What is ground state useful for?
Many textbooks and research are oriented to computing or finding the ground state of a quantum mechanical system. Why is it such a big deal? What can be done once one has the ground state?
quantum-mechanics ground-state
asked Dec 28 '18 at 19:44
Vladimir Vargas
670412
|
show 1 more comment
Many textbooks and research are oriented to computing or finding the ground state of a quantum mechanical system. Why is it such a big deal? What can be done once one has the ground state?
quantum-mechanics ground-state
asked Dec 28 '18 at 19:44
Vladimir Vargas
670412
It is sometimes easier to get the full spectrum once a ground state is known. Also super symmetric partner potentials are a function of the ground state wave function. This is a seed for a lot of useful information.
– ggcg
Dec 28 '18 at 20:09
@ggcg That seems like an answer, rather than a comment.
– rob♦
Dec 28 '18 at 20:13
I'll give it a try.
– ggcg
Dec 28 '18 at 20:28
1
@ggcg It seems to me that you put answers as comments pretty frequently. Comments are intended to either ask clarifying questions or to suggest improvements to the question. I would say if what you have to say doesn't fall into either of these two categories then you should either post it as an answer or not say anything at all.
– Aaron Stevens
Dec 29 '18 at 4:31
@AaronStevens, I hope you're policing every comment because I frequently see answers as comments from others as well. This gives the impression that it is standard practice.
– ggcg
Dec 29 '18 at 11:54
|
show 1 more comment
Many textbooks and research are oriented to computing or finding the ground state of a quantum mechanical system. Why is it such a big deal? What can be done once one has the ground state?
quantum-mechanics ground-state
asked Dec 28 '18 at 19:44
Vladimir Vargas
670412
Many textbooks and research are oriented to computing or finding the ground state of a quantum mechanical system. Why is it such a big deal? What can be done once one has the ground state?
quantum-mechanics ground-state
quantum-mechanics ground-state
asked Dec 28 '18 at 19:44
Vladimir Vargas
670412
asked Dec 28 '18 at 19:44
Vladimir Vargas
670412
asked Dec 28 '18 at 19:44
Vladimir Vargas
670412
asked Dec 28 '18 at 19:44
Vladimir Vargas
670412
asked Dec 28 '18 at 19:44
Vladimir Vargas
670412
670412
It is sometimes easier to get the full spectrum once a ground state is known. Also super symmetric partner potentials are a function of the ground state wave function. This is a seed for a lot of useful information.
– ggcg
Dec 28 '18 at 20:09
@ggcg That seems like an answer, rather than a comment.
– rob♦
Dec 28 '18 at 20:13
I'll give it a try.
– ggcg
Dec 28 '18 at 20:28
1
@ggcg It seems to me that you put answers as comments pretty frequently. Comments are intended to either ask clarifying questions or to suggest improvements to the question. I would say if what you have to say doesn't fall into either of these two categories then you should either post it as an answer or not say anything at all.
– Aaron Stevens
Dec 29 '18 at 4:31
@AaronStevens, I hope you're policing every comment because I frequently see answers as comments from others as well. This gives the impression that it is standard practice.
– ggcg
Dec 29 '18 at 11:54
|
show 1 more comment
It is sometimes easier to get the full spectrum once a ground state is known. Also super symmetric partner potentials are a function of the ground state wave function. This is a seed for a lot of useful information.
– ggcg
Dec 28 '18 at 20:09
@ggcg That seems like an answer, rather than a comment.
– rob♦
Dec 28 '18 at 20:13
I'll give it a try.
– ggcg
Dec 28 '18 at 20:28
1
@ggcg It seems to me that you put answers as comments pretty frequently. Comments are intended to either ask clarifying questions or to suggest improvements to the question. I would say if what you have to say doesn't fall into either of these two categories then you should either post it as an answer or not say anything at all.
– Aaron Stevens
Dec 29 '18 at 4:31
@AaronStevens, I hope you're policing every comment because I frequently see answers as comments from others as well. This gives the impression that it is standard practice.
– ggcg
Dec 29 '18 at 11:54
It is sometimes easier to get the full spectrum once a ground state is known. Also super symmetric partner potentials are a function of the ground state wave function. This is a seed for a lot of useful information.
– ggcg
Dec 28 '18 at 20:09
It is sometimes easier to get the full spectrum once a ground state is known. Also super symmetric partner potentials are a function of the ground state wave function. This is a seed for a lot of useful information.
– ggcg
Dec 28 '18 at 20:09
@ggcg That seems like an answer, rather than a comment.
– rob♦
Dec 28 '18 at 20:13
@ggcg That seems like an answer, rather than a comment.
– rob♦
Dec 28 '18 at 20:13
I'll give it a try.
– ggcg
Dec 28 '18 at 20:28
I'll give it a try.
– ggcg
Dec 28 '18 at 20:28
1
1
@ggcg It seems to me that you put answers as comments pretty frequently. Comments are intended to either ask clarifying questions or to suggest improvements to the question. I would say if what you have to say doesn't fall into either of these two categories then you should either post it as an answer or not say anything at all.
– Aaron Stevens
Dec 29 '18 at 4:31
@ggcg It seems to me that you put answers as comments pretty frequently. Comments are intended to either ask clarifying questions or to suggest improvements to the question. I would say if what you have to say doesn't fall into either of these two categories then you should either post it as an answer or not say anything at all.
– Aaron Stevens
Dec 29 '18 at 4:31
@AaronStevens, I hope you're policing every comment because I frequently see answers as comments from others as well. This gives the impression that it is standard practice.
– ggcg
Dec 29 '18 at 11:54
@AaronStevens, I hope you're policing every comment because I frequently see answers as comments from others as well. This gives the impression that it is standard practice.
– ggcg
Dec 29 '18 at 11:54
|
show 1 more comment
3 Answers
3
active
oldest
votes
Here is my comment in the form of an answer.
It is sometimes easier to get the full spectrum once a ground state is known. For example the full Schrodinger equation may not be solvable exactly but the ground state may be calculable, along with an energy value. It may also happen that a recursion relation among the higher energy states might be discoverable w/o solving the equation for the eigen functions. If this occurs then the entire spectrum can be determined if the ground state is known. There are a couple famous examples that are in most text books (and serve as a primary reason for focusing on ground states). I suspect your text has these examples in it. I do not recall them off the top of my head but I recall using variation of parameters to get a ground state energy level.
Also super symmetric partner potentials are a function of the ground state wave function. In my opinion this is a better reason to care about ground states. The SHO is an example of a self similar potential so you can generate the entire spectrum from the ground state. In general if you have the spectrum of one system you can define a generalization of the creation and annihilation operators, with a product that gives the Hamiltonian. Reversing the order gives an entirely new Hamiltonian. You can generate the entire set of eigen functions and eigen values for the new Hamiltonian by applying the step operators to the old set of functions, and the spectra line up. All but the ground state. So you need to find the ground state of the new system "by hand".
The SUSY QM is a special case of SUSY QFT from what I recall. This is a seed for a lot of useful information. You can look up a Physics Reports from the 80s on SUSY QM by Cooper, Sukhatme, et al for more details.
answered Dec 28 '18 at 20:39
ggcg
1,06913
Is there any use to supersymmetry to a general audience? My (maybe wrong) impression was that it's an old idea that's been mostly falsified by recent experiments. I would be surprised if supersymmetry is the most useful consequence of such a fundamental thing.
– Steven Sagona
Dec 29 '18 at 6:00
I think you are referring to SUSY particles. SUSY QM has all but doubled the number of exactly solvable systems and has found use in acoustics and solid state physics in periodic potentials. No experiment has proven falsehood in that application. There are some excellent QM texts include SUSY potentials in a later chapter or appendix.
– ggcg
Dec 29 '18 at 11:56
add a comment |
The ground state is often the state of lowest energy, i.e. the state that nature will most probably occupy, and if not, there must be a reason why not, a reason why the system should be excited.
In addition, the ground state is very often the zero-order approximation, e.g. of perturbation theory or mean-field approximation, so the ground state gives the highest contribution to your solution and the higher orders will give certain, maybe in your case even negligible, contributions or corrections to the calculated energy level.
It gives you a raw estimate of whatever solution you are looking for.
answered Dec 28 '18 at 19:56
kalle
396114
1
Sorry I don't see how the ground state is a zeroth-order approximation? Approximation to what? It is certainly not the zeroth-order approximation to any excited state...
– ZeroTheHero
Dec 28 '18 at 20:04
@ZeroTheHero If the excited state is brought in to existence by breaking the degeneracy of the ground state with a small perturbation, as is the case when a proton is subjected to a small external magnetic field, the excited state will not be very different from the original ground state.
– Display Name
Dec 29 '18 at 2:18
@DisplayName Are you saying the first excited state of - say - the 3D harmonic can be obtained from the perturbation of the ground state? How do you do perturbation using only the ground state?
– ZeroTheHero
Dec 29 '18 at 2:44
@ZeroTheHero You might actually have a change of getting it if the spring is weak enough, not from the 3D SHO ground state, but from the free particle ground state. I'm not sure if the harmonic oscillator potential will let you do this, but it works sometimes.
– Display Name
Dec 29 '18 at 2:49
@DisplayName I would looooove to see this.
– ZeroTheHero
Dec 29 '18 at 2:51
|
show 1 more comment
If you are looking for the eigenfunctions of the state's energy, the fact that they are all perpendicular can help. First you have the ground state, but if you remove the ground state from the Hilbert space you are working in (by requiring everything to be perpendicular to it), you can find the second lowest-energy state by computing the "ground state" of the new, constrained system. So in a silly mathematical sense, every state of your system is a ground state of a constrained version of it, so all you ever do is find ground states.
answered Dec 29 '18 at 4:16
Display Name
37818
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
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oldest
votes
Here is my comment in the form of an answer.
It is sometimes easier to get the full spectrum once a ground state is known. For example the full Schrodinger equation may not be solvable exactly but the ground state may be calculable, along with an energy value. It may also happen that a recursion relation among the higher energy states might be discoverable w/o solving the equation for the eigen functions. If this occurs then the entire spectrum can be determined if the ground state is known. There are a couple famous examples that are in most text books (and serve as a primary reason for focusing on ground states). I suspect your text has these examples in it. I do not recall them off the top of my head but I recall using variation of parameters to get a ground state energy level.
Also super symmetric partner potentials are a function of the ground state wave function. In my opinion this is a better reason to care about ground states. The SHO is an example of a self similar potential so you can generate the entire spectrum from the ground state. In general if you have the spectrum of one system you can define a generalization of the creation and annihilation operators, with a product that gives the Hamiltonian. Reversing the order gives an entirely new Hamiltonian. You can generate the entire set of eigen functions and eigen values for the new Hamiltonian by applying the step operators to the old set of functions, and the spectra line up. All but the ground state. So you need to find the ground state of the new system "by hand".
The SUSY QM is a special case of SUSY QFT from what I recall. This is a seed for a lot of useful information. You can look up a Physics Reports from the 80s on SUSY QM by Cooper, Sukhatme, et al for more details.
answered Dec 28 '18 at 20:39
ggcg
1,06913
Is there any use to supersymmetry to a general audience? My (maybe wrong) impression was that it's an old idea that's been mostly falsified by recent experiments. I would be surprised if supersymmetry is the most useful consequence of such a fundamental thing.
– Steven Sagona
Dec 29 '18 at 6:00
I think you are referring to SUSY particles. SUSY QM has all but doubled the number of exactly solvable systems and has found use in acoustics and solid state physics in periodic potentials. No experiment has proven falsehood in that application. There are some excellent QM texts include SUSY potentials in a later chapter or appendix.
– ggcg
Dec 29 '18 at 11:56
add a comment |
Here is my comment in the form of an answer.
It is sometimes easier to get the full spectrum once a ground state is known. For example the full Schrodinger equation may not be solvable exactly but the ground state may be calculable, along with an energy value. It may also happen that a recursion relation among the higher energy states might be discoverable w/o solving the equation for the eigen functions. If this occurs then the entire spectrum can be determined if the ground state is known. There are a couple famous examples that are in most text books (and serve as a primary reason for focusing on ground states). I suspect your text has these examples in it. I do not recall them off the top of my head but I recall using variation of parameters to get a ground state energy level.
Also super symmetric partner potentials are a function of the ground state wave function. In my opinion this is a better reason to care about ground states. The SHO is an example of a self similar potential so you can generate the entire spectrum from the ground state. In general if you have the spectrum of one system you can define a generalization of the creation and annihilation operators, with a product that gives the Hamiltonian. Reversing the order gives an entirely new Hamiltonian. You can generate the entire set of eigen functions and eigen values for the new Hamiltonian by applying the step operators to the old set of functions, and the spectra line up. All but the ground state. So you need to find the ground state of the new system "by hand".
The SUSY QM is a special case of SUSY QFT from what I recall. This is a seed for a lot of useful information. You can look up a Physics Reports from the 80s on SUSY QM by Cooper, Sukhatme, et al for more details.
answered Dec 28 '18 at 20:39
ggcg
1,06913
Is there any use to supersymmetry to a general audience? My (maybe wrong) impression was that it's an old idea that's been mostly falsified by recent experiments. I would be surprised if supersymmetry is the most useful consequence of such a fundamental thing.
– Steven Sagona
Dec 29 '18 at 6:00
I think you are referring to SUSY particles. SUSY QM has all but doubled the number of exactly solvable systems and has found use in acoustics and solid state physics in periodic potentials. No experiment has proven falsehood in that application. There are some excellent QM texts include SUSY potentials in a later chapter or appendix.
– ggcg
Dec 29 '18 at 11:56
add a comment |
Here is my comment in the form of an answer.
It is sometimes easier to get the full spectrum once a ground state is known. For example the full Schrodinger equation may not be solvable exactly but the ground state may be calculable, along with an energy value. It may also happen that a recursion relation among the higher energy states might be discoverable w/o solving the equation for the eigen functions. If this occurs then the entire spectrum can be determined if the ground state is known. There are a couple famous examples that are in most text books (and serve as a primary reason for focusing on ground states). I suspect your text has these examples in it. I do not recall them off the top of my head but I recall using variation of parameters to get a ground state energy level.
Also super symmetric partner potentials are a function of the ground state wave function. In my opinion this is a better reason to care about ground states. The SHO is an example of a self similar potential so you can generate the entire spectrum from the ground state. In general if you have the spectrum of one system you can define a generalization of the creation and annihilation operators, with a product that gives the Hamiltonian. Reversing the order gives an entirely new Hamiltonian. You can generate the entire set of eigen functions and eigen values for the new Hamiltonian by applying the step operators to the old set of functions, and the spectra line up. All but the ground state. So you need to find the ground state of the new system "by hand".
The SUSY QM is a special case of SUSY QFT from what I recall. This is a seed for a lot of useful information. You can look up a Physics Reports from the 80s on SUSY QM by Cooper, Sukhatme, et al for more details.
answered Dec 28 '18 at 20:39
ggcg
1,06913
Here is my comment in the form of an answer.
It is sometimes easier to get the full spectrum once a ground state is known. For example the full Schrodinger equation may not be solvable exactly but the ground state may be calculable, along with an energy value. It may also happen that a recursion relation among the higher energy states might be discoverable w/o solving the equation for the eigen functions. If this occurs then the entire spectrum can be determined if the ground state is known. There are a couple famous examples that are in most text books (and serve as a primary reason for focusing on ground states). I suspect your text has these examples in it. I do not recall them off the top of my head but I recall using variation of parameters to get a ground state energy level.
Also super symmetric partner potentials are a function of the ground state wave function. In my opinion this is a better reason to care about ground states. The SHO is an example of a self similar potential so you can generate the entire spectrum from the ground state. In general if you have the spectrum of one system you can define a generalization of the creation and annihilation operators, with a product that gives the Hamiltonian. Reversing the order gives an entirely new Hamiltonian. You can generate the entire set of eigen functions and eigen values for the new Hamiltonian by applying the step operators to the old set of functions, and the spectra line up. All but the ground state. So you need to find the ground state of the new system "by hand".
The SUSY QM is a special case of SUSY QFT from what I recall. This is a seed for a lot of useful information. You can look up a Physics Reports from the 80s on SUSY QM by Cooper, Sukhatme, et al for more details.
answered Dec 28 '18 at 20:39
ggcg
1,06913
answered Dec 28 '18 at 20:39
ggcg
1,06913
answered Dec 28 '18 at 20:39
ggcg
1,06913
answered Dec 28 '18 at 20:39
ggcg
1,06913
1,06913
Is there any use to supersymmetry to a general audience? My (maybe wrong) impression was that it's an old idea that's been mostly falsified by recent experiments. I would be surprised if supersymmetry is the most useful consequence of such a fundamental thing.
– Steven Sagona
Dec 29 '18 at 6:00
I think you are referring to SUSY particles. SUSY QM has all but doubled the number of exactly solvable systems and has found use in acoustics and solid state physics in periodic potentials. No experiment has proven falsehood in that application. There are some excellent QM texts include SUSY potentials in a later chapter or appendix.
– ggcg
Dec 29 '18 at 11:56
add a comment |
Is there any use to supersymmetry to a general audience? My (maybe wrong) impression was that it's an old idea that's been mostly falsified by recent experiments. I would be surprised if supersymmetry is the most useful consequence of such a fundamental thing.
– Steven Sagona
Dec 29 '18 at 6:00
I think you are referring to SUSY particles. SUSY QM has all but doubled the number of exactly solvable systems and has found use in acoustics and solid state physics in periodic potentials. No experiment has proven falsehood in that application. There are some excellent QM texts include SUSY potentials in a later chapter or appendix.
– ggcg
Dec 29 '18 at 11:56
Is there any use to supersymmetry to a general audience? My (maybe wrong) impression was that it's an old idea that's been mostly falsified by recent experiments. I would be surprised if supersymmetry is the most useful consequence of such a fundamental thing.
– Steven Sagona
Dec 29 '18 at 6:00
Is there any use to supersymmetry to a general audience? My (maybe wrong) impression was that it's an old idea that's been mostly falsified by recent experiments. I would be surprised if supersymmetry is the most useful consequence of such a fundamental thing.
– Steven Sagona
Dec 29 '18 at 6:00
I think you are referring to SUSY particles. SUSY QM has all but doubled the number of exactly solvable systems and has found use in acoustics and solid state physics in periodic potentials. No experiment has proven falsehood in that application. There are some excellent QM texts include SUSY potentials in a later chapter or appendix.
– ggcg
Dec 29 '18 at 11:56
I think you are referring to SUSY particles. SUSY QM has all but doubled the number of exactly solvable systems and has found use in acoustics and solid state physics in periodic potentials. No experiment has proven falsehood in that application. There are some excellent QM texts include SUSY potentials in a later chapter or appendix.
– ggcg
Dec 29 '18 at 11:56
add a comment |
The ground state is often the state of lowest energy, i.e. the state that nature will most probably occupy, and if not, there must be a reason why not, a reason why the system should be excited.
In addition, the ground state is very often the zero-order approximation, e.g. of perturbation theory or mean-field approximation, so the ground state gives the highest contribution to your solution and the higher orders will give certain, maybe in your case even negligible, contributions or corrections to the calculated energy level.
It gives you a raw estimate of whatever solution you are looking for.
answered Dec 28 '18 at 19:56
kalle
396114
1
Sorry I don't see how the ground state is a zeroth-order approximation? Approximation to what? It is certainly not the zeroth-order approximation to any excited state...
– ZeroTheHero
Dec 28 '18 at 20:04
@ZeroTheHero If the excited state is brought in to existence by breaking the degeneracy of the ground state with a small perturbation, as is the case when a proton is subjected to a small external magnetic field, the excited state will not be very different from the original ground state.
– Display Name
Dec 29 '18 at 2:18
@DisplayName Are you saying the first excited state of - say - the 3D harmonic can be obtained from the perturbation of the ground state? How do you do perturbation using only the ground state?
– ZeroTheHero
Dec 29 '18 at 2:44
@ZeroTheHero You might actually have a change of getting it if the spring is weak enough, not from the 3D SHO ground state, but from the free particle ground state. I'm not sure if the harmonic oscillator potential will let you do this, but it works sometimes.
– Display Name
Dec 29 '18 at 2:49
@DisplayName I would looooove to see this.
– ZeroTheHero
Dec 29 '18 at 2:51
|
show 1 more comment
The ground state is often the state of lowest energy, i.e. the state that nature will most probably occupy, and if not, there must be a reason why not, a reason why the system should be excited.
In addition, the ground state is very often the zero-order approximation, e.g. of perturbation theory or mean-field approximation, so the ground state gives the highest contribution to your solution and the higher orders will give certain, maybe in your case even negligible, contributions or corrections to the calculated energy level.
It gives you a raw estimate of whatever solution you are looking for.
answered Dec 28 '18 at 19:56
kalle
396114
1
Sorry I don't see how the ground state is a zeroth-order approximation? Approximation to what? It is certainly not the zeroth-order approximation to any excited state...
– ZeroTheHero
Dec 28 '18 at 20:04
@ZeroTheHero If the excited state is brought in to existence by breaking the degeneracy of the ground state with a small perturbation, as is the case when a proton is subjected to a small external magnetic field, the excited state will not be very different from the original ground state.
– Display Name
Dec 29 '18 at 2:18
@DisplayName Are you saying the first excited state of - say - the 3D harmonic can be obtained from the perturbation of the ground state? How do you do perturbation using only the ground state?
– ZeroTheHero
Dec 29 '18 at 2:44
@ZeroTheHero You might actually have a change of getting it if the spring is weak enough, not from the 3D SHO ground state, but from the free particle ground state. I'm not sure if the harmonic oscillator potential will let you do this, but it works sometimes.
– Display Name
Dec 29 '18 at 2:49
@DisplayName I would looooove to see this.
– ZeroTheHero
Dec 29 '18 at 2:51
|
show 1 more comment
The ground state is often the state of lowest energy, i.e. the state that nature will most probably occupy, and if not, there must be a reason why not, a reason why the system should be excited.
In addition, the ground state is very often the zero-order approximation, e.g. of perturbation theory or mean-field approximation, so the ground state gives the highest contribution to your solution and the higher orders will give certain, maybe in your case even negligible, contributions or corrections to the calculated energy level.
It gives you a raw estimate of whatever solution you are looking for.
answered Dec 28 '18 at 19:56
kalle
396114
The ground state is often the state of lowest energy, i.e. the state that nature will most probably occupy, and if not, there must be a reason why not, a reason why the system should be excited.
In addition, the ground state is very often the zero-order approximation, e.g. of perturbation theory or mean-field approximation, so the ground state gives the highest contribution to your solution and the higher orders will give certain, maybe in your case even negligible, contributions or corrections to the calculated energy level.
It gives you a raw estimate of whatever solution you are looking for.
answered Dec 28 '18 at 19:56
kalle
396114
answered Dec 28 '18 at 19:56
kalle
396114
answered Dec 28 '18 at 19:56
kalle
396114
answered Dec 28 '18 at 19:56
kalle
396114
396114
1
Sorry I don't see how the ground state is a zeroth-order approximation? Approximation to what? It is certainly not the zeroth-order approximation to any excited state...
– ZeroTheHero
Dec 28 '18 at 20:04
@ZeroTheHero If the excited state is brought in to existence by breaking the degeneracy of the ground state with a small perturbation, as is the case when a proton is subjected to a small external magnetic field, the excited state will not be very different from the original ground state.
– Display Name
Dec 29 '18 at 2:18
@DisplayName Are you saying the first excited state of - say - the 3D harmonic can be obtained from the perturbation of the ground state? How do you do perturbation using only the ground state?
– ZeroTheHero
Dec 29 '18 at 2:44
@ZeroTheHero You might actually have a change of getting it if the spring is weak enough, not from the 3D SHO ground state, but from the free particle ground state. I'm not sure if the harmonic oscillator potential will let you do this, but it works sometimes.
– Display Name
Dec 29 '18 at 2:49
@DisplayName I would looooove to see this.
– ZeroTheHero
Dec 29 '18 at 2:51
|
show 1 more comment
1
Sorry I don't see how the ground state is a zeroth-order approximation? Approximation to what? It is certainly not the zeroth-order approximation to any excited state...
– ZeroTheHero
Dec 28 '18 at 20:04
@ZeroTheHero If the excited state is brought in to existence by breaking the degeneracy of the ground state with a small perturbation, as is the case when a proton is subjected to a small external magnetic field, the excited state will not be very different from the original ground state.
– Display Name
Dec 29 '18 at 2:18
@DisplayName Are you saying the first excited state of - say - the 3D harmonic can be obtained from the perturbation of the ground state? How do you do perturbation using only the ground state?
– ZeroTheHero
Dec 29 '18 at 2:44
@ZeroTheHero You might actually have a change of getting it if the spring is weak enough, not from the 3D SHO ground state, but from the free particle ground state. I'm not sure if the harmonic oscillator potential will let you do this, but it works sometimes.
– Display Name
Dec 29 '18 at 2:49
@DisplayName I would looooove to see this.
– ZeroTheHero
Dec 29 '18 at 2:51
1
1
Sorry I don't see how the ground state is a zeroth-order approximation? Approximation to what? It is certainly not the zeroth-order approximation to any excited state...
– ZeroTheHero
Dec 28 '18 at 20:04
Sorry I don't see how the ground state is a zeroth-order approximation? Approximation to what? It is certainly not the zeroth-order approximation to any excited state...
– ZeroTheHero
Dec 28 '18 at 20:04
@ZeroTheHero If the excited state is brought in to existence by breaking the degeneracy of the ground state with a small perturbation, as is the case when a proton is subjected to a small external magnetic field, the excited state will not be very different from the original ground state.
– Display Name
Dec 29 '18 at 2:18
@ZeroTheHero If the excited state is brought in to existence by breaking the degeneracy of the ground state with a small perturbation, as is the case when a proton is subjected to a small external magnetic field, the excited state will not be very different from the original ground state.
– Display Name
Dec 29 '18 at 2:18
@DisplayName Are you saying the first excited state of - say - the 3D harmonic can be obtained from the perturbation of the ground state? How do you do perturbation using only the ground state?
– ZeroTheHero
Dec 29 '18 at 2:44
@DisplayName Are you saying the first excited state of - say - the 3D harmonic can be obtained from the perturbation of the ground state? How do you do perturbation using only the ground state?
– ZeroTheHero
Dec 29 '18 at 2:44
@ZeroTheHero You might actually have a change of getting it if the spring is weak enough, not from the 3D SHO ground state, but from the free particle ground state. I'm not sure if the harmonic oscillator potential will let you do this, but it works sometimes.
– Display Name
Dec 29 '18 at 2:49
@ZeroTheHero You might actually have a change of getting it if the spring is weak enough, not from the 3D SHO ground state, but from the free particle ground state. I'm not sure if the harmonic oscillator potential will let you do this, but it works sometimes.
– Display Name
Dec 29 '18 at 2:49
@DisplayName I would looooove to see this.
– ZeroTheHero
Dec 29 '18 at 2:51
@DisplayName I would looooove to see this.
– ZeroTheHero
Dec 29 '18 at 2:51
|
show 1 more comment
If you are looking for the eigenfunctions of the state's energy, the fact that they are all perpendicular can help. First you have the ground state, but if you remove the ground state from the Hilbert space you are working in (by requiring everything to be perpendicular to it), you can find the second lowest-energy state by computing the "ground state" of the new, constrained system. So in a silly mathematical sense, every state of your system is a ground state of a constrained version of it, so all you ever do is find ground states.
answered Dec 29 '18 at 4:16
Display Name
37818
add a comment |
If you are looking for the eigenfunctions of the state's energy, the fact that they are all perpendicular can help. First you have the ground state, but if you remove the ground state from the Hilbert space you are working in (by requiring everything to be perpendicular to it), you can find the second lowest-energy state by computing the "ground state" of the new, constrained system. So in a silly mathematical sense, every state of your system is a ground state of a constrained version of it, so all you ever do is find ground states.
answered Dec 29 '18 at 4:16
Display Name
37818
add a comment |
If you are looking for the eigenfunctions of the state's energy, the fact that they are all perpendicular can help. First you have the ground state, but if you remove the ground state from the Hilbert space you are working in (by requiring everything to be perpendicular to it), you can find the second lowest-energy state by computing the "ground state" of the new, constrained system. So in a silly mathematical sense, every state of your system is a ground state of a constrained version of it, so all you ever do is find ground states.
answered Dec 29 '18 at 4:16
Display Name
37818
If you are looking for the eigenfunctions of the state's energy, the fact that they are all perpendicular can help. First you have the ground state, but if you remove the ground state from the Hilbert space you are working in (by requiring everything to be perpendicular to it), you can find the second lowest-energy state by computing the "ground state" of the new, constrained system. So in a silly mathematical sense, every state of your system is a ground state of a constrained version of it, so all you ever do is find ground states.
answered Dec 29 '18 at 4:16
Display Name
37818
answered Dec 29 '18 at 4:16
Display Name
37818
answered Dec 29 '18 at 4:16
Display Name
37818
answered Dec 29 '18 at 4:16
Display Name
37818
37818
add a comment |
add a comment |
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It is sometimes easier to get the full spectrum once a ground state is known. Also super symmetric partner potentials are a function of the ground state wave function. This is a seed for a lot of useful information.
– ggcg
Dec 28 '18 at 20:09
@ggcg That seems like an answer, rather than a comment.
– rob♦
Dec 28 '18 at 20:13
I'll give it a try.
– ggcg
Dec 28 '18 at 20:28
1
@ggcg It seems to me that you put answers as comments pretty frequently. Comments are intended to either ask clarifying questions or to suggest improvements to the question. I would say if what you have to say doesn't fall into either of these two categories then you should either post it as an answer or not say anything at all.
– Aaron Stevens
Dec 29 '18 at 4:31
@AaronStevens, I hope you're policing every comment because I frequently see answers as comments from others as well. This gives the impression that it is standard practice.
– ggcg
Dec 29 '18 at 11:54