Is there only radial motion in the Hydrogen ground state?

The ground state of the Hydrogen atom is spherically symmetric. In other words, the wave function Psi depends only on the distance r of the electron from the nucleus.

As a consequence all derivatives of Psi with respect to angles theta and phi yield zero.

Does this imply that the average kinetic energy in the ground state [which can be calculated without difficulty from the wave function] is determined exclusively by the radial motion of the electron?

If so, that would be a rather odd result. Let us say the electron is at position (x, 0, 0). Then the kinetic energy would be the result of motion either away from the nucleus (direction +x) or towards the nucleus (-x), but not from motion perpendicular to the x-axis. So in essence the motion of the electron would be 1-dimensional, like a pendulum.

• The eigenfunctions of the hydrogen atom are time independant. There is no motion at all, or at least not in the sense of a tiny billiard ball ricocheting around. – John Rennie Sep 29 '14 at 7:14
• Well, except for the overall $\exp(-i Et/\hbar)$. While there is no motion in the sense of a tiny ball bouncing around, there is a non-zero expectation value for the kinetic energy. Maybe his question is how does one resolve these two notions of motion? – CuriousKev Sep 29 '14 at 7:47
• Excellent question and excellent answer by @CuriousKev, but I am not happy yet. Obviously the electron is in a standing wave with zero angular momentum, so all the expectation values you can think of will be zero. That doesn't mean the electron isn't moving - we know it is. Is it possible to write the wavefunction in terms of a superposition of plane waves (momentum states). If so, won't this give you answer? A Compton scattering experiment should tell you what the initial momentum of the electron was before the photon hit it. – akrasia Sep 29 '14 at 10:29
• @akrasia Alright, I'll add something about momentum decomposition. – CuriousKev Sep 29 '14 at 11:20
• I thought I was going to post the momentum space wave function, but turns out it can't be done easily! It was brought up here physics.stackexchange.com/q/63004 Ideally we would clearly see in that function what the values are for the theta and phi momentum components. For the present question, it would easily suffice to prove that that form is/isn't a Dirac delta function. So far no one has even managed that task. So this question is entirely up for grabs. – Alan Rominger Nov 5 '14 at 17:48

The probability density of the ground state is time independent, so there is no motion in this sense. Yet the expectation value of the kinetic energy is non-zero, so there is movement in this sense. How are these notions of movement reconciled?

First off, classically, if we had a particle in a $1/r$ potential and released it from rest, it would indeed bob back and forth like a pendulum as you describe. But in quantum mechanics we can't say the electron is taking any specific path around the proton. As there is no specific path, we cannot fully reconcile these notions of movement with any classical preconceptions.

Let's discuss some various notions of motion in quantum mechanics, that may help you here.

The ground state of the hydrogen atom is $$\psi(r,\theta,\phi,t) = A\, \exp\left(-\frac{r}{a}-i\frac{E t}{\hbar}\right)$$ Where $A =\sqrt{\frac{1}{\pi a^3}},a=\frac{\hbar^2}{me^2},E=\frac{-m e^4}{8 h^2 \varepsilon_0^2}$

The radial momentum operator in this basis is: $$\vec{p}_r = -i\hbar\hat{r}\frac{\partial}{\partial r}$$ where $\hat{r}$ is the radial unit vector (not an operator).

In calculating the expectation value of this: $$\langle \psi|\vec{p}_r|\psi\rangle = \int \psi^* (- i\hbar\hat{r}\frac{\partial}{\partial r} \psi) \sin(\theta) r^2 dr\,d\theta\,d\phi = \int \psi^* (- i\hbar\hat{r}\frac{-1}{a} \psi) \sin(\theta) r^2 dr\,d\theta\,d\phi$$

Due to symmetry, this will of course be zero. But the density term in the integral is

$$\psi^*(- i\hbar\hat{r}\frac{-1}{a} \psi) = i\hbar\frac{1}{a}(\psi^*\psi)\hat{r}$$

This might be what you want to interpret as 'motion', but since $(\psi^*\psi)\ge 0$ this is purely imaginary and doesn't have a directly physical interpretation as motion. As its imaginary, it's neither towards or away from the center.

Another notion of motion is the probability current:

$$\vec{j} = \frac{\hbar}{2mi}\left(\psi^* \vec\nabla \psi - \psi \vec\nabla \psi^{*} \right)$$

This is related to conservation of probability by:

$$\rho = \left|\psi\right|^2,\quad \frac{\partial \rho}{\partial t} + \vec\nabla \cdot \vec{j} = 0$$

For the hydrogen ground state we have:

$$\vec{j} = \frac{\hbar}{2mi}\hat{r}\left(\psi^* \left(\frac{\partial}{\partial r}\psi\right) - \psi \left(\frac{\partial}{\partial r}\psi^{*}\right) \right)$$ $$= \frac{\hbar}{2mi}\hat{r}\left(\psi^* \left(\frac{-1}{a}\psi\right) - \psi \left(\frac{-1}{a}\psi^{*}\right) \right) = 0$$

There is no probability current at any point. So any sense in which there is motion at some location, the net current into/out of this point is still zero. Which takes me to the only remaining way I know to discuss "motion" here. We are writing the state in the position basis, let me make this more clear, and also use the cartesian basis for a bit: $$|\psi\rangle = \int \phi(x,y,z,t)|x,y,z\rangle$$ The state $|\phi\rangle$ is but one vector in the infinite vector space, that is the Hilbert space for the electron here. When we write $\phi(x,y,z,t)$ these are really time dependent components for each basis element $|x,y,x\rangle$ in the chosen basis for this vector space. The state $|1,0,0\rangle$ by itself is the closest interpretation of your idea of starting with the electron at say x=1,y=0,z=0 and dropping it to see how it moves.

We can start with this pure position state and watch how it evolves according to the Hamiltonian operator. Since this state is not an energy eigenstate, it will spread (evolve to a state that now needs to be written as a super-position of many of our $|x,y,x\rangle$ basis states). It however will not swing like a pendulum though the origin like you imagine. It will spread out in all directions (since by the uncertainty principle, a pure position state is completely spread in momentum space).

The magic of the ground state is that if we consider this special weighted super position of an enormous (infinite) number of position states individually spreading, they spread exactly such that the super position of states remains the same and the net current is zero at every point. You could view this a bit like equilibrium with the principle of detailed balance: the position states will evolve into each other, but the amount that is "leaving" a pure position state must be replaced with exactly the same amount "entering" that state from other position states in this super position.

So in a sense, there is movement (kinetic energy is non-zero, the time evolution operator (Hamiltonian) is constantly evolving pure position states at each point to spread out), but the "net movement" of the wavefunction is zero (probability current is zero) and the probability density is time independent.

Consider this section an extended comment:

Akrasia suggested another way of looking at motion here: momentum decomposition.

Basically we can also write the state in terms of momentum basis in Hilbert space. $$|\psi\rangle = \int \phi(k_x,k_y,k_z,t)|k_x,k_y,k_z\rangle$$

These basis states are spread over all space (uniformly so). So they can't tell us about motion in some region. But we can get a probability density on this space, giving a notion of motion for parts of the state. And for the hydrogen ground state, it will be built up as standing waves of opposing momentum basis states. Since these cover all space, the momentum of a plain wave state is not just in the radial direction. So in this sense, the "motion" is not just in the radial direction.

Here is a related stack exchange question:
Hydrogen wave function in momentum space
And here is a paper which claims to work out the hydrogen wave function in spherical momentum space.

They find the ground state to be: $$\phi_{1,0,0} = \frac{1}{(p_r - i p_0)^2} \frac{1}{p_\theta^{1/2}}J_{1/2}(p_\theta)\delta(p_\phi)$$ Which means there is no contribution from basis elements with non-zero $p_\phi$ momentum, but there is for non-zero $p_\theta$. That is surprising to me, and I don't have time to read through the paper right now. So it would be best if someone else wrote up an answer covering this portion. If that paper works, then that seems to be a very nice notion of motion for which to answer the question here "Is there only radial motion in the Hydrogen ground state?".

• Thank you very much for your excellent reply! May I point out that there is a small error in your wave function. The position dependence goes with r/a (instead of a/r). This also affects the subsequent calculation. – M. Wind Sep 29 '14 at 21:24
• The Lombardi result for the wave function in momentum representation is very strange. It is straightforward to Fourier transform the groundstate Psi(r) = exp(-r/a), with the result: Phi(k) = C/(k^2 + 1/a^2)^2. As one would expect, Phi only depends on the absolute value of the momentum. No angular dependence manifests itself! – M. Wind Sep 29 '14 at 22:29
• I think there are fundamental reasons why the Lombardi result should not be interpreted too literally as answering the question. There are fundamental problems, which Lombardi discusses, with introducing angle operators in quantum mechanics: physics.stackexchange.com/q/63228/4552 . Lombardi is forced to use nonhermitian operators, but one of the usual axioms of quantum mechanics is that every observable is a hermitian operator. Therefore I don't think Lombardi's expansion tells us what we would actually measure. If you relax the requirement that observables be hermitian, [...] – user4552 Nov 7 '14 at 17:42
• I don't see any obvious reason why expanding a wavefunction in the basis of eigenstates tells us anything meaningful. Lombardi takes a spherically symmetric function and writes it as a sum of functions that lack that symmetry. This is not surprising, e.g., I can write the even function $x^2$ as a sum of $x^2+e^x$ and $-e^x$, neither of which is an even function. This doesn't reveal anything interesting about $x^2$. – user4552 Nov 7 '14 at 17:45

A few days ago I suddenly realized that the answer to the question that I posted here nearly 3 years ago is actually quite simple. "Is the kinetic energy of the electron in the Hydrogen groundstate the result of radial motion only, or are there also contributions from non-radial motion?"

To find the total kinetic energy one starts with the lenght of the momentum vector squared, p.p. This is turned into the differential operator - h-bar^2 grad.grad. When this operator acts on a radial wave function (as is the case in the ground state), we obtain the result that is well-known, e.g. from the Laplace operator in spherical coordinates:- h-bar^2 {d^2/dr^2 + (2/r)d/dr}.

Now we consider the kinetic energy associated with radial motion only. To eliminate non-radial terms we take the projection of the momentum vector on a unit vector from the origin to the point (x, y, z). This radial unit vector is given by u = (x/r, y/r, z/r). The radial momentum squared is now given by (u.p)^2. We convert this into the differential operator - h-bar^2 (u.grad)(u.grad). If this operator acts on a radial wave function, the result reduces to: - h-bar^2 d^2/dr^2.

Comparing these results, we see that there are indeed angular (non-radial) contributions to the kinetic energy. They are given by: - h-bar^2 (2/r)d/dr.

It is straightforward to evaluate the expectation values of these three operators in the Hydrogen ground state. The total kinetic energy is +1 (in Rydberg energy units), the radial kinetic energy is -1 and the angular kinetic energy is +2.

On average there is no motion at all, i.e., there is no systematic displacements. But there are "fluctuations" with non zero squares averaged. Classically speaking, it is like a Brownian motion in a limited space. But let us set aside a classical picture. Apart from momentum representation of the wave function, there is a simple proof that the electron may have unlimited velocity in the ground state.

Let us consider a scattering process in the first Born approximation. The projectile is heavy (proton, for example). From kinematic reasoning, a still electron cannot scatter a heavy proton to large angles, there is a limiting angle determined with the ratio $m_e/m_p$. However the scattering cross section is not zero for larger angles. Although small, the cross section is never zero. It is because the electron may have high instant velocity at the moment of scattering and this may push a heavy projectile back. The latter effect is described with the atomic form-factor $|F(\mathbf{q})|>0$ for any scattering angle.

• Very little of the material in this answer actually addresses the question. – user4552 Nov 9 '14 at 19:52
• Yes, it answers the question. On average there is no even a radial motion. That should be clear. And "fluctuations" are in all directions (non zero average squares). To scatter at $\pi/2$ from an electron, the proton should be pushed strongly by the electron. It is not possible if the electron does not move at high instant speed. – Vladimir Kalitvianski Nov 10 '14 at 23:41
• And I did not mentioned it, but the atomic nucleus also "moves around" the atomic center of mass (rotates) and creates a "positively charged cloud". It is not possible for the radial motion only. The nucleus does not stay at the atomic center! arxiv.org/abs/0806.2635 – Vladimir Kalitvianski Nov 10 '14 at 23:47
• @Kalivianski - The electron starts off with low velocity in a region far from the nucleus, gets accelerated by the Coulomb force, passes the nucleus with great velocity, and gets decelerated as it moves towards the far regions. Agreed! That is EXACTLY what my question was about. I referred to this (radial) motion as pendulum-like. Others called it a standing wave (= sum of two waves, representing ingoing and outgoing radial motion). – M. Wind Nov 12 '14 at 5:54
• @M.Wind: Yes, but it is not a pure pendulum motion in the radial direction. There are movements in the angular directions too since we deal with 3D "waves", not with particles. Everything happens there and it is described with a wave function giving the probability amplitude of this or that event. – Vladimir Kalitvianski Nov 12 '14 at 8:44

You have two questions which are not quite the same. One is

Is there only radial motion in the hydrogen ground state?

and the other is

If so, that would be a rather odd result. Let us say the electron is at position $(x, 0, 0)$. Then the kinetic energy would be the result of motion either away from the nucleus (direction $+x$) or towards the nucleus ($-x$), but not from motion perpendicular to the $x$-axis.

It's important to understand why these are different scenarios: an electron "at" the position $(x,0,0)$ can't be in the ground state. An electron the hydrogen ground state has the position distribution $$\psi_\text{ground}(r) = A e^{-r/a}$$ where CuriousKev is correct that the normalization $A = 1/\sqrt{\pi a^3}$ and the length scale $a = \hbar^2/me^2$ depend only on the electron$^1$ mass $m$, the unit charge $e$, and the unit of angular momentum $\hbar$. However, an electron at the position $(x,0,0)$ has the position distribution $$\psi_\text{localized}(r) = \delta^{(3)}(\vec r - (x,0,0)).$$ Since the hydrogen electronic orbitals form a complete orthonormal set of functions, you can use the techniques of Fourier analysis to write $\psi_\text{localized}$ as a superposition of all the ordinary $\psi_{n\ell m}$; the superposition will have contributions from states with all $n$ and $\ell$, but only with the projections $m=0$. (All of the angular momentum wavefunctions with projection $m\neq0$ vanish in the $x$-$y$ plane.)

This $\psi_\text{localized}$ is not a stationary state. The position of the electron will evolve as the different stationary components $\psi_n$ evolve with their different frequencies $E_n/\hbar$. Without having done the simulation I would expect the "most probable" position for the electron to move initially towards the nucleus, but for the probability density to spread out both on the $x$-axis and in the $y$-$z$ plane. So your scenario has both radial kinetic energy, due to the motion along the $x$-axis, and transverse kinetic energy, due to the spreading of the packet in the $y$-$z$-plane.

$^1$ Well, actually it's the reduced mass $m = m_e / (1 + m_e/m_p)$, but the difference is small.

• Since the hydrogen electronic orbitals form a complete orthonormal set of functions, you can use the techniques of Fourier analysis to write ψlocalized as a superposition of all the ordinary ψnℓm This sounds wrong to me. The ψnℓm are bound states, so any linear combination of them will have an energy with a negative expectation value. The energy of the delta-function state has infinite positive expectation value. – user4552 Nov 10 '14 at 23:34
• I would expect the "most probable" position for the electron to move initially towards the nucleus I don't think this is right either. Part of the Ehrenfest theorem is that $d\langle x\rangle/dt=\langle p\rangle /m$. But $\langle p\rangle$ diverges for the delta-function state. So your scenario has both radial kinetic energy, due to the motion along the x-axis, and transverse kinetic energy, due to the spreading of the packet in the y-z-plane. This is simpler to see on elementary grounds. The Heisenberg uncertainty principle guarantees infinite uncertainty in $p_x$, $p_y$, and $p_z$. – user4552 Nov 10 '14 at 23:42
• The only question I raised is whether the kinetic energy in the ground state is completely radial, or whether there are angular contributions. The purpose of the last paragraph in my initial post was to point out that without angular terms, the electron's motion is essentially one-dimensional (of course in a 3-dimensional standing wave version). – M. Wind Nov 11 '14 at 1:02
• @BenCrowell So uncharitable. Would you be happier with a Gaussian approximation to a delta function, so that $\left<p\right>$ is small enough that it's safe to neglect the unbound parts of the wavefunction? – rob Nov 11 '14 at 1:22
• @M.Wind But it's not one-dimensional, because the frequency of the back-and-forth motion along the $x$-axis is different from the frequency for spreading/contracting in the $y$-$z$ plane. After just a handful of oscillations along $x$ the probability distribution would be essentially spherical. It'd be a cute thing to model. – rob Nov 11 '14 at 1:26