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

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 how well your prediction is will depend on how much the two wavefunctions are overlapping.

But how well does a prediction do..?

Additionally, notice that: the probability $P(x_1|1s)$ is high if the wave function of $1s$ is very peaked in $x_1$ and if the probability of finding any particle in $x_1$ (the denominator) is very low, i.e. if the two states are very separate radially. 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.

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?

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..

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.

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 $\hat{x}$ 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 how well your prediction is will depend on how much the two wavefunctions are overlapping.

But how well does a prediction do..?

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 $x_1$ and if the probability of finding any particle in $x_1$ (the denominator) is very low, i.e. if the two states are very separate radially. 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.

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 $\hat{x}$ 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 how well your prediction is will depend on how much the two wavefunctions are overlapping.

But how well does a prediction do..?

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 $x_1$ and if the probability of finding any particle in $x_1$ (the denominator) is very low, i.e. if the two states are very separate radially. 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.

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 $\hat{x}$ 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 $1p$ states but how well your prediction is will depend on how much the two wavefunctions are overlapping.

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 $1p$ summed together appropriately.

If that value is high enough, you can assume if you measure $x_1$ then that particle is in $1s$ but you can not tell which of the two ones, as they are indistinguishable unless you measure spin.

Additionally, notice that: the probability $P(x_1|1s)$ is high if the wave function of $1s$ is very peaked in $x_1$ and if the probability of finding any particle in $x_1$ (the denominator) is very low, i.e. if the two states are very separate radially. 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.

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 $\hat{x}$ 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 how well your prediction is will depend on how much the two wavefunctions are overlapping.

But how well does a prediction do..?

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 $x_1$ and if the probability of finding any particle in $x_1$ (the denominator) is very low, i.e. if the two states are very separate radially. 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.