- #1

RedMech

- 13

- 0

Show that if the operator relation

e^(ipa/ħ)xe^(-ipa/ħ) = x+a

holds. The operator e^A is defined to the

∞

e^A= Ʃ(A^n)/n!

n=0

[Hint: Calculate e^(ipa/ħ)xe^(-ipa/ħ)f(p) where f(p)is any function of p, and use the representation x=iħd/dp]

2. Homework Equations :

I am not entirely sure but I think those presented in the problem statement are sufficient.

3. The Attempt at a Solution :

I calculated e^(ipa/ħ)xe^(-ipa/ħ)f(p) by replacing x=iħd/dp to get;

e^(ipa/ħ)iħd/dp[e^(-ipa/ħ)f(p)]= e^(ipa/ħ){iħ[(-ia/ħ)e^(-ipa/ħ)f(p)+e^(-ipa/ħ)f'(p)]}

= e^(ipa/ħ){ae^(-ipa/ħ)f(p)+e^(-ipa/ħ)f'(p)]}

= af(p)+f'(p).

∴ af(p)+f'(p) = x + a.

From this point I become entirely confused, I don't know how everything is suppose to tie into the power series.