# Few particle fermion system wavefuction

Suppose I have 3 fermions($\left|\psi_1\right\rangle$, $\left|\psi_2\right\rangle$, $\left|\psi_3\right\rangle$) and a system with 3 states ( $\left|1\right\rangle$, $\left|2\right\rangle$, $\left|3\right\rangle$)

By exclusion principle we can say that the total wave function of the system, $\left|\Psi\right\rangle$ , can only have terms like $\left|ijk\right\rangle$, where ijk is some permutation of $123$

Suppose the wave function of the particle $1$ is prepared as $\left|\psi_1\right\rangle= a \left|2\right\rangle+ b\left|3\right\rangle$ Then will it be true that the total wave function, $\left|\Psi\right\rangle$ will not have terms $\left|123\right\rangle$ and $\left|132\right\rangle$?

In general is it possible to prepare particle wave functions of a system of fermions to make sure that the total wave function of the system admits only certain configurations of the states?

• If you're dealing with this stuff you will probably (hopefully!) soon learn that there's a much easier way to handle multi-particle systems. Read about "second quatization". Here is an SE post about it: physics.stackexchange.com/questions/122570/…. Commented Oct 21, 2014 at 8:23

## 1 Answer

Once you consider all three fermions, their wave function must be completely antisymmetric which, in your case, means that it must be a multiple of $$\frac{ |123\rangle+ |231\rangle+ |312\rangle- |132\rangle- |321\rangle- |213\rangle }{\sqrt{6}}$$ You may imagine that at the very beginning when you only know about the single fermion, it has the wave function you wrote down. But once you consider all three, the wave function of all of them has to be the antisymmetric part of the tensor product, and not just the tensor product itself, which guarantees that the terms starting with "1" are represented just like all others.

Wave functions that are not fully antisymmetric are simply not allowed for fermions.

• So the anti symmetry requires always terms. There is no way by using fermions to get a wave function which has less than that number of terms? Commented Oct 21, 2014 at 8:19
• No. Maybe imagining that it's "many terms" - namely $3!=6$ here and $n!$, n factorial, more generally – is a bit misleading way to describe the situation. The vector I wrote down, a superposition of 6 other terms, may be viewed as "six terms", but it's still "one vector", and what is variable is the coefficient in front of this vector (the complex probability amplitude), so the vector describes "one" possibility for 3 fermions in 3 states. For $N$ fermions in $k$ (fewer) states, you get more than one possible states but if the number of "boxes" is the same as the number of f., there's 1 option Commented Oct 21, 2014 at 9:42
• I think you mean $k$ is larger than $N$ Commented Oct 21, 2014 at 10:22
• Yup, sorry. The number of states -- the dimension of the HIlbert space - is "k choose N" then. Commented Oct 21, 2014 at 12:24