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I want to find some formula related with \begin{align} [a^n, \bar{a}^m]=? \end{align} with $[a,\bar{a}]=d$ for some constant number $d$.

Note here $[A,B]= AB-BA$. $i.e$, $[ , ]$ is usual commutator.

First after some computation i obtain \begin{align} &[a,\bar{a}^m] = m \bar{a}^{m-1} [ a ,\bar{a}] \\ & [ a^n, \bar{a}] = n a^{n-1} [ a,\bar{a}] \end{align} By applying some formula like $BAC-CAB$ rule.

I want to obtain some formula related with $[a^n, \bar{a}^m]$, but it seems my formula dosen't seem nice,

Do you have any idea or guess for the form of $[a^n, \bar{a}^m]=$?

Consider the case for \begin{align} [a^2, \bar{a}^2] &= [a,\bar{a}^2]a + a[a,\bar{a}^2] \\ &= ([a,\bar{a}]\bar{a} + \bar{a}[a,\bar{a}])a + a( [a,\bar{a}]\bar{a} + \bar{a}[a,\bar{a}]) \\ & = [a,\bar{a}]2(\bar{a}a+a\bar{a}) = [a,\bar{a}] 2( [a,\bar{a}] +2\bar{a}a) = 2 [a,\bar{a}]^2 + 4 [a,\bar{a}] \bar{a}a \end{align}

phy_math
  • 3,782

2 Answers2

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Here's a slightly nicer way of thinking about it. The relation $[a, \bar{a}] = d$ can be rephrased as saying, "you can move a $\bar{a}$ to the left of an $a$, but you pick up an extra $d$ term".

Now, we want to compute the difference of $a^n \bar{a}^m$ and $\bar{a}^m a^n$. To convert the former into the latter, we need to move a $\bar{a}$ past an $a$, $nm$ times (that is, $n$ times for each of the $m$ $\bar{a}$'s), picking up a factor of $d a^{n-1} \bar{a}^{m-1}$ every time. Then the answer is $$[a^n, \bar{a}^m] = mnd a^{n-1} \bar{a}^{m-1}.$$

knzhou
  • 107,105
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You can use these relations $$\begin{align}[a,bc] &=[a,b]c+b[a,c]\\ [ab,c] &=a[b,c]+[a,c]b\\ [a,\bar{a}^m] &=[a,\bar{a}]\bar{a}^{m-1}+\bar{a}\left[a, \bar{a}^{m-1}\right]\end{align}$$

This is a recursive relationship $$=d\bar{a}^{m-1}+\bar{a}\left(d\bar{a}^{m-2}+\bar{a}\left[a, \bar{a}^{m-2}\right]\right)$$

And so on.. after $m$ recursions

$$=\sum_{i=0}^{m-1} \left(\bar{a}^{m-1}d\right)=m\bar{a}^{m-1}d$$

Similarly $$[a^n,\bar{a}]=na^{n-1}d$$

Now consider $$\begin{align}[a^n, \bar{a}^{m}] &=a^{n-1}\left[a,\bar{a}^{m}\right]+\left[a^{n-1}, \bar{a}^m\right]a\\ &=a^{n-1}m\bar{a}^{m-1}d+\left[a^{n-1}, \bar{a}^m\right]a\end{align}$$

After $n$ recursions $$\begin{align} &=\sum_{i=0}^{n-1}mda^{n-1}\bar{a}^{m-1}\\ &=mnda^{n-1}\bar{a}^{m-1}\,.\end{align}$$

Courage
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