Commutator relation identity for Quantum mechanics

Question

The question asked to show that the identity holds

$\left [ \hat{a}_{-},\left ( \hat{a}_{+} \right ) \right ]=n\left ( \hat{a}_{+} \right )^{n-1}$ between operators.

The solution that came with the tutorial were 'ugly' and a stark deviation in the flavour in my attempt. However, my attempt did not result in showing the identity despite the fact that-to the best of my knowledge-no error has been committed.

Attempt:

$\left [ \hat{a}_{-},\left ( \hat{a}_{+} \right ) \right ]=\hat{a}_{-}\left ( \hat{a}_{+} \right )^{n}-\left ( \hat{a}_{+} \right )^{n}\hat{a}_{-} =\hat{a}_{-}\hat{a}_{+}\cdot \cdot \cdot \hat{a}_{+}-\hat{a}_{-}\cdot \cdot \cdot \hat{a}_{-}\hat{a}_{+} $

$\left ( n\text{ factor for } \hat{a}_{+} \right )$

$=\left ( \hat{a}_{-}\hat{a}_{+} \right )\left ( \hat{a}_{+}\cdot \cdot \cdot \hat{a}_{+} \right )-\left ( \hat{a}_{+}\cdot \cdot \cdot \hat{a}_{+} \right )\left ( \hat{a}_{+}\hat{a}_{-} \right )$

$\left ( n-1\text{ factor for } \hat{a}_{+} \right )$

$=\left ( \left ( \hat{a}_{-}\hat{a}_{+} \right )-\left ( \hat{a}_{+}\hat{a}_{-} \right ) \right )\left ( \hat{a}_{+}\cdot \cdot \cdot \hat{a}_{+} \right )$

=$\left ( \hat{a}_{+} \right )^{n-1}$

Where's the $n$?

Answer 1

you mistke is after the line $$(a_-a_+)(a_+\dots a_+)-(a_+\dots a_+)(a_+a_-)$$ which is not equal to $(a_-a_+ - a_+a_-)(a_+\dots a_+)$, because $a_+a_-$ does not commute with $a_+$.

Hint for the correct solution: You need to use induction over $n$, i.e. reduce $[a_-,a_+^n]$ to some term involving $[a_-,a_+^{n-1}]$

Answer 2

As said, $a_+^n$ doesn't commute with $a_-$. Assuming that $[a_-,a_+]=1$, so is $a_+a_-=a_-a_+-1$, as suggested (anyway, it must be so, done we succeed proving the intial claim), we can proceed by induction. The first step and the induction step are as suit.

$a_+^na_-=a_+^{n-1}a_+a_-=a_+^{n-1}(a_-a_+-1)=a_+^{n-1}a_-a_+-a_+^{n-1}=$

$=a_+^{n-2}a_+a_-a_+-a_+^{n-1}=a_+^{n-2}(a_-a_+-1)a_+-a_+^{n-1}=$

$=a_+^{n-2}a_-a_+^2-2a_+^{n-1}$

Suppose, $a_+^na_-=a_+^{n-k}a_-a_+^k-ka_+^{n-1}$, then

$a_+^na_-=a_+^{n-k-1}a_+a_-a_+^k-ka_+^{n-1}=a_+^{n-k-1}(a_-a_+-1)a_+^k-ka_+^{n-1}=$

$=a_+^{n-(k+1)}a_-a_+^{k+1}-a_+^{n-(k+1)}a_+^k-ka_+^{n-1}=$

$=a_+^{n-(k+1)}a_-a_+^{k+1}-(k+1)a_+^{n-1}$

Finally,

$a_+^na_-=a_+^{n-n}a_-a_+^{n}-(n)a_+^{n}=$

$=a_-a_+^{n}-na_+^{n-1}$

Then

$[a_-,a_+^n]=a_-a_+^n-a_+^na_-=a_-a_+^n-a_-a_+^n+na_+^{n-1}=na_+^{n-1}$

Comment Maybe it's the same derivation, or very similar, from your manual. If so, I cannot agree with the ugliness of the proof. :)