Why is perturbation theory used in quantum mechanics?Perturbation theoryPerturbation theory in quantum...
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Why is perturbation theory used in quantum mechanics?
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Why is perturbation theory used in quantum mechanics?
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I don't seem to understand what perturbation theory really is, and what it is needed for. Can someone please provide an explanation for what it is, and why it is needed in QM?
quantum-mechanics perturbation-theory
edited 24 mins ago
knzhou
44.3k11121214
asked 6 hours ago
Claus KlausenClaus Klausen
164
New contributor
Claus Klausen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I don't seem to understand what perturbation theory really is, and what it is needed for. Can someone please provide an explanation for what it is, and why it is needed in QM?
quantum-mechanics perturbation-theory
edited 24 mins ago
knzhou
44.3k11121214
asked 6 hours ago
Claus KlausenClaus Klausen
164
New contributor
Claus Klausen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Perturbation theory allows you to calculate the response of a system to a small perturbation. For example, it can tell you, how energy levels shift under the influence of the perturbation, how energy level degeneracies are lifted, and it even allows you to calculate linear response functions. You should be able to find more detailed answers to your question in standard QM books and on the web.
$endgroup$
– flaudemus
5 hours ago
1
$begingroup$
This seems to be a side effect of an incomplete physics education. Perturbation theory is an incredibly broad set of techniques useful in all subfields of physics and much of applied mathematics at large. But often people are not introduced to it until a quantum mechanics course, probably because there's not enough time to look at nontrivial (i.e. no simple closed form solution) problems earlier in the curriculum, which would motivate introducing perturbation theory.
$endgroup$
– knzhou
22 mins ago
$begingroup$
As such this question ends up sounding like... "what is calculus and why is it introduced in quantum field theory class?", or "what is the quadratic formula and why is it useful for statistical mechanics?"
$endgroup$
– knzhou
21 mins ago
add a comment |
$begingroup$
I don't seem to understand what perturbation theory really is, and what it is needed for. Can someone please provide an explanation for what it is, and why it is needed in QM?
quantum-mechanics perturbation-theory
edited 24 mins ago
knzhou
44.3k11121214
asked 6 hours ago
Claus KlausenClaus Klausen
164
New contributor
Claus Klausen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I don't seem to understand what perturbation theory really is, and what it is needed for. Can someone please provide an explanation for what it is, and why it is needed in QM?
quantum-mechanics perturbation-theory
quantum-mechanics perturbation-theory
edited 24 mins ago
knzhou
44.3k11121214
asked 6 hours ago
Claus KlausenClaus Klausen
164
New contributor
Claus Klausen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 24 mins ago
knzhou
44.3k11121214
asked 6 hours ago
Claus KlausenClaus Klausen
164
New contributor
Claus Klausen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 24 mins ago
knzhou
44.3k11121214
edited 24 mins ago
knzhou
44.3k11121214
edited 24 mins ago
knzhou
44.3k11121214
44.3k11121214
asked 6 hours ago
Claus KlausenClaus Klausen
164
New contributor
Claus Klausen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
Claus KlausenClaus Klausen
164
asked 6 hours ago
Claus KlausenClaus Klausen
164
164
New contributor
Claus Klausen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Claus Klausen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Claus Klausen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Perturbation theory allows you to calculate the response of a system to a small perturbation. For example, it can tell you, how energy levels shift under the influence of the perturbation, how energy level degeneracies are lifted, and it even allows you to calculate linear response functions. You should be able to find more detailed answers to your question in standard QM books and on the web.
$endgroup$
– flaudemus
5 hours ago
1
$begingroup$
This seems to be a side effect of an incomplete physics education. Perturbation theory is an incredibly broad set of techniques useful in all subfields of physics and much of applied mathematics at large. But often people are not introduced to it until a quantum mechanics course, probably because there's not enough time to look at nontrivial (i.e. no simple closed form solution) problems earlier in the curriculum, which would motivate introducing perturbation theory.
$endgroup$
– knzhou
22 mins ago
$begingroup$
As such this question ends up sounding like... "what is calculus and why is it introduced in quantum field theory class?", or "what is the quadratic formula and why is it useful for statistical mechanics?"
$endgroup$
– knzhou
21 mins ago
add a comment |
$begingroup$
Perturbation theory allows you to calculate the response of a system to a small perturbation. For example, it can tell you, how energy levels shift under the influence of the perturbation, how energy level degeneracies are lifted, and it even allows you to calculate linear response functions. You should be able to find more detailed answers to your question in standard QM books and on the web.
$endgroup$
– flaudemus
5 hours ago
1
$begingroup$
This seems to be a side effect of an incomplete physics education. Perturbation theory is an incredibly broad set of techniques useful in all subfields of physics and much of applied mathematics at large. But often people are not introduced to it until a quantum mechanics course, probably because there's not enough time to look at nontrivial (i.e. no simple closed form solution) problems earlier in the curriculum, which would motivate introducing perturbation theory.
$endgroup$
– knzhou
22 mins ago
$begingroup$
As such this question ends up sounding like... "what is calculus and why is it introduced in quantum field theory class?", or "what is the quadratic formula and why is it useful for statistical mechanics?"
$endgroup$
– knzhou
21 mins ago
$begingroup$
Perturbation theory allows you to calculate the response of a system to a small perturbation. For example, it can tell you, how energy levels shift under the influence of the perturbation, how energy level degeneracies are lifted, and it even allows you to calculate linear response functions. You should be able to find more detailed answers to your question in standard QM books and on the web.
$endgroup$
– flaudemus
5 hours ago
$begingroup$
Perturbation theory allows you to calculate the response of a system to a small perturbation. For example, it can tell you, how energy levels shift under the influence of the perturbation, how energy level degeneracies are lifted, and it even allows you to calculate linear response functions. You should be able to find more detailed answers to your question in standard QM books and on the web.
$endgroup$
– flaudemus
5 hours ago
1
1
$begingroup$
This seems to be a side effect of an incomplete physics education. Perturbation theory is an incredibly broad set of techniques useful in all subfields of physics and much of applied mathematics at large. But often people are not introduced to it until a quantum mechanics course, probably because there's not enough time to look at nontrivial (i.e. no simple closed form solution) problems earlier in the curriculum, which would motivate introducing perturbation theory.
$endgroup$
– knzhou
22 mins ago
$begingroup$
This seems to be a side effect of an incomplete physics education. Perturbation theory is an incredibly broad set of techniques useful in all subfields of physics and much of applied mathematics at large. But often people are not introduced to it until a quantum mechanics course, probably because there's not enough time to look at nontrivial (i.e. no simple closed form solution) problems earlier in the curriculum, which would motivate introducing perturbation theory.
$endgroup$
– knzhou
22 mins ago
$begingroup$
As such this question ends up sounding like... "what is calculus and why is it introduced in quantum field theory class?", or "what is the quadratic formula and why is it useful for statistical mechanics?"
$endgroup$
– knzhou
21 mins ago
$begingroup$
As such this question ends up sounding like... "what is calculus and why is it introduced in quantum field theory class?", or "what is the quadratic formula and why is it useful for statistical mechanics?"
$endgroup$
– knzhou
21 mins ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are two main reasons.
The first is practical, and it is that QM is hard. Very few systems are solvable, particularly in the presence of interactions. There's a wide variety of situations where perturbation theory is the only available way to make any headway at all.
The second one is conceptual and it is basically that it allows us to establish a hierarchy in the strengths of the different components of a system, and it allows us to have a clearer picture of how this strengths scale with the parameters of the system. Thus, even if you can solve the full system, if one component is weak, it often makes sense to treat it explicitly as a perturbation, so that you can keep a clearer track of how the different aspects of the solution (energies, eigenfunctions, dipole moments, what have you) scale.
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.5k22205420
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Claus Klausen
5 hours ago
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
4 hours ago
1
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
3 hours ago
$begingroup$
I was thinking schemes where we have a small parameter but we are not doing perturbation theory, for example, the adiabatic approximation. I will try to formulate a separate question.
$endgroup$
– Dvij Mankad
3 hours ago
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
There are two main reasons.
The first is practical, and it is that QM is hard. Very few systems are solvable, particularly in the presence of interactions. There's a wide variety of situations where perturbation theory is the only available way to make any headway at all.
The second one is conceptual and it is basically that it allows us to establish a hierarchy in the strengths of the different components of a system, and it allows us to have a clearer picture of how this strengths scale with the parameters of the system. Thus, even if you can solve the full system, if one component is weak, it often makes sense to treat it explicitly as a perturbation, so that you can keep a clearer track of how the different aspects of the solution (energies, eigenfunctions, dipole moments, what have you) scale.
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.5k22205420
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Claus Klausen
5 hours ago
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
4 hours ago
1
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
3 hours ago
$begingroup$
I was thinking schemes where we have a small parameter but we are not doing perturbation theory, for example, the adiabatic approximation. I will try to formulate a separate question.
$endgroup$
– Dvij Mankad
3 hours ago
add a comment |
$begingroup$
There are two main reasons.
The first is practical, and it is that QM is hard. Very few systems are solvable, particularly in the presence of interactions. There's a wide variety of situations where perturbation theory is the only available way to make any headway at all.
The second one is conceptual and it is basically that it allows us to establish a hierarchy in the strengths of the different components of a system, and it allows us to have a clearer picture of how this strengths scale with the parameters of the system. Thus, even if you can solve the full system, if one component is weak, it often makes sense to treat it explicitly as a perturbation, so that you can keep a clearer track of how the different aspects of the solution (energies, eigenfunctions, dipole moments, what have you) scale.
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.5k22205420
$endgroup$
$begingroup$
Thank you very much!
$endgroup$
– Claus Klausen
5 hours ago
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
4 hours ago
1
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
3 hours ago
$begingroup$
I was thinking schemes where we have a small parameter but we are not doing perturbation theory, for example, the adiabatic approximation. I will try to formulate a separate question.
$endgroup$
– Dvij Mankad
3 hours ago
add a comment |
$begingroup$
There are two main reasons.
The first is practical, and it is that QM is hard. Very few systems are solvable, particularly in the presence of interactions. There's a wide variety of situations where perturbation theory is the only available way to make any headway at all.
The second one is conceptual and it is basically that it allows us to establish a hierarchy in the strengths of the different components of a system, and it allows us to have a clearer picture of how this strengths scale with the parameters of the system. Thus, even if you can solve the full system, if one component is weak, it often makes sense to treat it explicitly as a perturbation, so that you can keep a clearer track of how the different aspects of the solution (energies, eigenfunctions, dipole moments, what have you) scale.
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.5k22205420
$endgroup$
There are two main reasons.
The first is practical, and it is that QM is hard. Very few systems are solvable, particularly in the presence of interactions. There's a wide variety of situations where perturbation theory is the only available way to make any headway at all.
The second one is conceptual and it is basically that it allows us to establish a hierarchy in the strengths of the different components of a system, and it allows us to have a clearer picture of how this strengths scale with the parameters of the system. Thus, even if you can solve the full system, if one component is weak, it often makes sense to treat it explicitly as a perturbation, so that you can keep a clearer track of how the different aspects of the solution (energies, eigenfunctions, dipole moments, what have you) scale.
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.5k22205420
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.5k22205420
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.5k22205420
answered 5 hours ago
Emilio PisantyEmilio Pisanty
83.5k22205420
83.5k22205420
$begingroup$
Thank you very much!
$endgroup$
– Claus Klausen
5 hours ago
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
4 hours ago
1
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
3 hours ago
$begingroup$
I was thinking schemes where we have a small parameter but we are not doing perturbation theory, for example, the adiabatic approximation. I will try to formulate a separate question.
$endgroup$
– Dvij Mankad
3 hours ago
add a comment |
$begingroup$
Thank you very much!
$endgroup$
– Claus Klausen
5 hours ago
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
4 hours ago
1
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
3 hours ago
$begingroup$
I was thinking schemes where we have a small parameter but we are not doing perturbation theory, for example, the adiabatic approximation. I will try to formulate a separate question.
$endgroup$
– Dvij Mankad
3 hours ago
$begingroup$
Thank you very much!
$endgroup$
– Claus Klausen
5 hours ago
$begingroup$
Thank you very much!
$endgroup$
– Claus Klausen
5 hours ago
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
4 hours ago
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
4 hours ago
1
1
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
3 hours ago
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
3 hours ago
$begingroup$
I was thinking schemes where we have a small parameter but we are not doing perturbation theory, for example, the adiabatic approximation. I will try to formulate a separate question.
$endgroup$
– Dvij Mankad
3 hours ago
$begingroup$
I was thinking schemes where we have a small parameter but we are not doing perturbation theory, for example, the adiabatic approximation. I will try to formulate a separate question.
$endgroup$
– Dvij Mankad
3 hours ago
add a comment |
Claus Klausen is a new contributor. Be nice, and check out our Code of Conduct.
Claus Klausen is a new contributor. Be nice, and check out our Code of Conduct.
Claus Klausen is a new contributor. Be nice, and check out our Code of Conduct.
Claus Klausen is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Perturbation theory allows you to calculate the response of a system to a small perturbation. For example, it can tell you, how energy levels shift under the influence of the perturbation, how energy level degeneracies are lifted, and it even allows you to calculate linear response functions. You should be able to find more detailed answers to your question in standard QM books and on the web.
$endgroup$
– flaudemus
5 hours ago
1
$begingroup$
This seems to be a side effect of an incomplete physics education. Perturbation theory is an incredibly broad set of techniques useful in all subfields of physics and much of applied mathematics at large. But often people are not introduced to it until a quantum mechanics course, probably because there's not enough time to look at nontrivial (i.e. no simple closed form solution) problems earlier in the curriculum, which would motivate introducing perturbation theory.
$endgroup$
– knzhou
22 mins ago
$begingroup$
As such this question ends up sounding like... "what is calculus and why is it introduced in quantum field theory class?", or "what is the quadratic formula and why is it useful for statistical mechanics?"
$endgroup$
– knzhou
21 mins ago