Are the eigenstates of an electron in the hydrogen atom coordinate dependent?

Question

This is a simple question, but I cannot do the calculations necessary to answer it as it seems too complicated to me. I hope there is a easy way to answer this. When writing down the potential of the hydrogen atom, we take the origin to be at the nucleus. Suppose we took the origin to be slightly off center. Then the Hamiltonian will no longer be spherically symmetric. Does that mean the allowed states are no longer eigenstates of energy and angular momentum? Are the states of hydrogen atom coordinate dependent?

Answer 1

The system is still spherically symmetric about the position of the hydrogen nucleus. The wave functions will still be the same as they are, centered on the hydrogen nucleus. The only difference is that this won't be the origin in the new coordinate system. If the hydrogen atom is at position $\vec{r}_0$ in the new coordinate system, then the eigenstates of the Hamiltonian will just be given by $\psi_{nlm}(\vec{r} - \vec{r}_0)$. These are exactly the same hydrogen orbitals, centered on the same point in space. We're just calling that point in space $\vec{r}_0$ instead of 0.

This is equivalent to solving a particle-in-a-box system on the interval $[0,L]$ instead of $[-L/2,L/2]$. Using the latter interval, it's clear that the system has parity symmetry, and so the eigenstates will be even and odd functions (sines and cosines). If we shift the origin so that we are using the interval $[0,L]$ instead, we still get the same functions, just shifted so that they are centered at $L/2$ instead. The symmetry is not as obvious, but it's still true that the eigenstates are even or odd about the center of the box ($L/2$).

Answer 2

There is an elegant way to check how will be the new eigenfunctions of the changed Hamiltonian.

A change of origin is equivalent to a displacement of the system. Such transformation could be represented by an unitary operator $D(\vec a)$ such that

$$ D^{-1}(\vec a)\vec r D(\vec a) = \vec r + \vec a $$

and

$$ D^{-1}(\vec a)\vec p D(\vec a) = \vec p $$

It is easy to check that $D(\vec a) = e^{i\vec p \cdot \vec a/\hbar}$. Now, this operator acts similarly in any function of position operator

$$ D^{-1}(\vec a)F(\vec r)D(\vec a) = F(\vec r+ \vec a). $$

and the eigenvectors of position operators transform like $$ D(\vec a) |\vec r\rangle = |\vec r - \vec a\rangle $$

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When we change the coordinate system, the Hamiltonian changes to (where $D(\vec a) \equiv D$)

$$ H' = \frac{1}{2m} p^2 + V (\vec r + \vec a) = D^{-1}HD. $$

And the same for angular momentum $L' = D^{-1}LD$. Now, let's suppose we have a solution for the eigensystem

$$ \begin{cases} H |\psi_{nlm}\rangle = E|\psi_{nlm}\rangle\\ L|\psi_{nlm}\rangle = l(l+1)|\psi_{nlm}\rangle \end{cases} $$ By applying $D^{-1}$ in the equations above, and using the fact that $D^{-1}D=I$,

$$ \begin{cases} (D^{-1} H D)D^{-1}|\psi_{nlm}\rangle = E_n D^{-1}|\psi_{nlm}\rangle\\ (D^{-1}LD)D^{-1}|\psi_{nlm}\rangle = l(l+1)D^{-1}|\psi_{nlm}\rangle \end{cases} $$

So the new solutions are $D^{-1}|\psi_{mln}\rangle$. In the position basis

$$ \langle \vec r|D^{-1}|\psi_{nlm}\rangle = \langle \vec r+\vec a |\psi_{nlm}\rangle = \psi_{nlm}(\vec r +\vec a) $$

So the displaced wave function solves both the eigensystems, for the displaced Hamiltonian and displaced angular momentum operators.

Answer 3

The answer to your question is no, yes, and maybe. This is because the question needs clarification.

First, we're talking about a non-relativistic$^1$ scalar$^2$ charged particle in a static Coulomb$^3$ potential from a point$^4$ like nucleus with infinite mass$^5$. The 5 idealization allow me to ignore:

[1] 1st order relativistic perturbations:

$$ H' = \frac{\hat p^4}{8m^3_ec^2} $$

[2] Spin-Orbit:

$$ H_{SO} = \Big( \frac{Ze^2}{4\pi\epsilon_0} \Big) \Big( \frac{g_s-1}{2M_e^2c^2} \Big) \frac{\vec L \cdot \vec S}{r^3} $$

[3] A static field means no photon interactions, which keeps excited eigenstates as eigenstates. IRL, they are metastable states with finite width.

[4] Just ignoring the finite nuclear potential b/c the charge distribution is 0.88 fm wide, not $e\delta(\vec r)$.

[5] Now per your question where you state we normally take $r=0$ to be the nucleus. This is actually not the case. The full hydrogen atom hamiltonian is:

$$ H = \frac{\hat p_p^2}{2M_p} + \frac{\hat p_e^2}{2m_e} - k\frac{e^2}{|\vec r_p - \vec r_e|}$$

but we can split that into a center-of-mass hamiltonian around

$$ \vec R = \frac{M_p\vec r_p + m_e\vec r_e} {M_p + m_e} $$

that we ignore, and a reduced mass hamiltonian:

$$ H = -\frac{\hbar^2}{2\mu}\nabla^2 - k\frac{e^2}r $$

where the reduced mass is:

$$ \mu = \frac 1{\frac 1{M_p} + \frac 1{m_e}}$$

and its coordinate is:

$$ \vec r = \vec r_e - \vec r_p $$

which in a sense is centered around the nucleus, but irl the nucleus is also in an orbital wave function that we ignore. With $M_p \rightarrow \infty$, then we can say: the electron orbits the center of the nucleus.

So now we can address the question. The hamiltonian is spherical coordinates is separable, so the wave functions can be written as:

$$ \psi(\vec r) = R(r)Y_l^m(\theta, \phi) $$

where the $Y_l^m(\theta, \phi)$ result from spherical symmetry, and are eigenstates of $L^2$ and $L_z$. The solutions can be written as $|nlm\rangle$ with energy:

$$ E_{nlm} =\frac 1 2 m_ec^2\frac {\alpha^2}{n^2}$$

Of course, that does not depend on $m$, since the direction of angular momentum doesn't matter: that is clear from the choice of coordinates: $\vec L$ is conserved.

Note that is also doesn't depend on $l$, which is not clear from the coordinates. This is because the hamiltonian conserves the Laplace-Runge-Lenz vector:

$$ \vec A = \vec p \times \vec L -mk\hat r $$

which is clear if you write the hamiltonian in parabolic coordinates, where it is also separable:

$$ H= [TBD]^{(6)} $$

The point is, you get the LRL-vector conservation naturally in parabolic coordinates, but it's still there in spherical coordinates, it's just not manifestly obvious.

So: if you pick terrible coordinates, your solutions will still respect the symmetries of the problem (and conserve angular momentum), it just won't be obvious based on your solutions.

[6] So I don't have the parabolic form memorized, and idk if anyone noticed, now that Google uses AI in the search, the results for really technical things such as "Hydrogen atom in parabolic coordinates" just produces far poorer results than it did a few years ago.