What is ground state useful for?












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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?










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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
    2 hours ago










  • @ggcg That seems like an answer, rather than a comment.
    – rob
    2 hours ago










  • I'll give it a try.
    – ggcg
    2 hours ago
















3














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?










share|cite|improve this question






















  • 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
    2 hours ago










  • @ggcg That seems like an answer, rather than a comment.
    – rob
    2 hours ago










  • I'll give it a try.
    – ggcg
    2 hours ago














3












3








3


1





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?










share|cite|improve this question













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






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asked 3 hours ago









Vladimir Vargas

669412




669412












  • 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
    2 hours ago










  • @ggcg That seems like an answer, rather than a comment.
    – rob
    2 hours ago










  • I'll give it a try.
    – ggcg
    2 hours ago


















  • 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
    2 hours ago










  • @ggcg That seems like an answer, rather than a comment.
    – rob
    2 hours ago










  • I'll give it a try.
    – ggcg
    2 hours ago
















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
2 hours ago




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
2 hours ago












@ggcg That seems like an answer, rather than a comment.
– rob
2 hours ago




@ggcg That seems like an answer, rather than a comment.
– rob
2 hours ago












I'll give it a try.
– ggcg
2 hours ago




I'll give it a try.
– ggcg
2 hours ago










2 Answers
2






active

oldest

votes


















1














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.






share|cite|improve this answer





















  • 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
    2 hours ago



















1














Here is my comment in the form of an answer.




  1. 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.


  2. 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.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

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    active

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    votes






    active

    oldest

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    1














    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.






    share|cite|improve this answer





















    • 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
      2 hours ago
















    1














    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.






    share|cite|improve this answer





















    • 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
      2 hours ago














    1












    1








    1






    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.






    share|cite|improve this answer












    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 3 hours ago









    kalle

    381114




    381114












    • 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
      2 hours ago


















    • 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
      2 hours ago
















    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
    2 hours ago




    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
    2 hours ago











    1














    Here is my comment in the form of an answer.




    1. 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.


    2. 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.






    share|cite|improve this answer


























      1














      Here is my comment in the form of an answer.




      1. 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.


      2. 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.






      share|cite|improve this answer
























        1












        1








        1






        Here is my comment in the form of an answer.




        1. 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.


        2. 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.






        share|cite|improve this answer












        Here is my comment in the form of an answer.




        1. 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.


        2. 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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        ggcg

        1,01613




        1,01613






























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