Represent an arbitrary number state \(n\) in terms of creation operators and the vacuum state

for bosons

 * 1) Normalise the creation and annihilation operators.
 * 2) Calculate \(\|\hat{a}^{\dagger}|\alpha\rangle\|\): \(|\alpha+1\rangle\ = \frac{\hat{a}^{\dagger}|\alpha\rangle}{\|\hat{a}^{\dagger}|\alpha\rangle\|}\) \(\|\hat{a}^{\dagger}|\alpha\rangle\| = \sqrt{\langle \alpha|\hat{a}\hat{a}^{\dagger}|\alpha\rangle}\) \(\|\hat{a}^{\dagger}|\alpha\rangle\| = \sqrt{\langle \alpha|\left(1+\hat{a}^{\dagger}\hat{a}\right)|\alpha\rangle}=\sqrt{1+\alpha}\)
 * 3) Calculate \(\|\hat{a}|\alpha\rangle\|\): \(|\alpha-1\rangle\ = \frac{\hat{a}|\alpha\rangle}{\|\hat{a}|\alpha\rangle\|}\) \(\|\hat{a}|\alpha\rangle\| = \sqrt{\langle \alpha|\hat{a}^{\dagger}\hat{a}|\alpha\rangle}=\sqrt{\alpha}\)
 * 4) Set up a system in state \(|n\rangle\), where \(|n\rangle\) is a normalised eigenstate of the number operator \(\hat{a}^{\dagger}\hat{a}\).
 * 5) From the excited state, lower the state once: \(\hat{a}|n\rangle = \sqrt{n}|n-1\rangle\)
 * 6) Multiply on the left by \(\hat{a}^{\dagger}\): \(\hat{a}^{\dagger}\hat{a}|n\rangle = \sqrt{n}\hat{a}^{\dagger}|n-1\rangle\)
 * 7) Substitute the number operator \(\hat{n}\) for \(\hat{a}^{\dagger}\hat{a}\) and calculate: \(n|n\rangle = \sqrt{n}\hat{a}^{\dagger}|n-1\rangle\)
 * 8) Divide through by the coefficient of \(|n\rangle\): \(|n\rangle = \frac{1}{\sqrt{n}}\hat{a}^{\dagger}|n-1\rangle\)
 * 9) If the ket on the right hand side is not the vacuum state \(|\Omega\rangle\), go to step 3. Otherwise, continue to step 8. \(|n\rangle = \frac{1}{\sqrt{n}}\frac{1}{\sqrt{n-1}}\frac{1}{\sqrt{n-2}}\cdots\left(\hat{a}^{\dagger}\right)^n|\Omega\rangle\)
 * 10) Simplify: \(|n\rangle = \frac{1}{\sqrt{n!}}\left(\hat{a}^{\dagger}\right)^n|\Omega\rangle\)

for fermions

 * 1) Normalise the creation and annihilation operators.
 * 2) Calculate \(\|\hat{a}^{\dagger}|\alpha\rangle\|\): \(|\alpha+1\rangle\ = \frac{\hat{a}^{\dagger}|\alpha\rangle}{\|\hat{a}^{\dagger}|\alpha\rangle\|}\) \(\|\hat{a}^{\dagger}|\alpha\rangle\| = \sqrt{\langle \alpha|\hat{a}\hat{a}^{\dagger}|\alpha\rangle}\) \(\|\hat{a}^{\dagger}|\alpha\rangle\| = \sqrt{\langle \alpha|\left(1-\hat{a}^{\dagger}\hat{a}\right)|\alpha\rangle}=\sqrt{1-\alpha}\)
 * 3) Calculate \(\|\hat{a}|\alpha\rangle\|\): \(|\alpha-1\rangle\ = \frac{\hat{a}|\alpha\rangle}{\|\hat{a}|\alpha\rangle\|}\) \(\|\hat{a}|\alpha\rangle\| = \sqrt{\langle \alpha|\hat{a}^{\dagger}\hat{a}|\alpha\rangle}=\sqrt{\alpha}\)
 * 4) Set up a system in state \(|n\rangle\), where \(|n\rangle\) is a normalised eigenstate of the number operator \(\hat{a}^{\dagger}\hat{a}\).
 * 5) Find the value of \(\hat{a}^{\dagger}\hat{a}^{\dagger}\).
 * 6) Write expand the anticommutator\([\hat{a}^{\dagger},\hat{a}^{\dagger}]_{+} = 0\) \(\hat{a}^{\dagger}\hat{a}^{\dagger}+\hat{a}^{\dagger}\hat{a}^{\dagger} = 0\)
 * 7) Solve: \(\hat{a}^{\dagger}\hat{a}^{\dagger}=0\)
 * 8) Since \(\hat{a}^{\dagger}\hat{a}^{\dagger}\), disregard all states higher than \(n=1\). The only excited state that remains is \(|1\rangle = \frac{1}{\sqrt{1}}\hat{a}^{\dagger}|\Omega\rangle = \hat{a}^{\dagger}|0\rangle\)