How does Ehrenfest's theorem apply to the quantum harmonic oscillator?












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Ehrenfest's theorem, to my level of understanding, says that expectation values for quantum mechanical observables obey their Newtonian mechanics counterparts, which means that we can use newton's laws on expectation values. However, in the case of the quantum harmonic oscillator, this clearly does not look newtonian because the expectation value of the position does not oscillate like the newtonian $sinomega t$.



These states are of the form $psi=K(n, xi)e^{-xi^2/2}$. Why do they not obey Ehrenfest's theorem? They don't give a harmonic oscillator, imo.



oscillator










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  • $begingroup$
    Maybe this question of mine can answer your question: physics.stackexchange.com/questions/267835/…
    $endgroup$
    – Quantumwhisp
    15 hours ago










  • $begingroup$
    Though it does not address the question asked, I used to have students plot the position probability distribution for the quantum and classical cases over a range of energies corresponding to, say, $n = 1,2,3,4,10$ and a "big" number like $20$. I'll post a figure in chat: chat.stackexchange.com/transcript/message/49407141#49407141.
    $endgroup$
    – dmckee
    2 hours ago


















10












$begingroup$


Ehrenfest's theorem, to my level of understanding, says that expectation values for quantum mechanical observables obey their Newtonian mechanics counterparts, which means that we can use newton's laws on expectation values. However, in the case of the quantum harmonic oscillator, this clearly does not look newtonian because the expectation value of the position does not oscillate like the newtonian $sinomega t$.



These states are of the form $psi=K(n, xi)e^{-xi^2/2}$. Why do they not obey Ehrenfest's theorem? They don't give a harmonic oscillator, imo.



oscillator










share|cite|improve this question









New contributor




Soda is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • $begingroup$
    Maybe this question of mine can answer your question: physics.stackexchange.com/questions/267835/…
    $endgroup$
    – Quantumwhisp
    15 hours ago










  • $begingroup$
    Though it does not address the question asked, I used to have students plot the position probability distribution for the quantum and classical cases over a range of energies corresponding to, say, $n = 1,2,3,4,10$ and a "big" number like $20$. I'll post a figure in chat: chat.stackexchange.com/transcript/message/49407141#49407141.
    $endgroup$
    – dmckee
    2 hours ago
















10












10








10


3



$begingroup$


Ehrenfest's theorem, to my level of understanding, says that expectation values for quantum mechanical observables obey their Newtonian mechanics counterparts, which means that we can use newton's laws on expectation values. However, in the case of the quantum harmonic oscillator, this clearly does not look newtonian because the expectation value of the position does not oscillate like the newtonian $sinomega t$.



These states are of the form $psi=K(n, xi)e^{-xi^2/2}$. Why do they not obey Ehrenfest's theorem? They don't give a harmonic oscillator, imo.



oscillator










share|cite|improve this question









New contributor




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







$endgroup$




Ehrenfest's theorem, to my level of understanding, says that expectation values for quantum mechanical observables obey their Newtonian mechanics counterparts, which means that we can use newton's laws on expectation values. However, in the case of the quantum harmonic oscillator, this clearly does not look newtonian because the expectation value of the position does not oscillate like the newtonian $sinomega t$.



These states are of the form $psi=K(n, xi)e^{-xi^2/2}$. Why do they not obey Ehrenfest's theorem? They don't give a harmonic oscillator, imo.



oscillator







quantum-mechanics classical-mechanics wavefunction harmonic-oscillator






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Soda is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 10 hours ago









knzhou

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asked 16 hours ago









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  • $begingroup$
    Maybe this question of mine can answer your question: physics.stackexchange.com/questions/267835/…
    $endgroup$
    – Quantumwhisp
    15 hours ago










  • $begingroup$
    Though it does not address the question asked, I used to have students plot the position probability distribution for the quantum and classical cases over a range of energies corresponding to, say, $n = 1,2,3,4,10$ and a "big" number like $20$. I'll post a figure in chat: chat.stackexchange.com/transcript/message/49407141#49407141.
    $endgroup$
    – dmckee
    2 hours ago




















  • $begingroup$
    Maybe this question of mine can answer your question: physics.stackexchange.com/questions/267835/…
    $endgroup$
    – Quantumwhisp
    15 hours ago










  • $begingroup$
    Though it does not address the question asked, I used to have students plot the position probability distribution for the quantum and classical cases over a range of energies corresponding to, say, $n = 1,2,3,4,10$ and a "big" number like $20$. I'll post a figure in chat: chat.stackexchange.com/transcript/message/49407141#49407141.
    $endgroup$
    – dmckee
    2 hours ago


















$begingroup$
Maybe this question of mine can answer your question: physics.stackexchange.com/questions/267835/…
$endgroup$
– Quantumwhisp
15 hours ago




$begingroup$
Maybe this question of mine can answer your question: physics.stackexchange.com/questions/267835/…
$endgroup$
– Quantumwhisp
15 hours ago












$begingroup$
Though it does not address the question asked, I used to have students plot the position probability distribution for the quantum and classical cases over a range of energies corresponding to, say, $n = 1,2,3,4,10$ and a "big" number like $20$. I'll post a figure in chat: chat.stackexchange.com/transcript/message/49407141#49407141.
$endgroup$
– dmckee
2 hours ago






$begingroup$
Though it does not address the question asked, I used to have students plot the position probability distribution for the quantum and classical cases over a range of energies corresponding to, say, $n = 1,2,3,4,10$ and a "big" number like $20$. I'll post a figure in chat: chat.stackexchange.com/transcript/message/49407141#49407141.
$endgroup$
– dmckee
2 hours ago












3 Answers
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19












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It is actually true, in an almost trivial way. The Ehrenfest theorem states that,
begin{equation}
frac{d}{dt}langle xrangle=langle prangle,quad frac{d}{dt}langle prangle =- langle V'(x)rangle
end{equation}

However for all eigenfunctions for the Harmonic oscillator $langle xrangle=0$ (and therefore $langle V'(x)rangle=0$) and $langle prangle=0$. So the Ehrenfest theorem on the eigenstates reduces to $0=0$.



You can see that the general version of the Ehrenfest theorem works trivially for all eigenstates. It states that for the arbitrary observable $A$ its expectation value satisfies the equation,
begin{equation}
frac{d}{dt} langle Arangle=frac{1}{ihbar}langle [A,H]rangle+langle frac{partial A}{partial t}rangle
end{equation}

However on the eigenstates,
begin{equation}
langlepsi_n| [A,H]|psi_nrangle=langlepsi_n|AH-HA|psi_nrangle=E_nlanglepsi_n|A-A|psi_nrangle=0
end{equation}

So the expectation value of the observable that doesn't explicitly depend on time, doesn't evolve on the eigenstates which is what you would expect.



So where does the Ehrenfest theorem lead to the classical dynamics? You need to consider the localized wavepackets. The simplest example would be the coherent state of the Harmonic oscillator that is the Gaussian wavepacket that follows the classical trajectory
Coherent state evolution



For the Harmonic oscillator the Ehrenfest theorem is always "classical" if only in a trivial way (as in case of the eigenstates). However in general the Ehrenfest theorem reduces to the classical equation of motion only on such localized wavepackets that concentrate near the classical trajectory as $hbar$ goes to zero. The key point happens to be the interchange $langle V'(x)rangle mapsto V'(langle xrangle)$ that on general states can't be done. So if you want to recover some classical dynamics from the quantum theory look on the localized wavepackets.






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    6












    $begingroup$


    1. Your version of $psi$ is (I'm sure you know) derived from the Time Independent Schrodinger equation, $hat{H}psi=Epsi$. To find time-dependent solutions, we solve $$ifrac{partial}{partial t}psi=hat{H}psi.$$ You were trying to solve for stationary states, and the entire point of those is that $|psi(x)|^2$ does not change over time. Still, for these stationary solutions solutions,$$frac{mathrm{d}}{mathrm{d}t}left<xright>=left<pright>=0; frac{mathrm{d}}{mathrm{d}t}left<pright>=-left<frac{mathrm{d}}{mathrm{d}x}V (x)right>=0,$$ which is in accordance with the Ehrenfest theorem (albeit uninformatively).


    2. Be careful about the time dependence of your reported $psi$: you're actually dealing with $Psi(x, t)=psi(x)e^{-itE/hbar}$ for those stationary states. Of course, this doesn't relate to the Ehrenfest theorem, but it's something worth mentioning: the complex and real parts are oscillating, as shown by the pink and blue lines in this diagram (from the wikipedia page on QHO):complex and real parts of some QHO states



      Do not make the mistake of suggesting that all derivatives with respect to time are necessarily automatically equal to zero because the wavefunction is time-independent. Contrary to what the shorthand notation suggests, we do have a (separable) time-dependent bit.



    3. Referring to the same diagram, observe parts G and H: these represent coherent states, which can be understood using Ehrenfest's theorem because $|psi|^2$ looks like a Gaussian which follows a classical $sin$ or $cos$ function.



    See




    Kanasugi, H., and H. Okada. “Systematic Treatment of General Time-Dependent Harmonic Oscillator in Classical and Quantum Mechanics.” Progress of Theoretical Physics, vol. 93, no. 5, 1995, pp. 949–960., doi:10.1143/ptp/93.5.949.







    share|cite|improve this answer











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      0












      $begingroup$

      What you wrote down is just a complete set of solutions of the time independent Schrödinger equation.
      begin{align}
      [- frac{hbar²}{2m} Delta + V(x) ] Psi(x) = E Psi(x)
      end{align}

      Of course this solutions don't carry any time dependence, because the time-independent Schrödinger equation (in this representation) only makes statements about functions that depend on the spatial coordinates only (in this case, this is only x).



      How to get to the time-dependent solutions?
      A particular property of the solutions $Psi_{E}$ of the time-independent Schrödinger Equation is that $Psi_{E}(x) e^{-i frac{E}{hbar}t}$ is a solution to the time-dependent Schrödinger equation.



      If you would apply this to your suggested set of solutions of the harmonic oscillator, you would arrive at time dependent solutions. THOSE are the ones that the Ehrenfest-Theorem makes a statement about.
      You would then calculate that the expectation values $<X>$ and $<P>$ do not change. But you would as well calculate that both of this values are 0. This is in perfect agreement with the Theorems of Ehrenfest, and the classical analogon would be a particle resting at the deepest spot of the harmonic potential.






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        3 Answers
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        19












        $begingroup$

        It is actually true, in an almost trivial way. The Ehrenfest theorem states that,
        begin{equation}
        frac{d}{dt}langle xrangle=langle prangle,quad frac{d}{dt}langle prangle =- langle V'(x)rangle
        end{equation}

        However for all eigenfunctions for the Harmonic oscillator $langle xrangle=0$ (and therefore $langle V'(x)rangle=0$) and $langle prangle=0$. So the Ehrenfest theorem on the eigenstates reduces to $0=0$.



        You can see that the general version of the Ehrenfest theorem works trivially for all eigenstates. It states that for the arbitrary observable $A$ its expectation value satisfies the equation,
        begin{equation}
        frac{d}{dt} langle Arangle=frac{1}{ihbar}langle [A,H]rangle+langle frac{partial A}{partial t}rangle
        end{equation}

        However on the eigenstates,
        begin{equation}
        langlepsi_n| [A,H]|psi_nrangle=langlepsi_n|AH-HA|psi_nrangle=E_nlanglepsi_n|A-A|psi_nrangle=0
        end{equation}

        So the expectation value of the observable that doesn't explicitly depend on time, doesn't evolve on the eigenstates which is what you would expect.



        So where does the Ehrenfest theorem lead to the classical dynamics? You need to consider the localized wavepackets. The simplest example would be the coherent state of the Harmonic oscillator that is the Gaussian wavepacket that follows the classical trajectory
        Coherent state evolution



        For the Harmonic oscillator the Ehrenfest theorem is always "classical" if only in a trivial way (as in case of the eigenstates). However in general the Ehrenfest theorem reduces to the classical equation of motion only on such localized wavepackets that concentrate near the classical trajectory as $hbar$ goes to zero. The key point happens to be the interchange $langle V'(x)rangle mapsto V'(langle xrangle)$ that on general states can't be done. So if you want to recover some classical dynamics from the quantum theory look on the localized wavepackets.






        share|cite|improve this answer









        $endgroup$


















          19












          $begingroup$

          It is actually true, in an almost trivial way. The Ehrenfest theorem states that,
          begin{equation}
          frac{d}{dt}langle xrangle=langle prangle,quad frac{d}{dt}langle prangle =- langle V'(x)rangle
          end{equation}

          However for all eigenfunctions for the Harmonic oscillator $langle xrangle=0$ (and therefore $langle V'(x)rangle=0$) and $langle prangle=0$. So the Ehrenfest theorem on the eigenstates reduces to $0=0$.



          You can see that the general version of the Ehrenfest theorem works trivially for all eigenstates. It states that for the arbitrary observable $A$ its expectation value satisfies the equation,
          begin{equation}
          frac{d}{dt} langle Arangle=frac{1}{ihbar}langle [A,H]rangle+langle frac{partial A}{partial t}rangle
          end{equation}

          However on the eigenstates,
          begin{equation}
          langlepsi_n| [A,H]|psi_nrangle=langlepsi_n|AH-HA|psi_nrangle=E_nlanglepsi_n|A-A|psi_nrangle=0
          end{equation}

          So the expectation value of the observable that doesn't explicitly depend on time, doesn't evolve on the eigenstates which is what you would expect.



          So where does the Ehrenfest theorem lead to the classical dynamics? You need to consider the localized wavepackets. The simplest example would be the coherent state of the Harmonic oscillator that is the Gaussian wavepacket that follows the classical trajectory
          Coherent state evolution



          For the Harmonic oscillator the Ehrenfest theorem is always "classical" if only in a trivial way (as in case of the eigenstates). However in general the Ehrenfest theorem reduces to the classical equation of motion only on such localized wavepackets that concentrate near the classical trajectory as $hbar$ goes to zero. The key point happens to be the interchange $langle V'(x)rangle mapsto V'(langle xrangle)$ that on general states can't be done. So if you want to recover some classical dynamics from the quantum theory look on the localized wavepackets.






          share|cite|improve this answer









          $endgroup$
















            19












            19








            19





            $begingroup$

            It is actually true, in an almost trivial way. The Ehrenfest theorem states that,
            begin{equation}
            frac{d}{dt}langle xrangle=langle prangle,quad frac{d}{dt}langle prangle =- langle V'(x)rangle
            end{equation}

            However for all eigenfunctions for the Harmonic oscillator $langle xrangle=0$ (and therefore $langle V'(x)rangle=0$) and $langle prangle=0$. So the Ehrenfest theorem on the eigenstates reduces to $0=0$.



            You can see that the general version of the Ehrenfest theorem works trivially for all eigenstates. It states that for the arbitrary observable $A$ its expectation value satisfies the equation,
            begin{equation}
            frac{d}{dt} langle Arangle=frac{1}{ihbar}langle [A,H]rangle+langle frac{partial A}{partial t}rangle
            end{equation}

            However on the eigenstates,
            begin{equation}
            langlepsi_n| [A,H]|psi_nrangle=langlepsi_n|AH-HA|psi_nrangle=E_nlanglepsi_n|A-A|psi_nrangle=0
            end{equation}

            So the expectation value of the observable that doesn't explicitly depend on time, doesn't evolve on the eigenstates which is what you would expect.



            So where does the Ehrenfest theorem lead to the classical dynamics? You need to consider the localized wavepackets. The simplest example would be the coherent state of the Harmonic oscillator that is the Gaussian wavepacket that follows the classical trajectory
            Coherent state evolution



            For the Harmonic oscillator the Ehrenfest theorem is always "classical" if only in a trivial way (as in case of the eigenstates). However in general the Ehrenfest theorem reduces to the classical equation of motion only on such localized wavepackets that concentrate near the classical trajectory as $hbar$ goes to zero. The key point happens to be the interchange $langle V'(x)rangle mapsto V'(langle xrangle)$ that on general states can't be done. So if you want to recover some classical dynamics from the quantum theory look on the localized wavepackets.






            share|cite|improve this answer









            $endgroup$



            It is actually true, in an almost trivial way. The Ehrenfest theorem states that,
            begin{equation}
            frac{d}{dt}langle xrangle=langle prangle,quad frac{d}{dt}langle prangle =- langle V'(x)rangle
            end{equation}

            However for all eigenfunctions for the Harmonic oscillator $langle xrangle=0$ (and therefore $langle V'(x)rangle=0$) and $langle prangle=0$. So the Ehrenfest theorem on the eigenstates reduces to $0=0$.



            You can see that the general version of the Ehrenfest theorem works trivially for all eigenstates. It states that for the arbitrary observable $A$ its expectation value satisfies the equation,
            begin{equation}
            frac{d}{dt} langle Arangle=frac{1}{ihbar}langle [A,H]rangle+langle frac{partial A}{partial t}rangle
            end{equation}

            However on the eigenstates,
            begin{equation}
            langlepsi_n| [A,H]|psi_nrangle=langlepsi_n|AH-HA|psi_nrangle=E_nlanglepsi_n|A-A|psi_nrangle=0
            end{equation}

            So the expectation value of the observable that doesn't explicitly depend on time, doesn't evolve on the eigenstates which is what you would expect.



            So where does the Ehrenfest theorem lead to the classical dynamics? You need to consider the localized wavepackets. The simplest example would be the coherent state of the Harmonic oscillator that is the Gaussian wavepacket that follows the classical trajectory
            Coherent state evolution



            For the Harmonic oscillator the Ehrenfest theorem is always "classical" if only in a trivial way (as in case of the eigenstates). However in general the Ehrenfest theorem reduces to the classical equation of motion only on such localized wavepackets that concentrate near the classical trajectory as $hbar$ goes to zero. The key point happens to be the interchange $langle V'(x)rangle mapsto V'(langle xrangle)$ that on general states can't be done. So if you want to recover some classical dynamics from the quantum theory look on the localized wavepackets.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 15 hours ago









            OONOON

            4,5792816




            4,5792816























                6












                $begingroup$


                1. Your version of $psi$ is (I'm sure you know) derived from the Time Independent Schrodinger equation, $hat{H}psi=Epsi$. To find time-dependent solutions, we solve $$ifrac{partial}{partial t}psi=hat{H}psi.$$ You were trying to solve for stationary states, and the entire point of those is that $|psi(x)|^2$ does not change over time. Still, for these stationary solutions solutions,$$frac{mathrm{d}}{mathrm{d}t}left<xright>=left<pright>=0; frac{mathrm{d}}{mathrm{d}t}left<pright>=-left<frac{mathrm{d}}{mathrm{d}x}V (x)right>=0,$$ which is in accordance with the Ehrenfest theorem (albeit uninformatively).


                2. Be careful about the time dependence of your reported $psi$: you're actually dealing with $Psi(x, t)=psi(x)e^{-itE/hbar}$ for those stationary states. Of course, this doesn't relate to the Ehrenfest theorem, but it's something worth mentioning: the complex and real parts are oscillating, as shown by the pink and blue lines in this diagram (from the wikipedia page on QHO):complex and real parts of some QHO states



                  Do not make the mistake of suggesting that all derivatives with respect to time are necessarily automatically equal to zero because the wavefunction is time-independent. Contrary to what the shorthand notation suggests, we do have a (separable) time-dependent bit.



                3. Referring to the same diagram, observe parts G and H: these represent coherent states, which can be understood using Ehrenfest's theorem because $|psi|^2$ looks like a Gaussian which follows a classical $sin$ or $cos$ function.



                See




                Kanasugi, H., and H. Okada. “Systematic Treatment of General Time-Dependent Harmonic Oscillator in Classical and Quantum Mechanics.” Progress of Theoretical Physics, vol. 93, no. 5, 1995, pp. 949–960., doi:10.1143/ptp/93.5.949.







                share|cite|improve this answer











                $endgroup$


















                  6












                  $begingroup$


                  1. Your version of $psi$ is (I'm sure you know) derived from the Time Independent Schrodinger equation, $hat{H}psi=Epsi$. To find time-dependent solutions, we solve $$ifrac{partial}{partial t}psi=hat{H}psi.$$ You were trying to solve for stationary states, and the entire point of those is that $|psi(x)|^2$ does not change over time. Still, for these stationary solutions solutions,$$frac{mathrm{d}}{mathrm{d}t}left<xright>=left<pright>=0; frac{mathrm{d}}{mathrm{d}t}left<pright>=-left<frac{mathrm{d}}{mathrm{d}x}V (x)right>=0,$$ which is in accordance with the Ehrenfest theorem (albeit uninformatively).


                  2. Be careful about the time dependence of your reported $psi$: you're actually dealing with $Psi(x, t)=psi(x)e^{-itE/hbar}$ for those stationary states. Of course, this doesn't relate to the Ehrenfest theorem, but it's something worth mentioning: the complex and real parts are oscillating, as shown by the pink and blue lines in this diagram (from the wikipedia page on QHO):complex and real parts of some QHO states



                    Do not make the mistake of suggesting that all derivatives with respect to time are necessarily automatically equal to zero because the wavefunction is time-independent. Contrary to what the shorthand notation suggests, we do have a (separable) time-dependent bit.



                  3. Referring to the same diagram, observe parts G and H: these represent coherent states, which can be understood using Ehrenfest's theorem because $|psi|^2$ looks like a Gaussian which follows a classical $sin$ or $cos$ function.



                  See




                  Kanasugi, H., and H. Okada. “Systematic Treatment of General Time-Dependent Harmonic Oscillator in Classical and Quantum Mechanics.” Progress of Theoretical Physics, vol. 93, no. 5, 1995, pp. 949–960., doi:10.1143/ptp/93.5.949.







                  share|cite|improve this answer











                  $endgroup$
















                    6












                    6








                    6





                    $begingroup$


                    1. Your version of $psi$ is (I'm sure you know) derived from the Time Independent Schrodinger equation, $hat{H}psi=Epsi$. To find time-dependent solutions, we solve $$ifrac{partial}{partial t}psi=hat{H}psi.$$ You were trying to solve for stationary states, and the entire point of those is that $|psi(x)|^2$ does not change over time. Still, for these stationary solutions solutions,$$frac{mathrm{d}}{mathrm{d}t}left<xright>=left<pright>=0; frac{mathrm{d}}{mathrm{d}t}left<pright>=-left<frac{mathrm{d}}{mathrm{d}x}V (x)right>=0,$$ which is in accordance with the Ehrenfest theorem (albeit uninformatively).


                    2. Be careful about the time dependence of your reported $psi$: you're actually dealing with $Psi(x, t)=psi(x)e^{-itE/hbar}$ for those stationary states. Of course, this doesn't relate to the Ehrenfest theorem, but it's something worth mentioning: the complex and real parts are oscillating, as shown by the pink and blue lines in this diagram (from the wikipedia page on QHO):complex and real parts of some QHO states



                      Do not make the mistake of suggesting that all derivatives with respect to time are necessarily automatically equal to zero because the wavefunction is time-independent. Contrary to what the shorthand notation suggests, we do have a (separable) time-dependent bit.



                    3. Referring to the same diagram, observe parts G and H: these represent coherent states, which can be understood using Ehrenfest's theorem because $|psi|^2$ looks like a Gaussian which follows a classical $sin$ or $cos$ function.



                    See




                    Kanasugi, H., and H. Okada. “Systematic Treatment of General Time-Dependent Harmonic Oscillator in Classical and Quantum Mechanics.” Progress of Theoretical Physics, vol. 93, no. 5, 1995, pp. 949–960., doi:10.1143/ptp/93.5.949.







                    share|cite|improve this answer











                    $endgroup$




                    1. Your version of $psi$ is (I'm sure you know) derived from the Time Independent Schrodinger equation, $hat{H}psi=Epsi$. To find time-dependent solutions, we solve $$ifrac{partial}{partial t}psi=hat{H}psi.$$ You were trying to solve for stationary states, and the entire point of those is that $|psi(x)|^2$ does not change over time. Still, for these stationary solutions solutions,$$frac{mathrm{d}}{mathrm{d}t}left<xright>=left<pright>=0; frac{mathrm{d}}{mathrm{d}t}left<pright>=-left<frac{mathrm{d}}{mathrm{d}x}V (x)right>=0,$$ which is in accordance with the Ehrenfest theorem (albeit uninformatively).


                    2. Be careful about the time dependence of your reported $psi$: you're actually dealing with $Psi(x, t)=psi(x)e^{-itE/hbar}$ for those stationary states. Of course, this doesn't relate to the Ehrenfest theorem, but it's something worth mentioning: the complex and real parts are oscillating, as shown by the pink and blue lines in this diagram (from the wikipedia page on QHO):complex and real parts of some QHO states



                      Do not make the mistake of suggesting that all derivatives with respect to time are necessarily automatically equal to zero because the wavefunction is time-independent. Contrary to what the shorthand notation suggests, we do have a (separable) time-dependent bit.



                    3. Referring to the same diagram, observe parts G and H: these represent coherent states, which can be understood using Ehrenfest's theorem because $|psi|^2$ looks like a Gaussian which follows a classical $sin$ or $cos$ function.



                    See




                    Kanasugi, H., and H. Okada. “Systematic Treatment of General Time-Dependent Harmonic Oscillator in Classical and Quantum Mechanics.” Progress of Theoretical Physics, vol. 93, no. 5, 1995, pp. 949–960., doi:10.1143/ptp/93.5.949.








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                    edited 11 hours ago

























                    answered 15 hours ago









                    ChairChair

                    4,33772240




                    4,33772240























                        0












                        $begingroup$

                        What you wrote down is just a complete set of solutions of the time independent Schrödinger equation.
                        begin{align}
                        [- frac{hbar²}{2m} Delta + V(x) ] Psi(x) = E Psi(x)
                        end{align}

                        Of course this solutions don't carry any time dependence, because the time-independent Schrödinger equation (in this representation) only makes statements about functions that depend on the spatial coordinates only (in this case, this is only x).



                        How to get to the time-dependent solutions?
                        A particular property of the solutions $Psi_{E}$ of the time-independent Schrödinger Equation is that $Psi_{E}(x) e^{-i frac{E}{hbar}t}$ is a solution to the time-dependent Schrödinger equation.



                        If you would apply this to your suggested set of solutions of the harmonic oscillator, you would arrive at time dependent solutions. THOSE are the ones that the Ehrenfest-Theorem makes a statement about.
                        You would then calculate that the expectation values $<X>$ and $<P>$ do not change. But you would as well calculate that both of this values are 0. This is in perfect agreement with the Theorems of Ehrenfest, and the classical analogon would be a particle resting at the deepest spot of the harmonic potential.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          What you wrote down is just a complete set of solutions of the time independent Schrödinger equation.
                          begin{align}
                          [- frac{hbar²}{2m} Delta + V(x) ] Psi(x) = E Psi(x)
                          end{align}

                          Of course this solutions don't carry any time dependence, because the time-independent Schrödinger equation (in this representation) only makes statements about functions that depend on the spatial coordinates only (in this case, this is only x).



                          How to get to the time-dependent solutions?
                          A particular property of the solutions $Psi_{E}$ of the time-independent Schrödinger Equation is that $Psi_{E}(x) e^{-i frac{E}{hbar}t}$ is a solution to the time-dependent Schrödinger equation.



                          If you would apply this to your suggested set of solutions of the harmonic oscillator, you would arrive at time dependent solutions. THOSE are the ones that the Ehrenfest-Theorem makes a statement about.
                          You would then calculate that the expectation values $<X>$ and $<P>$ do not change. But you would as well calculate that both of this values are 0. This is in perfect agreement with the Theorems of Ehrenfest, and the classical analogon would be a particle resting at the deepest spot of the harmonic potential.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            What you wrote down is just a complete set of solutions of the time independent Schrödinger equation.
                            begin{align}
                            [- frac{hbar²}{2m} Delta + V(x) ] Psi(x) = E Psi(x)
                            end{align}

                            Of course this solutions don't carry any time dependence, because the time-independent Schrödinger equation (in this representation) only makes statements about functions that depend on the spatial coordinates only (in this case, this is only x).



                            How to get to the time-dependent solutions?
                            A particular property of the solutions $Psi_{E}$ of the time-independent Schrödinger Equation is that $Psi_{E}(x) e^{-i frac{E}{hbar}t}$ is a solution to the time-dependent Schrödinger equation.



                            If you would apply this to your suggested set of solutions of the harmonic oscillator, you would arrive at time dependent solutions. THOSE are the ones that the Ehrenfest-Theorem makes a statement about.
                            You would then calculate that the expectation values $<X>$ and $<P>$ do not change. But you would as well calculate that both of this values are 0. This is in perfect agreement with the Theorems of Ehrenfest, and the classical analogon would be a particle resting at the deepest spot of the harmonic potential.






                            share|cite|improve this answer









                            $endgroup$



                            What you wrote down is just a complete set of solutions of the time independent Schrödinger equation.
                            begin{align}
                            [- frac{hbar²}{2m} Delta + V(x) ] Psi(x) = E Psi(x)
                            end{align}

                            Of course this solutions don't carry any time dependence, because the time-independent Schrödinger equation (in this representation) only makes statements about functions that depend on the spatial coordinates only (in this case, this is only x).



                            How to get to the time-dependent solutions?
                            A particular property of the solutions $Psi_{E}$ of the time-independent Schrödinger Equation is that $Psi_{E}(x) e^{-i frac{E}{hbar}t}$ is a solution to the time-dependent Schrödinger equation.



                            If you would apply this to your suggested set of solutions of the harmonic oscillator, you would arrive at time dependent solutions. THOSE are the ones that the Ehrenfest-Theorem makes a statement about.
                            You would then calculate that the expectation values $<X>$ and $<P>$ do not change. But you would as well calculate that both of this values are 0. This is in perfect agreement with the Theorems of Ehrenfest, and the classical analogon would be a particle resting at the deepest spot of the harmonic potential.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 15 hours ago









                            QuantumwhispQuantumwhisp

                            2,748724




                            2,748724






















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