# Physical kets

So far we have seen that there exist a unique ket which properly describe the state of a system of identical particles. This ket is a completely symmetric ket for bosons and a completely antisymmetric ket for fermions. The problem is now to construct these kets. The procedure is theoretically simple:

• number the particles and construct the ket ${\displaystyle \left|u\right\rangle }$ corresponding to the physical state and to the number given to the particles
• Apply ${\displaystyle S}$ or ${\displaystyle A}$ depending on the nature of the particles
• Normalize

This process allows us to construct easily the physical ket if the number of the particles is low. On the contrary for a system composed of a large number of particles we need a mechanical method in order to simplify calculations.

Example (Two identical particles)

Suppose we are dealing with a system with two identical particles,labeled with ${\displaystyle \left(1\right),\left(2\right)}$. Assume that each particle can be in the orthogonal states ${\displaystyle \left|\varphi \right\rangle }$ or ${\displaystyle \left|\chi \right\rangle }$. The state of the system can be:

{\displaystyle {\begin{aligned}\left|\varphi \right\rangle _{1}\left|\chi \right\rangle _{2}=\left|\varphi \left(1\right),\chi \left(2\right)\right\rangle \\\left|\varphi \right\rangle _{2}\left|\chi \right\rangle _{1}=\left|\varphi \left(2\right),\chi \left(1\right)\right\rangle \end{aligned}}}
. We chose ${\displaystyle \left|u\right\rangle =\left|\varphi \left(1\right),\chi \left(2\right)\right\rangle }$, then:

• if the particles are bosons we apply ${\displaystyle S}$ on ${\displaystyle \left|u\right\rangle }$ and normalize.We get the symmetric ket:
${\displaystyle S\mid u\rangle ={\frac {1}{\sqrt {2}}}\left(\mid \varphi \left(1\right),\chi \left(2\right)\rangle +\mid \varphi \left(2\right),\chi \left(1\right)\rangle \right)}$
• if the particles are fermions we apply ${\displaystyle A}$ on ${\displaystyle \left|u\right\rangle }$ and normalize. We get the antisymmetric ket:
${\displaystyle S\left|u\right\rangle ={\frac {1}{\sqrt {2}}}\left(\left|\varphi \left(1\right),\chi \left(2\right)\right\rangle -\left|\varphi \left(2\right),\chi \left(1\right)\right\rangle \right)}$

We can easily see that if ${\displaystyle \left|\chi \right\rangle =\left|\varphi \right\rangle }$ there isn't any antisymmetric ket able to describe two fermions in the same state: this is a first insight in the Pauli exclusion principle,according to which two identical fermions cannot be in the same individual state.

## Slater determinant

For a system of ${\displaystyle N}$ identical particles there is a simple way to construct the symmetric and antisymmetric ket, based on the properties of a matrix determinant (remember that the definition of determinant is strictly linked to permutations). We summarize the procedure:

• number the particles ${\displaystyle 1,2,\dots N}$
• find the individual states accessible to the particles: ${\displaystyle \left|a\right\rangle ,\left|b\right\rangle ,\left|c\right\rangle \dots \left|z\right\rangle }$. There are ${\displaystyle N}$ states,not necessary different.
• construct a squared matrix with dimension ${\displaystyle N\times N}$ as follows:
{\displaystyle {\begin{aligned}{\begin{bmatrix}\left|a\left(1\right)\right\rangle &\cdots &\left|z\left(1\right)\right\rangle \\\vdots &\ddots &\vdots \\\left|a\left(N\right)\right\rangle &\cdots &\left|z\left(N\right)\right\rangle \end{bmatrix}}\end{aligned}}}
• write down the expression for determinant. If you're dealing with bosons change each ${\displaystyle -}$ with ${\displaystyle +}$,otherwise leave the determinant unchanged.
• normalize the expression found with the factor ${\displaystyle {\frac {1}{\sqrt {N!}}}}$
• You have found the completely symmetric or completely antisymmetric ket describing the system!

## When are antisymmetrization and symmetrization necessary?

If the application of the symmetrization postulate were always indispensable we should consider every electron in the universe even in the study of a single atom, since all electrons are identical. Clearly this is not the case. In fact under special conditions we can neglect the indistiguishability of the particles. Let us discuss when this approximation is appropriate.

Consider two identical particles ${\displaystyle (1),(2)}$, one being in the individual state ${\displaystyle \left|\psi \right\rangle }$ and the other in the individual state ${\displaystyle \left|\varphi \right\rangle }$. Assume that ${\displaystyle \left\langle x\mid \psi \right\rangle =\psi (x)}$ vanishes outside a region ${\displaystyle \mathbb {D} }$ and ${\displaystyle \left\langle x\mid \varphi \right\rangle =\varphi (x)}$ vanishes outside a region ${\displaystyle \mathbb {B} }$, with ${\displaystyle \mathbb {D} \cap \mathbb {B} =\mathbb {0} }$. The symmetric and antisymmetric waves function for the system are:

{\displaystyle {\begin{aligned}{\frac {1}{\sqrt {2}}}\left(\psi (x_{1})\varphi (x_{2})+\psi (x_{2})\varphi (x_{1})\right)\\{\frac {1}{\sqrt {2}}}\left(\psi (x_{1})\varphi (x_{2})-\psi (x_{2})\varphi (x_{1})\right)\end{aligned}}}
Since the individual wave functions do not overlap, if ${\displaystyle x_{1}\in \mathbb {D} }$ and ${\displaystyle x_{2}\in \mathbb {B} }$ we have:
{\displaystyle {\begin{aligned}\psi (x_{1})\neq 0\quad {\mbox{since }}x_{1}\in \mathbb {D} \\\varphi (x_{2})\neq 0\quad {\mbox{since }}x_{2}\in \mathbb {B} \\\psi (x_{2})=0\quad {\mbox{since }}x_{2}\notin \mathbb {D} \\\varphi (x_{1})=0\quad {\mbox{since }}x_{1}\notin \mathbb {B} \end{aligned}}}
The system wave functions then become:
{\displaystyle {\begin{aligned}{\frac {1}{\sqrt {2}}}\left(\psi (x_{1})\varphi (x_{2})+0\right)\\{\frac {1}{\sqrt {2}}}\left(\psi (x_{1})\varphi (x_{2})-0\right)\end{aligned}}}
We can conclude that if wave functions do not overlap symmetrization or antisymmetrization aren't necessary