# Expectation value of following function

1. Jun 25, 2005

### Kruger

I need to find the momentum expectation value of the function in the attached picture. It is the function of the harmonic oscillator (first excited state).

I know that the expectation value is the value that we measure with the highest probability if we measure the system. But what the hell does the standard deviation d(p)=<p^2>-<p>^2 mean for the harmonic oscillator? Does it mean that if we make 1000 times the same experiment and measure with every experiment the momentum that we result in this d(p) if we always measure p?

And another question is can we measure in one experiment <p> and <x> (the two expectation values) exactly at the same time? I think no, because the uncertainty principle would not be sadisfied.

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2. Jun 25, 2005

### dextercioby

There's an error there:

$$\langle\hat{p}_{x}\rangle_{\psi (x)}=-i\hbar\int_{\mathbb{R}} \psi^{*}(x) \frac{d\psi (x)}{dx} \ dx$$

Daniel.

3. Jun 26, 2005

### Edgardo

Hello Kruger,

standard deviation is the spread around your mean value. Say you measure your momentum 1000 times.

From the data you collected you can calculate the mean value <p>. In addition, you can also calculate the standard deviation often denoted as $\sigma$.

What does $<p> \pm \sigma$ mean?
In case of a normal distribution 68% of your data, that is 680 data points will lie within $<p> \pm \sigma$.

Or to give you a better feeling, say you have a friend who asks you about your measurements. He wants to know, what momentum he will measure if he conducted the same experiment. Then you can tell him: Well, the mean value is <p>, but you will measure with 68% probability a value within $<p> \pm \sigma$.

The standard deviation moreover tells you how "spread" your values will be around your mean value, the greater $\sigma$ the greater your spread.

By typing "Standard deviation" into google, I found the following websites:
http://en.wikipedia.org/wiki/Standard_deviation
http://www.robertniles.com/stats/stdev.shtml

Last edited: Jun 26, 2005
4. Jun 26, 2005

### Staff: Mentor

Correction: you need to take the square root:

$$\Delta p = \sqrt {<p^2> - <p>^2}$$

That is approximately correct. Think of the $\Delta p$ that you can calculate from the wave function as an "ideal" or "exact" value. The $\Delta p$ that you calculate from actual experimental measurements is an approximation, which improves as the number of data points increases.

In one experiment (trial), you cannot meaningfully measure the expectation value of either x or p because you have only one data point to work with.

If you repeat the experiment many many times (identically prepared each time, or course), then measure both x and p each time, you can then estimate $<x>$ and $<p>$ from your data, also $<x^2>$ and $<p^2>$. Now you can calculate $\Delta p$ as described above, and $\Delta x$ similarly. You will find that those two quantities always satisfy Heisenberg's uncertainty relation, in the limit of an infinite number of trials. If your measured values of p lie within a small range, then the values of x will spread out over a large range, and vice versa. (For some suitable definition of "small" and "large", of course.) That's the meaning of the Heisenberg uncertainty principle.

Note that the HUP is a statement about the probability distributions that underlie the measured values of x and p. In a finite number of trials, you can test the HUP only approximately, and there is a chance that your particular data may actually appear to violate the HUP! As the number of trials becomes larger this becomes less likely to happen.