hydrogenquantum mechanics
Expectation values of $(x,y,z)$ in the $| n\ell m\rangle$ state of hydrogen?
Does anyone know of a quick way of finding this (if there is even one)? Can I somehow use the relation that:
$$\langle r\rangle ~=~ \frac{a_0}{2}(3n^2 – \ell(\ell+1)),$$
or do I just have to brute force and use properties of Laguerre polynomials and spherical harmonics and what not?
Let $\alpha \in {\Bbb C}$, and let $\vert{n}\rangle $ be the harmonic oscillator state with energy $(n+\textstyle\frac{1}{2})\hbar\omega$. At $t=0$, the coherent state $\vert {\alpha(0)}\rangle $ is defined by $$ \vert{\alpha(0)}\rangle= e^{-\vert \alpha \vert^2/2}\,\left( \sum_{n=0}^{\infty} \displaystyle{\alpha^n\over \sqrt{n!}}\,\vert{n}\rangle\right) \tag{1} $$
What is $\vert{\alpha(t)}\rangle$, the coherent state at time $t$? Start with (1). Since $\left\vert n\right\rangle $ is an eigenstate of the harmonic oscillator hamiltonian $\hat{H}=\left( \hat a^{\dagger }\hat a+\frac{1}{2}\right) \hbar \omega $ with eigenvalue $\left( n+\frac{1}{2}\right) \hbar \omega ,$ the time evolution of $\left\vert n\right\rangle $ is simply $\left\vert n(t)\right\rangle =e^{-i(n+\frac{1}{2})\omega t}\left\vert n\right\rangle $ and thus \begin{equation} \left\vert \alpha (t)\right\rangle =e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n}}{\sqrt{n!}}e^{-i(n+\frac{% 1}{2})\omega t}\left\vert n\right\rangle \right) . \end{equation} It is easy to show that $\left\vert \alpha (t)\right\rangle $ is normalized.
Now we first need to show that $a\vert{\alpha(t)}\rangle=\alpha e^{i\hbar \omega t}\vert{\alpha(t)}\rangle$. Recall that $\hat{a}\left\vert n\right\rangle =\sqrt{n}\left\vert n-1\right\rangle .$ \ Then, since $\hat{a}$ is linear, \begin{eqnarray} \hat{a}\left\vert \alpha (t)\right\rangle &=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n}}{\sqrt{n!}}% e^{-i(n+\frac{1}{2})\omega t}\hat{a}\left\vert n\right\rangle \right) , \\ &=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{% \alpha ^{n}}{\sqrt{n!}}e^{-i(n+\frac{1}{2})\omega t}\sqrt{n}\left\vert n-1\right\rangle \right) , \\ &=&e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{% \alpha ^{n}}{\sqrt{\left( n-1\right) !}}e^{-i(n+\frac{1}{2})\omega t}\left\vert n-1\right\rangle \right) , \\ &=&\alpha e^{-i\omega t}e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{n=0}^{\infty }\frac{\alpha ^{n-1}}{\sqrt{\left( n-1\right) !}}% e^{-i(n-1+\frac{1}{2})\omega t}\left\vert n-1\right\rangle \right) .\quad \end{eqnarray} The sum properly starts at $n=1$ since the $n=0$ term does not exist. Thus, setting $m=n-1,$ we can rewrite this sum in terms of $m,$ with $m$ starting at $m=0.$ Hence \begin{eqnarray} \hat{a}\left\vert \alpha (t)\right\rangle &=&\alpha e^{-i\omega t}\left[ e^{-\left\vert \alpha \right\vert ^{2}/2}\left( \sum_{m=0}^{\infty }\frac{% \alpha ^{m}}{\sqrt{m!}}e^{-i(m+\frac{1}{2})\omega t}\left\vert m\right\rangle \right) \right] \\ &=&\alpha e^{-i\omega t}\left\vert \alpha (t)\right\rangle . \end{eqnarray} A useful secondary result, which follows immediately from above, is \begin{eqnarray} \left[ \hat{a}\left\vert \alpha (t)\right\rangle \right] ^{\dagger } &=&\left\langle \alpha (t)\right\vert \hat{a}^{\dagger } \\ &=&\left[ \alpha e^{-i\omega t}\left\vert \alpha (t)\right\rangle \right] ^{\dagger }=\alpha ^{\ast }e^{i\omega t}\left\langle \alpha (t)\right\vert \end{eqnarray}
Now $\langle \hat p(t) \rangle$ and $\langle \hat x(t)\rangle$ for $\vert{\alpha(t)}\rangle$. Starting from the definitions $$ \hat{a} =\sqrt{\frac{m\omega }{2\hbar }}\left( \hat{x}+\frac{i}{% m\omega }\hat{p}\right) , \qquad \hat{a}^{\dagger } =\sqrt{\frac{m\omega }{2\hbar }}\left( \hat{x}-% \frac{i}{m\omega }\hat{p}\right) , $$ we have $$ \hat{x} =\sqrt{\frac{\hbar }{2m\omega }}\left( \hat{a}^{\dagger }+% \hat{a}\right) , \qquad \hat{p} =i\sqrt{\frac{m\omega \hbar }{2}}\left( \hat{a}^{\dagger }-% \hat{a}\right) , $$ and thus \begin{eqnarray} \left\langle x(t)\right\rangle &=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\left\vert \alpha (t)\right\rangle +\left\langle \alpha (t)\right\vert \hat{a}\left\vert \alpha (t)\right\rangle \right]\, , \\ &=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \alpha ^{\ast }e^{i\omega t}+\alpha e^{-i\omega t}\right] \left\langle \alpha (t)\right. \left\vert \alpha (t)\right\rangle \\ &=&\sqrt{\frac{\hbar }{2m\omega }}\left[ \alpha ^{\ast }e^{i\omega t}+\alpha e^{-i\omega t}\right] , \end{eqnarray} which is real, as expected. We can clean this up by writing $\alpha =\left\vert \alpha \right\vert e^{i\theta }$ to obtain% \begin{equation} \left\langle x(t)\right\rangle =\sqrt{\frac{2\hbar }{m\omega }}\left\vert \alpha \right\vert \cos \left( \omega t-\theta \right) \tag{2} \end{equation}
Likewise, \begin{eqnarray} \left\langle p(t)\right\rangle &=&i\sqrt{\frac{m\omega \hbar }{2}}\left[ \left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\left\vert \alpha (t)\right\rangle -\left\langle \alpha (t)\right\vert \hat{a}\left\vert \alpha (t)\right\rangle \right] \\ &=&i\sqrt{\frac{m\omega \hbar }{2}}\left[ \alpha ^{\ast }e^{i\omega t}-\alpha e^{-i\omega t}\right] \left\langle \alpha (t)\right. \left\vert \alpha (t)\right\rangle \\ &=&-\sqrt{2m\omega \hbar }\left\vert \alpha \right\vert \sin \left( \omega t-\theta \right) \tag{3} \end{eqnarray} which is again real.
In your specific case you are starting with a coherent state for which, at $t=0$, we have $$ \langle x(0)\rangle= b\sqrt{2}x_0\, ,\qquad \langle p(0)\rangle=0 $$ so this implies from (2) and (3) evaluated at $t=0$ that $$ b\sqrt{2}x_0=\sqrt{\frac{2\hbar }{m\omega }}\left\vert \alpha \right\vert \cos \left(\theta \right)\, , \qquad 0= \sqrt{2m\omega \hbar }\left\vert \alpha \right\vert \sin \left(\theta \right) $$ Comparing with your initial conditions gives $\theta=0$ and $b\sqrt{2}x_0=\sqrt{\frac{2\hbar }{m\omega }} \alpha $ with $\alpha$ real.
Finally, $\hat{x}^{2}$ and $\hat{p}^{2}.$ From $\hat{x}$ and $\hat{p},$ we find \begin{eqnarray} \hat{x}^{2} &=&\frac{\hbar }{2m\omega }\left( \hat{a}^{\dagger }+ \hat{a}\right) ^{2}=\frac{\hbar }{2m\omega }\left( \left( \hat{a} ^{\dagger }\right) ^{2}+\hat{a}^{\dagger }\hat{a}+\hat{a}\hat{a} ^{\dagger }+\left( \hat{a}\right) ^{2}\right) , \\ &=&\frac{\hbar }{2m\omega }\left( \left( \hat{a}^{\dagger }\right) ^{2}+2 \hat{a}^{\dagger }\hat{a}+1+\left( \hat{a}\right) ^{2}\right) , \\ \hat{p}^{2} &=&-\frac{m\omega \hbar }{2}\left( \hat{a}-\hat{a} ^{\dagger }\right) ^{2}=-\frac{m\omega \hbar }{2}\left( \left( \hat{a} ^{\dagger }\right) ^{2}-\hat{a}^{\dagger }\hat{a}-\hat{a}\hat{a}% ^{\dagger }+\left( \hat{a}\right) ^{2}\right) , \\ &=&-\frac{m\omega \hbar }{2}\left( \left( \hat{a}^{\dagger }\right) ^{2}-2% \hat{a}^{\dagger }\hat{a}-1+\left( \hat{a}\right) ^{2}\right) , \end{eqnarray} where \begin{equation} \hat{a}\hat{a}^{\dagger }=\hat{a}\hat{a}^{\dagger }-\hat{a}% ^{\dagger }\hat{a}+\hat{a}^{\dagger }\hat{a}=\left[ \hat{a},% \hat{a}^{\dagger }\right] +\hat{a}^{\dagger }\hat{a}=1+\hat{a}% ^{\dagger }\hat{a} \end{equation} has been used. Thus, \begin{eqnarray} \left\langle x^{2}(t)\right\rangle &=&\frac{\hbar }{2m\omega }\left[ \left\langle \alpha (t)\right\vert \left( \hat{a}^{\dagger }\right) ^{2}\left\vert \alpha (t)\right\rangle +2\left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\hat{a}\left\vert \alpha (t)\right\rangle\right.\nonumber \\ &&\left.\qquad\qquad\qquad\qquad\qquad\quad +1+\left\langle \alpha (t)\right\vert \hat{a}^{2}\left\vert \alpha (t)\right\rangle \right] , \\ &=&\frac{\hbar }{2m\omega }\left[ \left( \alpha ^{\ast }e^{i\omega t}\right) ^{2}+2\alpha ^{\ast }\alpha +1+\left( \alpha e^{-i\omega t}\right) ^{2}% \right]\, ,\\ &=&\frac{\hbar }{2m\omega }\left[ \left( \alpha ^{\ast }e^{i\omega t}+\alpha e^{-i\omega t}\right) ^{2}+1\right] , \\ &=&\frac{\hbar }{2m\omega }\left[ 4\left\vert \alpha \right\vert ^{2}\cos ^{2}\left( \omega t-\theta \right) +1\right] . \\ \left\langle p^{2}(t)\right\rangle &=&-\frac{m\omega \hbar }{2}\left[ \left\langle \alpha (t)\right\vert \left( \hat{a}^{\dagger }\right) ^{2}\left\vert \alpha (t)\right\rangle -2\left\langle \alpha (t)\right\vert \hat{a}^{\dagger }\hat{a}\left\vert \alpha (t)\right\rangle\right.\nonumber \\ &&\left.\qquad\qquad\qquad\qquad\qquad\quad -1+\left\langle \alpha (t)\right\vert \hat{a}^{2}\left\vert \alpha (t)\right\rangle \right] , \\ &=&-\frac{m\omega \hbar }{2}\left[ \left( \alpha ^{\ast }e^{i\omega t}\right) ^{2}-2\alpha ^{\ast }\alpha -1+\left( \alpha e^{-i\omega t}\right) ^{2}\right]\, ,\\ &=&-\frac{m\omega \hbar }{2}\left[ \left( \alpha ^{\ast }e^{i\omega t}-\alpha e^{-i\omega t}\right) ^{2}-1\right] , \\ &=&-\frac{m\omega \hbar }{2}\left[ -4\left\vert \alpha \right\vert ^{2}\sin ^{2}\left( \omega t-\theta \right) -1\right]\, ,\\ &=&\frac{m\omega \hbar }{2}\left[ 4\left\vert \alpha \right\vert ^{2}\sin ^{2}\left( \omega t-\theta \right) +1% \right] . \end{eqnarray}
This is actually not surprising. Even an electron orbit in the Bohr atom has a zero expectation value for the momentum if you average over whole orbits.
Think of what it would mean for there to be a finite expectation value for momentum: you'd see an electron that's escaping from the nucleus!
Perhaps what you want to calculate is the root mean square speed of the electron. In that case, you want to find the expectation value of $\vec{p}^2$, not $\vec{p}$.
Best Answer
You're confusing the expectation values of the vector $\mathbf{r}=(x,y,z)$ and its magnitude $r=\sqrt{x^2+y^2+z^2}$. Because the hydrogen hamiltonian is parity invariant, all its eigenfunctions are chosen to have a definite parity. This means that the expectation value $\langle\mathbf{r}\rangle$ will always be zero because each component is the integral of an odd function times an even probability distribution function.
The second expectation value you mention, $$\langle n,l,m|r|n,l,m\rangle=\int \mathrm{d}x \mathrm{d}y \mathrm{d}z \psi_{n,l,m}^\ast(x,y,z)\sqrt{x^2+y^2+z^2}\psi_{n,l,m}(x,y,z),$$ is quite different, since you're not counting the direction of the distances you add. This is the same as comparing the averages of $x$ and $|x|$ over some 1D probability distribution function $p(x)$. The average of $x$ will be zero if $p(x)$ is even, while the average of $|x|$ will only vanish is $p$ is a delta function, concentrated at the origin. Thus $\langle|x|\rangle$ can be quite large while $\langle x\rangle$ is zero because of cancellations.
This is not to say that expectation values of $\mathbf{r}$ are not interesting, but one must simply be more careful. The fact that the diagonal matrix elements vanish says that the eigenstates have no permanent dipole moment - which of course they can't as they are eigenstates of an isotropic system. What doesn't vanish, however, are the transition matrix elements, $$\langle n',l',m'|\mathbf{r}|n,l,m\rangle,$$ which play an important role in the hydrogen atom's interaction with radiation. (Specifically, they control the leading-order interaction energy, which is the dipole coupling $-q\mathbf{r}\cdot\mathbf{E}$.) Because the components of $\mathbf{r}$ are themselves spherical harmonics of degree 1 (or linear combinations of them), the matrix element above will involve the spherical integral $$\int\mathrm{d}\Omega \, Y_{l'm'}^\ast Y_{1,\mu} Y_{lm}$$ for $\mu=-1,0,1$.
Because the spherical harmonics have a rich orthogonality structure, only a few of these integrals will survive. Specifically, you need $\Delta l=|l-l'|=1$ and $m'=m+\mu$, which has the physical content of the dipole selection rules: for a dipole coupling to radiation, only S-P, P-D, D-F, ... transitions are allowed. Further, only a few spatial orientations are allowed: $m$ can increase or decrease by one (which happens if $|\mu|=1$, and corresponds to circularly polarized light in either direction, with the electric field rotating in the $x,y$ plane) or stay the same (when $m=0$ and the light is linearly polarized along $z$).