Why is perturbation theory used in quantum mechanics?
Clash Royale CLAN TAG#URR8PPP
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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 Feb 22 at 16:18
knzhou
45.2k11122219
asked Feb 22 at 10:30
Claus KlausenClaus Klausen
366
$endgroup$
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 Feb 22 at 16:18
knzhou
45.2k11122219
asked Feb 22 at 10:30
Claus KlausenClaus Klausen
366
$endgroup$
1
$begingroup$
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
Feb 22 at 11:24
5
$begingroup$
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
Feb 22 at 16:19
1
$begingroup$
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
Feb 22 at 16:21
1
$begingroup$
@knzhou I don't see where is the OP is implying that perturbation theory is useful only for QM. Also, the question "why is perturbation theory needed in QM?" is very on point, IMO.
$endgroup$
– pppqqq
Feb 22 at 19:14
$begingroup$
wikipedia has a fairly good explanation of this, have you read it?
$endgroup$
– niels nielsen
Feb 22 at 23:53
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 Feb 22 at 16:18
knzhou
45.2k11122219
asked Feb 22 at 10:30
Claus KlausenClaus Klausen
366
$endgroup$
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 Feb 22 at 16:18
knzhou
45.2k11122219
asked Feb 22 at 10:30
Claus KlausenClaus Klausen
366
edited Feb 22 at 16:18
knzhou
45.2k11122219
asked Feb 22 at 10:30
Claus KlausenClaus Klausen
366
edited Feb 22 at 16:18
knzhou
45.2k11122219
edited Feb 22 at 16:18
knzhou
45.2k11122219
edited Feb 22 at 16:18
knzhou
45.2k11122219
45.2k11122219
asked Feb 22 at 10:30
Claus KlausenClaus Klausen
366
asked Feb 22 at 10:30
Claus KlausenClaus Klausen
366
asked Feb 22 at 10:30
Claus KlausenClaus Klausen
366
366
1
$begingroup$
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
Feb 22 at 11:24
5
$begingroup$
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
Feb 22 at 16:19
1
$begingroup$
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
Feb 22 at 16:21
1
$begingroup$
@knzhou I don't see where is the OP is implying that perturbation theory is useful only for QM. Also, the question "why is perturbation theory needed in QM?" is very on point, IMO.
$endgroup$
– pppqqq
Feb 22 at 19:14
$begingroup$
wikipedia has a fairly good explanation of this, have you read it?
$endgroup$
– niels nielsen
Feb 22 at 23:53
add a comment |
1
$begingroup$
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
Feb 22 at 11:24
5
$begingroup$
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
Feb 22 at 16:19
1
$begingroup$
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
Feb 22 at 16:21
1
$begingroup$
@knzhou I don't see where is the OP is implying that perturbation theory is useful only for QM. Also, the question "why is perturbation theory needed in QM?" is very on point, IMO.
$endgroup$
– pppqqq
Feb 22 at 19:14
$begingroup$
wikipedia has a fairly good explanation of this, have you read it?
$endgroup$
– niels nielsen
Feb 22 at 23:53
1
1
$begingroup$
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
Feb 22 at 11:24
$begingroup$
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
Feb 22 at 11:24
5
5
$begingroup$
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
Feb 22 at 16:19
$begingroup$
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
Feb 22 at 16:19
1
1
$begingroup$
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
Feb 22 at 16:21
$begingroup$
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
Feb 22 at 16:21
1
1
$begingroup$
@knzhou I don't see where is the OP is implying that perturbation theory is useful only for QM. Also, the question "why is perturbation theory needed in QM?" is very on point, IMO.
$endgroup$
– pppqqq
Feb 22 at 19:14
$begingroup$
@knzhou I don't see where is the OP is implying that perturbation theory is useful only for QM. Also, the question "why is perturbation theory needed in QM?" is very on point, IMO.
$endgroup$
– pppqqq
Feb 22 at 19:14
$begingroup$
wikipedia has a fairly good explanation of this, have you read it?
$endgroup$
– niels nielsen
Feb 22 at 23:53
$begingroup$
wikipedia has a fairly good explanation of this, have you read it?
$endgroup$
– niels nielsen
Feb 22 at 23:53
add a comment |
1 Answer
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oldest
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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 Feb 22 at 10:54
Emilio PisantyEmilio Pisanty
85.7k23211430
$endgroup$
1
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
Feb 22 at 12:16
2
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
Feb 22 at 12:50
$begingroup$
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
Feb 22 at 13:10
add a comment |
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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 Feb 22 at 10:54
Emilio PisantyEmilio Pisanty
85.7k23211430
$endgroup$
1
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
Feb 22 at 12:16
2
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
Feb 22 at 12:50
$begingroup$
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
Feb 22 at 13:10
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 Feb 22 at 10:54
Emilio PisantyEmilio Pisanty
85.7k23211430
$endgroup$
1
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
Feb 22 at 12:16
2
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
Feb 22 at 12:50
$begingroup$
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
Feb 22 at 13:10
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 Feb 22 at 10:54
Emilio PisantyEmilio Pisanty
85.7k23211430
$endgroup$
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 Feb 22 at 10:54
Emilio PisantyEmilio Pisanty
85.7k23211430
answered Feb 22 at 10:54
Emilio PisantyEmilio Pisanty
85.7k23211430
answered Feb 22 at 10:54
Emilio PisantyEmilio Pisanty
85.7k23211430
answered Feb 22 at 10:54
Emilio PisantyEmilio Pisanty
85.7k23211430
85.7k23211430
1
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
Feb 22 at 12:16
2
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
Feb 22 at 12:50
$begingroup$
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
Feb 22 at 13:10
add a comment |
1
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
Feb 22 at 12:16
2
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
Feb 22 at 12:50
$begingroup$
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
Feb 22 at 13:10
1
1
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
Feb 22 at 12:16
$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
Feb 22 at 12:16
2
2
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
Feb 22 at 12:50
$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
Feb 22 at 12:50
$begingroup$
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
Feb 22 at 13:10
$begingroup$
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
Feb 22 at 13:10
add a comment |
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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
Feb 22 at 11:24
5
$begingroup$
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
Feb 22 at 16:19
1
$begingroup$
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
Feb 22 at 16:21
1
$begingroup$
@knzhou I don't see where is the OP is implying that perturbation theory is useful only for QM. Also, the question "why is perturbation theory needed in QM?" is very on point, IMO.
$endgroup$
– pppqqq
Feb 22 at 19:14
$begingroup$
wikipedia has a fairly good explanation of this, have you read it?
$endgroup$
– niels nielsen
Feb 22 at 23:53