03 | May | 2015 | Physics Town

A First Course in String Theory

$\sum_{s_1, s_2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$

Prove that “only the part of the wavefunction $\psi$ that is antisymmetric under the simultaneous exchange $p_1 \leftrightarrow p_2$ and $s_1 \leftrightarrow s_2$ contributes…”

~~~

We divide the proof into two parts. In the first part, we prove that being antisymmetric under $p_1 \leftrightarrow p_2$ is a must. Then, in the second part, we use the same method to prove that being antisymmetric under $s_1 \leftrightarrow s_2$ is also a must.

First, we separate $\psi$ into two parts:

$\psi (p_1, s_1; p_2, s_2) = \psi_{sp} + \psi_{ap}$ ,

where $\psi_{sp}$ is symmetric under $p_1 \leftrightarrow p_2$ and $\psi_{ap}$ is anti-symmetric.

Wavefunctions $\psi_{sp}$ and $\psi_{ap}$ can be defined as

$\psi_{sp} = \frac{1}{2} \left[ \psi (p_1, s_1; p_2, s_2) + \psi (p_2, s_1; p_1, s_2) \right]$
$\psi_{ap} = \frac{1}{2} \left[ \psi (p_1, s_1; p_2, s_2) - \psi (p_2, s_1; p_1, s_2) \right]$

Now, we consider the contribution of the symmetric part, $\psi_{sp} (p_1, s_1; p_2, s_2)$ .

$\int d \vec p \psi_{sp} (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$
$= \frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$
$+ \frac{1}{2} \int d \vec p \psi (p_2, s_1; p_1, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$

By interchanging the dummy variables $p_1$ and $p_2$ in the second term, we have

$...$
$= \frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger | \Omega \rangle$
$+ \frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) f_{p_2, s_1}^\dagger f_{p_1, s_2}^\dagger | \Omega \rangle$
$=\frac{1}{2} \int d \vec p \psi (p_1, s_1; p_2, s_2) \left[ f_{p_1, s_1}^\dagger f_{p_2, s_2}^\dagger + f_{p_2, s_1}^\dagger f_{p_1, s_2}^\dagger \right] | \Omega \rangle$
$= 0$