Show that if the lowering operator acts on a non-zero eigenstate of the number operator, the result is also an eigenstate of the number operator

for a boson

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

for a fermion

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