# Do atomic orbitals “pulse” in time?

I understand that atomic orbitals are solutions to the time-independent Schrödinger equation, and that they are are analogous to standing waves ("stationary states"). However, even a standing wave has motion, in the sense that (at points other than the nodes) the amplitude varies with time. My question is, do atomic orbitals or spherical harmonics, as standing waves in 3D space, also have such motion? Intuitively, do they "pulse" or "breathe"?

(More precisely, my question is in regard to the behavior of an isosurface of an orbital, the often-pictured shell containing some arbitrary probability, as atomic orbitals themselves have infinite spatial extent.)

## 2 Answers

Short answer: yes, but only the phase factor has the time dependence. The spatial profile is constant in time because the eigenstates of the Hamiltonian are Stationary states.

Maths:

The time dependent Schroedinger equation looks like this:

$$i\hbar \frac{\partial \Psi}{\partial t} = H \Psi = \left ( -\frac{\hbar^2 }{2 m}\frac{\partial^2}{\partial x^2} + V(x,t) \right ) \Psi(x,t) ,$$

you attempt a solution via separation of variables: $$\Psi(x,t) = \psi(x) T(t)$$, plug it in.

If the potential $$V$$ is time independent such that $$V(x,t) = V(x)$$, then the equation above splits into two independent equations:

$$\left ( -\frac{\hbar^2 }{2 m}\frac{\mathrm{d}^2}{\mathrm{d} x^2} + V(x) \right ) \psi(x) = E \psi(x), \quad \text{Time independent Schroedinger equation}$$

and:

$$i\hbar\frac{\mathrm{d} T}{\mathrm{d} t} = ET \implies T(t) \propto e^{-iEt/\hbar} = e^{-i\omega t}.$$

With $$E$$ a constant identified with energy.

Hence the full solution will be $$\Psi(x,t) = \psi(x) e^{-i\omega t}$$, with the time dependence only in the phase factor.

Any physical observable depends on $$|\Psi|^2 \propto |\psi(x)|^2$$ so the time dependence of the phase factor does not affect any of the physics.

• "so the time dependence of the phase factor does not affect any of the physics." I understand that but then what is the phase factor 'good for'? Thanks for the nice answer. +1 from me. – Gert May 11 at 15:27
• Well the phase is irrelevant only if you are in a definite state. IF you are in a superposition of states $E_1$ and $E_2$, then the observables will have a time dependence $(E_1 -E_2)t/\hbar$. This affects the dyanmics. And will also affect the chemical bonds, see for instance here: chemistry.stackexchange.com/questions/35212/… – SuperCiocia May 11 at 15:33
• Ok, that's very clear. – Gert May 11 at 17:15
• I think in $L\ne 0$ states, the phase factors lead to probability current "orbiting" the nucleus, though nothing is changing. – JEB May 11 at 19:35
• No any stationary state has no probability current, no matter its quantum numbers. You need a superposition in order to have a non zero probability current – SuperCiocia May 11 at 19:54

No, no observable quantity changes over time for a stationary state. The analogy between stationary states in quantum mechanics and standing waves isn't very close.

Whether or not a stationary quantum state has any time dependence at all depends on which mathematical formalism you're using. In the "pure state" or "state vector" formalism, which I suspect you're using at this stage in your physics education, the state vector does formally oscillate in time through its complex phase, which is vaguely similar to a standing wave (although unlike with a standing wave, the actual frequency of oscillation is completely unmeasurable). In the "density matrix" formalism, the state operator is completely time-independent and doesn't have any oscillating phases, which corresponds to the fact that nothing physically measurable is changing over time.

In my opinion, the state-vector formalism is mathematically simpler but the density-matrix formalism is conceptually simpler (you don't have to "remember to forget about the phase factor"), so which one is better depends on the use case and personal taste.

That's for pure states; for mixed states, the density-matrix formalism is both mathematically and conceptually simpler, and the state-vector formalism is only useful for rather esoteric applications. (At this stage in your education, you probably haven't learned about pure and mixed states yet. Suffice it to say that all the states you've probably studied so far have been pure states, and there's a slightly more general notion called a "mixed state" that you may eventually learn about.)