Show that if the raising operator acts on an eigenstate of the number operator, the result is also an eigenstate of the number operator

for a boson

 * 1) Multiply the state \(\hat{a}^{\dagger}|\alpha\rangle\) by the number operator: \(\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger}|\alpha\rangle\)
 * 2) Add \(\hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a} - \hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a} = 0\): \(\left(\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger} - \hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a} + \hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a}\right)|\alpha\rangle\)
 * 3) Simplify with a commutator: \(\left([\hat{a}^{\dagger}\hat{a},\hat{a}^{\dagger}] + \hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a}\right)|\alpha\rangle\)
 * 4) Calculate the commutator.
 * 5) Use \([AB,C] = A[B,C] + [A,C]B\) to expand the commutator: \([\hat{a}^{\dagger}\hat{a},\hat{a}^{\dagger}]=\hat{a}^{\dagger}[\hat{a},\hat{a}^{\dagger}]+[\hat{a}^{\dagger},\hat{a}^{\dagger}]\hat{a}\)
 * 6) Solve: \([\hat{a}^{\dagger}\hat{a},\hat{a}^{\dagger}]=\hat{a}^{\dagger}\cdot 1-0\cdot\hat{a} = \hat{a}^{\dagger}\)
 * 7) Substitute \([\hat{a}^{\dagger}\hat{a},\hat{a}^{\dagger}] = \hat{a}^{\dagger}\) into the expression: \(\left(\hat{a}^{\dagger} + \hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a}\right)|\alpha\rangle\)
 * 8) Simplify: \(\hat{a}^{\dagger}\left(1 + \hat{a}^{\dagger}\hat{a}\right)|\alpha\rangle\)
 * 9) Substitute in the number operator eigenstate equation \(\hat{a}^{\dagger}\hat{a}|\alpha\rangle = \alpha|\alpha\rangle \): \(\hat{a}^{\dagger}\left(1 + \alpha\right)|\alpha\rangle\)
 * 10) Rearrange: \(\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger}|\alpha\rangle = \left(1 + \alpha\right)\hat{a}^{\dagger}|\alpha\rangle\)

for a fermion

 * 1) Multiply the state \(\hat{a}^{\dagger}|\alpha\rangle\) by the number operator: \(\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger}|\alpha\rangle\)
 * 2) Add \(\hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a} - \hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a} = 0\): \(\left(\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger} - \hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a} + \hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a}\right)|\alpha\rangle\)
 * 3) Simplify with an anticommutator: \(\left([\hat{a}^{\dagger}\hat{a},\hat{a}^{\dagger}]_{+} - \hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a}\right)|\alpha\rangle\)
 * 4) Calculate the anticommutator.
 * 5) Use \([A,B]_{+} = [B,A]_{+}\) to solve the anticommutator: \([\hat{a}^{\dagger}\hat{a},\hat{a}^{\dagger}]_{+}=\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger} + \hat{a}\hat{a}^{\dagger}\hat{a}^{\dagger} = [\hat{a}^{\dagger},\hat{a}]_{+}\hat{a}^{\dagger}\)
 * 6) Solve: \([\hat{a}^{\dagger}\hat{a},\hat{a}^{\dagger}]_{+}= \hat{a}^{\dagger}\)
 * 7) Substitute \([\hat{a}^{\dagger}\hat{a},\hat{a}^{\dagger}]_{+} = \hat{a}^{\dagger}\) into the expression: \(\left(\hat{a}^{\dagger} - \hat{a}^{\dagger}\hat{a}^{\dagger}\hat{a}\right)|\alpha\rangle\)
 * 8) Simplify: \(\hat{a}^{\dagger}\left(1 - \hat{a}^{\dagger}\hat{a}\right)|\alpha\rangle\)
 * 9) Substitute in the number operator eigenstate equation \(\hat{a}^{\dagger}\hat{a}|\alpha\rangle = \alpha|\alpha\rangle \): \(\hat{a}^{\dagger}\left(1 - \alpha\right)|\alpha\rangle\)
 * 10) Rearrange: \(\hat{a}^{\dagger}\hat{a}\hat{a}^{\dagger}|\alpha\rangle = \left(1 - \alpha\right)\hat{a}^{\dagger}|\alpha\rangle\)