Question 1.3: Consider the gaussian distribution ρ(x) = A e^-λ(x-a)^2 , wh...

(a)

$1=\int_{-\infty }^{\infty }{Ae^{-\lambda (x-a)^2}} dx$.    Let u ≡ x-a , du = dx , $u : -\infty → \infty$.

$1=A\int_{-\infty }^{\infty } {e^{-λu^2}}du = A \sqrt{\frac{π}{λ} }$  ⇒ $A = \sqrt{\frac{λ}{π} }$.

(b)

$\left\langle x\right\rangle = \int_{-\infty }^{\infty } {x^{-λ(x-a)^2}}dx = A \int_{-\infty }^{\infty } {(u+a)e^{-λu^2}}du$.

$= A \biggl[ \int_{-\infty }^{\infty }{ue^{-λu^2}}du + a \int_{-\infty }^{\infty }{e^{-λu^2}}du\biggr] = A \biggl(0+a \sqrt{\frac{π}{λ} } \biggr) = a$.

$\left\langle x^2\right\rangle = A\int_{-\infty }^{\infty }{x^2 e^{-λ(x-a)^2}}dx$.

$= A \left\{\int_{-\infty }^{\infty }{u^2 e^{-λu^2}} du + 2a\int_{-\infty }^{\infty }{u e^{-λu^2}} du + a^2 \int_{-\infty }^{\infty }{ e^{-λu^2}} du\right\}$.

$= A \biggl[\frac{1}{2λ} \sqrt{\frac{π}{λ} } + 0 + a^2 \sqrt{\frac{π}{λ} } \biggr] = a^2 + \frac{1}{2λ}$.

$σ^2 = \left\langle x^2\right\rangle – \left\langle x\right\rangle^2  = a^2 + \frac{1}{2λ} – a^2 = \frac{1}{2λ}$.

$σ = \frac{1}{ \sqrt{2λ}}$.

(c)