# What does the $\langle \rangle$ notation mean?

I have what I think is a very simple question, basically, what does the notation "$$\langle \rangle$$" stand for?

My background is in math and I am not familiar with physics notations. I am reading the following:

"we assume that $$\epsilon_k$$ can be approximated as zero mean Gaussian measurement noise with $$\langle \epsilon_j \epsilon_k \rangle=\sigma^2\delta_{jk}$$."

From what I have found it seems like the $$\langle \epsilon_k \rangle$$ notation would indicate the mean, following the example above $$\langle \epsilon_k \rangle=0$$. Yet I am not clear what to make of the $$\langle \epsilon_j \epsilon_k \rangle$$. How is it defined?

Context

The $$\{\epsilon_k\}$$ with $$k$$ in $$\{1,..,.n\}$$ would refer to noise at each timepoint in the measurement of a particle trajectory. And it is assumed that

$$\epsilon_k \sim Normal(0,\sigma^2)$$

Based on the answers received I understand the $$\langle \rangle$$ notation to represent the expectation. And in this case:

$$\langle \epsilon_k \rangle=0$$ (the first moment) and,

$$\langle \epsilon^2_k \rangle=\sigma^2$$ (the second moment)

If the noise is not correlated between timepoints then:

$$\langle \epsilon_i \epsilon_j\rangle=0$$ for $$i \neq j$$

What I find confusing is that I could see this if we were talking about consecutive values:

$$\langle \epsilon_i \epsilon_{i+1} \rangle= \frac{1}{n-1} \sum_{i=1}^{n-1}(\epsilon_i \epsilon_{i+1})$$

But does the notation indicate the product at all possible intervals? What does it mean exactly to do, i.e. how would one calculate:

$$\langle \epsilon_i \epsilon_j \rangle$$

I guess if one wanted to calculate it one would need to know the distance between the $$i$$'s and $$j$$'s.

• This needs more context. Often, an expectation value. Commented Feb 5, 2021 at 19:22
• This is the notation of expectation, in probability theory. Commented Feb 5, 2021 at 19:26
• @AbdelmalekAbdesselam I'll note that this is common notation for probability theory within physics. Within mathematics, this is not a standard notation that I've seen. Commented Feb 5, 2021 at 19:37
• @RichardMyers: what I meant is the notion of expectation is the one "in probability theory", not that the notation $\langle\cdots\rangle$ is used in probability theory, where one would write $E(\cdots)$. Commented Feb 5, 2021 at 19:38
• In mathematics, in particular in probability theory, you would typically denote that average by $E$, e.g. $E\{\epsilon_k\}=0$. They are the same exact thing, just with a different notation. Commented Feb 5, 2021 at 20:39

$$\langle x\rangle$$ refers to the expectation value of $$x$$.
$$\delta_{jk}$$ is the Kronecker delta, defined as:
\delta_{jk}=\left\{\begin{align}0 && j\ne k \\ 1 && j=k\end{align}\right.
So this is a shorthand way of saying that for any $$j$$, $$\langle e_j^2\rangle=\sigma^2$$ and that if $$j\ne k$$, $$\langle e_je_k\rangle=0$$. In other words, the RMS value of the noise is $$\sigma$$ for all time, and the value of the noise is uncorrelated between any two time points $$j$$ and $$k$$.
Chris answer is correct. To put it into a math context think of $$\epsilon_i$$ as independent identically distributed random variables. Two random variables $$\epsilon_i$$ and $$\epsilon_j$$ differ if $$i\ne k$$. Since they are independent we get $$\langle \epsilon_i \epsilon_j \rangle =:E[\epsilon_i \epsilon_j] = E[\epsilon_i] E[\epsilon_j] = 0$$ where $$E[.]$$ is the expectation value. However, if $$i=j$$ we get $$\langle \epsilon_i \epsilon_i \rangle =:E[\epsilon_i^2] = E[\epsilon_i^2] - \underbrace{E[\epsilon_i]^2}_{=0} = Var[\epsilon_i] =: \sigma^2$$ where $$Var[.]$$ is the variance.