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Einsteins non interacting gas.


Sorcerer

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Consider a collection of N noninteracting particles, which can each be in one of two quantum states, \scriptstyle|0\rangle and \scriptstyle|1\rangle. If the two states are equal in energy, each different configuration is equally likely.

 

If we can tell which particle is which, there are 2^N different configurations, since each particle can be in \scriptstyle|0\rangle or \scriptstyle|1\rangle independently. In almost all of the configurations, about half the particles are in \scriptstyle|0\rangle and the other half in \scriptstyle|1\rangle. The balance is a statistical effect: the number of configurations is largest when the particles are divided equally.

 

If the particles are indistinguishable, however, there are only N+1 different configurations. If there are K particles in state \scriptstyle|1\rangle, there are N − K particles in state \scriptstyle|0\rangle. Whether any particular particle is in state \scriptstyle|0\rangle or in state \scriptstyle|1\rangle cannot be determined, so each value of K determines a unique quantum state for the whole system.

This confuses me, why is a particle being distinguishable from another a factor in determining the number of configurations? Surely the configurations exist independent of our ability to distinguish them.

 

As a thought experiment I visualised a room with a billiards table in it. The table is packed in a lattice with cue balls. Two people are in the room, Bill and Ben. Bill leaves and Ben, switches as many balls as he likes or none at all and then asks Bill to return. This process is repeated hundreds of times over the course of the day.

 

Bill could apply the quoted logic to the cue balls and Ben would know he was completely wrong.

 

Since for us, particles can move and "Ben" is always present, we're almost always "out of the room", why do we assume we are right?

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Swapping two identical cue balls doesn't change anything. The table is the same before and after. (If you think it isn't, you need to say how you think it is objectively [measurably] different.)

 

It is only if you swap a green and a red ball that there will be a difference.

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Surely being swapped changed the state of the table. There would be a permutation of cue ball movements, why is only the state before and after important?

 

Also, then why is the number of states N+1 , if we remove the energy level by your logic the number of states is 1 for any value of N.

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Surely being swapped changed the state of the table.

 

How has it changed? What tests could you do to see if it has changed?

 

 

There would be a permutation of cue ball movements, why is only the state before and after important?

 

Ummm... Because that is what "change" means? :)

 

 

Also, then why is the number of states N+1 , if we remove the energy level by your logic the number of states is 1 for any value of N.

 

Each particle can be in one of two states. So if there are 3 particles, then the possibilities are:

 

1. All in low energy state.

2. One particle in high energy state (it doesn't matter which because they are indistinguishable)

3. Two particles in high energy state (it doesn't matter ... )

4. All three particles in high energy state (ditto)

 

So, 3 particles = 4 states.

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You could observe it as it changes. Do these particles all occupy the same space?

 

My definition of change differs, change doesn't require a different static configuration, before and after, merely a movement between 2 points in time. Say you took the two cue balls played a game of 1 ball, then racked them as before while Bill was out of the room. He would think nothing changed. But that's only because he didn't observe the event, which occurs between 2 points in time.

 

I don't understand if there are 3 particles, let's represent them { ... } low energy and { ::: } high energy. How can they not be distinguished by their position in space?

 

{ ::. } {.::} and {:.:} all differ.

 

I understand bosons can all occupy the same space, even so, just because the change appeared the same, doesn't make it the same. Also doesn't uncertainty allow each independent probability of position. Why does our ignorance effect reality, just because we can't know something does it objectively mean it doesn't exist. In Bills reality did the cue ball really not move?

 

I noticed Einstein assumed non interacting gas, is this because interaction allows the particles to each be an observer?

Also, if they all occupy the same space. How in your example are 2 and 3 observably different. 1 particle would appear the same as 2 particles. Shouldn't N only have 3 possible states regardless of its value? Up down or both. How are identical particles counted, how could you know you haven't counted twice?

__________

 

OK I get it, only 2 states all in one place. There is no intermediate position, no transition to observe. But then how can there be any change, there's only 1 possible configuration. It can't change, also it doesn't exist, because it can't be observed, because it's non interacting.

Edited by Sorcerer
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You could observe it as it changes. Do these particles all occupy the same space?

 

My definition of change differs, change doesn't require a different static configuration, before and after, merely a movement between 2 points in time. Say you took the two cue balls played a game of 1 ball, then racked them as before while Bill was out of the room. He would think nothing changed. But that's only because he didn't observe the event, which occurs between 2 points in time.

 

I don't understand if there are 3 particles, let's represent them { ... } low energy and { ::: } high energy. How can they not be distinguished by their position in space?

 

{ ::. } {.::} and {:.:} all differ.

 

I understand bosons can all occupy the same space, even so, just because the change appeared the same, doesn't make it the same. Also doesn't uncertainty allow each independent probability of position. Why does our ignorance effect reality, just because we can't know something does it objectively mean it doesn't exist. In Bills reality did the cue ball really not move?

 

I noticed Einstein assumed non interacting gas, is this because interaction allows the particles to each be an observer?

Also, if they all occupy the same space. How in your example are 2 and 3 observably different. 1 particle would appear the same as 2 particles. Shouldn't N only have 3 possible states regardless of its value? Up down or both. How are identical particles counted, how could you know you haven't counted twice?

__________

 

OK I get it, only 2 states all in one place. There is no intermediate position, no transition to observe. But then how can there be any change, there's only 1 possible configuration. It can't change, also it doesn't exist, because it can't be observed, because it's non interacting.

 

You can't observe it as it changes, though. Not in quantum systems.

 

Einstein assumed a non-interacting gas, meaning the particles don't interact with each other. Interactions between them give rise to more complicated behaviors. There is no assumption that you can't observe the system.

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  • 2 weeks later...

 

You can't observe it as it changes, though. Not in quantum systems.

 

Einstein assumed a non-interacting gas, meaning the particles don't interact with each other. Interactions between them give rise to more complicated behaviors. There is no assumption that you can't observe the system.

To be indistinguishable all the particles would have to occupy the same space right? So bosons only?

 

If they didn't then shouldn't this example be correct?

 

 

I don't understand if there are 3 particles, let's represent them { ... } low energy and { ::: } high energy. How can they not be distinguished by their position in space?

 

{ ::. } {.::} and {:.:} all differ.

 

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So I'm confused as to why they're not different.

Each particle can be in one of two states. So if there are 3 particles, then the possibilities are:

 

1. All in low energy state.

2. One particle in high energy state (it doesn't matter which because they are indistinguishable)

3. Two particles in high energy state (it doesn't matter ... )

4. All three particles in high energy state (ditto)

 

So, 3 particles = 4 states.

But the particles permuations represented spatially are clearly distinguishable.

 

::. Mirrors .::

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So I'm confused as to why they're not different.

 

But the particles permuations represented spatially are clearly distinguishable.

 

::. Mirrors .::

 

 

You have identical particles. One is at point A, the other at point B. You turn your back, and I may or may not switch them. How can you tell if I did?

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You have identical particles. One is at point A, the other at point B. You turn your back, and I may or may not switch them. How can you tell if I did?

Yes I understand that, but the gas includes 2 energy states. If one was up the other down they wouldnt be identical right?

 

Is it, if I turn my back and you transfer the energy, how could I tell if you switched them or transferred the energy?

 

Or am I unable to measure individual particles energy, only the total system, there's something missing.

 

I could ask you

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Yes I understand that, but the gas includes 2 energy states. If one was up the other down they wouldnt be identical right?

 

Is it, if I turn my back and you transfer the energy, how could I tell if you switched them or transferred the energy?

 

Or am I unable to measure individual particles energy, only the total system, there's something missing.

 

I could ask you

 

Yes, adding a second energy state increases the number of configurations. Strange covered this earlier.

 

Measuring the individual particle energy gains you nothing, since the particles are indistinguishable.

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Yes, adding a second energy state increases the number of configurations. Strange covered this earlier.

 

Measuring the individual particle energy gains you nothing, since the particles are indistinguishable.

Huh? both particles have a different set of co-ordinates. I can almost grasp the bosonic form, but fermionic doesn't seem possible.

Does uncertainty prevent me from doing this?

Edited by Sorcerer
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Huh? both particles have a different set of co-ordinates. I can almost grasp the bosonic form, but fermionic doesn't seem possible.

Does uncertainty prevent me from doing this?

 

 

But how do you know which particle is at location 1 vs location 2? What is the measurable difference if they swapped positions?

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But how do you know which particle is at location 1 vs location 2? What is the measurable difference if they swapped positions?

If they had different energy levels. Why are you reducing it to 2 and removing the energy states, the example was given with 3 with 2 states.

Being able to determine their position in space and measure 1 of 2 energy levels makes them distinguishable, therefore only Bosons can be indistinguishable. Otherwise there are permutations formed by variations in energy level.

Each particle can be in one of two states. So if there are 3 particles, then the possibilities are:

 

1. All in low energy state.

2. One particle in high energy state (it doesn't matter which because they are indistinguishable)

3. Two particles in high energy state (it doesn't matter ... )

4. All three particles in high energy state (ditto)

 

So, 3 particles = 4 states.

1. ...

2. :.. .:. ..: <------spatially distinguishable, it does matter

3. .:: :.: ::.

4. :::

How do we observe a non-interacting gas?

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If they had different energy levels. Why are you reducing it to 2 and removing the energy states, the example was given with 3 with 2 states.

 

I gave an example with two, hoping it would make the situation simpler. The original problem said nothing about location. It's not a good quantum number.

 

I'm not removing the energy states. If there are two energy states, the energy state is what differentiates the particles. But you keep going back to coordinates, which has no bearing. Switching the coordinates does not allow you to identify the particles. Again, Strange covered this concept already.

 

Here, I'll quote it so you don't have to scroll back

 

 

Each particle can be in one of two states. So if there are 3 particles, then the possibilities are:

 

1. All in low energy state.

2. One particle in high energy state (it doesn't matter which because they are indistinguishable)

3. Two particles in high energy state (it doesn't matter ... )

4. All three particles in high energy state (ditto)

 

So, 3 particles = 4 states.

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*Facedesk*

3 positions with 2 energy states to choose from, because the order matters spatially, is the permutation 2x2x2=8.

 

I can distinguish between each energy state in each position.

Edited by Sorcerer
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*Facedesk*

3 positions with 2 energy states to choose from, because the order matters spatially is the permutation 2x2x2=8.

 

I can distinguish between each energy state in each position.

 

Where does position come into play in Einstein's derivation? This is a gas. There is no fixed position. Position is not a good quantum number. It's not being used to describe any of the particles. Energy is.

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Fermions must all be assigned a seperate spatial coordinate. If the uncertainty principle prevented me however from knowing both which state and which position though, the possibilities reduce.

 

If we use pool balls though, one black one white to show the 2 states =n. And assign them 3 seperate positions =r. Then it's the permutation nPr.

 

Bosons also aren't excluded from having seperate positions, but are allowed to have the same. So only in one special case is the number of ways reduced to 4. If we use only 3 of the 1D coordinates of a line on the table surface, this is when the balls are stacked vertically.

 

And I haven't even accounted for a change in coordinates, this is only when position swaps exactly.

 

This was useful, I understand his assumptions now. I'll read up further to see how they model reality.

 

Got it! NO INTERNAL DEGREES OF FREEDOM

Edited by Sorcerer
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