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In my introduction to Quantum Physics class we're learning about wave functions and uncertainties (with a great deal of it being shown through graphical means). However, in the notes from my lecture I came about something that was rather confusing:

enter image description here

In part b) of my professor's notes, he wrote, "the expectation value is independent of sigma (standard deviation)....". Now, I do understand why the expectation value is 0 here, however, I'm confused on it's relationship to the standard deviation and uncertainty:

  1. Correct me if I'm wrong, but aren't the standard deviation (sigma) and the uncertainty in x (∆x) the same thing (correlating to the width of the

  2. If so, isn't sigma related to the expectation value through the following:

enter image description here

(where sigma and ∆x are the same)

And if the aforementioned statement is true, then wouldn't increasing the expectation value, in turn, increase the standard deviation?

Again, I'd really appreciate any guidance in helping me figure this out, as I'm rather confused on how to even go about this.

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2 Answers 2

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Correct me if I'm wrong, but aren't the standard deviation (sigma) and the uncertainty in x (∆x) the same thing (correlating to the width of the

You can certainly define the uncertainty to be the standard deviation. Personally I would tend to say that "the uncertainty" is a more general concept, and the standard deviation is one (common and useful) metric that can be used to quantify the uncertainty. Another example of a metric that's sometimes used to quantify uncertainty (although not often in quantum mechanics) is a 90% confidence interval.

If so, isn't sigma related to the expectation value through the following: $$ \Delta x = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} $$

The equation you wrote down is correct. However, it is also true that $\Delta x$ is independent of $\langle x\rangle$.

It is actually a good exercise to prove this -- if you add a constant $a$ to every data point $x_i$ and compute the standard deviation $\Delta x$, you will find that $a$ drops out of the final answer.

However, we can also give a little bit of intuition.

The idea is that shifting the mean $\langle x \rangle$ will itself shift the average of $x^2$, $\langle x^2 \rangle$. This is because every data point has been made bigger (or smaller), so an average of any monotonic function of $x$ will get bigger (or smaller). However, the standard deviation is a combination of the two possible "averages" that are of order $x^2$ which is independent of the mean. Therefore, it provides independent information characterizing the distribution.

In fact, you can generalize the idea of the standard deviation, to look at higher exponents. For example, there is a combination of "averages of $x^3$" (namely, a combination of $\langle x^3\rangle, \langle x^2\rangle \langle x \rangle, \langle x\rangle^3$) which are independent of the mean and give additional information about the shape of the distribution. These are known as standardized moments and include the skewness (information in averages of order $x^3$) and kurtosis (information in averages of order $x^4$). Compared to the standard deviation, higher order moments tend to be more sensitive to the tails of the distribution.

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Given the list $\mathbf x=(1,2,3)$ the expectation value and standard deviation are $\langle x\rangle =2$ and $\Delta x=\sqrt{2/3}$. If I shift everything to the right, e.g. $\mathbf x=(11,12,13)$ we now have $\langle x\rangle =12$, but the expectation value is still $\Delta x=\sqrt{2/3}$. The standard deviation is just the spread of something and it can be changed separately from the expectation value. If I scale the entire vector with a constant however, both quanties will change.

To answer your questions:

  1. For quantum mechanics these quantities mean the same thing. For other fields these terminologies might change slightly but they will often correspond to the same thing
  2. Yes
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