I need help!

I'm trying to calculate the Power Spectral Density of a quantum operator ($\delta \hat{n}(t)$) given by:

$$ \delta \hat{n}(t) = A(t)\delta\hat{a}(t)+A^{*}(t)\delta\hat{a}^{\dagger}(t) $$

$\delta\hat{a}(t)$ is a wide-sense stationary process and $A(t)$ is a complex time-dependent deterministic factor. To calculate the PSD ($S_{nn}(\omega)$) I'm employing the following definition:

$$ S_{nn}(\omega) = \int_{-\infty}^{\infty} d\tau \overline{\langle \delta\hat{n}^{\dagger}(t+\tau)\delta\hat{n}(t)\rangle} e^{i \omega \tau}$$

where the bar indicates a time average. When calculating the autocorrelation I'm left with:

$$\langle \delta\hat{n}^{\dagger}(t+\tau)\delta\hat{n}(t)\rangle = A^{*}(t+\tau)A(t) \langle \delta\hat{a}^{\dagger}(t+\tau)\delta\hat{a}(t)\rangle + A(t+\tau)A^{∗}(t) \langle \delta\hat{a}(t+\tau)\delta\hat{a}^{\dagger}(t)\rangle+ A(t+\tau)A(t)\langle \delta\hat{a}(t+\tau)\delta\hat{a}(t)\rangle+ A(t+\tau)^{∗}A^{∗}(t)\langle \delta\hat{a}^{\dagger}(t+\tau)\delta\hat{a}^{\dagger}(t)\rangle$$

For reference, $A(t)=a_{0} + a_{1} e^{-i\delta t}$, where both $a_0$ and $a_1$ are complex. My problem arises when calculating the third and fourth terms on top. I expect the autocorrelation to be real (and hence the PSD), but calculating the time-average of the third and fourth terms of the autocorrelation gives:

$$\overline{A(t+\tau)A(t)} = (a_{0}^{*})^2$$

$$\overline{A(t+\tau)^{∗}A^{∗}(t)} = a_{0}^2$$

Therefore, the only way for the autocorrelation of $\delta\hat{n}(t)$ to be real is if $\langle \delta\hat{a}^{\dagger}(t+\tau)\delta\hat{a}^{\dagger}(t)\rangle$ and $\langle \delta\hat{a}(t+\tau)\delta\hat{a}(t)\rangle$ are both zero, but I don't see why. I'm trying to replicate the calculations made in this article: https://arxiv.org/abs/1402.6929v2 (eq. 32, pg.5).

So basically:

  1. Should $\langle \delta\hat{a}^{\dagger}(t+\tau)\delta\hat{a}^{\dagger}(t)\rangle = \langle \delta\hat{a}(t+\tau)\delta\hat{a}(t)\rangle = 0$?
  2. or is the something fundamentally wrong with my calculations?
  • 1
    $\begingroup$ I'm not following all of the symbols involved so only commenting, but a primary reason the autocorrelation incorrectly doesn't come out to be real is due to forgetting a complex conjugate. For example, if at transpose operation is involved, then a conjugate transpose should be used. $\endgroup$ Commented Jan 11 at 21:56
  • $\begingroup$ the dagger symbol is, as expected, the hermitian conjugate of the operator $\endgroup$ Commented Jan 11 at 22:18

2 Answers 2



It seems to me that you maybe have some misconceptions. The PSD is a real non-negative valued function, and the correlation sequence is a conjugate symmetric sequence that is only real if its inputs are real. The following provides a typical proof followed by (in my opinion) a more meaningful proof of these concepts.

Typical Derivation

I think there is some confusion here. The PSD is a real and non-negative valued function since it is a power density. Since in your definition you use a Hermitian transpose notation, I am assuming that your signal is complex.

The autocorrelation sequence is defined as

\begin{equation} r(k) = E\{x(t)x^{*}(t-k)\} \end{equation}

From this definition, it is clear that $r(k)$ is only guaranteed to be real at $k=0$ for all non-constant complex phase.

$r(k)$ has some useful properties, such as

\begin{equation} r(k) = r^{*}(-k) \end{equation}


\begin{equation} r(0) \geq |r(k)| \; \forall \; k \end{equation}

Additionally, we assume our signal is wide sense stationary in order for it to have finite average power. Because of this assumption, we can say that

\begin{equation} r(t,k) = r(t-k,0) \end{equation}

The power spectral density is defined as

\begin{equation}\phi(\omega) = \sum_{k=-\infty}^{\infty}r(k)e^{-j\omega k} \end{equation}

From these properties, we can see that the signal has an even symmetric magnitude and an odd symmetric phase, which means that the Fourier transform of this signal will be real. This coincides with the idea that a power density must be real since it contains power information.

Matrix Derivation

It goes deeper than that though. Letting $\underline{y}$ be a column vector of length $m$, we can define the correlation matrix as

\begin{equation} R = E\{\underline{y}\,\underline{y}^{H}\} \end{equation}

where $^{H}$ represents the Hermitian transpose. This means that $R(t,s) = E\{\underline{y}(t)\underline{y}^{*}(s)\}$.

The first important property of the correlation matrix is that it is positive semi-definite. This means that $\underline{w}^{H}R\underline{w} \geq 0$ for all non-zero vectors $\underline{w}$. However, we are interested in the power spectral density. By letting

\begin{equation} \underline{a}(\omega) = \begin{bmatrix}1 & e^{j\omega} & \cdots & e^{j\omega(m-1)}\end{bmatrix}^{T}\end{equation}

the estimate of the power spectral density can be written as (to within a scale factor)

\begin{equation}\phi(\omega) = \underline{a}^{H}(\omega)R\,\underline{a}(\omega)\end{equation}

This is pretty awesome since, because $R$ is positive semi-definite, it proves again that $\phi(\omega)$ is non-negative.

Now, a second important property of the power-spectral density is that it is Hermitian. This means that $A_{i,j} = A^{*}_{j,i}$. Additionally, we've assumed that our signal is wide-sense stationary, which means that the correlation matrix is Toeplitz. This coincides with the idea that

\begin{equation} r(k) = r^{*}(-k) \end{equation}

With some simple algebraic manipulation, $\underline{a}^{H}(\omega)R\,\underline{a}(\omega)$ can be computed via a Fourier transform (FFT). This means that the result is also expected to be real, as was the case in the previous derivation. These conclusions, namely that the correlation function is conjugate symmetric and that the PSD is real and non-negative, are not new, but they are now expressed in matrix form.


The Fourier Transform of the autocorrelation function $R_{xx}(\tau)$ for a Wide Sense Stationary process $x(t)$ is the Power Spectral Density (PSD). $x(t)$ can be real or complex, and for complex $x(t)$, $R_{xx}(\tau)$ will also be complex (however $R_{xx}(0)$ will always be real). This is easy to confirm by considering when we cross multiply the complex function with itself delayed in time ($x(t)x^*(t-\tau)$) as part of the autocorrelation computation, the phase at $x(t)$ is not necessarily the same as the phase $x(t-\tau)$. The generalized complex product $x(t)y^*(t)$ can only be real when the phase of $x(t)$ is the same as the phase of $y(t)$ such that the result has zero phase (real).

But the autocorrelation result, $R_{xx}(\tau)$, will be complex conjugate symmetric (Hermitian) given the symmetry of the autocorrelation process: $R_{xx}(-\tau) = R_{xx}^*(\tau)$, so it's Fourier Transform (as the PSD) will be real.

  • If a time domain signal is real, then it's Fourier Transform will be conjugate symmetric.

  • Similarly is a time domain signal is conjugate symmetric, then it's
    Fourier Transform will be real.


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