- #1
orentago
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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?
