0

The question asked to show that the identity holds

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

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}_{-} $

$\left ( n-factorial \space for \space \hat{a}_{+} \right )$

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

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

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

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

Where's the $n$?

Qmechanic
  • 220,844
Physkid
  • 889

2 Answers2

1

There is quite an elegant method which is based on the observation that the operators $\hat a_+$ and $\hat a_-$ have the same commutations relations as $\xi$ and $\partial_\xi$. Explicitly: $$ \hat a_+\mapsto\xi\, ,\qquad \hat a_-\mapsto \partial_\xi. $$
Since the proof of the results only involves commutators, proving that $[a_-,a_+^n]=a_+^{n-1}$ on an arbitrary ket $\vert p \rangle\mapsto\xi^{p}/\sqrt{p!}$ is the same as proving \begin{align} \partial_\xi \left(\xi^n \frac{\xi^p}{\sqrt{p!}}\right)-\xi^n\partial_\xi \left(\frac{\xi^p}{\sqrt{p!}}\right)= &(n+p) \frac{\xi^{n+p-1}}{\sqrt{p!}} - \xi^n\,p\,\frac{\xi^{p-1}}{\sqrt{p!}}\\ &= n\frac{\xi^{n+p-1}}{\sqrt{p!}}\mapsto na_+^{n-1}\vert p\rangle\, , \end{align}

ZeroTheHero
  • 49,168
  • 21
  • 71
  • 148
0

I can't quite follow your notation with (n-1 factorial for a_+) etc. So I can't pinpoint your error but this is the correct derivation:

$$[a, (a^\dagger)^n] = a (a^\dagger)^n - (a^\dagger)^na$$

The plan is to commute the a in the second term to the front such that both terms cancel. In doing so we will find some extra terms that will add up to the sought $(a^\dagger)^n$

$$= a(a^\dagger)^n - (a^\dagger)^{n-1}a^\dagger a$$

But $a a^\dagger - a^\dagger a = 1$ such that

$$= a(a^\dagger)^n - (a^\dagger)^{n-1}(aa^\dagger -1)$$

$$= a(a^\dagger)^n - (a^\dagger)^{n-1}aa^\dagger + (a^\dagger)$$

If we want the first term to cancel the second we need to redo this "trick" an additional n-1 times. Every application gives us a rest term $a^\dagger$. such that we find:

$$= a(a^\dagger)^n - a(a^\dagger)^n + n\cdot a^\dagger$$

I think that this is more or less what you were going for isn't it ? If not please tell me and I'll try to help you better.

gertian
  • 1,333