What is the difference between Statistical Mechanics and Quantum Mechanics [closed] The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What is the difference between thermodynamics and statistical mechanics?Relation between statistical mechanics and quantum field theoryWhat are the key properties of and differences between classical and quantum statistical mechanics?Statistical mechanics: What is a “microscopic realization” of a system?What is the difference between classical thermodynamics and statistical mechanics?Statistical Mechanics deals with the same systems that Thermodynamics does?Is kinetic theory part of statistical mechanics?Density matrix in Quantum Statistical MechanicsDensity matrix in quantum computation and quantum statistical mechanicsIn the context of statistical mechanics, how should I visualize a quantum state?
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What is the difference between Statistical Mechanics and Quantum Mechanics [closed]
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)What is the difference between thermodynamics and statistical mechanics?Relation between statistical mechanics and quantum field theoryWhat are the key properties of and differences between classical and quantum statistical mechanics?Statistical mechanics: What is a “microscopic realization” of a system?What is the difference between classical thermodynamics and statistical mechanics?Statistical Mechanics deals with the same systems that Thermodynamics does?Is kinetic theory part of statistical mechanics?Density matrix in Quantum Statistical MechanicsDensity matrix in quantum computation and quantum statistical mechanicsIn the context of statistical mechanics, how should I visualize a quantum state?
$begingroup$
What is the difference between Statistical and Quantum Mechanics?
In both we try to study the property of small particles using probability and hence apply to macroscopic systems.
quantum-mechanics statistical-mechanics
asked Mar 31 at 9:59
Sawan KumawatSawan Kumawat
465
$endgroup$
closed as too broad by By Symmetry, my2cts, Jon Custer, stafusa, valerio Apr 3 at 7:35
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
What is the difference between Statistical and Quantum Mechanics?
In both we try to study the property of small particles using probability and hence apply to macroscopic systems.
quantum-mechanics statistical-mechanics
asked Mar 31 at 9:59
Sawan KumawatSawan Kumawat
465
$endgroup$
closed as too broad by By Symmetry, my2cts, Jon Custer, stafusa, valerio Apr 3 at 7:35
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
What is the difference between Statistical and Quantum Mechanics?
In both we try to study the property of small particles using probability and hence apply to macroscopic systems.
quantum-mechanics statistical-mechanics
asked Mar 31 at 9:59
Sawan KumawatSawan Kumawat
465
$endgroup$
What is the difference between Statistical and Quantum Mechanics?
In both we try to study the property of small particles using probability and hence apply to macroscopic systems.
quantum-mechanics statistical-mechanics
quantum-mechanics statistical-mechanics
asked Mar 31 at 9:59
Sawan KumawatSawan Kumawat
465
asked Mar 31 at 9:59
Sawan KumawatSawan Kumawat
465
asked Mar 31 at 9:59
Sawan KumawatSawan Kumawat
465
asked Mar 31 at 9:59
Sawan KumawatSawan Kumawat
465
asked Mar 31 at 9:59
Sawan KumawatSawan Kumawat
465
465
closed as too broad by By Symmetry, my2cts, Jon Custer, stafusa, valerio Apr 3 at 7:35
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by By Symmetry, my2cts, Jon Custer, stafusa, valerio Apr 3 at 7:35
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
In statistical mechanics the system at any time is in a definite microstate (e.g. positions and velocities of all the particles in a gas), yet we don't know what this state is. Instead, we define certain global properties of the system that are defined on longer time scales (like total energy, entropy, temperature, volume) that are useful in many processes and try to predict them (in equilibrium) from the microscopic degrees of freedom.
In quantum mechanics, on the other hand, there are many options for what we mean by "states". The most intuitive definition is to define them in terms of things we can measure, like positions or velocities of particles, etc. However, the state is actually a wave in the space of these states and any given particle can actually spread out in state space and occupy many of these states simultaneously, with a different "amplitude" $psi$, just as a wave can spread out over space with a different amplitude at any point. (There are also restrictions on these wave functions, such as the fact that it must spread both in position $Delta x$ and in momentum $Delta p$ such that $Delta x Delta p geq hbar /2$, and similarly for other variables.) But basically the system can spread over what we would call "measurable" states.
The probabilities come in in Quantum mechanics, for example, when you try to measure the position of a particle that is spread over many different positions. This is where quantum mechanics gets confusing and leads to endless discussions about reality, but in short, the wave function "collapses" and you only measure one position, with probability $|psi(x)|^2$.
answered Mar 31 at 12:14
Eric David KramerEric David Kramer
1,036311
$endgroup$
$begingroup$
Shouldn't the Heisenberg uncertainty be $Delta x Delta p gte hbar / 2$?
$endgroup$
– michi7x7
Mar 31 at 14:50
$begingroup$
Sorry, typo. Yes, of course.
$endgroup$
– Eric David Kramer
Mar 31 at 15:06
add a comment |
$begingroup$
Since you are specifically focusing on the probability aspect, this is what I will talk about. Otherwise the question and answer will be way too broad.
In classical statistical mechanics probabilities arise due to limited knowledge of the system. This is usually due to the fact that our systems are made up of so many particles that it would be impossible to keep track of everything. Therefore, we use statistical methods to describe the system and use those statistics to determine macroscopic properties of the system.
In terms of limited knowledge, the typical example given is a fair coin toss. We say we have a $0.5$ probability of getting heads and a $0.5$ probability of getting tails, but really this is just due to our limited knowledge of the system. If we knew the exact initial conditions of the coin flip, the interaction of the coin with the air, how the coin is caught/landed on the floor, etc. then we wouldn't need probability. We could know with exactly certainty what the result of the coin flip would be.
The same is true for statistical mechanics. If we could know the position and momentum of every particle, how each particle interacts with each other particle, external effects, etc. then we wouldn't need statistical mechanics. We would know exactly how the entire system would behave and evolve over time. You'll notice that this and the previous example are very unreasonable though, hence we use probabilities.
And then we have quantum mechanics. The difference here is that we can know everything there is to know about the system, yet the result of a measurement of that system will still not have a predicable outcome. All we can predict is the probability of a certain outcome. Probability seems to be an inherent property of QM that cannot be taken away like in the statistical mechanics examples above.
Of course this doesn't mean we can't make predictions about properties of our system. Like I said above, QM does great at determining what the probabilities should be. But we can't "dig deeper", collect more system information, etc. to remove these probabilities and make each measurement of a quantum system deterministic.
answered Mar 31 at 11:57
Aaron StevensAaron Stevens
15.4k42555
$endgroup$
$begingroup$
Sorry Aaron this website as well as you provide so long answers that it seems that we are studying English literature. We Indians do not have so much time to read so much long answers. I am not blaming you please don't take it personally friend but provide very compact answer which can be understood in limited time.😍😍
$endgroup$
– Shreyansh
Mar 31 at 14:10
7
$begingroup$
@Shreyansh First, I timed myself, and it look me about one minute to read this answer at an average pace from start to finish. If one minute isn't "limited time", then I don't know what is. Second, I highly doubt this is the case for all Indians, so you should probably just speak for yourself.
$endgroup$
– Aaron Stevens
Mar 31 at 14:41
$begingroup$
Hi Aaron I was in great trouble yesterday so sorry for all the bad I wrote.Let's delete our comments.
$endgroup$
– Shreyansh
Apr 1 at 5:45
add a comment |
$begingroup$
The microscopic particles themselves in classical statistical mechanics follow classical mechanics laws.
Elementary particles follow the laws of quantum mehanics.. Quantum mechanics was invented because elementary particles did not obey classical mechanics, but the new postulates of quantum mechanics. This, for large ensembles of quantum mechanical particles leads to quantum statistical mechanics, with differing average behaviors than the ones expected from classical statistical mechanics.
answered Mar 31 at 11:57
anna vanna v
162k8153456
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In statistical mechanics the system at any time is in a definite microstate (e.g. positions and velocities of all the particles in a gas), yet we don't know what this state is. Instead, we define certain global properties of the system that are defined on longer time scales (like total energy, entropy, temperature, volume) that are useful in many processes and try to predict them (in equilibrium) from the microscopic degrees of freedom.
In quantum mechanics, on the other hand, there are many options for what we mean by "states". The most intuitive definition is to define them in terms of things we can measure, like positions or velocities of particles, etc. However, the state is actually a wave in the space of these states and any given particle can actually spread out in state space and occupy many of these states simultaneously, with a different "amplitude" $psi$, just as a wave can spread out over space with a different amplitude at any point. (There are also restrictions on these wave functions, such as the fact that it must spread both in position $Delta x$ and in momentum $Delta p$ such that $Delta x Delta p geq hbar /2$, and similarly for other variables.) But basically the system can spread over what we would call "measurable" states.
The probabilities come in in Quantum mechanics, for example, when you try to measure the position of a particle that is spread over many different positions. This is where quantum mechanics gets confusing and leads to endless discussions about reality, but in short, the wave function "collapses" and you only measure one position, with probability $|psi(x)|^2$.
answered Mar 31 at 12:14
Eric David KramerEric David Kramer
1,036311
$endgroup$
$begingroup$
Shouldn't the Heisenberg uncertainty be $Delta x Delta p gte hbar / 2$?
$endgroup$
– michi7x7
Mar 31 at 14:50
$begingroup$
Sorry, typo. Yes, of course.
$endgroup$
– Eric David Kramer
Mar 31 at 15:06
add a comment |
$begingroup$
In statistical mechanics the system at any time is in a definite microstate (e.g. positions and velocities of all the particles in a gas), yet we don't know what this state is. Instead, we define certain global properties of the system that are defined on longer time scales (like total energy, entropy, temperature, volume) that are useful in many processes and try to predict them (in equilibrium) from the microscopic degrees of freedom.
In quantum mechanics, on the other hand, there are many options for what we mean by "states". The most intuitive definition is to define them in terms of things we can measure, like positions or velocities of particles, etc. However, the state is actually a wave in the space of these states and any given particle can actually spread out in state space and occupy many of these states simultaneously, with a different "amplitude" $psi$, just as a wave can spread out over space with a different amplitude at any point. (There are also restrictions on these wave functions, such as the fact that it must spread both in position $Delta x$ and in momentum $Delta p$ such that $Delta x Delta p geq hbar /2$, and similarly for other variables.) But basically the system can spread over what we would call "measurable" states.
The probabilities come in in Quantum mechanics, for example, when you try to measure the position of a particle that is spread over many different positions. This is where quantum mechanics gets confusing and leads to endless discussions about reality, but in short, the wave function "collapses" and you only measure one position, with probability $|psi(x)|^2$.
answered Mar 31 at 12:14
Eric David KramerEric David Kramer
1,036311
$endgroup$
$begingroup$
Shouldn't the Heisenberg uncertainty be $Delta x Delta p gte hbar / 2$?
$endgroup$
– michi7x7
Mar 31 at 14:50
$begingroup$
Sorry, typo. Yes, of course.
$endgroup$
– Eric David Kramer
Mar 31 at 15:06
add a comment |
$begingroup$
In statistical mechanics the system at any time is in a definite microstate (e.g. positions and velocities of all the particles in a gas), yet we don't know what this state is. Instead, we define certain global properties of the system that are defined on longer time scales (like total energy, entropy, temperature, volume) that are useful in many processes and try to predict them (in equilibrium) from the microscopic degrees of freedom.
In quantum mechanics, on the other hand, there are many options for what we mean by "states". The most intuitive definition is to define them in terms of things we can measure, like positions or velocities of particles, etc. However, the state is actually a wave in the space of these states and any given particle can actually spread out in state space and occupy many of these states simultaneously, with a different "amplitude" $psi$, just as a wave can spread out over space with a different amplitude at any point. (There are also restrictions on these wave functions, such as the fact that it must spread both in position $Delta x$ and in momentum $Delta p$ such that $Delta x Delta p geq hbar /2$, and similarly for other variables.) But basically the system can spread over what we would call "measurable" states.
The probabilities come in in Quantum mechanics, for example, when you try to measure the position of a particle that is spread over many different positions. This is where quantum mechanics gets confusing and leads to endless discussions about reality, but in short, the wave function "collapses" and you only measure one position, with probability $|psi(x)|^2$.
answered Mar 31 at 12:14
Eric David KramerEric David Kramer
1,036311
$endgroup$
In statistical mechanics the system at any time is in a definite microstate (e.g. positions and velocities of all the particles in a gas), yet we don't know what this state is. Instead, we define certain global properties of the system that are defined on longer time scales (like total energy, entropy, temperature, volume) that are useful in many processes and try to predict them (in equilibrium) from the microscopic degrees of freedom.
In quantum mechanics, on the other hand, there are many options for what we mean by "states". The most intuitive definition is to define them in terms of things we can measure, like positions or velocities of particles, etc. However, the state is actually a wave in the space of these states and any given particle can actually spread out in state space and occupy many of these states simultaneously, with a different "amplitude" $psi$, just as a wave can spread out over space with a different amplitude at any point. (There are also restrictions on these wave functions, such as the fact that it must spread both in position $Delta x$ and in momentum $Delta p$ such that $Delta x Delta p geq hbar /2$, and similarly for other variables.) But basically the system can spread over what we would call "measurable" states.
The probabilities come in in Quantum mechanics, for example, when you try to measure the position of a particle that is spread over many different positions. This is where quantum mechanics gets confusing and leads to endless discussions about reality, but in short, the wave function "collapses" and you only measure one position, with probability $|psi(x)|^2$.
answered Mar 31 at 12:14
Eric David KramerEric David Kramer
1,036311
edited Mar 31 at 15:05
answered Mar 31 at 12:14
Eric David KramerEric David Kramer
1,036311
answered Mar 31 at 12:14
Eric David KramerEric David Kramer
1,036311
answered Mar 31 at 12:14
Eric David KramerEric David Kramer
1,036311
1,036311
$begingroup$
Shouldn't the Heisenberg uncertainty be $Delta x Delta p gte hbar / 2$?
$endgroup$
– michi7x7
Mar 31 at 14:50
$begingroup$
Sorry, typo. Yes, of course.
$endgroup$
– Eric David Kramer
Mar 31 at 15:06
add a comment |
$begingroup$
Shouldn't the Heisenberg uncertainty be $Delta x Delta p gte hbar / 2$?
$endgroup$
– michi7x7
Mar 31 at 14:50
$begingroup$
Sorry, typo. Yes, of course.
$endgroup$
– Eric David Kramer
Mar 31 at 15:06
$begingroup$
Shouldn't the Heisenberg uncertainty be $Delta x Delta p gte hbar / 2$?
$endgroup$
– michi7x7
Mar 31 at 14:50
$begingroup$
Shouldn't the Heisenberg uncertainty be $Delta x Delta p gte hbar / 2$?
$endgroup$
– michi7x7
Mar 31 at 14:50
$begingroup$
Sorry, typo. Yes, of course.
$endgroup$
– Eric David Kramer
Mar 31 at 15:06
$begingroup$
Sorry, typo. Yes, of course.
$endgroup$
– Eric David Kramer
Mar 31 at 15:06
add a comment |
$begingroup$
Since you are specifically focusing on the probability aspect, this is what I will talk about. Otherwise the question and answer will be way too broad.
In classical statistical mechanics probabilities arise due to limited knowledge of the system. This is usually due to the fact that our systems are made up of so many particles that it would be impossible to keep track of everything. Therefore, we use statistical methods to describe the system and use those statistics to determine macroscopic properties of the system.
In terms of limited knowledge, the typical example given is a fair coin toss. We say we have a $0.5$ probability of getting heads and a $0.5$ probability of getting tails, but really this is just due to our limited knowledge of the system. If we knew the exact initial conditions of the coin flip, the interaction of the coin with the air, how the coin is caught/landed on the floor, etc. then we wouldn't need probability. We could know with exactly certainty what the result of the coin flip would be.
The same is true for statistical mechanics. If we could know the position and momentum of every particle, how each particle interacts with each other particle, external effects, etc. then we wouldn't need statistical mechanics. We would know exactly how the entire system would behave and evolve over time. You'll notice that this and the previous example are very unreasonable though, hence we use probabilities.
And then we have quantum mechanics. The difference here is that we can know everything there is to know about the system, yet the result of a measurement of that system will still not have a predicable outcome. All we can predict is the probability of a certain outcome. Probability seems to be an inherent property of QM that cannot be taken away like in the statistical mechanics examples above.
Of course this doesn't mean we can't make predictions about properties of our system. Like I said above, QM does great at determining what the probabilities should be. But we can't "dig deeper", collect more system information, etc. to remove these probabilities and make each measurement of a quantum system deterministic.
answered Mar 31 at 11:57
Aaron StevensAaron Stevens
15.4k42555
$endgroup$
$begingroup$
Sorry Aaron this website as well as you provide so long answers that it seems that we are studying English literature. We Indians do not have so much time to read so much long answers. I am not blaming you please don't take it personally friend but provide very compact answer which can be understood in limited time.😍😍
$endgroup$
– Shreyansh
Mar 31 at 14:10
7
$begingroup$
@Shreyansh First, I timed myself, and it look me about one minute to read this answer at an average pace from start to finish. If one minute isn't "limited time", then I don't know what is. Second, I highly doubt this is the case for all Indians, so you should probably just speak for yourself.
$endgroup$
– Aaron Stevens
Mar 31 at 14:41
$begingroup$
Hi Aaron I was in great trouble yesterday so sorry for all the bad I wrote.Let's delete our comments.
$endgroup$
– Shreyansh
Apr 1 at 5:45
add a comment |
$begingroup$
Since you are specifically focusing on the probability aspect, this is what I will talk about. Otherwise the question and answer will be way too broad.
In classical statistical mechanics probabilities arise due to limited knowledge of the system. This is usually due to the fact that our systems are made up of so many particles that it would be impossible to keep track of everything. Therefore, we use statistical methods to describe the system and use those statistics to determine macroscopic properties of the system.
In terms of limited knowledge, the typical example given is a fair coin toss. We say we have a $0.5$ probability of getting heads and a $0.5$ probability of getting tails, but really this is just due to our limited knowledge of the system. If we knew the exact initial conditions of the coin flip, the interaction of the coin with the air, how the coin is caught/landed on the floor, etc. then we wouldn't need probability. We could know with exactly certainty what the result of the coin flip would be.
The same is true for statistical mechanics. If we could know the position and momentum of every particle, how each particle interacts with each other particle, external effects, etc. then we wouldn't need statistical mechanics. We would know exactly how the entire system would behave and evolve over time. You'll notice that this and the previous example are very unreasonable though, hence we use probabilities.
And then we have quantum mechanics. The difference here is that we can know everything there is to know about the system, yet the result of a measurement of that system will still not have a predicable outcome. All we can predict is the probability of a certain outcome. Probability seems to be an inherent property of QM that cannot be taken away like in the statistical mechanics examples above.
Of course this doesn't mean we can't make predictions about properties of our system. Like I said above, QM does great at determining what the probabilities should be. But we can't "dig deeper", collect more system information, etc. to remove these probabilities and make each measurement of a quantum system deterministic.
answered Mar 31 at 11:57
Aaron StevensAaron Stevens
15.4k42555
$endgroup$
$begingroup$
Sorry Aaron this website as well as you provide so long answers that it seems that we are studying English literature. We Indians do not have so much time to read so much long answers. I am not blaming you please don't take it personally friend but provide very compact answer which can be understood in limited time.😍😍
$endgroup$
– Shreyansh
Mar 31 at 14:10
7
$begingroup$
@Shreyansh First, I timed myself, and it look me about one minute to read this answer at an average pace from start to finish. If one minute isn't "limited time", then I don't know what is. Second, I highly doubt this is the case for all Indians, so you should probably just speak for yourself.
$endgroup$
– Aaron Stevens
Mar 31 at 14:41
$begingroup$
Hi Aaron I was in great trouble yesterday so sorry for all the bad I wrote.Let's delete our comments.
$endgroup$
– Shreyansh
Apr 1 at 5:45
add a comment |
$begingroup$
Since you are specifically focusing on the probability aspect, this is what I will talk about. Otherwise the question and answer will be way too broad.
In classical statistical mechanics probabilities arise due to limited knowledge of the system. This is usually due to the fact that our systems are made up of so many particles that it would be impossible to keep track of everything. Therefore, we use statistical methods to describe the system and use those statistics to determine macroscopic properties of the system.
In terms of limited knowledge, the typical example given is a fair coin toss. We say we have a $0.5$ probability of getting heads and a $0.5$ probability of getting tails, but really this is just due to our limited knowledge of the system. If we knew the exact initial conditions of the coin flip, the interaction of the coin with the air, how the coin is caught/landed on the floor, etc. then we wouldn't need probability. We could know with exactly certainty what the result of the coin flip would be.
The same is true for statistical mechanics. If we could know the position and momentum of every particle, how each particle interacts with each other particle, external effects, etc. then we wouldn't need statistical mechanics. We would know exactly how the entire system would behave and evolve over time. You'll notice that this and the previous example are very unreasonable though, hence we use probabilities.
And then we have quantum mechanics. The difference here is that we can know everything there is to know about the system, yet the result of a measurement of that system will still not have a predicable outcome. All we can predict is the probability of a certain outcome. Probability seems to be an inherent property of QM that cannot be taken away like in the statistical mechanics examples above.
Of course this doesn't mean we can't make predictions about properties of our system. Like I said above, QM does great at determining what the probabilities should be. But we can't "dig deeper", collect more system information, etc. to remove these probabilities and make each measurement of a quantum system deterministic.
answered Mar 31 at 11:57
Aaron StevensAaron Stevens
15.4k42555
$endgroup$
Since you are specifically focusing on the probability aspect, this is what I will talk about. Otherwise the question and answer will be way too broad.
In classical statistical mechanics probabilities arise due to limited knowledge of the system. This is usually due to the fact that our systems are made up of so many particles that it would be impossible to keep track of everything. Therefore, we use statistical methods to describe the system and use those statistics to determine macroscopic properties of the system.
In terms of limited knowledge, the typical example given is a fair coin toss. We say we have a $0.5$ probability of getting heads and a $0.5$ probability of getting tails, but really this is just due to our limited knowledge of the system. If we knew the exact initial conditions of the coin flip, the interaction of the coin with the air, how the coin is caught/landed on the floor, etc. then we wouldn't need probability. We could know with exactly certainty what the result of the coin flip would be.
The same is true for statistical mechanics. If we could know the position and momentum of every particle, how each particle interacts with each other particle, external effects, etc. then we wouldn't need statistical mechanics. We would know exactly how the entire system would behave and evolve over time. You'll notice that this and the previous example are very unreasonable though, hence we use probabilities.
And then we have quantum mechanics. The difference here is that we can know everything there is to know about the system, yet the result of a measurement of that system will still not have a predicable outcome. All we can predict is the probability of a certain outcome. Probability seems to be an inherent property of QM that cannot be taken away like in the statistical mechanics examples above.
Of course this doesn't mean we can't make predictions about properties of our system. Like I said above, QM does great at determining what the probabilities should be. But we can't "dig deeper", collect more system information, etc. to remove these probabilities and make each measurement of a quantum system deterministic.
answered Mar 31 at 11:57
Aaron StevensAaron Stevens
15.4k42555
edited Mar 31 at 12:04
answered Mar 31 at 11:57
Aaron StevensAaron Stevens
15.4k42555
answered Mar 31 at 11:57
Aaron StevensAaron Stevens
15.4k42555
answered Mar 31 at 11:57
Aaron StevensAaron Stevens
15.4k42555
15.4k42555
$begingroup$
Sorry Aaron this website as well as you provide so long answers that it seems that we are studying English literature. We Indians do not have so much time to read so much long answers. I am not blaming you please don't take it personally friend but provide very compact answer which can be understood in limited time.😍😍
$endgroup$
– Shreyansh
Mar 31 at 14:10
7
$begingroup$
@Shreyansh First, I timed myself, and it look me about one minute to read this answer at an average pace from start to finish. If one minute isn't "limited time", then I don't know what is. Second, I highly doubt this is the case for all Indians, so you should probably just speak for yourself.
$endgroup$
– Aaron Stevens
Mar 31 at 14:41
$begingroup$
Hi Aaron I was in great trouble yesterday so sorry for all the bad I wrote.Let's delete our comments.
$endgroup$
– Shreyansh
Apr 1 at 5:45
add a comment |
$begingroup$
Sorry Aaron this website as well as you provide so long answers that it seems that we are studying English literature. We Indians do not have so much time to read so much long answers. I am not blaming you please don't take it personally friend but provide very compact answer which can be understood in limited time.😍😍
$endgroup$
– Shreyansh
Mar 31 at 14:10
7
$begingroup$
@Shreyansh First, I timed myself, and it look me about one minute to read this answer at an average pace from start to finish. If one minute isn't "limited time", then I don't know what is. Second, I highly doubt this is the case for all Indians, so you should probably just speak for yourself.
$endgroup$
– Aaron Stevens
Mar 31 at 14:41
$begingroup$
Hi Aaron I was in great trouble yesterday so sorry for all the bad I wrote.Let's delete our comments.
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– Shreyansh
Apr 1 at 5:45
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Sorry Aaron this website as well as you provide so long answers that it seems that we are studying English literature. We Indians do not have so much time to read so much long answers. I am not blaming you please don't take it personally friend but provide very compact answer which can be understood in limited time.😍😍
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– Shreyansh
Mar 31 at 14:10
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Sorry Aaron this website as well as you provide so long answers that it seems that we are studying English literature. We Indians do not have so much time to read so much long answers. I am not blaming you please don't take it personally friend but provide very compact answer which can be understood in limited time.😍😍
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– Shreyansh
Mar 31 at 14:10
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@Shreyansh First, I timed myself, and it look me about one minute to read this answer at an average pace from start to finish. If one minute isn't "limited time", then I don't know what is. Second, I highly doubt this is the case for all Indians, so you should probably just speak for yourself.
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– Aaron Stevens
Mar 31 at 14:41
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@Shreyansh First, I timed myself, and it look me about one minute to read this answer at an average pace from start to finish. If one minute isn't "limited time", then I don't know what is. Second, I highly doubt this is the case for all Indians, so you should probably just speak for yourself.
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– Aaron Stevens
Mar 31 at 14:41
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Hi Aaron I was in great trouble yesterday so sorry for all the bad I wrote.Let's delete our comments.
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– Shreyansh
Apr 1 at 5:45
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Hi Aaron I was in great trouble yesterday so sorry for all the bad I wrote.Let's delete our comments.
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– Shreyansh
Apr 1 at 5:45
add a comment |
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The microscopic particles themselves in classical statistical mechanics follow classical mechanics laws.
Elementary particles follow the laws of quantum mehanics.. Quantum mechanics was invented because elementary particles did not obey classical mechanics, but the new postulates of quantum mechanics. This, for large ensembles of quantum mechanical particles leads to quantum statistical mechanics, with differing average behaviors than the ones expected from classical statistical mechanics.
answered Mar 31 at 11:57
anna vanna v
162k8153456
$endgroup$
add a comment |
$begingroup$
The microscopic particles themselves in classical statistical mechanics follow classical mechanics laws.
Elementary particles follow the laws of quantum mehanics.. Quantum mechanics was invented because elementary particles did not obey classical mechanics, but the new postulates of quantum mechanics. This, for large ensembles of quantum mechanical particles leads to quantum statistical mechanics, with differing average behaviors than the ones expected from classical statistical mechanics.
answered Mar 31 at 11:57
anna vanna v
162k8153456
$endgroup$
add a comment |
$begingroup$
The microscopic particles themselves in classical statistical mechanics follow classical mechanics laws.
Elementary particles follow the laws of quantum mehanics.. Quantum mechanics was invented because elementary particles did not obey classical mechanics, but the new postulates of quantum mechanics. This, for large ensembles of quantum mechanical particles leads to quantum statistical mechanics, with differing average behaviors than the ones expected from classical statistical mechanics.
answered Mar 31 at 11:57
anna vanna v
162k8153456
$endgroup$
The microscopic particles themselves in classical statistical mechanics follow classical mechanics laws.
Elementary particles follow the laws of quantum mehanics.. Quantum mechanics was invented because elementary particles did not obey classical mechanics, but the new postulates of quantum mechanics. This, for large ensembles of quantum mechanical particles leads to quantum statistical mechanics, with differing average behaviors than the ones expected from classical statistical mechanics.
answered Mar 31 at 11:57
anna vanna v
162k8153456
answered Mar 31 at 11:57
anna vanna v
162k8153456
answered Mar 31 at 11:57
anna vanna v
162k8153456
answered Mar 31 at 11:57
anna vanna v
162k8153456
162k8153456
add a comment |
add a comment |
