Why is perturbation theory used in quantum mechanics?












4












$begingroup$


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?










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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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  • $begingroup$
    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
    16 hours ago








  • 4




    $begingroup$
    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
    12 hours ago








  • 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
    11 hours ago










  • $begingroup$
    @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
    9 hours ago










  • $begingroup$
    wikipedia has a fairly good explanation of this, have you read it?
    $endgroup$
    – niels nielsen
    4 hours ago
















4












$begingroup$


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?










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
    16 hours ago








  • 4




    $begingroup$
    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
    12 hours ago








  • 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
    11 hours ago










  • $begingroup$
    @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
    9 hours ago










  • $begingroup$
    wikipedia has a fairly good explanation of this, have you read it?
    $endgroup$
    – niels nielsen
    4 hours ago














4












4








4


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?










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






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









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




share|cite|improve this question








edited 12 hours ago









knzhou

44.3k11121214




44.3k11121214






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asked 17 hours ago









Claus KlausenClaus Klausen

294




294




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
    16 hours ago








  • 4




    $begingroup$
    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
    12 hours ago








  • 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
    11 hours ago










  • $begingroup$
    @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
    9 hours ago










  • $begingroup$
    wikipedia has a fairly good explanation of this, have you read it?
    $endgroup$
    – niels nielsen
    4 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
    16 hours ago








  • 4




    $begingroup$
    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
    12 hours ago








  • 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
    11 hours ago










  • $begingroup$
    @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
    9 hours ago










  • $begingroup$
    wikipedia has a fairly good explanation of this, have you read it?
    $endgroup$
    – niels nielsen
    4 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
16 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
16 hours ago






4




4




$begingroup$
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
12 hours 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
12 hours ago






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
11 hours 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
11 hours ago












$begingroup$
@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
9 hours ago




$begingroup$
@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
9 hours ago












$begingroup$
wikipedia has a fairly good explanation of this, have you read it?
$endgroup$
– niels nielsen
4 hours ago




$begingroup$
wikipedia has a fairly good explanation of this, have you read it?
$endgroup$
– niels nielsen
4 hours ago










2 Answers
2






active

oldest

votes


















13












$begingroup$

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
    17 hours ago






  • 1




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






  • 2




    $begingroup$
    @DvijMankad I have no idea what scope you're thinking of. Ask separately.
    $endgroup$
    – Emilio Pisanty
    15 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
    15 hours ago



















-1












$begingroup$

Perturbation Theory in Quantum Mechanics is probably deep rooted with the kind of the mathematical formulation of Quantum Mechanics in terms of Self Adjoint Operators (Von Neumann formulation )and specially the nature of analytical functions expected to describe Physical parameters on quantum physics :The energy bound state levels of a given quantum Hamiltonian is always expected to be at worse a meromorphic function of the "tuning" parameters on the theory (essentially those coupling constants mediating the interaction of the systems interacting to made up a bigger one (the full Hamiltonian is always constructed mathematically rigorously from PERTURBATION KATO-RELICH like self adjoint theorems -The general Atomic Hmiltonian for instance ) .Even if one wishes to handle non perturbative phenomena in Quantum Physics ,in much of the cases , one must reformulate the problem in news variables where now the non perturbative scheme becomes a perturbative scheme .However ,computer oriented calculations are somewhat non perturbative in their mathematical nature , but they remains used in a perturbative scheme .So , concluding :Our Differential Integral calculus and Functional Analysis thinking on Quantum Mechanics and Classical Physics remains almost complete only for sort of "Analytical " functions and their close variants .The reader should compare with the mathematical thinking need in probability theory problems , where "analytical functions thinking-perturbation theory is almost zero !(All sample function in space functions of Measurable sets are at best continuous functions like those made up Wienner trajectories ) .






share|cite|improve this answer








New contributor




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






$endgroup$









  • 1




    $begingroup$
    This answer seems quite unclear to me, and I cannot even tell, if it is a real answer to the question.
    $endgroup$
    – flaudemus
    7 hours ago













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









13












$begingroup$

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
    17 hours ago






  • 1




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






  • 2




    $begingroup$
    @DvijMankad I have no idea what scope you're thinking of. Ask separately.
    $endgroup$
    – Emilio Pisanty
    15 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
    15 hours ago
















13












$begingroup$

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
    17 hours ago






  • 1




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






  • 2




    $begingroup$
    @DvijMankad I have no idea what scope you're thinking of. Ask separately.
    $endgroup$
    – Emilio Pisanty
    15 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
    15 hours ago














13












13








13





$begingroup$

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 17 hours ago









Emilio PisantyEmilio Pisanty

83.5k22206421




83.5k22206421












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






  • 1




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






  • 2




    $begingroup$
    @DvijMankad I have no idea what scope you're thinking of. Ask separately.
    $endgroup$
    – Emilio Pisanty
    15 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
    15 hours ago


















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






  • 1




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






  • 2




    $begingroup$
    @DvijMankad I have no idea what scope you're thinking of. Ask separately.
    $endgroup$
    – Emilio Pisanty
    15 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
    15 hours ago
















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




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




1




1




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




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




2




2




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




$begingroup$
@DvijMankad I have no idea what scope you're thinking of. Ask separately.
$endgroup$
– Emilio Pisanty
15 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
15 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
15 hours ago











-1












$begingroup$

Perturbation Theory in Quantum Mechanics is probably deep rooted with the kind of the mathematical formulation of Quantum Mechanics in terms of Self Adjoint Operators (Von Neumann formulation )and specially the nature of analytical functions expected to describe Physical parameters on quantum physics :The energy bound state levels of a given quantum Hamiltonian is always expected to be at worse a meromorphic function of the "tuning" parameters on the theory (essentially those coupling constants mediating the interaction of the systems interacting to made up a bigger one (the full Hamiltonian is always constructed mathematically rigorously from PERTURBATION KATO-RELICH like self adjoint theorems -The general Atomic Hmiltonian for instance ) .Even if one wishes to handle non perturbative phenomena in Quantum Physics ,in much of the cases , one must reformulate the problem in news variables where now the non perturbative scheme becomes a perturbative scheme .However ,computer oriented calculations are somewhat non perturbative in their mathematical nature , but they remains used in a perturbative scheme .So , concluding :Our Differential Integral calculus and Functional Analysis thinking on Quantum Mechanics and Classical Physics remains almost complete only for sort of "Analytical " functions and their close variants .The reader should compare with the mathematical thinking need in probability theory problems , where "analytical functions thinking-perturbation theory is almost zero !(All sample function in space functions of Measurable sets are at best continuous functions like those made up Wienner trajectories ) .






share|cite|improve this answer








New contributor




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






$endgroup$









  • 1




    $begingroup$
    This answer seems quite unclear to me, and I cannot even tell, if it is a real answer to the question.
    $endgroup$
    – flaudemus
    7 hours ago


















-1












$begingroup$

Perturbation Theory in Quantum Mechanics is probably deep rooted with the kind of the mathematical formulation of Quantum Mechanics in terms of Self Adjoint Operators (Von Neumann formulation )and specially the nature of analytical functions expected to describe Physical parameters on quantum physics :The energy bound state levels of a given quantum Hamiltonian is always expected to be at worse a meromorphic function of the "tuning" parameters on the theory (essentially those coupling constants mediating the interaction of the systems interacting to made up a bigger one (the full Hamiltonian is always constructed mathematically rigorously from PERTURBATION KATO-RELICH like self adjoint theorems -The general Atomic Hmiltonian for instance ) .Even if one wishes to handle non perturbative phenomena in Quantum Physics ,in much of the cases , one must reformulate the problem in news variables where now the non perturbative scheme becomes a perturbative scheme .However ,computer oriented calculations are somewhat non perturbative in their mathematical nature , but they remains used in a perturbative scheme .So , concluding :Our Differential Integral calculus and Functional Analysis thinking on Quantum Mechanics and Classical Physics remains almost complete only for sort of "Analytical " functions and their close variants .The reader should compare with the mathematical thinking need in probability theory problems , where "analytical functions thinking-perturbation theory is almost zero !(All sample function in space functions of Measurable sets are at best continuous functions like those made up Wienner trajectories ) .






share|cite|improve this answer








New contributor




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






$endgroup$









  • 1




    $begingroup$
    This answer seems quite unclear to me, and I cannot even tell, if it is a real answer to the question.
    $endgroup$
    – flaudemus
    7 hours ago
















-1












-1








-1





$begingroup$

Perturbation Theory in Quantum Mechanics is probably deep rooted with the kind of the mathematical formulation of Quantum Mechanics in terms of Self Adjoint Operators (Von Neumann formulation )and specially the nature of analytical functions expected to describe Physical parameters on quantum physics :The energy bound state levels of a given quantum Hamiltonian is always expected to be at worse a meromorphic function of the "tuning" parameters on the theory (essentially those coupling constants mediating the interaction of the systems interacting to made up a bigger one (the full Hamiltonian is always constructed mathematically rigorously from PERTURBATION KATO-RELICH like self adjoint theorems -The general Atomic Hmiltonian for instance ) .Even if one wishes to handle non perturbative phenomena in Quantum Physics ,in much of the cases , one must reformulate the problem in news variables where now the non perturbative scheme becomes a perturbative scheme .However ,computer oriented calculations are somewhat non perturbative in their mathematical nature , but they remains used in a perturbative scheme .So , concluding :Our Differential Integral calculus and Functional Analysis thinking on Quantum Mechanics and Classical Physics remains almost complete only for sort of "Analytical " functions and their close variants .The reader should compare with the mathematical thinking need in probability theory problems , where "analytical functions thinking-perturbation theory is almost zero !(All sample function in space functions of Measurable sets are at best continuous functions like those made up Wienner trajectories ) .






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Perturbation Theory in Quantum Mechanics is probably deep rooted with the kind of the mathematical formulation of Quantum Mechanics in terms of Self Adjoint Operators (Von Neumann formulation )and specially the nature of analytical functions expected to describe Physical parameters on quantum physics :The energy bound state levels of a given quantum Hamiltonian is always expected to be at worse a meromorphic function of the "tuning" parameters on the theory (essentially those coupling constants mediating the interaction of the systems interacting to made up a bigger one (the full Hamiltonian is always constructed mathematically rigorously from PERTURBATION KATO-RELICH like self adjoint theorems -The general Atomic Hmiltonian for instance ) .Even if one wishes to handle non perturbative phenomena in Quantum Physics ,in much of the cases , one must reformulate the problem in news variables where now the non perturbative scheme becomes a perturbative scheme .However ,computer oriented calculations are somewhat non perturbative in their mathematical nature , but they remains used in a perturbative scheme .So , concluding :Our Differential Integral calculus and Functional Analysis thinking on Quantum Mechanics and Classical Physics remains almost complete only for sort of "Analytical " functions and their close variants .The reader should compare with the mathematical thinking need in probability theory problems , where "analytical functions thinking-perturbation theory is almost zero !(All sample function in space functions of Measurable sets are at best continuous functions like those made up Wienner trajectories ) .







share|cite|improve this answer








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Luiz C L Botelho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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answered 10 hours ago









Luiz C L BotelhoLuiz C L Botelho

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    This answer seems quite unclear to me, and I cannot even tell, if it is a real answer to the question.
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    – flaudemus
    7 hours ago
















  • 1




    $begingroup$
    This answer seems quite unclear to me, and I cannot even tell, if it is a real answer to the question.
    $endgroup$
    – flaudemus
    7 hours ago










1




1




$begingroup$
This answer seems quite unclear to me, and I cannot even tell, if it is a real answer to the question.
$endgroup$
– flaudemus
7 hours ago






$begingroup$
This answer seems quite unclear to me, and I cannot even tell, if it is a real answer to the question.
$endgroup$
– flaudemus
7 hours ago












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










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