Consider two systems, each made of $N$ particles. In both systems particles interact pairwise and the interaction is given by the same Hamiltonian for both systems. Any other constraints and/or requirements you'd like to add should be the same. Except for one -- the only difference between these systems is that the first particles are bosons and the second particles are fermions.

I'm interested in ground states of these systems. My intuition tells me that the ground state energy of bosons should always be lower than the ground state of fermions -- no matter what kind of interactions or other external properties we've chosen. But I cannot think of any reasonably general proof for that statement.

Maybe, someone knows how to prove that?

Or, maybe I'm wrong and there is a counterexample?

Edit: If you worry about spins of those particles, then you are free to make this difference too. I'll even give you all the following degeneracies, but the interaction shoud be the same -- spin-independent.

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    $\begingroup$ Not that the Question isn't interesting, but how can the interaction Hamiltonian have the same structure, given that fermions must be spinor fields, whereas bosons must be not spinor fields? $\endgroup$ Mar 8, 2012 at 22:57
  • $\begingroup$ @PeterMorgan Just thought that forgot to comment on spins. I thought about something like "you cannot make interacion the same", but that is not an answer -- just a way to avoid the quesion. Also don't really need the QFT decsription -- I'd happy with with basic quantum mechanics. $\endgroup$
    – Kostya
    Mar 8, 2012 at 23:43
  • $\begingroup$ Maybe an explicit example would be the Bose-Hubbard Hamiltonian vs. the Hubbard model for spinless fermions. Both have a simple tight-binding single-particle part and an on-site interaction term. Is this what you have in mind? $\endgroup$
    – Lagerbaer
    Mar 9, 2012 at 5:02
  • $\begingroup$ Not 100% sure, but what about the Bose-Hubbard model at half filling and infinite repulsion? In that case the ground state has, basically, energy $0$. The corresponding fermionic Hubbard model would also have that energy. $\endgroup$
    – Lagerbaer
    Mar 9, 2012 at 5:07
  • $\begingroup$ @Lagerbaer Actually I'd like to see an example, when energy of fermions is exactly lower that that of bosons. $\endgroup$
    – Kostya
    Mar 9, 2012 at 9:42

2 Answers 2


My intuition tells me that the ground state energy of bosons should always be lower than the ground state of fermions -- no matter what kind of interactions or other external properties we've chosen.

I think your intuition is usually correct; but it's possible to define systems where the ground state fermions would have less energy than the ground state bosons. First, a note on why fermions tend to have higher energies, than a note on how one can arrange for the bosons to have more energy than the fermions.

(1) Assume no particle interactions.

Under the assumption that the Hamiltonian for the boson and fermion are identical, the energies for single particles (i.e. $N=1$) will also be identical. This follows from the fact that the Schroedinger wave equation applies equally to a boson or fermion. In particular, under this assumption, the ground state energies are identical, call that energy $E_1$.

The simplest assumption for particle interactions is that there is none. In this case, the ground state energy for the boson is easy, it's just $N\; E_1$ as all the bosons are in the same state.

The ground state for the fermion will be a higher energy (due to the Pauli exclusion principle) except in the case that the ground state is $N$-fold degenerate in which case the bosons and fermions will have the same energy.

To some people, the above might be obvious in itself. For others, they might want a little less hand waving and a little more mathematics. So let the $N$ lowest energy eigenstates be $\psi_n(x)$ for $n=1,2,3,...,N$ with no degeneracy so that $H\psi_n = E_n\psi_n$. In the boson case the ground state has all the particles in this state so the wave function is a symmetrization of $\psi(x_1,x_2,...,x_n) = \psi_1(x_1)\psi_1(x_2)...\psi_1(x_n)$. But no matter how the positions are permuted, the energy of this state is $E_1+E_1+...+E_1 = N\;E_1$. Similarly, the energy for any permutation of the fermi ground state $\psi_1(x_1)\psi_2(x_2)...\psi_n(x_n)$ is $E_1+E_2+...E_N > N\;E_1$.

(1) Assume arbitrary particle interactions.

With arbitrary particle interactions, it's easy to create a situation where bosons have the same ground state energy as fermions. One simply adds energy to the bose wave functions without adding energy to the fermi states. Let's do this explicitly for a 2-particle wave function. To do this, we need to define the wave functions for $\psi_1 \times \psi_2$, that is, we need to define the tensor product. Let's use the simplest possible wave functions, spinors, and define the combined wave function $\psi_1\otimes \psi_2$ by:
$$|1\rangle\otimes |2\rangle=\left(\begin{array}{c}a_1a_2\\b_1a_2\\a_1b_2\\b_1b_2\end{array}\right).$$
Now (ignoring an unimportant factor of $\sqrt{1/2}$ from here on) a fermi symmetrization looks like this:
$$|1\rangle\otimes|2\rangle-|2\rangle\otimes|1\rangle=\left(\begin{array}{c}a_1a_2-a_2a_1\\b_1a_2-b_2a_1\\a_1b_2-a_2b_1\\b_1b_2-b_2b_1\end{array}\right)= \left(\begin{array}{c}0\\b_1a_2-b_2a_1\\a_1b_2-a_2b_1\\0\end{array}\right),$$
while a bose symmetrization looks like this: $$|1\rangle\otimes|2\rangle+|2\rangle\otimes|1\rangle=\left(\begin{array}{c}a_1a_2+a_2a_1\\b_1a_2+b_2a_1\\a_1b_2+a_2b_1\\b_1b_2+b_2b_1\end{array}\right)= \left(\begin{array}{c}2a_1a_2\\b_1a_2+b_2a_1\\a_1b_2+a_2b_1\\2b_1b_2\end{array}\right).$$
So to add energy only to the bose symmetry case, all we have to do is to include a potential term that looks like:
$$V = \left(\begin{array}{cccc}E&0&0&0\\0&0&0&0\\0&0&0&0\\0&0&0&E\end{array}\right).$$

Note that the above matrix has eigenvalues of 0 and $E$, but the $E$ eigenvalues are only accessible to a boson, the fermions automatically have an energy 0 for this potential. For a bose state to avoid the energy addition, it needs to have $a_1a_2=b_1b_2=0$. There are two ways to do this; either have $a_1=0$ or $a_2=0$. Since you can't have both $a_1$ and $a_2$ zero, or have both $b_1$ and $b_2$ zero, a little algebra will show that there is only one bose state that avoids the $E$:
The above is the bose symmetrization of (1,0) x (0,1). On the other hand, the fermi symmetrization of these two states is: $$\left(\begin{array}{c}0\\+1\\-1\\0\end{array}\right).$$
So to add an energy to the bose symmetrization but not the fermi symmetrization, use a potential of:
$$V = \left(\begin{array}{cccc}E&0&0&0\\0&E/2&E/2&0\\0&E/2&E/2&0\\0&0&0&E\end{array}\right).$$ The above has three eigenvectors with eigenvalue $E$ and only a single eigenvector with eigenvalue 0; this eigenvalue is accessible only to fermions.

  • $\begingroup$ Whilst I think this answers the OP, I do wonder if this method generalises to large $N$? The problem is the restriction to pair-wise interactions --- which imposes constraints on the allowed $V$. I can't shake off the feeling that there should be a simple field theory way to do this... $\endgroup$
    – genneth
    Mar 12, 2012 at 18:24
  • $\begingroup$ Wow. This was a lot of work for only two up votes. $\endgroup$ Jun 23, 2012 at 2:22
  • $\begingroup$ Looks strange that you take a wavefunction of boson as a rank 1 spinor. I'd rather just call it a wavefunction of a bosonic two-level system, not a spinor. $\endgroup$
    – Ruslan
    Dec 17, 2014 at 10:51

The question makes sense in a nonrelativistic setting, where either symmetry or antisymmetry can be imposed on the wave function.

The symmetric ground state always has lower energy, as it is also a ground state of the non-symmetrized system. (Proof: The symmetrization of an arbitrary ground state is again a ground state.)

  • $\begingroup$ This holds only under some caveats; see chapter 3.2.4 in the book by Lieb and Seiringer: The stability of matter in quantum mechanics. $\endgroup$ May 4 at 8:47

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