmilioPisanty I'm not sure I understand your argument. How are tensor products of hermitian operators relevant here? It is true that the permutation operator can be written as a composition of transposition operators, but even so, as pointed out above, such a composition is not necessarily hermitian. I think there's a bit more to this for $N>2$ than you're giving it credit for... $\endgroup$ Mar 21, 2013 at 16:39
  • $\begingroup$ Yes, you're right. I'll update my answer to account for that. $\endgroup$ Mar 21, 2013 at 16:52

Lets start with a simple case where we consider only two particles. Take a complete set of one particle state such that a basis of two particle states can be obtained as a product


with the permutation operator that exchanges particles


Applying twice this operator we recover the original state, thus $P^{2}=1$. $P$ is showed to be hermitian considering any matrix element

$\langle u_{i}^{(1)},u_{j}^{(2)}|P|u_{i'}^{(1)},u_{j'}^{(2)}\rangle=\langle u_{i}^{(1)},u_{j}^{(2)}|u_{i'}^{(2)},u_{j'}^{(1)}\rangle=\delta_{ij'} \delta_{ji'}$

Since all the matrix elements are real you get $P^{\dagger}=P$. And since $P^{2}=1$ we also have $PP=P^{\dagger}P=1$, thus $P$ is unitary. This can be easily generalized for $N$ particles. This implies $S^{\dagger}=S$, that the operator is hermitian. This result is also true for the antisymmetrization operator. To convince yourself, take states that are eigenstates of all permutation operators in a system with $N$ particles. Index with $p$ the number of all possible $N!$ permutation and let $|\psi\rangle$ such that $P_{p}|\psi\rangle=c_{p}|\psi\rangle$. Its easily seen that the eigenvalue for a permutation should be $c_{p}=(\pm 1)^{n_p}$ where $n_{p}$ is the number of transpositions in which the $p$-th permutation can be splitted. Now you find two case, the totally symmetric and antisymmetric case. which leads you to the symmetrisation and antisymmetrisation operators.


For $N>2$ I generalize the case for two particles; construct a basis for the $N$-particles state as a product of one particle states


In general there are $N!$ possible permutations. A general permutation is denoted by

$P_{l_{1}\dots{l_N}}$ where $l_{j}=1\dots N$ and $j=1\dots N$

However, not to carry to many indices, I'll discuss the case for $N=3$. Obviously the are $6$ possible permutation $P_{123},P_{231},P_{312},P_{213},P_{321},P_{132}$. The permutation operators form a group since

  1. the product of two permutation operators is a permutation operator $P_{312}P_{213}=P_{132}$
  2. the identity is also a permutation operator. $P_{123}$ in this case
  3. the inverse of a permutation is also a permutation. $P_{231}^{-1}=P_{312}$, etc.

But, in general permutation do not commute. Here it can be seen that $P_{213}P_{312}=P_{321}\ne P_{132}$. As you know, a permutation where only two particles are permuted is called a transposition. For $N=3$ these are $P_{132}, P_{321},P_{213}$. As I've showed above for $N=2$ the transpositions are hermitian and unitary. Also, each permutation can be expressed as product of transpositions, $P_{312}=P_{132}P_{213}$. The parity of a permutation is given by the parity of the number of transpositions in which a permutation can be splitted. Since a permutation can be expressed as a product of transpositions and transpositions are unitary, permutations are unitary. But, as above, permutations do not commute in general, although transpositions are hermitian, a general permutation is not.

Now, with this lets pass to symmetrization and antisymmetrization operators. As stated in the last part of my comment before the edit we look for states that are eigenstates of all permutation operators in a system with $N$ particles. The states must satisfy $P_{p}|\psi\rangle=c_{p}|\psi\rangle$ ($p$ is the same as stated above before the edit). Now, the easiest case is that of a transposition, where $c_{p}^{transpose}=(\pm 1)^{n_{p}}$. Since $|\psi\rangle$ is assumed to be an eigenvector of all permutations, it should be an eigenvector of all transpositions. Also, if the particles are indistinguishable, the eigenvalue of the transposition cannot depend on the particular transposition, thus the eigenvalue should be the same for all transpositions. The eigenvalue for a permutation is then $c_{p}=(\pm 1)^{n_{p}}$. It is even or odd depending on the parity of the permutation. This gives us two cases

Total symmetric case $P_{p}|\psi_{S}\rangle=|\psi_{S}\rangle$, $c_{p}=1$

Total antisymmetric case $P_{p}|\psi_{A}\rangle=a_{p}|\psi_{A}\rangle$ where $a_{p}=\pm 1$ for odd and even permutations

Now, the totally symmetric and antisymmetric states spans a subspace of the total Hilbert space which we denote $\mathcal{H}_{S}$ and $\mathcal{H}_{A}$. Its easily proven that the states of these two subspaces are orthogonal to each other


Here I assumed that the parity of $P_{p}^{-1}$ is odd. Finally, we build the projection operators on the subspaces $\mathcal{H}_{S}$ and $\mathcal{H}_{A}$ s.t. we obtain elements of either subspaces from any given states

$S=\frac{1}{N!}\sum_{p}P_{p}$ symmetrization op.

$A=\frac{1}{N!}\sum_{p}a_{p}P_{p}$ antisymmetrization op.

Since $P_{p}$ is unitary, $P^{\dagger}=P^{-1}$ is also a permutation operator. All of this, implies that $S^{\dagger}=S$ and $A^{\dagger}=A$, thus both operators are hermitian. Also, given a random permutation operator $P_{r}$ we find


with this you can show that $S^{2}=S$. This can help you in a more detailed proof. Hope this helps.

  • $\begingroup$ Thanks for your answer. But can you elaborate on "This can be easily generalized for N particles."? Since I seem to be unable to find that easy way (see my Soulution Edit) $\endgroup$
    – Haatschii
    Mar 21, 2013 at 12:16
  • $\begingroup$ @Haatschii I've edited my answer, hope it helps you. $\endgroup$ Mar 21, 2013 at 13:44
  • $\begingroup$ Thanks for you edit, that helped. Still the tricky part for me is this line: $\langle\psi_{S}|\psi_{A}\rangle=\langle\psi_{S}|P_{p}^{\dagger}|\psi_{A}\rangle=\langle\psi_{S}|P_{p}^{-1}|\psi_{A}\rangle=-\langle\psi_{S}|\psi_{A}\rangle$, because at this Point I don't know that $\hat P_p$ is hermitian, do I? $\endgroup$
    – Haatschii
    Mar 21, 2013 at 14:19

First let me denote the basis state $|\Psi\rangle$ that you have defined as $$ |k_1, \dots, k_N\rangle $$ and note that what you said at first is not exactly write; these states form a basis for $\mathcal H_N$ so every state $|\psi\rangle\in\mathcal H_N$ can be written as some linear combination of these basis states; $$ |\psi\rangle = \sum_{k_1, \dots, k_N}c_{k_1\cdots k_N}|k_1, \dots, k_N\rangle $$ To make our notation even more compact, let's define $$ |\mathbf k\rangle = |k_1, \dots, k_N\rangle $$ The matrix elements of the permutation operator in this basis are $$ P_{\mathbf k', \mathbf k}=\langle \mathbf k'|\hat P|\mathbf k\rangle $$ Now the question is, is it the case that $\hat P^\dagger = \hat P$, which in terms of matrix elements reads $$ P^*_{\mathbf k, \mathbf k'} = P_{\mathbf k', \mathbf k}\qquad ? $$ I'll leave it to you to move on from here :)

Edit 1. Comments and proof.

So, as pointed out above, the permutation operator is not hermitian. However, here is a short, elegant proof for the fact that $\hat S$ is hermitian. To write the (one line) proof, let me introduce the following notation: Let $S_N$ denote the symmetric group which is just the set of bijections of the set $\{1, 2, \dots N\}$. For each $\sigma\in S_N$, and for each $\mathbf k$ let $$ \mathbf k_\sigma = (k_{\sigma(1)}, k_{\sigma(2)}, \dots, k_{\sigma(N)}) $$ and therefore the action of a permutation on states can be written as $$ \hat P_\sigma|\mathbf k\rangle = |\mathbf k_\sigma\rangle $$ Proof. \begin{align} \hat S_{\mathbf k', \mathbf k} &=\langle\mathbf k'|\hat S|\mathbf k\rangle = \frac{1}{N!}\sum_{\sigma\in S_N} \langle\mathbf k'|\hat P_\sigma|\mathbf k\rangle = \frac{1}{N!}\sum_{\sigma\in S_N} \langle \mathbf k'|\mathbf k_\sigma\rangle = \frac{1}{N!}\sum_{\sigma\in S_N} \langle \mathbf k'_{\sigma^{-1}}|\mathbf k\rangle \\ &= \frac{1}{N!}\sum_{\sigma^{-1}\in S_N} \langle \mathbf k'_{\sigma^{-1}}|\mathbf k\rangle = \frac{1}{N!}\sum_{\sigma\in S_N} \langle \mathbf k'_{\sigma}|\mathbf k\rangle = \frac{1}{N!}\sum_{\sigma\in S_N} \langle \mathbf k|\mathbf k'_{\sigma}\rangle = \frac{1}{N!}\sum_{\sigma\in S_N} \langle \mathbf k|\hat P_\sigma|\mathbf k'\rangle \\ &= \hat S_{\mathbf k, \mathbf k'} \\ \end{align} This shows that $\hat S$ is symmetric, but since its entries are real, it is therefore hermitian.

Note. The crucial step here was to observe that $$ \langle \mathbf k'|\mathbf k_\sigma\rangle = \langle \mathbf k'_{\sigma^{-1}}|\mathbf k\rangle $$ This follows from the fact that $$ \langle \mathbf k'|\mathbf k\rangle = \delta_{\mathbf k', \mathbf k} $$ and the fact that $\mathbf k'=\mathbf k_\sigma$ if and only if $\mathbf k' = \mathbf k_\sigma$.

  • $\begingroup$ I did come up with a soulution (See my edit), however I doubt it is what you intended? $\endgroup$
    – Haatschii
    Mar 21, 2013 at 12:17
  • $\begingroup$ I posted a pretty short proof; I think user Nijankowski's answer somewhat obscures the issue; just my humble opinion. $\endgroup$ Mar 21, 2013 at 16:34

I think P is unitary because it maps an ortonormal basis to an ortonormal basis. PP=1 so P^-1=P and being unitary P*=P. So S*=S, from SS=S you have S=S^-1=S* Marco


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