I Most Likely Position of an Electron

[bob012345]

TL;DR Summary

What is the most likely position of an electron in a Hydrogen atom in the 1s state if one considers the space broken into equal cartesian differential volumes. By position, I do not mean distance from the nucleus, I mean {x,y,z}position.

The peak of the radial probability density for the 1s state is at the Bohr radius $a_0$. But if we consider equal volumes as cartesian boxes, can one fairly say the electron has a higher probability of occupying a box near the nucleus than elsewhere because the peak of $\psi^* \psi$ is at zero?

 

[anuttarasammyak]

The peak is not at center nucleus where the probability is zero.

 

[BvU]

The peak is not at center nucleus where the probability is zero.

The OP asks for probability density, which does peak at zero.

And expectation values for $x, y, z$ do come out at zero.

$\$

 

[PeterDonis]

near the nucleus

This phrase means that you are using the radial probability density, not the "cartesian boxes" one. The radial probability density is the probability density as a function of how near the nucleus you are. In "cartesian boxes" terms, roughly speaking, the radial probability density combines the probability density "per cartesian box" with the number of boxes at a given radius (distance from the nucleus), which goes to zero at $r = 0$, and which gets larger as $r$ increases. The combination of these two effects tells you that the most likely distance from the nucleus to find the electron is $a_0$; that means that at distances nearer the nucleus, you are less likely to find the electron there, because the reduced number of boxes more than outweighs the increased value of $\psi^* \psi$ in each box.

 

[bob012345]

This phrase means that you are using the radial probability density, not the "cartesian boxes" one. The radial probability density is the probability density as a function of how near the nucleus you are. In "cartesian boxes" terms, roughly speaking, the radial probability density combines the probability density "per cartesian box" with the number of boxes at a given radius (distance from the nucleus), which goes to zero at $r = 0$, and which gets larger as $r$ increases. The combination of these two effects tells you that the most likely distance from the nucleus to find the electron is $a_0$; that means that at distances nearer the nucleus, you are less likely to find the electron there, because the reduced number of boxes more than outweighs the increased value of $\psi^* \psi$ in each box.

I agree, Thanks. This was an exercise in debate with someone who argued it does hang around the nucleus which I disagree with. Trying to come up with different arguments because they seem to think the radial probability density as usually constructed is misleading. They seem to think that because $\psi^*\psi$ peaks at zero it means the electron is most likely there.

 

[bob012345]

The OP asks for probability density, which does peak at zero.

And expectation values for $x, y, z$ do come out at zero.

$\$

Yes, but clearly to go from these to thinking the electron is mostly around {$0,0,0$} would be misleading.

 

[Vanadium 50]

First, using PF to argue with someone by proxy is bad form.

Second, the electron does spend measurable time in the nucleus. Is it "significant"? That's linguistics, not physics. It certainly shows up in spectra.

If P is the probability, dP/dr tells you something, but since the volume element is r² this may not be what you want. But more importantly, "most probable" does not mean "probabe". If I flip 1000 coins, the most probable outcome is 500 heads, but the probability of another outcome is above 95%.

And nuclei are small.

 

[bob012345]

First, using PF to argue with someone by proxy is bad form.

Sorry, not meant to mean an argument but about technical arguments people can argue.

Second, the electron does spend measurable time in the nucleus. Is it "significant"? That's linguistics, not physics. It certainly shows up in spectra.

If P is the probability, dP/dr tells you something, but since the volume element is r₂ this may not be what you want. But more importantly, "most probable" does not mean "probabe". If I flip 1000 coins, the most probable outcome is 500 heads, but the probability of another outcome is above 95%.

And nuclei are small.

I agree. I like your statement that "most probable" does not mean "probable". That's what I was trying to express. I like the coin reference also. I had thought of coming up with such an example. I was thinking of a circle where one throws pennies to cover the whole circle but with a bit higher rate of hitting nearer the center. There might be more pennies near the center on a per area basis but overall, there are a lot more pennies further away from the center than near the center.

 

[Vanadium 50]

One other complication - the hydrogen atom Schroedinger equation separates into R, θ and φ pieces. It does not separate into x,y and x pieces. In a very real sense you cannot talk at all about an electrons x, y and z distribution. It doesn't have one.

 

[bob012345]

One other complication - the hydrogen atom Schroedinger equation separates into R, θ and φ pieces. It does not separate into x,y and x pieces. In a very real sense you cannot talk at all about an electrons x, y and z distribution. It doesn't have one.

Non-separable in terms of cartesian coordinates does not mean you cannot describe it in terms of x, y and z. The radial wavefunction could be written as $exp(-\sqrt{x^2+y^2+z^2}/a_0)$. Higher wave functions can be expressed in cartesian coordinates as well since the spherical harmonics can be.

 

[Vanadium 50]

You can't even say "it is in a 1S state" in cartesian coordinates, If you constrain the electron to be somewhere in, say, x, you are then no longer in a 1S state.

The best you can do is to look at the expectation values of x, y and z from your 1S state. They are all zero, by symmetry.

 

[bob012345]

You can't even say "it is in a 1S state" in cartesian coordinates, If you constrain the electron to be somewhere in, say, x, you are then no longer in a 1S state.

The best you can do is to look at the expectation values of x, y and z from your 1S state. They are all zero, by symmetry.

But I do believe one can ask what is the probability a 1s electron is in a box $dxdydz$ centered around the point {$x,y,z$} rather than in a spherical shell at radius r. Then you can compare that to other boxes. We are not constraining the electron, just computing probabilities.

 

[Vanadium 50]

X and y (and z) do not commute, just like p and x do not commute. While you can always evaluate an integral (at least numerically), you cannot interpret it as a probability. You can't even interpret it as a well-defined state.

When you say "1S state" you are saying it is in an eigenstate of n, l and m. These states do not commute with x, y and z. You can say you start with an atom in the 1S state, and calculate the probability you will find it in a volume dV. However, if you find it, you are no longer in the 1S state. If you don't find it, you are also not in the 1S state.

This may be unsatisfactory - but it's as good as you will get in QM.

 

[PeterDonis]

the probability you will find it in a volume dV

If x, y, and z do not commute, is there even an operator that corresponds to this?

 

[Vanadium 50]

What is calculable is \int \Psi^* \Psi dV over some restricted volume. Here the interval is taken over a restricted volume. That's a number. Interpreting it as a probability is, for the reasons I discussed, problematic.

You can also calculate \int \Psi^* x \Psi dV over all space. You will get zero. (Likewise for y and z). But I don't think this tells you where the electron is, other than "somewhere near the nucleus".

I suppose you can calculate \int \Psi^* x^2 \Psi dV again taken over all space. I have a hard time interpreting this as a position, or even a coordinate. The units are not right, for starters.

I don't think one can do any better than what the OP doesn't like, project the r axis onto x, y and z.

 

[renormalize]

X and y (and z) do not commute, just like p and x do not commute.

Huh?

(Merzbacher, Quantum Mechanics Second Edition, pg. 339)

 

[bob012345]

When you say "1S state" you are saying it is in an eigenstate of n, l and m. These states do not commute with x, y and z. You can say you start with an atom in the 1S state, and calculate the probability you will find it in a volume dV. However, if you find it, you are no longer in the 1S state. If you don't find it, you are also not in the 1S state.

Of course it is in an eigenstate of n,l and m. We can calculate probabilities or how likely the electron will be found in a spherical shell around r. We can also calculate how likely it will be found in some cartesian box. That doesn't change the state. I don't think QM calculations are dependent on only certain coordinate systems.

 

[vanhees71]

One other complication - the hydrogen atom Schroedinger equation separates into R, θ and φ pieces. It does not separate into x,y and x pieces. In a very real sense you cannot talk at all about an electrons x, y and z distribution. It doesn't have one.

Why not? Of course
$$|\psi(x,y,z)|^2 \equiv |\psi(r,\vartheta,\varphi)|^2$$
is the probability density function for the position $(x,y,z)$. Why should this distribution be separable in $(x,y,z)$. For the hydrogon energy-eigenfunctions that's of course not the case.

 

[vanhees71]

If x, y, and z do not commute, is there even an operator that corresponds to this?

The position components $(x,y,z)$ of course always commute. It's part of the basic realizations of the Galileo group (or rather it's quantum extension, but that doesn't matter so much here) in terms of the Heisenberg algebra. Writing $(x_1,x_2,x_3)$ it's given by
$$[\hat{x}_j,\hat{x}_k]=0, \quad [\hat{p}_j,\hat{x}_k]=0, \quad [\hat{x}_j,\hat{p}_k]=\mathrm{i} \delta_{jk} \hat{1}.$$
The hydrogen energy eigenfunctions are of course no eigenstates of $\hat{\vec{x}}$, because $\hat{\vec{x}}$ doesn't commute with the Hamiltonian, of which you want eigenstates. Due to rotational symmetry to get unique energy eigenstates you need to determine the common eigenstates of a complete set of compatible observables, which (from rotational symmetry again) can be chosen as $(H,\vec{L}^2,L_3)$, leading to the here discussed energy eigenfunctions.

The probability density for position is for any wave function, i.e., also for these energy-eigenfunctions
$$P(\vec{x})=|\psi(\vec{x})|^2.$$
In this case these eigenfunctions take the form
$$u_{n,l,m}(r,\vartheta,\varphi)=R_{nl}(r') \text{Y}_{lm}(\vartheta,\varphi).$$
Of course, the probability distribution for the distance, $r$ of the electron from the proton is given by
$$\tilde{P}(r)=\int_{\mathbb{R}^3} \mathrm{d}^3 x P(\vec{x}) \delta(|\vec{x}|-r),$$
as for any probability distribution following from "coarse graining". You get
$$\tilde{P}(r)=\int_{0}^{\infty} \mathrm{d} r' \int_{S_2} \mathrm{d}^2 \Omega r^{\prime 2} \delta(r'-r) |R_n(r)|^2 |Y_{lm}(\vartheta,\varphi)|^2 = r^2 |R_{nl}(r)|^2.$$

 

[vanhees71]

Of course it is in an eigenstate of n,l and m. We can calculate probabilities or how likely the electron will be found in a spherical shell around r. We can also calculate how likely it will be found in some cartesian box. That doesn't change the state. I don't think QM calculations are dependent on only certain coordinate systems.

That's of course right. QM is invariant under arbitrary coordinate transformations. In its form as wave mechanics (i.e., the position representation) formulated entirely in terms of the standard 3D Euclidean tensor analysis. This is no surprise since QM by contruction is compatible with Newtonian spacetime symmetries.

 

[Vanadium 50]

My point is that they don't commute with the variables you have already specified. Maybe that was unclear.

And while you can calculate what the probability was, you cant calculate what it is. "The electron is in a 1S state" and "the electron is over here" (or even "not over here") are incompatible.

 

[bob012345]

My point is that they don't commute with the variables you have already specified. Maybe that was unclear.

And while you can calculate what the probability was, you cant calculate what it is. "The electron is in a 1S state" and "the electron is over here" (or even "not over here") are incompatible.

What do you mean by 'was' vs. 'is' in this context?

 

[renormalize]

I am confused about what's being debated in this thread: calculations or measurements? Consider a hydrogen atom in a 1s state:

• You can calculate the probability of finding the electron in a given cartesian box by computing the integral of the 1s probability-density over the volume of the box, numerically if necessary.
• With a suitable apparatus, you can measure the probability of finding the electron in a given cartesian box by repeated measurements using an ensemble of such atoms, but immediately after each such measurement the atom observed will no longer be in a 1s state.

Is @bob012345 talking about the former and @Vanadium 50 the latter?

 

[Vanadium 50]

It's clear that I am making it more confusing, not less, so I will shut up now.

 

[PeterDonis]

What do you mean by 'was' vs. 'is' in this context?

He means this:

With a suitable apparatus, you can measure the probability of finding the electron in a given cartesian box by repeated measurements using an ensemble of such atoms, but immediately after each such measurement the atom observed will no longer be in a 1s state.

 

[PeterDonis]

Is @bob012345 talking about the former and @Vanadium 50 the latter?

@Vanadium 50 is talking about the latter. @bob012345 may be talking about the former but I'm not entirely sure; he might want to clarify.

 

[bob012345]

@Vanadium 50 is talking about the latter. @bob012345 may be talking about the former but I'm not entirely sure; he might want to clarify.

I was only talking about computations and not measurements.

 

[bob012345]

It's clear that I am making it more confusing, not less, so I will shut up now.

I want you to know I appreciate your input and perspective as well as everyone else who participates. Thanks.

 

[PeterDonis]

I was only talking about computations and not measurements.

Ok. Then the key caveat is that you have to be careful attributing physical significance to computations.

 

[bob012345]

Ok. Then the key caveat is that you have to be careful attributing physical significance to computations.

Always! Thanks to all for input and perspectives.

 

[vanhees71]

My point is that they don't commute with the variables you have already specified. Maybe that was unclear.

And while you can calculate what the probability was, you cant calculate what it is. "The electron is in a 1S state" and "the electron is over here" (or even "not over here") are incompatible.

Sigh. Of course you can, in principle, ask about the position of an electron in the 1S state of a hydrogen atom. Nothing forbids to measure observables of a system only because the system is not in an eigenstate of the measured observable. It's only that due to the preparation in this state the position is indetermined, and you'll find a probability density when repeating the same measurement on many electrons prepared in this 1S state.