# Expectation Value <x^2>

1. Oct 15, 2011

### teroenza

1. The problem statement, all variables and given/known data
1. What is <x$^{2}$>, in terms of position and expectation values.
2. How can I use the correspondence principal to explain the quantum vs classical results (below).

My textbook (Serway, Modern Physics) uses <x> as the expectation value, meaning the average position of a particle. It also uses <x$^{2}$> in one example, and wants me to calculate it. I have calculated it by using the formula for <x>, but have no idea what it represents. The book does not define it, and I can't find it elsewhere.

The problem is to compare my calculated classical <x> and <x$^{2}$> values vs the books example calculating them for a quantum situation (particle in infinite square well of length L).

2. Relevant equations
<x> = $\int$x (1/L)dx from 0 to L

<x$^{2}$> = = $\int$x$^{}$ (1/L)dx from 0 to L

3. The attempt at a solution
The probability density (classical) is given as 1/L.
My classical <x> agrees with that of the book's quantum = L/2

My classical <x$^{2}$> is L$^{2}$/3 while the book's quantum value is L$^{2}$/3 -L$^{3}$/(2 pi$^{2}$).

I am also to use the correspondence principal to discuss the findings, but I cannot see how the quantum value would be altered (no n factor to increase) as the quantum world approaches the macro.

2. Oct 15, 2011

### vela

Staff Emeritus
The expectation value of a function g(X) of a continuous random variable X is given by
$$\langle g(x)\rangle = \int g(x) f(x)\,dx$$where f(x) is the probability density function for X.

For the classical case, you have f(x)=1/L, so
\begin{align*}
\langle x \rangle &= \int_0^L x \frac{1}{L}\,dx \\
\langle x^2 \rangle &= \int_0^L x^2 \frac{1}{L}\,dx
\end{align*}
The only difference for the quantum case is that you now have f(x)=|ψ(x)|2.

3. Oct 16, 2011

### teroenza

O.k. thank you. I used the equation with x$^{2}$ to arrive at my answer for the <x$^{2}$> part of the question. So if <x> is the average position, <x$^{2}$> is just the average of the position squared? I don't seem to understand the reason for the squaring. I assume <x$^{2}$> means the same thing for the classical and quantum, like <x> = average position, in both.

4. Oct 16, 2011

### vela

Staff Emeritus
Yes, you got it. As for the reason why you'd calculate <x2>, it's simply because the math calls for it. For example, the characteristic width of the wave function is $\Delta x = \sqrt{\langle (x-\langle x \rangle)^2 \rangle} = \sqrt{\langle x^2 \rangle - \langle x \rangle^2}$. In plain old probability and statistics, this is how you'd calculate the standard deviation of a random distribution.

5. Oct 16, 2011

### teroenza

Great, thanks for the help.

I am still puzzled as to how I can use these results to "discuss" Bohr's correspondence principal. I see the classical and quantum results agree for the <x> value, which makes sense, because in in infinite well they are both forever confined within. However, I cannot explain the discrepancy in the <x$^{2}$>value. There is no "n" quantum number for me to increase ad thus see what happens as the quantum scale approaches the classical.

6. Oct 16, 2011

### vela

Staff Emeritus
Did your book calculate the expectation value for a particular value of n? If not, it's a typo. There should be an n in the answer.

7. Oct 16, 2011

### teroenza

You are correct, the books quantum expression was for the ground state (n=1).
There was an n in every answer, just hidden as n=1, in the denominator. So as n$\rightarrow$infinity, both quantum terms go to zero. Thus predicting a zero average position and average position squared. Does this mean the particle is now localized at the origin?

8. Oct 16, 2011

### vela

Staff Emeritus
That's not correct. In the limit as n goes to infinity, the quantum mechanical result should go to the classical result.

9. Oct 17, 2011

### teroenza

I made a mistake in following the n through the books calculations. You are correct, the 2nd term drops out (in the <x^2>, so that it approaches macroscopic reality as n gets large.