Does the Wavefunction of an Electron Collapse When Captured by a Proton?

[syfry]

TL;DR Summary

If the wavefunction of a free electron is in a superposition, will it collapse when the electron binds to a proton to create a hydrogen atom?

And then, if the wavefunction does collapse as the electron binds to the proton, is that collapse temporary, or will the wavefunction remain collapsed until the electron escapes and is free again?

Or, will the wavefunction of a freshly bound electron eventually return to a max superposition while still bound?

Searches online led to only general info on wavefunction and not about the above specific scenario with bound electrons.

 

[PeterDonis]

If the wavefunction of a free electron is in a superposition

A superposition of what? "Superposition" is basis dependent.

will it collapse when the electron binds to a proton to create a hydrogen atom?

Collapse to what? Obviously the electron's quantum state will change, but why would you view this as a "collapse" to anything?

The rest of your post just extends these same confusions.

 

[Haborix]

Following on from Peter, I would suggest setting aside all talk of collapse and try to first understand the time evolution of the wavefunction. The time evolution of the wavefunction is already a very challenging problem and will quickly expand to issues of photons and QFT.

 

[PeroK]

Or, will the wavefunction of a freshly bound electron ...

"Freshly bound electrons" sounds like something you might find at a physicists' market!

 

[syfry]

A superposition of what? "Superposition" is basis dependent.

For example, the reply by this person at stack exchange.

Sounds like they're saying a wavefunction is in a superposition, and I don't know nearly enough to properly gauge the accuracy of my interpretation or of their reply.

Collapse to what? Obviously the electron's quantum state will change, but why would you view this as a "collapse" to anything?

Collapse from a distribution pattern of where the electron is likely to be found, to a specific location when it interacts.

My question is from wondering if the wavefunction is collapsed because the bound electron is continually interacting with the proton electromagnetically. So wouldn't that interaction collapse the wavefunction, and keep it collapsed?

"Freshly bound electrons" sounds like something you might find at a physicists' market!

To rephrase: very recently bound!

 

[PeroK]

For example, the reply by this person at stack exchange.

Sounds like they're saying a wavefunction is in a superposition, and I don't know nearly enough to properly gauge the accuracy of my interpretation or of their reply.

I didn't find that thread very helpful. Let's take the ground state of the hydrogen atom. That is an energy eigenstate - and can be described by a single wavefunction ($\psi_1$, say). But, that wavefunction is a superposition of position eigenstates. So, the atom has a definite energy, but the electron does not have a definite position.

Like all quantum states, it can be described as a single wavefunction or a superposition of wavefunctions. From that point of view, wave function collapse is a bit of a misnomer. It's actually that the wavefunction becomes a specifc eigenstate of the relevant observable.

Collapse from a distribution pattern of where the electron is likely to be found, to a specific location when it interacts.

This is just a jumble of words.

My question is from wondering if the wavefunction is collapsed because the bound electron is continually interacting with the proton electromagnetically. So wouldn't that interaction collapse the wavefunction, and keep it collapsed?

It's nothing like that.

 

[PeterDonis]

Sounds like they're saying a wavefunction is in a superposition

"a wavefunction is in a superposition", with no other information given, is meaningless; it tells you nothing.

The Stack Exchange thread correctly notes that, since QM is linear, you can add any two wave functions and get another wave function. The most general meaning of "superposition" is simply that, and that is more or less how the term is being used there.

However, that usage is so general that it conveys no useful information. To have any useful information, you need to be more specific. First you need to have some specific wave function in mind. For example, the wave function of a free electron. Typically wave functions of free particles are given as wave packets. A wave packet is basically a Gaussian in the position representation and a similar Gaussian in the momentum representation. That means it can be said to be in a superposition of many different position eigenstates, and to be in a superposition of many different momentum eigenstates. For a free electron, this also means it will be in a superposition of different energy eigenstates.

If we now suppose this electron is captured by a proton and forms a hydrogen atom in its ground state, the electron's wave function is now a different one, the 1s state. This state is also a superposition of many different position eigenstates, and of many different momentum eigenstates, but it's a different superposition than the free electron wave function. Also, unlike the free electron wave function, the 1s wave function for the electron is an eigenstate of energy; that is, it is not a superposition of different energy eigenstates.

In other words, if we adopt a more useful meaning for the term "superposition", we find that whether or not a wave function is a "superposition" depends both on the wave function itself, and on the basis of eigenstates we choose, which in turn will depend on what we intend to measure. If we want to measure the electron's position, then we represent the wave function in the basis of position eigenstates, and the wave function then tells us the probabilities of different possible positions. If we want to measure the electron's momentum, we use the basis of momentum eigenstates and we get the probabilities of different possible momenta. And similarly for energy. So it makes no sense to just say a wave function is a "superposition": you have to specify what observable you are using as your basis of eigenstates, and different choices of observable can give different answers.

 

[PeterDonis]

wouldn't that interaction collapse the wavefunction

No.

 

[PeterDonis]

Collapse from a distribution pattern of where the electron is likely to be found, to a specific location when it interacts.

You might be groping here towards using the position representation, i.e., representing the wave function in terms of position eigenstates. However, as my previous post will show you, both the free electron wave function and the 1s state wave function for an electron bound in a hydrogen atom are superpositions in the position representation. For that process to be a position measurement, it would have to result in the electron being in a position eigenstate. But it doesn't.

 

[anuttarasammyak]

@syfry An initial free or unbounded state evolves in time to a superposition state of bounded and unbounded states according to Hamiltonian of the system. Later when you obesrve it a bounded ( or unbonded ) state, it means that the superposition state collapsed.

 

[Nugatory]

@syfry in many other threads on this forum I have recommended Giancarlo Ghirardi's book "Sneaking a look at God's cards" as one of the very few B-level explanations of quantum mechanics - it's no substitute for a real textbook but it is a good alternative if you lack the necessary mathematical background for that.

 

[vanhees71]

You might be groping here towards using the position representation, i.e., representing the wave function in terms of position eigenstates. However, as my previous post will show you, both the free electron wave function and the 1s state wave function for an electron bound in a hydrogen atom are superpositions in the position representation. For that process to be a position measurement, it would have to result in the electron being in a position eigenstate. But it doesn't.

Also one should emphasize that position "eigenvectors" are not true Hilbert-space vectors. They are rather distributions and thus cannot represent a physically realizable state. That's why it can be "normalized" only to a $\delta$-distribution,
$$\langle \vec{x}_1|\vec{x}_2 \rangle=\delta(x_1-x_2).$$
A physically realizable state is is always given by something like
$$|\psi \rightarrow =\int_{\mathbb{R}} \mathrm{d} x \psi(x) |x \rangle \langle x|,$$
where $\psi(x)$ is an $\mathrm{L}^2(\mathbb{R}^2)$ function, i.e., it can be properly normalized
$$\langle \psi|\rangle = \int_{\mathbb{R}} \mathrm{d} x |\psi(x)|^2=1.$$
To be very precise, the pure state is then represented by the statistical operator,
$$\hat{\rho}=|\psi \rangle \langle \psi|.$$

 

[dx]

An interesting historical fact related this this conversation - the ratio of magnetization over angular momentum of a metallic rod is equal to e/m_e, and this is the value found by two experimenters sometime after 1915 (E. Beck, Arvdisson). Einstein and de Haas predicted the classical value to be e/2m_e.

Pauli wrote to Lande in 1923 - "The energy values for strong fields can be interpreted [only] by the assumption that the magnetic moment of the inner system [i.e. the atomic core] assumes twice the value it would have"

This factor two anomaly also shows up in the description of the anomalous Zeeman effect

 

[vanhees71]

This story has even another amusing twist: de Haas indeed had found indications for another gyro-factor than 1. Indeed they had two measurements, one giving $g=1.46$, the other $g=1.02$. They only published the 2nd result because it confirmed the classical theory for the magnetic moment, i.e., Amperian molecular currents as the cause of the magnetic moment and classical electrodynamics. One problem of the experiment was that they didn't meausure the magnetic field but used theoretical values. See A. Pais, Subtle is the Lord about more details on the history.

Indeed the Lande factor of nearly 2 of the electron was a big puzzle in the early 20th century, also in connection with the Zeeman effect and the Stern-Gerlach experiment. It's not clear to me, how Stern and Gerlach's results of a split of their Ag-atom beam into only 2 can be really satisfactorily explained within the old Bohr-Sommerfeld theory used at the time to explaine "Richtungsquantelung" ("quantization of direction") since there should also be a third un-deflected line due to the value $m=0$ which must occur when using only orbital angular momentum (that's also true in modern QT of course).

All this finally lead to the discovery of the spin of 1/2, which is indeed generically quantum as is the Lande factor close to 2 for an elementary Dirac fermion (a result of the minimal-coupling assumption of gauge theory applied to QED). The deviations of $g$ from 2 is of course also another great discovery, i.e., one of the most accurate confirmations of QED by experiment and with still some quibbles left (more for the muon than for the electron, however) due to the trouble with strong-interaction radiative effects, which are hard to calculate accurately, although more and more lattice-QCD calculations seem to indicate that there's after all no discrepancy to the expectation of the standard model.