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The expectation value of $r=\sqrt{x^2 + y^2 + z^2}$ for the electron in the ground state in hydrogen is $\frac{3a}{2}$ where a is the bohr radius.

I can easily see from the integration that the expectation values of $x$ , $y$, $z$ individually is $0$ because of the factor $\cos\phi$, $\sin\phi$ and $\cos\theta$(.$\sin\theta$) respectively for $x$ $y$, $z$ integrating out to $0$.

What does this mean? Is it an artifact of the spherically symmetric potential? Since $r$ has a non zero value but what makes up $r$ ie $x$, $y$, $z$ (or rather the squareroot of the sum of their squares) each has individually $0$ and i can simultaneously determine their values since they commute with each other, which means $x,y,z$ are all simultaneously $0$ and $r$ is not?!

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  • $\begingroup$ I could see how my question is related to the answer to this question, but i want to understand what it means physically $\endgroup$
    – Prasad Mani
    Commented Oct 21, 2016 at 7:03
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    $\begingroup$ It is possible for $\langle x \rangle$ to be zero but $\langle x^2 \rangle$ to be nonzero. For example, consider $x = \pm 1$ with equal probability. $\endgroup$
    – knzhou
    Commented Oct 21, 2016 at 7:08
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    $\begingroup$ The point is that $\langle x \rangle = 0$ really doesn't tell you anything about whether $\langle f(x) \rangle = 0$. It's not a coincidence, it follows from symmetry, but it has little to do with $\langle r \rangle$. $\endgroup$
    – knzhou
    Commented Oct 21, 2016 at 7:12
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    $\begingroup$ If you make a measurement, you've completely changed the state! And just because $\langle x \rangle = 0$ does not mean you'll always measure zero... Also, you measure $x$, not $\langle x \rangle$. A lot of this sounds very confused. $\endgroup$
    – knzhou
    Commented Oct 21, 2016 at 7:15
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    $\begingroup$ Again, it is possible for $\langle x \rangle$ to be zero and for $\langle f(x) \rangle$ to be nonzero. $\endgroup$
    – knzhou
    Commented Oct 21, 2016 at 7:18

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$x$, $y$ and $z$ can take any value from $-\infty$ to $\infty$, and for all three the wavefuntion is symmetric about the origin. So it's (hopefully) obvious that the expectation value of all three is zero.

By contrast $r \ge 0$ i.e. $r$ cannot be negative. So it's (hopefully) obvious that the expectation value of $r$ will lie somewhere in the range $0 \le r \le \infty$.

You would get a similar result if you calculated the expectation value of $|x|$ rather than $x$.

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  • $\begingroup$ I see that the expectation or the 'average' value is zero that way. I shouldnt worry about what it means physically? $\endgroup$
    – Prasad Mani
    Commented Oct 21, 2016 at 7:07
  • $\begingroup$ @PrasadMani: suppose you were whirling a stone around on a string of length $\ell$. The average distance of the stone from you is obviously $\ell$ because that's the length of the string. However the average value of $x$ and $y$ (assuming you're whirling the stone in the horizontal plane) would both be zero. $\endgroup$
    – John Rennie
    Commented Oct 21, 2016 at 7:13

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