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?


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

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

Reflection and question based on some the answers received

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.

  • 2
    $\begingroup$ This needs more context. Often, an expectation value. $\endgroup$ Commented Feb 5, 2021 at 19:22
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    $\begingroup$ This is the notation of expectation, in probability theory. $\endgroup$ Commented Feb 5, 2021 at 19:26
  • $\begingroup$ @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. $\endgroup$ Commented Feb 5, 2021 at 19:37
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    $\begingroup$ @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 $E(\cdots)$. $\endgroup$ Commented Feb 5, 2021 at 19:38
  • $\begingroup$ 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. $\endgroup$ Commented Feb 5, 2021 at 20:39

3 Answers 3


$\langle x\rangle$ 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 $k$.


It looks like it is the mean of the product of two components. Whatever these are. The relationship you show could mean that the noise in the components is not correlated so that the average of the noise in the products of different components cancel out whereas for the square of the components itself you get the standard deviation.


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.


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