Question 33.9: Derive an expression for the Johnson noise across a resistor......

Simple circuit theory yields

V + IR =\frac{Q}{C} .          (33.107)

The charge Q and voltage V are conjugate variables (their product has dimensions of energy) and so we write

\widetilde{Q} (ω)=\widetilde{χ} (ω)\widetilde{V} (ω),           (33.108)

where the response function \widetilde{χ}(ω) is given for this circuit by

\widetilde{χ} (ω)= \frac{1}{C^{−1} −iωR}.          (33.109)

Hence \widetilde{χ}^{\prime \prime }(ω) is given by

\widetilde{χ}^{\prime \prime } (ω)= \frac{ωR}{C^{−2} + ω^{2} R^{2} }.          (33.110)

At low frequency (ω \ll 1/RC, and since the capacitance will be small, 1/RC will be very high so that this is not a severe resistriction) we have that \widetilde{χ}^{\prime \prime }(ω) → ω RC². Thus the fluctuation–dissipation theorem(eqn 33.103) gives

\boxed{\widetilde{C} _{xx} (ω)=2k_{B} T\frac{\widetilde{χ}^{\prime \prime } (ω)}{ω } , }          (33.103)

\widetilde{C} _{QQ} (ω)=2k_{B} T\frac{\widetilde{χ}^{\prime \prime } (ω)}{ω }=2 k_{B} TRC^{2} .           (33.111)

Because Q = CV for a capacitor, correlations in Q and V are related by

\widetilde{C}_{VV} (ω)=\frac{\widetilde{C} _{QQ} (ω) }{C^{2} },          (33.112)

and hence

\widetilde{C} _{VV} (ω)=2 k_{B} TR.           (33.113)

Equation 33.84 implies that

\boxed{\left\langle x^{2} \right\rangle =\frac{1}{2\pi } \int_{-\infty }^{\infty }{}\widetilde{C}_{xx} (ω)  dω. }          (33.84)

\left\langle V^{2} \right\rangle =\frac{1}{2\pi } \int_{-\infty }^{\infty }{}\widetilde{C}_{VV} (ω)  dω.          (33.114)

and hence if this integral is carried out, not over all frequencies, but only in a small interval ∆f =∆ω/(2π) about some frequency ±ω_{0} (see Fig. 33.4),

\left\langle V^{2} \right\rangle = 2C_{VV} (ω) ∆f =4 k_{B} T R ∆f,            (33.115)

in agreement with eqn 33.28.

\boxed{\left\langle V^{2} \right\rangle = 4 k_{B} T R ∆f.}          (33.28)