Parity Transformation of Annihilation Operator

[orentago]

Homework Statement

Given that, for operators $$A$$ and $$B$$:

$$\mathrm{e}^{\mathrm{i} \alpha A} B \mathrm{e}^{-\mathrm{i} \alpha A} = \sum_{n=0}^{\infinity} {(\mathrm{i}\alpha)^n \over n!} B_n$$

where $$B_0 = B$$ and $$B_n = [A, B_{n-1}]$$ for n=1,2,...

show that:

$$P_1 a(\mathbf{k})P_1^{-1}=\mathrm{i}a(\mathbf{k})$$

and

$$P_2 a(\mathbf{k})P_2^{-1}=-\mathrm{i}\eta_P a(-\mathbf{k})$$

Where:

$$P_1=\exp\left[-\mathrm{i} {\pi \over 2} \sum_{\mathbf{k}} a^\dagger (\mathbf{k}) a(\mathbf{k})\right]$$

and

$$P_2=\exp\left[\mathrm{i} {\pi \over 2} \eta_P \sum_{\mathbf{k}} a^\dagger (\mathbf{k}) a(\mathbf{-k})\right]$$

Homework Equations

The Attempt at a Solution

First start by calculating $$B_1$$. From the question it is apparent that for $$P_1$$, $$B_0 = B = a(\mathrm{k})$$ and $$\alpha=-{\pi \over 2}$$, with $$A=\sum_{\mathbf{k}} a^\dagger (\mathbf{k}) a(\mathbf{k})$$.

Hence:

$$B_1=[A,B_0]=\left[\sum_{\mathbf{k}} a^\dagger (\mathbf{k}) a(\mathbf{k}), a(\mathbf{k'}) \right]=\sum_{\mathbf{k}} [a^\dagger (\mathbf{k}) a(\mathbf{k}), a(\mathbf{k'})]$$
$$=\sum_{\mathbf{k}} a^\dagger (\mathbf{k}) [a(\mathbf{k}), a(\mathbf{k'})]+\sum_{\mathbf{k}} [a^\dagger (\mathbf{k}), a(\mathbf{k'})] a(\mathbf{k})=\sum_{\mathbf{k}} \delta_{\mathbf{kk'}} a(\mathbf{k})=a(\mathbf{k'})$$

Since $$B_0=B_1$$, it is evident that all $$B_n=B=a(\mathbf{k})$$

Hence substituting this and the value for alpha into the summation over n, above, and using the fact that this summation is now simply the expansion of the exponential function, one obtains:

$$\sum_{n=0}^{\infinity} {(-\mathrm{i} \pi / 2)^n \over n!} a(\mathbf{k}) = \mathrm{e}^{-\mathrm{i} \pi / 2} a(\mathbf{k})= -\mathrm{i} a(\mathbf{k})$$

I could add a similar derivation for $$P_2$$, but it's practically the same so I won't. What I'd like to know is how I've picked up a minus sign in my derivation, and so have not obtained the same answer as the question gives.

Any ideas?

 

[orentago]

Ah no, problem solved. I evaluated my commutators wrongly (they should be the negative of the Kronecker delta, not the positive).