# Can two or more fermions be in the same mode?

I have always thought that only one fermion can occupy the same mode of Fock state. If two fermions are lying on the Fermi surface, it means that they have the same momentum. Doesn´t it mean that they are occupying the same mode if they have the same momentum? So my question is: Can two or more fermions be in the same mode?

• Momentum is a vector quantity. Two electrons on the Fermi surface may have the same magnitude of momentum, but they cannot have the same momentum (assuming, of course, there are no other degrees of freedom, such as spin) Jan 31, 2021 at 19:58
• I didn´t express myself clearly. Doesn´t it mean that they are occupying the same mode if they have the same magnitude of momentum? Jan 31, 2021 at 20:01
• No. Different vector momentum represent different modes, even if they have the same magnitude Jan 31, 2021 at 20:02

The story about fermions not being allowed to be in the same mode is a consequence of the algebra their creation and annihilation operators obey. Suppose that $$a(\mathbf{p},\sigma,n)$$ and $$a^\dagger(\mathbf{p},\sigma,n)$$ are annihilation and creation operators associated to fermions with momentum $$\mathbf{p}$$, spin/helicity $$\sigma$$ and species $$n$$. The species index includes, for example, possible internal symmetry quantum numbers, like color.
In that case we have $$\{a(\mathbf{p},\sigma,n),a^\dagger(\mathbf{p}',\sigma',n')\}=\delta_{\sigma\sigma'}\delta_{nn'}\delta^{(3)}(\mathbf{p}-\mathbf{p}')\\\{a^\dagger(\mathbf{p},\sigma,n),a^\dagger(\mathbf{p}',\sigma',n')\}=0,\quad \{a(\mathbf{p},\sigma,n),a(\mathbf{p}',\sigma',n')\}=0,\tag{1}$$
Now let $$|\psi\rangle$$ be an arbitrary state. Then you have, from the first equality in the second line $$a^\dagger(\mathbf{p},\sigma,n)a^\dagger(\mathbf{p'},\sigma',n')|\psi\rangle=-a^\dagger(\mathbf{p}',\sigma',n')a^\dagger(\mathbf{p},\sigma,n)|\psi\rangle\tag{2}.$$