1
$\begingroup$

Let's say we have an atom with three electrons in the lowest energy states. So two electrons are at energy level 1 and one at energy level 2. Now we measure the positions of those electrons, and found three positions $x_1,x_2,x_3$.

Can we know whether the electron found at a position, say $x_1$, was an electron with energy level 1 or 2?

I know about those antisymmetric state things, no need to explain from there.

Somewhat related:

Making indistinguishable particles distinguishable?

The electrons in atom are not far away so I think we may not be able to tell if a particle found was in lower or higher energy state.

Edit: Allow me to clarify. I saw someone wrote "the electron in 2s orbital might sometimes be closer to the nucleus that the electron in 1s orbital". How true is this statement? Can we even know if an electron was in 2s orbital and not 1s when we find that electron somewhere in space?

Edit 2: And yes, I know the whole thing about wave function and radial position probability. Don't start explaining this part. My real question is, upon finding an electron in this three-electron system, do we know if the electron we found was in 1s state or 2s state before we find it?

$\endgroup$

3 Answers 3

1
$\begingroup$

Well, short answer: no. You can't tell for sure. ... but that has nothing to do with indistinguishable particles!

If you have two wavefunctions $\psi(x, ...)$ and $\phi(x, ...)$ where $..$ indicates whatever quantum number you want (energy, spin, etc) and you only measure $\hat{x}$ (i.e. the position) and get $x_1$ and $x_2$ then how can you tell which one of the two wave functions gave which result?

You could do it if the two prob. distr. functions do not overlap substantially. Suppose the probability of finding $\psi$ in (approx) $x_1$ is 0.01 and that of finding $\phi$ in $x_1$ is $0.9$ then you might be tempted to assign $x_1$ to $\phi$ but.. it's still just a guess!

But never in this reasoning have we exploited the fact that the particles are bosons or fermions! Knowing that they are fermions is only telling us that $$\psi(x, ...)!=\phi(x, ... )$$ so that the two probabilities must be different when considering all quantum numbers!

In your example, the two particles in the $1s$ orbital have exactly the same probability distribution in space so you have no way of distinguishing them with a measurements that does not involve spin. You could distinguish $1s$ and $2s$ states but, unless you measure energy, how well your prediction is will depend on how much the two spatial wavefunctions are overlapping.

But what you can do is compute how well does a prediction do and what factors influence it..

Using Bayes theorem to know more...

We can understand better what factors influence your confidence in assigning a result to a given state using Bayes theorem [important theorem: google it if you don't know it!]

The best you can do is estimate the probability that the particle you find in e.g. $x_1$ is in a given state, i.e. for example, considering the state $1s$:

$$P(1s|x_1)$$ is the probability that the particle is in $1s$ given the fact that it was found in $x_1$

we can compute this using Bayes theorem as $$P(1s|x_1) = P(x_1|1s)P(1s)/P(x_1)$$

where

$$P(x_1|1s)=|\psi(x, 1s)|^2\Delta x$$ is the probability of finding the particle in $[x_1, x_1+\Delta x]$ if the particle is actually in $1s$, and you can find it easily using the position wave function $\psi(x, 1s)$ of a $1s$ particle and taking its squared value.

$$P(1s)=2/3$$ is the probability of having a particle in $1s$: two out f three particles are in $1s$ so that is $2/3$

$$P(x_1)$$ is the probability of finding any particle in $x_1$. You can find it from the individual wave functions of $1s$ and $2s$ summed together appropriately.

If that value is high enough (say, 0.82), you can assume if you measure $x_1$ then that particle is in $1s$ (with 82 % confidence) but you can not tell for sure.

Additionally, notice that:

  • the probability $P(x_1|1s)$ is high if the wave function of $1s$ is very peaked in $x_1$ (numerator), if you have a lot of particles in $1s$ (second part of the numerator) and/or if the probability of finding any particle in $x_1$ (the denominator) is very low, i.e. your prediction is good if the two states are very separate in space. Like, if $x_1$ is very close to the nucleus, you can probably assume it's not an $20 s$ particle. But not for sure.
  • your measurement, having 2 particles in $1s$ and only one in $2s$ is already biased towards $1s$ so that a random guess saying "this particle is in $1s$" is already pretty good, getting it right approx. 2/3 of the times ;)
  • if the particles were to be bosons, this prediction would not change.

However, if you do the same thing and look at relevant quantum measurements (e.g. energy or spin) then your prediction can be 100 % efficient. So the answer to your question is not general, it depends on whether you actually measure the position or something else.

$\endgroup$
0
$\begingroup$

I will assume that are familiar with the bra-ket notation (if not, I can edit the answer).

The atomic levels are the eigenstates of the Hamiltonian. When you say that a particle is the level 1, what you mean is that the particle is in a state of the eigenbasis, denoted by $|1\rangle$. The same can be applied to the particle in the level 2, $|2\rangle$, and so forth.

In this case, as you want to measure the position of the particle. The projection into the position basis is given by $$P_1(x) = \langle x| 1\rangle,$$ and should be interpreted as: $P_1(x)$ is the probability of finding an electron, which is at the level 1, at the position $x$. We can do the same for $P_2(x)$.

From this perspective yours question is: If I do not know in which level the particle is, but I found it at the position $x_1$, can I say what was the energy level before the measurement?

The answer is that, in general, you cannot. In particular, you can only do it when there is a common eigenbasis between the position operator and the Hamiltonian (which means they commute).

You can work out the probabilities though. If the particle was measured at $x_1$ and you know that it could only have been at level 1 or 2, then the probality that it was at level 1 is then $$\dfrac{P_1(x_1)}{P_1(x_1) + P_2(x_1)}$$ and the probability that it was level 2 then $$\dfrac{P_2(x_1)}{P_1(x_1) + P_2(x_1)}$$.

This is the question that you are actually making. However, you are missing an important point in your question. If the electrons are in different energy levels, then they are distinguishable. What characterizes an electron is the quantum numbers that described them. In this case, as they are in different energy levels they are distinguishable.

Then, why was my answer negative? Because you are using the wrong operator. What you want to actually measure is the energy. In this case, it would be clear that they are distinguishable.

$\endgroup$
-2
$\begingroup$

Well, the electron with more energy level would be the one closer to the nucleus, right?

Because E = -13.6 (z/n)^2

Where n is the shell number, so more the value of n, more the distance and less the energy.

So you should be able to find out where the electron is placed using its energy value.

$\endgroup$
1
  • $\begingroup$ Different shells have different mean radii, but they are all still continuous probability distributions. This is the essence of what the OP is asking about. $\endgroup$
    – Rococo
    Dec 26, 2020 at 18:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.