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Picking the different solutions to the time independent Schrodinger eqaution


Solving the time independent Schrodinger equation: Does a complex solution make sense?Change of variable in harmonic oscillator time independent Schrodinger equationQuantum mechanics in electric fieldConstructing solutions to the time-dependent Schrödinger's equationProtocol for solving time independent Schrodinger equationScaling the Time Independent Schrodinger EquationMomentum in time independent schrodinger equationOn Griffith Quantum example 2.1: normalization of wave function in time.How are the symmetry(ies) of the Hamiltonian reflected on the general solutions of the time-dependent Schrodinger equation?1D Time independent Schrodinger eq. with limit













1












$begingroup$


The time independent Schrodinger equation
$$-frachbar^22m fracd^2psidx^2+Vpsi = Epsi$$ can have many different solutions of $psi$ for a particular value of $E$.



For example, if we found a complex solution $psi(x)$ for a particular value of $E$, say $E_0$, we can write $psi(x)=a(x)+ib(x)$. Then $a(x)$ and $b(x)$ will also be solutions to the T.I.S.E with $E=E_0$. Furthermore $c_1a(x)+c_2b(x)$ will also be solutions with $c_1$ and $c_2$ being arbitrary constants.



I read that one can always choose any of these solutions as the solution for the stationary state with energy $E_0$. But does that mean all these different solutions represent the same physical state of a particle?



The expectation value of any dynamical variable $Q(x,p)$ is given by
$int psi^*Q(x,frachbarifracddx)psi. $ How do we know for sure that $int a(x)^*Q(x,frachbarifracddx)a(x) $ and $int b(x)^*Q(x,frachbarifracddx)b(x)$ gives the same expectation values?










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$endgroup$











  • $begingroup$
    The last two integrals don't seem to have an infinitesimal?
    $endgroup$
    – Gert
    49 mins ago















1












$begingroup$


The time independent Schrodinger equation
$$-frachbar^22m fracd^2psidx^2+Vpsi = Epsi$$ can have many different solutions of $psi$ for a particular value of $E$.



For example, if we found a complex solution $psi(x)$ for a particular value of $E$, say $E_0$, we can write $psi(x)=a(x)+ib(x)$. Then $a(x)$ and $b(x)$ will also be solutions to the T.I.S.E with $E=E_0$. Furthermore $c_1a(x)+c_2b(x)$ will also be solutions with $c_1$ and $c_2$ being arbitrary constants.



I read that one can always choose any of these solutions as the solution for the stationary state with energy $E_0$. But does that mean all these different solutions represent the same physical state of a particle?



The expectation value of any dynamical variable $Q(x,p)$ is given by
$int psi^*Q(x,frachbarifracddx)psi. $ How do we know for sure that $int a(x)^*Q(x,frachbarifracddx)a(x) $ and $int b(x)^*Q(x,frachbarifracddx)b(x)$ gives the same expectation values?










share|cite|improve this question









$endgroup$











  • $begingroup$
    The last two integrals don't seem to have an infinitesimal?
    $endgroup$
    – Gert
    49 mins ago













1












1








1





$begingroup$


The time independent Schrodinger equation
$$-frachbar^22m fracd^2psidx^2+Vpsi = Epsi$$ can have many different solutions of $psi$ for a particular value of $E$.



For example, if we found a complex solution $psi(x)$ for a particular value of $E$, say $E_0$, we can write $psi(x)=a(x)+ib(x)$. Then $a(x)$ and $b(x)$ will also be solutions to the T.I.S.E with $E=E_0$. Furthermore $c_1a(x)+c_2b(x)$ will also be solutions with $c_1$ and $c_2$ being arbitrary constants.



I read that one can always choose any of these solutions as the solution for the stationary state with energy $E_0$. But does that mean all these different solutions represent the same physical state of a particle?



The expectation value of any dynamical variable $Q(x,p)$ is given by
$int psi^*Q(x,frachbarifracddx)psi. $ How do we know for sure that $int a(x)^*Q(x,frachbarifracddx)a(x) $ and $int b(x)^*Q(x,frachbarifracddx)b(x)$ gives the same expectation values?










share|cite|improve this question









$endgroup$




The time independent Schrodinger equation
$$-frachbar^22m fracd^2psidx^2+Vpsi = Epsi$$ can have many different solutions of $psi$ for a particular value of $E$.



For example, if we found a complex solution $psi(x)$ for a particular value of $E$, say $E_0$, we can write $psi(x)=a(x)+ib(x)$. Then $a(x)$ and $b(x)$ will also be solutions to the T.I.S.E with $E=E_0$. Furthermore $c_1a(x)+c_2b(x)$ will also be solutions with $c_1$ and $c_2$ being arbitrary constants.



I read that one can always choose any of these solutions as the solution for the stationary state with energy $E_0$. But does that mean all these different solutions represent the same physical state of a particle?



The expectation value of any dynamical variable $Q(x,p)$ is given by
$int psi^*Q(x,frachbarifracddx)psi. $ How do we know for sure that $int a(x)^*Q(x,frachbarifracddx)a(x) $ and $int b(x)^*Q(x,frachbarifracddx)b(x)$ gives the same expectation values?







quantum-mechanics schroedinger-equation






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asked 2 hours ago









TaeNyFanTaeNyFan

62414




62414











  • $begingroup$
    The last two integrals don't seem to have an infinitesimal?
    $endgroup$
    – Gert
    49 mins ago
















  • $begingroup$
    The last two integrals don't seem to have an infinitesimal?
    $endgroup$
    – Gert
    49 mins ago















$begingroup$
The last two integrals don't seem to have an infinitesimal?
$endgroup$
– Gert
49 mins ago




$begingroup$
The last two integrals don't seem to have an infinitesimal?
$endgroup$
– Gert
49 mins ago










3 Answers
3






active

oldest

votes


















1












$begingroup$

Solutions can be degenerate with the same expectation value of the energy, but they can have different expectation values for other operators like angular momentum or it $z$ component.



For example the hydrogenic wave functions with $n=2$: $2s$ and $2p$. And then there are three different $2p$ wave functions that can be written as linear combinations of $2p_x, 2p_y, 2p_z$ or of the spherical harmonics $Y_ell,m$ with $ell=1$ and $m=-1,0,1$.






share|cite|improve this answer









$endgroup$




















    1












    $begingroup$


    ... one can always choose any of these solutions as the solution for the stationary state with energy $E_0$.




    That statement is misleading because of how it uses the word "the," suggesting uniqueness. If both occurrances of "the" were replaced by "a," then the statement would make sense.



    A given observable will typically have different expectation values in two different states with the same energy. For example, when $V=0$, the functions $exp(pm ipx/hbar)$ both have energy $E=p^2/2m$, but they have different eigenvalues ($pm p$) of the momentum operator $P=-ihbarpartial/partial x$.



    (I'm being relaxed here about using words like "eigenstate" for non-normalizable functions, but I think those mathematical technicalities are beside the point of this question.)






    share|cite|improve this answer









    $endgroup$




















      1












      $begingroup$

      1. In ordinary quantum mechanics, two wavefunctions represent the same physical state if and only if they are multiples of each other, that is $psi$ and $cpsi$ represent the same state for any $cinmathbbC$. If you insist on wavefunctions being normalized, then $c$ is restricted to complex numbers of absolute value 1, i.e. number of the form $mathrme^mathrmik$ for some $kin[0,2pi)$.



      2. If $psi(x) = a(x) + mathrmib(x)$ is a solution of the Schrödinger equation, then it is not automatically true that $a(x)$ and $b(x)$ are also solutions. It is "accidentally" true for the time-independent Schrödinger equation because applying complex conjugation shows us directly that $psi^ast(x)$ is a solution if $psi(x)$ is, and $a(x)$ and $b(x)$ can be obtained by linear combinations of $psi(x)$ and $psi^ast(x)$.



        There are now two cases: If $psi$ and $psi^ast$ are not linearly independent - i.e. one can be obtained from the other by multiplication with a complex constant - then the space of solutions for this energy is still one-dimensional, and there's only a single physical state. If they are linearly independent, then there are at least two distinct physical states with this energy.



        Note that already the free particle with $V=0$ gives a counter-example to the claim that all solutions for the same energy have the same values for all expectation values. There we have plane wave solutions $psi(x) = mathrme^mathrmipx$ and $psi^ast(x) = mathrme^-mathrmipx$ that are linearly-independent complex conjugates with the same energy that differ in the sign of their expectation value for the momentum operator $p$.







      share|cite|improve this answer









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        3 Answers
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        3 Answers
        3






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        active

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        active

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        1












        $begingroup$

        Solutions can be degenerate with the same expectation value of the energy, but they can have different expectation values for other operators like angular momentum or it $z$ component.



        For example the hydrogenic wave functions with $n=2$: $2s$ and $2p$. And then there are three different $2p$ wave functions that can be written as linear combinations of $2p_x, 2p_y, 2p_z$ or of the spherical harmonics $Y_ell,m$ with $ell=1$ and $m=-1,0,1$.






        share|cite|improve this answer









        $endgroup$

















          1












          $begingroup$

          Solutions can be degenerate with the same expectation value of the energy, but they can have different expectation values for other operators like angular momentum or it $z$ component.



          For example the hydrogenic wave functions with $n=2$: $2s$ and $2p$. And then there are three different $2p$ wave functions that can be written as linear combinations of $2p_x, 2p_y, 2p_z$ or of the spherical harmonics $Y_ell,m$ with $ell=1$ and $m=-1,0,1$.






          share|cite|improve this answer









          $endgroup$















            1












            1








            1





            $begingroup$

            Solutions can be degenerate with the same expectation value of the energy, but they can have different expectation values for other operators like angular momentum or it $z$ component.



            For example the hydrogenic wave functions with $n=2$: $2s$ and $2p$. And then there are three different $2p$ wave functions that can be written as linear combinations of $2p_x, 2p_y, 2p_z$ or of the spherical harmonics $Y_ell,m$ with $ell=1$ and $m=-1,0,1$.






            share|cite|improve this answer









            $endgroup$



            Solutions can be degenerate with the same expectation value of the energy, but they can have different expectation values for other operators like angular momentum or it $z$ component.



            For example the hydrogenic wave functions with $n=2$: $2s$ and $2p$. And then there are three different $2p$ wave functions that can be written as linear combinations of $2p_x, 2p_y, 2p_z$ or of the spherical harmonics $Y_ell,m$ with $ell=1$ and $m=-1,0,1$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 1 hour ago









            PieterPieter

            9,09231536




            9,09231536





















                1












                $begingroup$


                ... one can always choose any of these solutions as the solution for the stationary state with energy $E_0$.




                That statement is misleading because of how it uses the word "the," suggesting uniqueness. If both occurrances of "the" were replaced by "a," then the statement would make sense.



                A given observable will typically have different expectation values in two different states with the same energy. For example, when $V=0$, the functions $exp(pm ipx/hbar)$ both have energy $E=p^2/2m$, but they have different eigenvalues ($pm p$) of the momentum operator $P=-ihbarpartial/partial x$.



                (I'm being relaxed here about using words like "eigenstate" for non-normalizable functions, but I think those mathematical technicalities are beside the point of this question.)






                share|cite|improve this answer









                $endgroup$

















                  1












                  $begingroup$


                  ... one can always choose any of these solutions as the solution for the stationary state with energy $E_0$.




                  That statement is misleading because of how it uses the word "the," suggesting uniqueness. If both occurrances of "the" were replaced by "a," then the statement would make sense.



                  A given observable will typically have different expectation values in two different states with the same energy. For example, when $V=0$, the functions $exp(pm ipx/hbar)$ both have energy $E=p^2/2m$, but they have different eigenvalues ($pm p$) of the momentum operator $P=-ihbarpartial/partial x$.



                  (I'm being relaxed here about using words like "eigenstate" for non-normalizable functions, but I think those mathematical technicalities are beside the point of this question.)






                  share|cite|improve this answer









                  $endgroup$















                    1












                    1








                    1





                    $begingroup$


                    ... one can always choose any of these solutions as the solution for the stationary state with energy $E_0$.




                    That statement is misleading because of how it uses the word "the," suggesting uniqueness. If both occurrances of "the" were replaced by "a," then the statement would make sense.



                    A given observable will typically have different expectation values in two different states with the same energy. For example, when $V=0$, the functions $exp(pm ipx/hbar)$ both have energy $E=p^2/2m$, but they have different eigenvalues ($pm p$) of the momentum operator $P=-ihbarpartial/partial x$.



                    (I'm being relaxed here about using words like "eigenstate" for non-normalizable functions, but I think those mathematical technicalities are beside the point of this question.)






                    share|cite|improve this answer









                    $endgroup$




                    ... one can always choose any of these solutions as the solution for the stationary state with energy $E_0$.




                    That statement is misleading because of how it uses the word "the," suggesting uniqueness. If both occurrances of "the" were replaced by "a," then the statement would make sense.



                    A given observable will typically have different expectation values in two different states with the same energy. For example, when $V=0$, the functions $exp(pm ipx/hbar)$ both have energy $E=p^2/2m$, but they have different eigenvalues ($pm p$) of the momentum operator $P=-ihbarpartial/partial x$.



                    (I'm being relaxed here about using words like "eigenstate" for non-normalizable functions, but I think those mathematical technicalities are beside the point of this question.)







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    Chiral AnomalyChiral Anomaly

                    12.1k21540




                    12.1k21540





















                        1












                        $begingroup$

                        1. In ordinary quantum mechanics, two wavefunctions represent the same physical state if and only if they are multiples of each other, that is $psi$ and $cpsi$ represent the same state for any $cinmathbbC$. If you insist on wavefunctions being normalized, then $c$ is restricted to complex numbers of absolute value 1, i.e. number of the form $mathrme^mathrmik$ for some $kin[0,2pi)$.



                        2. If $psi(x) = a(x) + mathrmib(x)$ is a solution of the Schrödinger equation, then it is not automatically true that $a(x)$ and $b(x)$ are also solutions. It is "accidentally" true for the time-independent Schrödinger equation because applying complex conjugation shows us directly that $psi^ast(x)$ is a solution if $psi(x)$ is, and $a(x)$ and $b(x)$ can be obtained by linear combinations of $psi(x)$ and $psi^ast(x)$.



                          There are now two cases: If $psi$ and $psi^ast$ are not linearly independent - i.e. one can be obtained from the other by multiplication with a complex constant - then the space of solutions for this energy is still one-dimensional, and there's only a single physical state. If they are linearly independent, then there are at least two distinct physical states with this energy.



                          Note that already the free particle with $V=0$ gives a counter-example to the claim that all solutions for the same energy have the same values for all expectation values. There we have plane wave solutions $psi(x) = mathrme^mathrmipx$ and $psi^ast(x) = mathrme^-mathrmipx$ that are linearly-independent complex conjugates with the same energy that differ in the sign of their expectation value for the momentum operator $p$.







                        share|cite|improve this answer









                        $endgroup$

















                          1












                          $begingroup$

                          1. In ordinary quantum mechanics, two wavefunctions represent the same physical state if and only if they are multiples of each other, that is $psi$ and $cpsi$ represent the same state for any $cinmathbbC$. If you insist on wavefunctions being normalized, then $c$ is restricted to complex numbers of absolute value 1, i.e. number of the form $mathrme^mathrmik$ for some $kin[0,2pi)$.



                          2. If $psi(x) = a(x) + mathrmib(x)$ is a solution of the Schrödinger equation, then it is not automatically true that $a(x)$ and $b(x)$ are also solutions. It is "accidentally" true for the time-independent Schrödinger equation because applying complex conjugation shows us directly that $psi^ast(x)$ is a solution if $psi(x)$ is, and $a(x)$ and $b(x)$ can be obtained by linear combinations of $psi(x)$ and $psi^ast(x)$.



                            There are now two cases: If $psi$ and $psi^ast$ are not linearly independent - i.e. one can be obtained from the other by multiplication with a complex constant - then the space of solutions for this energy is still one-dimensional, and there's only a single physical state. If they are linearly independent, then there are at least two distinct physical states with this energy.



                            Note that already the free particle with $V=0$ gives a counter-example to the claim that all solutions for the same energy have the same values for all expectation values. There we have plane wave solutions $psi(x) = mathrme^mathrmipx$ and $psi^ast(x) = mathrme^-mathrmipx$ that are linearly-independent complex conjugates with the same energy that differ in the sign of their expectation value for the momentum operator $p$.







                          share|cite|improve this answer









                          $endgroup$















                            1












                            1








                            1





                            $begingroup$

                            1. In ordinary quantum mechanics, two wavefunctions represent the same physical state if and only if they are multiples of each other, that is $psi$ and $cpsi$ represent the same state for any $cinmathbbC$. If you insist on wavefunctions being normalized, then $c$ is restricted to complex numbers of absolute value 1, i.e. number of the form $mathrme^mathrmik$ for some $kin[0,2pi)$.



                            2. If $psi(x) = a(x) + mathrmib(x)$ is a solution of the Schrödinger equation, then it is not automatically true that $a(x)$ and $b(x)$ are also solutions. It is "accidentally" true for the time-independent Schrödinger equation because applying complex conjugation shows us directly that $psi^ast(x)$ is a solution if $psi(x)$ is, and $a(x)$ and $b(x)$ can be obtained by linear combinations of $psi(x)$ and $psi^ast(x)$.



                              There are now two cases: If $psi$ and $psi^ast$ are not linearly independent - i.e. one can be obtained from the other by multiplication with a complex constant - then the space of solutions for this energy is still one-dimensional, and there's only a single physical state. If they are linearly independent, then there are at least two distinct physical states with this energy.



                              Note that already the free particle with $V=0$ gives a counter-example to the claim that all solutions for the same energy have the same values for all expectation values. There we have plane wave solutions $psi(x) = mathrme^mathrmipx$ and $psi^ast(x) = mathrme^-mathrmipx$ that are linearly-independent complex conjugates with the same energy that differ in the sign of their expectation value for the momentum operator $p$.







                            share|cite|improve this answer









                            $endgroup$



                            1. In ordinary quantum mechanics, two wavefunctions represent the same physical state if and only if they are multiples of each other, that is $psi$ and $cpsi$ represent the same state for any $cinmathbbC$. If you insist on wavefunctions being normalized, then $c$ is restricted to complex numbers of absolute value 1, i.e. number of the form $mathrme^mathrmik$ for some $kin[0,2pi)$.



                            2. If $psi(x) = a(x) + mathrmib(x)$ is a solution of the Schrödinger equation, then it is not automatically true that $a(x)$ and $b(x)$ are also solutions. It is "accidentally" true for the time-independent Schrödinger equation because applying complex conjugation shows us directly that $psi^ast(x)$ is a solution if $psi(x)$ is, and $a(x)$ and $b(x)$ can be obtained by linear combinations of $psi(x)$ and $psi^ast(x)$.



                              There are now two cases: If $psi$ and $psi^ast$ are not linearly independent - i.e. one can be obtained from the other by multiplication with a complex constant - then the space of solutions for this energy is still one-dimensional, and there's only a single physical state. If they are linearly independent, then there are at least two distinct physical states with this energy.



                              Note that already the free particle with $V=0$ gives a counter-example to the claim that all solutions for the same energy have the same values for all expectation values. There we have plane wave solutions $psi(x) = mathrme^mathrmipx$ and $psi^ast(x) = mathrme^-mathrmipx$ that are linearly-independent complex conjugates with the same energy that differ in the sign of their expectation value for the momentum operator $p$.








                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 1 hour ago









                            ACuriousMindACuriousMind

                            72.9k18130322




                            72.9k18130322



























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