Algebra proof for quantum mechanics ladder operator

Question

My question contains the word quantum mechanics but is a purely algebraic problem. The so-called ladder operators $a_{-}$ and $a_{+}$ from quantum mechanics are operators that do not commute and the following relation holds between the operators: $$ a_+a_- = a_-a_+-1. \text{ (eq.1)} $$ Using the above, I must prove that: $$ [a_-,a_+^n]=na_+^{n-1} \text{ (eq.2)} $$ where $[a_-,a_+^n]$ denotes the commutator $a_-a_+^n - a_+^na_-$.
I know that the idea of the proof is to "shift" the $a_-$ from the second term to the left $n$ times as follows: $$ [a_-,a_+^n]=a_-a_+^n - a_+^na_- \\ [a_-,a_+^n]=a_-a_+^n - a_+^{n-1}a_+a_- \\ \text{Substitute (eq.1) : } \\ [a_-,a_+^n]=a_-a_+^n - a_+^{n-1}(a_-a_+-1) \\ [a_-,a_+^n]=a_-a_+^n - a_+^{n-1}a_-a_+ + a_+^{n-1} \\ ... \text{after } k \text{ iterations} \\ [a_-,a_+^n]=a_-a_+^n - a_+^{n-k}a_-a_+^k + ka_+^{n-1} \\ ... \text{after } n \text{ iterations} \\ [a_-,a_+^n]=a_-a_+^n - a_-a_+^n + na_+^{n-1} \\ [a_-,a_+^n]=0 + na_+^{n-1} \\ [a_-,a_+^n]=na_+^{n-1}. \\ $$ which corresponds to (eq.2).

I am looking for a way to write this proof formally, with less steps. The way I wrote it down is understandable and probably correct, but it feels like it could be done more rigorously, but I do not know how.

I would appreciate if someone knows if there is a more formal way of writing this proof. Thank you.

Answer 1

Hint:

You can write a proof using induction. The base case is $$ [a_-,a_+^2]=2a_+ $$ (easy to proof). And the induction step is: $$ [a_-,a_+^n]=na_+^{n-1} \Rightarrow [a_-,a_+^{n+1}]=(n+1)a_+^{n} $$

Answer 2

$$a_-a_+^n-a_+^na_-=\sum_{k=0}^{n-1}(a_+^ka_-a_+^{n-k}-a_+^{k+1}a_-a_+^{n-k-1}) =\sum_{k=0}^{n-1}a_+^k(a_+a_--a_-a_+)a_+^{n-k-1} =\sum_{k=0}^{n-1}a_+^k1a_+^{n-k-1} =na_+^{n-1}.$$

Answer 3

As a reaction to Emilio Novati's hint, here is the proof which I was looking for.

Proof by induction

Statement:
$[a_-,a_+^n]=na_+^{n-1} \ \forall \ n \in \mathbb{N}-\{0,1\}$ provided that $a_+$ and $a_-$ are operators that do not commute and that $a_+a_-=a_-a_+-1$. The operators are associative and distributive.
Base:
$n = 2$
According to the statement: $[a_-,a_+^2]=2a_+^{2-1}=2a_+$.
According to the given situation: $[a_-,a_+^2]=a_-a_+^2-a_+^2a_- = a_-a_+^2-a_+(a_-a_+-1)=a_-a_+^2-(a_-a_+-1)a_++a_+=2a_+$.
Induction:
Suppose for one $k\in\mathbb{N}-\{0,1\}$ that the statement is true. For $k+1$ it holds that: $$ [a_-,a_+^{k+1}] = a_-a_+^{k+1}-a_+^{k+1}a_-\\ [a_-,a_+^{k+1}] = a_-a_+^{k+1}-a_+^{k}a_+a_- \\ [a_-,a_+^{k+1}] = a_-a_+^{k+1}-a_+^{k}(a_-a_+-1) \\ [a_-,a_+^{k+1}] = a_-a_+^{k}a_+-a_+^{k}a_-a_++a_+ \\ $$ By distributivity: $$ [a_-,a_+^{k+1}] = (a_-a_+^{k}-a_+^{k}a_-)a_++a_+ \\ [a_-,a_+^{k+1}] = [a_-,a_+^{k}]a_++a_+ \\ $$ According to the induction hypothesis: $$ [a_-,a_+^{k+1}] = ka_+^{k-1}a_++a_+ \\ [a_-,a_+^{k+1}] = ka_+^{k} + a_+\\ [a_-,a_+^{k+1}] = (k+1)a_+^{k}. $$ It follows that the assumption is also true for $k+1$. By induction, the statement holds true.