= Thermal noise spectral convention
{title2=$S_v^{(1)}(f)=4k_{\rm B}TR$}
Model an ideal <resistor> of constant <electrical resistance> $R>0$ at uniform absolute <temperature> $T$ by its <Thévenin equivalent>: a zero-mean random <voltage source> in series with a noiseless $R$. In the classical range its one-sided, <open-circuit voltage> <power spectral density>, per <hertz>, is
$$
S_v^{(1)}(f)=4k_{\rm B}TR\quad(f\geq0),\qquad [S_v^{(1)}]=\mathrm{V^2/Hz},
$$
where $k_{\rm B}$ is the <Boltzmann constant>. Thus a positive-<frequency> interval of width $df$ contributes $4k_{\rm B}TR\,df$ to the mean square voltage. This is approximately <white noise> only where the <electrical resistance> is effectively constant and $hf\ll k_{\rm B}T$, with $h$ the <Planck constant>. The ideal constant spectrum cannot describe arbitrarily high <frequencies>.https://123.physics.ucdavis.edu/johnson_files/nyquist_1928.pdf[Nyquist (1928), p. 113, equations (4)–(8): classical spectrum and high-frequency thermal correction]{ref}
For a real voltage, an equivalent two-sided <power spectral density> is $S_v^{(2)}(f)=2k_{\rm B}TR$ over signed <frequencies>. With a real <linear time-invariant system> of <frequency response> $H(f)$, the same output mean square is
$$
\langle v_o^2\rangle=\int_0^\infty S_v^{(1)}(f)|H(f)|^2\,df
=\int_{-\infty}^{\infty}S_v^{(2)}(f)|H(f)|^2\,df.
$$
Angle brackets denote an <expected value>; for this zero-mean <noise> the mean square is its <variance>. Do not integrate $4k_{\rm B}TR$ over both signs of <frequency>: that counts the same fluctuations twice. All integrals here use <frequency> $f$ in <hertz>, not <angular frequency> $\omega=2\pi f$. The <amplitude spectral density> is $\sqrt{4k_{\rm B}TR}$ in $\mathrm V/\sqrt{\mathrm{Hz}}$, whereas the <root mean square> voltage requires a specified measurement <signal bandwidth>.
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