Can someone explain the need for perturbation theory in QM?












1












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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







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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
    2 hours ago


















1












$begingroup$


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







share|cite|improve this question









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
    2 hours ago
















1












1








1


1



$begingroup$


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







share|cite|improve this question









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






share|cite|improve this question









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.











share|cite|improve this question









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Claus Klausen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Qmechanic♦

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asked 3 hours ago









Claus KlausenClaus Klausen

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New contributor




Claus Klausen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
    2 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
    2 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
2 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
2 hours ago












1 Answer
1






active

oldest

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6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – Claus Klausen
    3 hours ago










  • $begingroup$
    Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
    $endgroup$
    – Dvij Mankad
    1 hour ago






  • 1




    $begingroup$
    @DvijMankad I have no idea what scope you're thinking of. Ask separately.
    $endgroup$
    – Emilio Pisanty
    1 hour 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
    53 mins ago











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1 Answer
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1 Answer
1






active

oldest

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active

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active

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6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – Claus Klausen
    3 hours ago










  • $begingroup$
    Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
    $endgroup$
    – Dvij Mankad
    1 hour ago






  • 1




    $begingroup$
    @DvijMankad I have no idea what scope you're thinking of. Ask separately.
    $endgroup$
    – Emilio Pisanty
    1 hour 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
    53 mins ago
















6












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you very much!
    $endgroup$
    – Claus Klausen
    3 hours ago










  • $begingroup$
    Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
    $endgroup$
    – Dvij Mankad
    1 hour ago






  • 1




    $begingroup$
    @DvijMankad I have no idea what scope you're thinking of. Ask separately.
    $endgroup$
    – Emilio Pisanty
    1 hour 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
    53 mins ago














6












6








6





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









Emilio PisantyEmilio Pisanty

83.5k22204420




83.5k22204420












  • $begingroup$
    Thank you very much!
    $endgroup$
    – Claus Klausen
    3 hours ago










  • $begingroup$
    Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
    $endgroup$
    – Dvij Mankad
    1 hour ago






  • 1




    $begingroup$
    @DvijMankad I have no idea what scope you're thinking of. Ask separately.
    $endgroup$
    – Emilio Pisanty
    1 hour 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
    53 mins ago


















  • $begingroup$
    Thank you very much!
    $endgroup$
    – Claus Klausen
    3 hours ago










  • $begingroup$
    Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
    $endgroup$
    – Dvij Mankad
    1 hour ago






  • 1




    $begingroup$
    @DvijMankad I have no idea what scope you're thinking of. Ask separately.
    $endgroup$
    – Emilio Pisanty
    1 hour 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
    53 mins ago
















$begingroup$
Thank you very much!
$endgroup$
– Claus Klausen
3 hours ago




$begingroup$
Thank you very much!
$endgroup$
– Claus Klausen
3 hours ago












$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
1 hour ago




$begingroup$
Does your answer work well for all controlled approximations, perturbative or otherwise? Thanks.
$endgroup$
– Dvij Mankad
1 hour ago




1




1




$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
1 hour ago




$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
1 hour 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
53 mins 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
53 mins ago










Claus Klausen is a new contributor. Be nice, and check out our Code of Conduct.










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