Assume we have a system with N particles, each particle in one of $\gamma$ different single particle states $|k^\gamma\rangle$. The state space of the multi-particle system is spanned by the basis $$\mathcal{B}=\{|k^{p_1},k^{p_2}...k^{p_\gamma}\rangle\}_{\textrm{All permutations } p_i\in\{1..\gamma\}}$$

where the following notation is used$$|k^{p_1},k^{p_2}...k^{p_\gamma}\rangle:=|k^{p_\gamma}\rangle\otimes...\otimes|\textrm k^{p_\gamma}\rangle$$ Let us consider Bosonic systems. Take as an axiom, that the multi-particle state of a Bosonic system have to be symmetric under any exchange of the states of two particles.

How to prove in general that there is only one multi-particle state $|\psi\rangle$ with $n_i$ particles in state $|k_i\rangle$? One can easily show this for 3 particles in a 3 level system, but the general case requires a bit more combinatorics and I'm struggeling. For now I would even be happy if someone can just assert that this is in fact true.

Motivation: I need to know that this is actually true in order to properly understand the derivation of the grand-canonical partition sum of the Bose gas. There one replaces the trace over all Bosonic multi-particle states with arbitrary particle number $N$ by a sum over all single-particle state occupation numbers.

  • $\begingroup$ I’m a little confused about your statement of the problem. Is your question how to prove that giving occupation numbers for every single-particle state uniquely determines your many-body state? $\endgroup$ – Chris Fechisin Jun 19 '20 at 15:11
  • $\begingroup$ @ChrisFechisin Yes sorry for the unclear notation, but this is exactly what I'm asking. I want to make sure that there aren't two quantum states with the same occupation numbers. I think that the anti/symmetrization condition can be used to show this. $\endgroup$ – user224659 Jun 19 '20 at 17:17
  • $\begingroup$ Is that state not just $|k_i \rangle \otimes \ldots \otimes |k_i \rangle$, with $n_i$ tensor products? What is there to show? $\endgroup$ – user1379857 Jun 19 '20 at 18:08
  • $\begingroup$ @user1379857 well consider a two energy system. $\epsilon_1$ and $\epsilon_2$. consider a state with $n_{\epsilon_1}=1$ and $n_{\epsilon_2}=1$. One could assume that all states $c_1|\epsilon_1\rangle+c_2|\epsilon_2\rangle$ fulfill this condition. But indeed only one state is symmetric or antisymmetric. Namely $\frac{1}{\sqrt{2}}(|\epsilon_1\rangle\pm |\epsilon_2\rangle)$. In this sense for both fermions and bosons the occupation representation is unique. But I want to prove this for an arbitrary particle number with an arbitrary number of energy levels. $\endgroup$ – user224659 Jun 19 '20 at 18:38

This follows because the fully symmetric representation of the permutation group $S_m$ of $m$ states is 1-dimensional, and because the symmetric state \begin{align} \sum_{\sigma\in S_\gamma} P(\sigma) |k^{p_1},k^{p_2}...k^{p_\gamma}\rangle\, , \end{align} where $$ P(\sigma)|k^{p_1},k^{p_2}...k^{p_\gamma}\rangle = |k^{p_{\sigma(1)}},k^{p_{\sigma(2)}}...k^{p_{\sigma(\gamma)}}\rangle $$ carries this representation of the permutation group. Here this group would be $S_\gamma$. For instance the combination $$ \frac{1}{\sqrt{6}} \left(|210\rangle + |201\rangle+ \vert 120\rangle + \vert 102\rangle + \vert 021\rangle+\vert 012\rangle \right) \tag{1} $$ is an example of such a fully symmetric state of $S_3$. The dimension of the representation is the dimension of the set $\{|210\rangle, |201\rangle, \vert 120\rangle, \vert 102\rangle ,\vert 021\rangle,\vert 012\rangle\}$ and is is obviously the number of permutations of three distinct numbers $\{0,1,2\}$, i.e. the three possible states of your system.

It is easy to see, by construction, that the symmetric irrep occurs at least once and is spanned precisely by the combination (1). But can the symmetric representation appear more than once? The answer is no by orthogonality: any state orthogonal to (1) or its generalization must ``traceless'', i.e. of the type $$ \frac{1}{\sqrt{6}} \left(a|210\rangle + b |201\rangle+ c \vert 120\rangle + \vert 102\rangle + d \vert 021\rangle+f \vert 012\rangle \right)\, , \tag{2} $$ with $a+b+c+d+e+f=0$. For a state like (2) to be symmetric under any permutation, one needs $a=b=c=d=e=f$, which is incompatible with the traceless condition.

Thus (up to an overall phase an normalization), symmetrization guarantees you necessarily are the a 1-dimensional subspace of your full Hilbert space, and that this representation occurs once.

  • $\begingroup$ I got a bit confused by the first part of your answer, but the part after But can the symmetric representation appear more than once? perfectly answered my question, thanks! $\endgroup$ – user224659 Jun 24 '20 at 14:01

When you’re writing down a fermionic or bosonic many-body wavefunction in terms of occupation numbers of single-particle states, you’re writing a Fock state. This is simply the many-body wavefunction written in the basis of occupation numbers of single-particle states, with an implicit underlying algebra (one follows different rules when changing occupation numbers in a fermionic versus a bosonic state).

For a given Fock state (read: list of occupation numbers), there is a unique way to write down the wavefunction taking into account the indistinguishability of quantum particles and the symmetry (antisymmetry) requirements of bosonic (fermionic) wavefunctions.

For fermions, we employ the fully antisymmetric Slater determinant. There is a similar mapping using the fully symmetric matrix permanent.

Consider a Fock state $$|\Psi\rangle_\text{Fock}=|n_1,n_2\rangle=|1,1\rangle$$ where $n_i$ gives the occupation numbers of the single-particle state $\psi_i$. If the particles are fermions, we write the many-body wavefunction: $$|\Psi\rangle_F=\frac{1}{\sqrt{2}}\left(\psi_1(x_1) \psi_2(x_2)- \psi_1(x_2) \psi_2(x_1)\right)$$ Notice that we can write $|\Psi\rangle_F$ as a determinant: $$|\Psi\rangle_F=\frac{1}{\sqrt{2}} \text{det}\left|\begin{array}. \psi_1(x_1) && \psi_2(x_1)\\ \psi_1(x_2)&&\psi_2(x_2)\end{array}\right|$$ This is the slater determinant, and it generalizes to arbitrarily many fermions $N$.

The bosonic case is only slightly different. For the same Fock state, if the particles are bosons, we write the many-body wavefunction: $$|\Psi\rangle_B=\frac{1}{\sqrt{2}}\left(\psi_1(x_1) \psi_2(x_2)+\psi_1(x_2) \psi_2(x_1)\right)$$ Notice that we can write $|\Psi\rangle_B$ as a permanent: $$|\Psi\rangle_B=\frac{1}{\sqrt{2}} \text{perm}\left|\begin{array}. \psi_1(x_1) && \psi_2(x_1)\\ \psi_1(x_2) &&\psi_2(x_2)\end{array}\right|$$ This too generalizes to arbitrarily many bosons $N$.

The determinant (permanent) is the simplest (and perhaps only?) totally antisymmetric (symmetric) operator mapping Fock space to the space of single-particle product states. In this way, a given Fock state specifies a single many-body wavefunction.

Edit: Uniqueness Proof

Assume $N$ identical particles with coordinates $x_1,\ldots,x_N$ and $M$ occupied single-particle states $\psi_1,\ldots,\psi_M$. For one Fock state with integer occupation numbers $n_1,\ldots,n_M$, the most general corresponding many-body state may be written: $$|\Psi\rangle=\sum_{\text{perms. of }x_{ij}}s_{ij}\otimes_{i=1}^M\otimes_{j=1}^{n_i}\psi_{i,j}(x_{ij})$$ where $\psi_{i,j}$ denotes the $j^\text{th}$ particle in state $\psi_i$ and $s_{ij}$ is a factor corresponding to each permutation of the coordinates $x_{ij}$. Because you are given occupation numbers for each wavefunction, you know which single-particle wave functions must appear in every term; your only remaining degree of freedom is the matching of wavefunctions to particle coordinates, thus the sum over all permutations thereof.

For the Fock state $|\Psi\rangle=|1,2,1\rangle$ this looks like: $$|\Psi\rangle=\sum_{\text{perms. of }x_{ij}}s_{ij}\psi_{1,1}(x_{1,1})\psi_{2,1}(x_{2,1})\psi_{2,2}(x_{2,2})\psi_{3,1}(x_{3,1})$$

This above general expression for $|\Psi\rangle$ is the immanant of the matrix of possible single-particle states:

$$M=\left[\begin{array}1 \psi_{1,1}(x_1) && \psi_{1,2}(x_1) && \cdots && \psi_{1,n_1}(x_1) &&\cdots&& \psi_{M,n_M}(x_1)\\ \vdots && && && && && \vdots\\ \psi_{1,1}(x_N) && \psi_{1,2}(x_N) && \cdots && \psi_{1,n_1}(x_N) && \cdots && \psi_{M,n_M}(x_N)\end{array}\right]$$

You can write down lots of funky immanants, but the permanent (determinant) is the unique fully symmetric (antisymmetric) immanant. We can therefore conclude a given Fock state maps to a unique many-particle state in the single-particle product state basis.

  • $\begingroup$ Okay thanks , now I know the term fock states and permanent, this is nice, never heard of this before. But I still believe that one should be able to show the uniqueness, this is what I was looking for. I wrote another comment to my question, to make it more clear. @Chris Fechisin $\endgroup$ – user224659 Jun 19 '20 at 18:42
  • $\begingroup$ thanks for introducing me to all of this terminology! I will try to find a good link to a proof of the theorem you stated once my exam is done. Theorem: The permanent (determinant) is the unique fully symmetric (antisymmetric) immanant $\endgroup$ – user224659 Jun 19 '20 at 22:22

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