# What is the physical meaning of the operators of the positive and negative frequency components?

In quantum optics, after performing the quantization of the radiation field, the field operator $$E$$ is often split into the positive- and negative-frequency parts as $$E(\mathbf{r},t) = E^{\left(+\right)}(\mathbf{r},t) + E^{\left(-\right)}(\mathbf{r},t),$$ where

$$E^{\left(+\right)}(\mathbf{r},t) = \sum_{\mathbf{k}} \mathcal{E}_{\mathbf{k}} \mathrm{e}^{-\mathrm{i}\omega\left(\mathbf{k}\right)t+\mathrm{i}\mathbf{k}\cdot\mathbf{r}} a(\mathbf{k})$$ and $$E^{\left(-\right)}(\mathbf{r},t) = \left[E^{\left(+\right)}(\mathbf{r},t)\right]^{\dagger}$$ Products of $$E^{\left(+\right)}(\mathbf{r},t)$$ and $$E^{\left(-\right)}(\mathbf{r},t)$$ appear, for example, in the definitions of the correlation functions: $$G^{(n)}(\mathbf{r}_1,\dots,\mathbf{r}_{2n};t_1,\dots,t_{2n}) = \langle E^{\left(-\right)}(\mathbf{r}_1,t_1) \dots E^{\left(-\right)}(\mathbf{r}_{n},t_{n}) E^{\left(+\right)}(\mathbf{r}_{n+1},t)\dots E^{\left(+\right)}(\mathbf{r}_{2n},t_{2n})\rangle$$

I am wondering about the physical meaning of these two operators. Do they have an intuitive interpretation?

The positive and negative frequency components are the energy-decreasing and -increasing parts (or conversely, depending on sign conventions), also called annihilation and creation operators.

To deduce this, let $$A(t)$$ be any operator in the Heisenberg picture, with time-dependence given by $$i\dot A(t) = \big[A(t),\,H\big].$$ Consider the operator $$A_\omega$$ defined by $$A_\omega\equiv \int_{-\infty}^\infty dt\ e^{-\epsilon t^2} e^{i\omega t} A(t),$$ where $$\epsilon>0$$ is an arbitrarily small positive coefficient that I'm including to help ensure that the integral is well-defined. This operator satisfies \begin{align*} [A_\omega,\,H] &= \int_{-\infty}^\infty dt\ e^{-\epsilon t^2} e^{i\omega t} \big[A(t),\,H\big] \\ &= \int_{-\infty}^\infty dt\ e^{-\epsilon t^2} e^{i\omega t} i\frac{d}{dt}A(t) \\ &= -i\int_{-\infty}^\infty dt\ A(t) \frac{d}{dt} e^{-\epsilon t^2} e^{i\omega t} \\ &= \omega A_\omega+O(\epsilon), \end{align*} which may also be written $$H A_\omega = A_\omega (H-\omega).$$ This shows that if $$\omega>0$$, then $$A_\omega$$ decreases the energy of any state on which it acts. The vacuum state is already the state of lowest energy, so $$A_\omega$$ annihilates the vacuum state if $$\omega>0$$. In other words, the positive-frequency part of $$A(t)$$ is an annihilation operator. Its adjoint, the negative-frequency part, is a creation operator (adds energy to the state on which it acts).

Here, I considered just a single frequency $$\omega$$. We can also write $$A(t)=A_+(t)+A_-(t)$$, where the two terms on the right-hand side contain all positive and negative frequencies, respectively, as shown in the question.

The derivation shown above is general. It is not specific to the electromagnetic field, or even to free fields. In the special case where $$A(t)$$ is a component of the electromagnetic field, its negative-frequency part creates a photon.

• @user135743 Suppose that in the Schrödinger (s) picture, the operator $A$ is indepenent of time. In the Heisenberg (h) picture, it becomes a time-dependent operator $A(t)$. Now consider $A_f\equiv\int_{-\infty}^\infty dt\ f(t)A(t)$, which is independent of $t$ because we integrated over $t$. Passing back to the s-picture can't give $A$ in general, because the relationship between $A_f$ and $A$ depends on $f$, but the relationship between the h- and s-pictures doesn't. In general, the s-picture version of $A_f$ will depend on $t$. Commented Sep 20, 2023 at 14:57
• @user135743 If that seems weird, it's because the concept of the s-picture itself is weird when we're dealing with observables smeared out in time, like $A_f$ is. As an example, consider a single harmonic oscillator, so that $A=a+a^\dagger$ in the s-picture and $A(t)=a e^{-i\omega t}+a^\dagger e^{i\omega t}$ in the h-picture, where $a$ and $a^\dagger$ are the usual annihilation and creation operators for that oscillator. Then $A_\omega = a$ in the h-picture, which becomes $a e^{i\omega t}$ (not $A$) in the s-picture. Commented Sep 20, 2023 at 14:57
• @user135743 This might seem less bothersome if we generalize the definition of $A_f$ to $A_f(t_0)\equiv \int_{-\infty}^\infty dt\ f(t-t_0)A(t)$. Changing the value of $t_0$ corresponds to changing the center of time about which $A_f$ is being smeared. When applied to the case in the answer, this gives $A_\omega(t_0)=A_\omega e^{-i\omega t_0}$ in the h-picture. If we choose to think of this as a time-dependent operator with respect to the time $t_0$, and if we use that as time when converting back to the s-picture, then we end up with just plain $A_\omega$ (still not $A$) in the s-picture. Commented Sep 20, 2023 at 14:58
• Great explanations. BUT I'm a huge fan of the S picture so it is sad that the expression there is uglier than the positive\negative frequency. Instead it's twist by time with hamiltonian then take positive\negative frequences. Thank you anyways! Commented Sep 22, 2023 at 11:01
• @user135743 Fans of the s-picture can still take comfort in the fact that the s-picture is at the heart of one of the most aesthetically pleasing attempts to axiomatize QFT. :) I'll come back and post a link or two when I'm not in such a hurry. Commented Sep 22, 2023 at 23:00

The represent the parts of the field that are responsible for emitting and absorbing photons. The usual convention (although this is not always followed), is that $$E^{(+)}$$ contains only annihilation operators, and $$E^{(-)}$$ contains only creation operators. So the quantum mechanical description of light intensity does not involve $$[E(\vec{x})]^{2}=E^{(+)}E^{(+)}+E^{(+)}E^{(-)}+E^{(-)}E^{(+)}+E^{(-)}E^{(-)}$$, but only the $$E^{(-)}E^{(+)}$$ term that represents the amplitude squared for a state $$\propto E^{(+)}|\psi_{0}\rangle$$ in which one photon is annihilated relative to the initial state $$|\psi_{0}\rangle$$. This makes the expression free from zero-point fluctuations—which would be included in $$\langle\psi_{0}|E^{(+)}E^{(-)}|\psi_{0}\rangle$$.

Then, since the intensity is formed from only $$E^{(+)}(\vec{x})E^{(-)}(\vec{x})$$, it makes sense to define the two-point correlation function with the same combination of fields, $$E^{(+)}(\vec{x})E^{(-)}(\vec{y})$$, so it becomes the intensity at $$\vec{x}=\vec{y}$$. For higher-order correlation functions, the same kind of arguments are involved.

Suppose we have a real function $$f(t)$$ of time. We can Fourier expand $$f(t)$$:

$$f(t) = \int_{\omega=-\infty}^{\infty}\tilde{f}(\omega)e^{-i\omega t}$$

If $$f(t)$$ is a real function then it can be proven that $$\tilde{f}(\omega) = \tilde{f}^*(-\omega)$$ so we can write

$$f(t) = \int_{\omega=0}^{\infty} \tilde{f}(\omega)e^{-i\omega t} + \int_{\omega=0}^{\infty}\tilde{f}^*(\omega)e^{i\omega t}$$

The part of this expression that goes like $$e^{-i\omega t}$$ is conventionally called the "positive" rotating part and the part of this expression that goes like $$e^{+i\omega t}$$ is called the "negative" rotating part. The positive (negative) rotating part rotates clockwise (counter clockwise) in the complex plane as $$t$$ increases. Note that some authors reverse the definitions of positive and negative rotating, see Fourier transform standard practice for physics.

We define

$$f^{(+)}(t) = \int_{\omega=0}^{\infty}\tilde{f}(\omega)e^{-i\omega t}$$

and $$f^{(-)}(t) = (f^{(+)}(t))^*$$ so that

$$f(t) = f^{(+)}(t) + f^{(-)}(t)$$

For quantum (Hermitian) operators the situation is the exact same except we replace complex conjugate with Hermitian conjugation. In the Heisenberg picture operators will have a temporal dependence which can be expressed as a function of time and that temporal dependence can be Fourier expanded as above and the operator can be split into components that rotate clockwise and counterclockwise in the complex plane. If you're uncomfortable with thinking of operators like we think of complex numbers then apply the comments in this paragraph to the expectation value of the operator.

This gives us the important insight that $$\hat{a}(\omega)$$ (or $$\hat{a}(\boldsymbol{k})$$ if we are working in multiple dimensions and must label our modes with $$\boldsymbol{k}$$ instead of $$\omega$$) is best thought of as the quantum analog of the complex amplitude of the spatial mode of the classical electric field corresponding to frequency $$\omega$$.

For the sake of brevity I've not spent time or space including the expansion into spatial modes. This is because it isn't strictly necessary to understand the positive and negative rotating components of something. The positive/negative rotating components have to do with exactly what I said above: they are the components of a decomposition of a real function which rotate in complex space at positive and negative frequencies.