Time evolution of a Gaussian wave packet, why convert to k-space? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) 2019 Moderator Election Q&A - Question CollectionWhy do we consider the evolution (usually in time) of a wave function?Time evolution of Gaussian wave packetTime evolution in Quantum Mechanics abstract state spaceTime reversal symmetry in the Schrodinger equation and evolution of wave packetIs the free particle Gaussian wavepacket in the Schwartz space?Convert time operator from momentum space to position spaceTransient simulation of a Gaussian wave packet using time dependent Liouville-von Neumann equation in center mass coordinates?Time evolution of squeezed statesHow big will the wave packet be?How to propagate Heller's model of the Gaussian Wave Packet?

What is the chair depicted in Cesare Maccari's 1889 painting "Cicerone denuncia Catilina"?

Is multiple magic items in one inherently imbalanced?

Converted a Scalar function to a TVF function for parallel execution-Still running in Serial mode

Crossing US/Canada Border for less than 24 hours

How many time has Arya actually used Needle?

Why do early math courses focus on the cross sections of a cone and not on other 3D objects?

Does the Mueller report show a conspiracy between Russia and the Trump Campaign?

Is there public access to the Meteor Crater in Arizona?

Misunderstanding of Sylow theory

Do wooden building fires get hotter than 600°C?

What does 丫 mean? 丫是什么意思?

How do I find out the mythology and history of my Fortress?

How many morphisms from 1 to 1+1 can there be?

Significance of Cersei's obsession with elephants?

Why are vacuum tubes still used in amateur radios?

Why does it sometimes sound good to play a grace note as a lead in to a note in a melody?

How do living politicians protect their readily obtainable signatures from misuse?

Has negative voting ever been officially implemented in elections, or seriously proposed, or even studied?

How can I set the aperture on my DSLR when it's attached to a telescope instead of a lens?

Sum letters are not two different

Would it be easier to apply for a UK visa if there is a host family to sponsor for you in going there?

What is an "asse" in Elizabethan English?

Why weren't discrete x86 CPUs ever used in game hardware?

Why is it faster to reheat something than it is to cook it?



Time evolution of a Gaussian wave packet, why convert to k-space?



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
2019 Moderator Election Q&A - Question CollectionWhy do we consider the evolution (usually in time) of a wave function?Time evolution of Gaussian wave packetTime evolution in Quantum Mechanics abstract state spaceTime reversal symmetry in the Schrodinger equation and evolution of wave packetIs the free particle Gaussian wavepacket in the Schwartz space?Convert time operator from momentum space to position spaceTransient simulation of a Gaussian wave packet using time dependent Liouville-von Neumann equation in center mass coordinates?Time evolution of squeezed statesHow big will the wave packet be?How to propagate Heller's model of the Gaussian Wave Packet?










3












$begingroup$


I'm trying to do a homework problem where I'm time evolving a Gaussian wave packet with a Hamiltonian of $ fracp^22m $



So if I have a Gaussian wave packet given by:



$$ Psi(x) = Ae^-alpha x^2 , .$$



I want to time evolve it, my first instinct would be to just tack on the time evolution term of $e^-fraciEthbar$.



However, in the solution it tells me that this is incorrect, and I first need to convert the wave function into k-space by using a Fourier transform due to the Hamiltonian being $ p^2/2m$. Can anyone tell me why I need to convert it to k-space first? In a finite well example with the same Hamiltonian we can just multiply the time evolution term to each term of the wave function. Why can't we can't do that to a Gaussian wave packet?










share|cite|improve this question









New contributor




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







$endgroup$







  • 1




    $begingroup$
    Ask yourself this: why do you think you can tack on the time dependence? What reason do you have to think that's correct?
    $endgroup$
    – DanielSank
    3 hours ago






  • 1




    $begingroup$
    And, more importantly, what value of the energy would you choose? Is your state an eigenstate of the hamiltonian, with a well-defined energy?
    $endgroup$
    – Emilio Pisanty
    3 hours ago















3












$begingroup$


I'm trying to do a homework problem where I'm time evolving a Gaussian wave packet with a Hamiltonian of $ fracp^22m $



So if I have a Gaussian wave packet given by:



$$ Psi(x) = Ae^-alpha x^2 , .$$



I want to time evolve it, my first instinct would be to just tack on the time evolution term of $e^-fraciEthbar$.



However, in the solution it tells me that this is incorrect, and I first need to convert the wave function into k-space by using a Fourier transform due to the Hamiltonian being $ p^2/2m$. Can anyone tell me why I need to convert it to k-space first? In a finite well example with the same Hamiltonian we can just multiply the time evolution term to each term of the wave function. Why can't we can't do that to a Gaussian wave packet?










share|cite|improve this question









New contributor




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







$endgroup$







  • 1




    $begingroup$
    Ask yourself this: why do you think you can tack on the time dependence? What reason do you have to think that's correct?
    $endgroup$
    – DanielSank
    3 hours ago






  • 1




    $begingroup$
    And, more importantly, what value of the energy would you choose? Is your state an eigenstate of the hamiltonian, with a well-defined energy?
    $endgroup$
    – Emilio Pisanty
    3 hours ago













3












3








3





$begingroup$


I'm trying to do a homework problem where I'm time evolving a Gaussian wave packet with a Hamiltonian of $ fracp^22m $



So if I have a Gaussian wave packet given by:



$$ Psi(x) = Ae^-alpha x^2 , .$$



I want to time evolve it, my first instinct would be to just tack on the time evolution term of $e^-fraciEthbar$.



However, in the solution it tells me that this is incorrect, and I first need to convert the wave function into k-space by using a Fourier transform due to the Hamiltonian being $ p^2/2m$. Can anyone tell me why I need to convert it to k-space first? In a finite well example with the same Hamiltonian we can just multiply the time evolution term to each term of the wave function. Why can't we can't do that to a Gaussian wave packet?










share|cite|improve this question









New contributor




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







$endgroup$




I'm trying to do a homework problem where I'm time evolving a Gaussian wave packet with a Hamiltonian of $ fracp^22m $



So if I have a Gaussian wave packet given by:



$$ Psi(x) = Ae^-alpha x^2 , .$$



I want to time evolve it, my first instinct would be to just tack on the time evolution term of $e^-fraciEthbar$.



However, in the solution it tells me that this is incorrect, and I first need to convert the wave function into k-space by using a Fourier transform due to the Hamiltonian being $ p^2/2m$. Can anyone tell me why I need to convert it to k-space first? In a finite well example with the same Hamiltonian we can just multiply the time evolution term to each term of the wave function. Why can't we can't do that to a Gaussian wave packet?







quantum-mechanics homework-and-exercises






share|cite|improve this question









New contributor




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











share|cite|improve this question









New contributor




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









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









DanielSank

17.9k45178




17.9k45178






New contributor




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









asked 4 hours ago









M-BM-B

182




182




New contributor




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





New contributor





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






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







  • 1




    $begingroup$
    Ask yourself this: why do you think you can tack on the time dependence? What reason do you have to think that's correct?
    $endgroup$
    – DanielSank
    3 hours ago






  • 1




    $begingroup$
    And, more importantly, what value of the energy would you choose? Is your state an eigenstate of the hamiltonian, with a well-defined energy?
    $endgroup$
    – Emilio Pisanty
    3 hours ago












  • 1




    $begingroup$
    Ask yourself this: why do you think you can tack on the time dependence? What reason do you have to think that's correct?
    $endgroup$
    – DanielSank
    3 hours ago






  • 1




    $begingroup$
    And, more importantly, what value of the energy would you choose? Is your state an eigenstate of the hamiltonian, with a well-defined energy?
    $endgroup$
    – Emilio Pisanty
    3 hours ago







1




1




$begingroup$
Ask yourself this: why do you think you can tack on the time dependence? What reason do you have to think that's correct?
$endgroup$
– DanielSank
3 hours ago




$begingroup$
Ask yourself this: why do you think you can tack on the time dependence? What reason do you have to think that's correct?
$endgroup$
– DanielSank
3 hours ago




1




1




$begingroup$
And, more importantly, what value of the energy would you choose? Is your state an eigenstate of the hamiltonian, with a well-defined energy?
$endgroup$
– Emilio Pisanty
3 hours ago




$begingroup$
And, more importantly, what value of the energy would you choose? Is your state an eigenstate of the hamiltonian, with a well-defined energy?
$endgroup$
– Emilio Pisanty
3 hours ago










1 Answer
1






active

oldest

votes


















3












$begingroup$

Tacking on a term $e^-iEt/hbar$ is the correct interpretation of the Schrödinger equation $$ihbar |partial_t Psirangle = hat H |Psirangle$$only for those eigenstates for which $$hat H |Psirangle = E|Psirangle,$$as otherwise you do not know what value of $E$ should be used to substitute. Hypothetically you can still do it, but you pay a very painful cost that the $E$ is in fact a full-fledged operator and you therefore need to exponentiate an operator, which is nontrivial.



If this is all sounding a bit complicated, please remember that QM is just linear algebra in funny hats, and so you could get an intuition for similar systems by just using some matrices and vectors, for example looking at $$ihbar beginbmatrix f'(t) \ g'(t) endbmatrix = epsilon beginbmatrix 0&1\1&0endbmatrix beginbmatrix f(t) \ g(t)endbmatrix.$$One can in fact express this as $$beginbmatrixf(t)\g(t)endbmatrix = e^-ihat H t/hbar beginbmatrix f_0\ g_0endbmatrix,$$ but one has to exponentiate this matrix. That is not hard because it squares to the identity matrix, causing a simple expansion, $$beginbmatrixf(t)\g(t)endbmatrix = cos(epsilon t/hbar) beginbmatrix f_0\ g_0endbmatrix - i sin(epsilon t/hbar) beginbmatrix g_0\ f_0endbmatrix. $$ One can then confirm that indeed this satisfies the Schrödinger equation given above. One can also immediately see that this does not directly have the form $e^-iepsilon t/hbar [f_0; g_0],$ but how could it? That would be a different Hamiltonian $hat H = epsilon I.$



But, with some creativity, one can see that if $f_0 = g_0$ those two remaining vectors would be parallel, or if $f_0 = -g_0$, and one can indeed rewrite this solution in terms of those eigenvectors of the original $hat H$ as $$beginbmatrixf(t)\g(t)endbmatrix = e^-iepsilon t/hbar alpha beginbmatrix 1\ 1endbmatrix + e^iepsilon t/hbar beta beginbmatrix -1\ 1endbmatrix. $$ So the trick to more easily finding general solutions is to find these eigenvectors first and then form a general linear combination of those eigenvectors once they have been multiplied individually by their time dependence. Then for a given initial state, we need to find the $alpha$ and $beta$ terms: in this case it is simple enough by looking at $t=0$ where $alpha - beta = f_0$ while $alpha + beta = g_0.$



Similarly for your Hamiltonian $hat H = hat p^2/(2m) = -frachbar^22mfracpartial^2~partial x^2,$ you know that the eigenvectors are plane waves, $$phi_k(x) = e^ikx.$$You know that you can then add time dependence to them in the obvious way, $$Phi_k(x, t) = e^i(k x - omega_k t),$$ where of course $$hbar omega_k = frachbar^2k^22m.$$ So the eigenvector story is just beautifully simple for you to do, all you need is the ability to take derivatives of exponentials.



The part of the story that is more complicated is assembling an arbitrary $psi(x)$ as a sum of these exponentials. However while it is complicated it is not impossible: you know from Fourier's theorem that $$psi(x) = frac12piint_-infty^infty dk ~e^i k x int_-infty^infty dxi ~e^-ikxi ~psi(xi).$$ Let your eyes glaze over the second integral and see it as just what it is, some $psi[k]$ coefficent in $k$-space. What we have here then is a sum—a continuous sum, but still a sum!—of coefficients times eigenfunctions:$$psi(x) = int_-infty^inftyfracdk~psi[k]2pi~phi_k(x).$$



And we know how to Schrödinger-ize such a sum, we just add $e^-iomega_k t$ terms to each of the eigenfunctions, turning $phi_k$ into $Phi_k.$ So we get,
$$Psi(x, t) = frac12piint_-infty^infty dk ~e^i (k x - omega_k t) ~psi[k].$$
You do not have to do it this way, you can try to do some sort of $$expleft[-i frachbar t2m fracpartial^2~partial x^2right] e^-a x^2$$
monstrosity, expanding the operator in a power series and then seeing whether there are patterns you can use among the $n^textth$ derivatives of Gaussians to simplify. But the operator expansion way looks really pretty difficult, while the eigenvector way is really easy.



The reason it is really easy is that both $hat H$ and $ihbar partial_t$ are linear operators: they distribute over sums. So if you are still feeling queasy about this procedure, convince yourself by just writing it out: calculate this value $$0 = left(ihbar fracpartial~partial t + frachbar^22mfracpartial^2~partial x^2right) frac12pi int_-infty^infty dk~psi[k] ~e^i (k x - omega_k t).$$ Notice that it holds with pretty much no restriction on the actual form of $psi[k]$ so that you only need to choose coefficients $psi[k]$ such that $Psi(x, 0) = psi(x).$






share|cite|improve this answer











$endgroup$













    Your Answer








    StackExchange.ready(function()
    var channelOptions =
    tags: "".split(" "),
    id: "151"
    ;
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function()
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled)
    StackExchange.using("snippets", function()
    createEditor();
    );

    else
    createEditor();

    );

    function createEditor()
    StackExchange.prepareEditor(
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader:
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    ,
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    );



    );






    M-B is a new contributor. Be nice, and check out our Code of Conduct.









    draft saved

    draft discarded


















    StackExchange.ready(
    function ()
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f473865%2ftime-evolution-of-a-gaussian-wave-packet-why-convert-to-k-space%23new-answer', 'question_page');

    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Tacking on a term $e^-iEt/hbar$ is the correct interpretation of the Schrödinger equation $$ihbar |partial_t Psirangle = hat H |Psirangle$$only for those eigenstates for which $$hat H |Psirangle = E|Psirangle,$$as otherwise you do not know what value of $E$ should be used to substitute. Hypothetically you can still do it, but you pay a very painful cost that the $E$ is in fact a full-fledged operator and you therefore need to exponentiate an operator, which is nontrivial.



    If this is all sounding a bit complicated, please remember that QM is just linear algebra in funny hats, and so you could get an intuition for similar systems by just using some matrices and vectors, for example looking at $$ihbar beginbmatrix f'(t) \ g'(t) endbmatrix = epsilon beginbmatrix 0&1\1&0endbmatrix beginbmatrix f(t) \ g(t)endbmatrix.$$One can in fact express this as $$beginbmatrixf(t)\g(t)endbmatrix = e^-ihat H t/hbar beginbmatrix f_0\ g_0endbmatrix,$$ but one has to exponentiate this matrix. That is not hard because it squares to the identity matrix, causing a simple expansion, $$beginbmatrixf(t)\g(t)endbmatrix = cos(epsilon t/hbar) beginbmatrix f_0\ g_0endbmatrix - i sin(epsilon t/hbar) beginbmatrix g_0\ f_0endbmatrix. $$ One can then confirm that indeed this satisfies the Schrödinger equation given above. One can also immediately see that this does not directly have the form $e^-iepsilon t/hbar [f_0; g_0],$ but how could it? That would be a different Hamiltonian $hat H = epsilon I.$



    But, with some creativity, one can see that if $f_0 = g_0$ those two remaining vectors would be parallel, or if $f_0 = -g_0$, and one can indeed rewrite this solution in terms of those eigenvectors of the original $hat H$ as $$beginbmatrixf(t)\g(t)endbmatrix = e^-iepsilon t/hbar alpha beginbmatrix 1\ 1endbmatrix + e^iepsilon t/hbar beta beginbmatrix -1\ 1endbmatrix. $$ So the trick to more easily finding general solutions is to find these eigenvectors first and then form a general linear combination of those eigenvectors once they have been multiplied individually by their time dependence. Then for a given initial state, we need to find the $alpha$ and $beta$ terms: in this case it is simple enough by looking at $t=0$ where $alpha - beta = f_0$ while $alpha + beta = g_0.$



    Similarly for your Hamiltonian $hat H = hat p^2/(2m) = -frachbar^22mfracpartial^2~partial x^2,$ you know that the eigenvectors are plane waves, $$phi_k(x) = e^ikx.$$You know that you can then add time dependence to them in the obvious way, $$Phi_k(x, t) = e^i(k x - omega_k t),$$ where of course $$hbar omega_k = frachbar^2k^22m.$$ So the eigenvector story is just beautifully simple for you to do, all you need is the ability to take derivatives of exponentials.



    The part of the story that is more complicated is assembling an arbitrary $psi(x)$ as a sum of these exponentials. However while it is complicated it is not impossible: you know from Fourier's theorem that $$psi(x) = frac12piint_-infty^infty dk ~e^i k x int_-infty^infty dxi ~e^-ikxi ~psi(xi).$$ Let your eyes glaze over the second integral and see it as just what it is, some $psi[k]$ coefficent in $k$-space. What we have here then is a sum—a continuous sum, but still a sum!—of coefficients times eigenfunctions:$$psi(x) = int_-infty^inftyfracdk~psi[k]2pi~phi_k(x).$$



    And we know how to Schrödinger-ize such a sum, we just add $e^-iomega_k t$ terms to each of the eigenfunctions, turning $phi_k$ into $Phi_k.$ So we get,
    $$Psi(x, t) = frac12piint_-infty^infty dk ~e^i (k x - omega_k t) ~psi[k].$$
    You do not have to do it this way, you can try to do some sort of $$expleft[-i frachbar t2m fracpartial^2~partial x^2right] e^-a x^2$$
    monstrosity, expanding the operator in a power series and then seeing whether there are patterns you can use among the $n^textth$ derivatives of Gaussians to simplify. But the operator expansion way looks really pretty difficult, while the eigenvector way is really easy.



    The reason it is really easy is that both $hat H$ and $ihbar partial_t$ are linear operators: they distribute over sums. So if you are still feeling queasy about this procedure, convince yourself by just writing it out: calculate this value $$0 = left(ihbar fracpartial~partial t + frachbar^22mfracpartial^2~partial x^2right) frac12pi int_-infty^infty dk~psi[k] ~e^i (k x - omega_k t).$$ Notice that it holds with pretty much no restriction on the actual form of $psi[k]$ so that you only need to choose coefficients $psi[k]$ such that $Psi(x, 0) = psi(x).$






    share|cite|improve this answer











    $endgroup$

















      3












      $begingroup$

      Tacking on a term $e^-iEt/hbar$ is the correct interpretation of the Schrödinger equation $$ihbar |partial_t Psirangle = hat H |Psirangle$$only for those eigenstates for which $$hat H |Psirangle = E|Psirangle,$$as otherwise you do not know what value of $E$ should be used to substitute. Hypothetically you can still do it, but you pay a very painful cost that the $E$ is in fact a full-fledged operator and you therefore need to exponentiate an operator, which is nontrivial.



      If this is all sounding a bit complicated, please remember that QM is just linear algebra in funny hats, and so you could get an intuition for similar systems by just using some matrices and vectors, for example looking at $$ihbar beginbmatrix f'(t) \ g'(t) endbmatrix = epsilon beginbmatrix 0&1\1&0endbmatrix beginbmatrix f(t) \ g(t)endbmatrix.$$One can in fact express this as $$beginbmatrixf(t)\g(t)endbmatrix = e^-ihat H t/hbar beginbmatrix f_0\ g_0endbmatrix,$$ but one has to exponentiate this matrix. That is not hard because it squares to the identity matrix, causing a simple expansion, $$beginbmatrixf(t)\g(t)endbmatrix = cos(epsilon t/hbar) beginbmatrix f_0\ g_0endbmatrix - i sin(epsilon t/hbar) beginbmatrix g_0\ f_0endbmatrix. $$ One can then confirm that indeed this satisfies the Schrödinger equation given above. One can also immediately see that this does not directly have the form $e^-iepsilon t/hbar [f_0; g_0],$ but how could it? That would be a different Hamiltonian $hat H = epsilon I.$



      But, with some creativity, one can see that if $f_0 = g_0$ those two remaining vectors would be parallel, or if $f_0 = -g_0$, and one can indeed rewrite this solution in terms of those eigenvectors of the original $hat H$ as $$beginbmatrixf(t)\g(t)endbmatrix = e^-iepsilon t/hbar alpha beginbmatrix 1\ 1endbmatrix + e^iepsilon t/hbar beta beginbmatrix -1\ 1endbmatrix. $$ So the trick to more easily finding general solutions is to find these eigenvectors first and then form a general linear combination of those eigenvectors once they have been multiplied individually by their time dependence. Then for a given initial state, we need to find the $alpha$ and $beta$ terms: in this case it is simple enough by looking at $t=0$ where $alpha - beta = f_0$ while $alpha + beta = g_0.$



      Similarly for your Hamiltonian $hat H = hat p^2/(2m) = -frachbar^22mfracpartial^2~partial x^2,$ you know that the eigenvectors are plane waves, $$phi_k(x) = e^ikx.$$You know that you can then add time dependence to them in the obvious way, $$Phi_k(x, t) = e^i(k x - omega_k t),$$ where of course $$hbar omega_k = frachbar^2k^22m.$$ So the eigenvector story is just beautifully simple for you to do, all you need is the ability to take derivatives of exponentials.



      The part of the story that is more complicated is assembling an arbitrary $psi(x)$ as a sum of these exponentials. However while it is complicated it is not impossible: you know from Fourier's theorem that $$psi(x) = frac12piint_-infty^infty dk ~e^i k x int_-infty^infty dxi ~e^-ikxi ~psi(xi).$$ Let your eyes glaze over the second integral and see it as just what it is, some $psi[k]$ coefficent in $k$-space. What we have here then is a sum—a continuous sum, but still a sum!—of coefficients times eigenfunctions:$$psi(x) = int_-infty^inftyfracdk~psi[k]2pi~phi_k(x).$$



      And we know how to Schrödinger-ize such a sum, we just add $e^-iomega_k t$ terms to each of the eigenfunctions, turning $phi_k$ into $Phi_k.$ So we get,
      $$Psi(x, t) = frac12piint_-infty^infty dk ~e^i (k x - omega_k t) ~psi[k].$$
      You do not have to do it this way, you can try to do some sort of $$expleft[-i frachbar t2m fracpartial^2~partial x^2right] e^-a x^2$$
      monstrosity, expanding the operator in a power series and then seeing whether there are patterns you can use among the $n^textth$ derivatives of Gaussians to simplify. But the operator expansion way looks really pretty difficult, while the eigenvector way is really easy.



      The reason it is really easy is that both $hat H$ and $ihbar partial_t$ are linear operators: they distribute over sums. So if you are still feeling queasy about this procedure, convince yourself by just writing it out: calculate this value $$0 = left(ihbar fracpartial~partial t + frachbar^22mfracpartial^2~partial x^2right) frac12pi int_-infty^infty dk~psi[k] ~e^i (k x - omega_k t).$$ Notice that it holds with pretty much no restriction on the actual form of $psi[k]$ so that you only need to choose coefficients $psi[k]$ such that $Psi(x, 0) = psi(x).$






      share|cite|improve this answer











      $endgroup$















        3












        3








        3





        $begingroup$

        Tacking on a term $e^-iEt/hbar$ is the correct interpretation of the Schrödinger equation $$ihbar |partial_t Psirangle = hat H |Psirangle$$only for those eigenstates for which $$hat H |Psirangle = E|Psirangle,$$as otherwise you do not know what value of $E$ should be used to substitute. Hypothetically you can still do it, but you pay a very painful cost that the $E$ is in fact a full-fledged operator and you therefore need to exponentiate an operator, which is nontrivial.



        If this is all sounding a bit complicated, please remember that QM is just linear algebra in funny hats, and so you could get an intuition for similar systems by just using some matrices and vectors, for example looking at $$ihbar beginbmatrix f'(t) \ g'(t) endbmatrix = epsilon beginbmatrix 0&1\1&0endbmatrix beginbmatrix f(t) \ g(t)endbmatrix.$$One can in fact express this as $$beginbmatrixf(t)\g(t)endbmatrix = e^-ihat H t/hbar beginbmatrix f_0\ g_0endbmatrix,$$ but one has to exponentiate this matrix. That is not hard because it squares to the identity matrix, causing a simple expansion, $$beginbmatrixf(t)\g(t)endbmatrix = cos(epsilon t/hbar) beginbmatrix f_0\ g_0endbmatrix - i sin(epsilon t/hbar) beginbmatrix g_0\ f_0endbmatrix. $$ One can then confirm that indeed this satisfies the Schrödinger equation given above. One can also immediately see that this does not directly have the form $e^-iepsilon t/hbar [f_0; g_0],$ but how could it? That would be a different Hamiltonian $hat H = epsilon I.$



        But, with some creativity, one can see that if $f_0 = g_0$ those two remaining vectors would be parallel, or if $f_0 = -g_0$, and one can indeed rewrite this solution in terms of those eigenvectors of the original $hat H$ as $$beginbmatrixf(t)\g(t)endbmatrix = e^-iepsilon t/hbar alpha beginbmatrix 1\ 1endbmatrix + e^iepsilon t/hbar beta beginbmatrix -1\ 1endbmatrix. $$ So the trick to more easily finding general solutions is to find these eigenvectors first and then form a general linear combination of those eigenvectors once they have been multiplied individually by their time dependence. Then for a given initial state, we need to find the $alpha$ and $beta$ terms: in this case it is simple enough by looking at $t=0$ where $alpha - beta = f_0$ while $alpha + beta = g_0.$



        Similarly for your Hamiltonian $hat H = hat p^2/(2m) = -frachbar^22mfracpartial^2~partial x^2,$ you know that the eigenvectors are plane waves, $$phi_k(x) = e^ikx.$$You know that you can then add time dependence to them in the obvious way, $$Phi_k(x, t) = e^i(k x - omega_k t),$$ where of course $$hbar omega_k = frachbar^2k^22m.$$ So the eigenvector story is just beautifully simple for you to do, all you need is the ability to take derivatives of exponentials.



        The part of the story that is more complicated is assembling an arbitrary $psi(x)$ as a sum of these exponentials. However while it is complicated it is not impossible: you know from Fourier's theorem that $$psi(x) = frac12piint_-infty^infty dk ~e^i k x int_-infty^infty dxi ~e^-ikxi ~psi(xi).$$ Let your eyes glaze over the second integral and see it as just what it is, some $psi[k]$ coefficent in $k$-space. What we have here then is a sum—a continuous sum, but still a sum!—of coefficients times eigenfunctions:$$psi(x) = int_-infty^inftyfracdk~psi[k]2pi~phi_k(x).$$



        And we know how to Schrödinger-ize such a sum, we just add $e^-iomega_k t$ terms to each of the eigenfunctions, turning $phi_k$ into $Phi_k.$ So we get,
        $$Psi(x, t) = frac12piint_-infty^infty dk ~e^i (k x - omega_k t) ~psi[k].$$
        You do not have to do it this way, you can try to do some sort of $$expleft[-i frachbar t2m fracpartial^2~partial x^2right] e^-a x^2$$
        monstrosity, expanding the operator in a power series and then seeing whether there are patterns you can use among the $n^textth$ derivatives of Gaussians to simplify. But the operator expansion way looks really pretty difficult, while the eigenvector way is really easy.



        The reason it is really easy is that both $hat H$ and $ihbar partial_t$ are linear operators: they distribute over sums. So if you are still feeling queasy about this procedure, convince yourself by just writing it out: calculate this value $$0 = left(ihbar fracpartial~partial t + frachbar^22mfracpartial^2~partial x^2right) frac12pi int_-infty^infty dk~psi[k] ~e^i (k x - omega_k t).$$ Notice that it holds with pretty much no restriction on the actual form of $psi[k]$ so that you only need to choose coefficients $psi[k]$ such that $Psi(x, 0) = psi(x).$






        share|cite|improve this answer











        $endgroup$



        Tacking on a term $e^-iEt/hbar$ is the correct interpretation of the Schrödinger equation $$ihbar |partial_t Psirangle = hat H |Psirangle$$only for those eigenstates for which $$hat H |Psirangle = E|Psirangle,$$as otherwise you do not know what value of $E$ should be used to substitute. Hypothetically you can still do it, but you pay a very painful cost that the $E$ is in fact a full-fledged operator and you therefore need to exponentiate an operator, which is nontrivial.



        If this is all sounding a bit complicated, please remember that QM is just linear algebra in funny hats, and so you could get an intuition for similar systems by just using some matrices and vectors, for example looking at $$ihbar beginbmatrix f'(t) \ g'(t) endbmatrix = epsilon beginbmatrix 0&1\1&0endbmatrix beginbmatrix f(t) \ g(t)endbmatrix.$$One can in fact express this as $$beginbmatrixf(t)\g(t)endbmatrix = e^-ihat H t/hbar beginbmatrix f_0\ g_0endbmatrix,$$ but one has to exponentiate this matrix. That is not hard because it squares to the identity matrix, causing a simple expansion, $$beginbmatrixf(t)\g(t)endbmatrix = cos(epsilon t/hbar) beginbmatrix f_0\ g_0endbmatrix - i sin(epsilon t/hbar) beginbmatrix g_0\ f_0endbmatrix. $$ One can then confirm that indeed this satisfies the Schrödinger equation given above. One can also immediately see that this does not directly have the form $e^-iepsilon t/hbar [f_0; g_0],$ but how could it? That would be a different Hamiltonian $hat H = epsilon I.$



        But, with some creativity, one can see that if $f_0 = g_0$ those two remaining vectors would be parallel, or if $f_0 = -g_0$, and one can indeed rewrite this solution in terms of those eigenvectors of the original $hat H$ as $$beginbmatrixf(t)\g(t)endbmatrix = e^-iepsilon t/hbar alpha beginbmatrix 1\ 1endbmatrix + e^iepsilon t/hbar beta beginbmatrix -1\ 1endbmatrix. $$ So the trick to more easily finding general solutions is to find these eigenvectors first and then form a general linear combination of those eigenvectors once they have been multiplied individually by their time dependence. Then for a given initial state, we need to find the $alpha$ and $beta$ terms: in this case it is simple enough by looking at $t=0$ where $alpha - beta = f_0$ while $alpha + beta = g_0.$



        Similarly for your Hamiltonian $hat H = hat p^2/(2m) = -frachbar^22mfracpartial^2~partial x^2,$ you know that the eigenvectors are plane waves, $$phi_k(x) = e^ikx.$$You know that you can then add time dependence to them in the obvious way, $$Phi_k(x, t) = e^i(k x - omega_k t),$$ where of course $$hbar omega_k = frachbar^2k^22m.$$ So the eigenvector story is just beautifully simple for you to do, all you need is the ability to take derivatives of exponentials.



        The part of the story that is more complicated is assembling an arbitrary $psi(x)$ as a sum of these exponentials. However while it is complicated it is not impossible: you know from Fourier's theorem that $$psi(x) = frac12piint_-infty^infty dk ~e^i k x int_-infty^infty dxi ~e^-ikxi ~psi(xi).$$ Let your eyes glaze over the second integral and see it as just what it is, some $psi[k]$ coefficent in $k$-space. What we have here then is a sum—a continuous sum, but still a sum!—of coefficients times eigenfunctions:$$psi(x) = int_-infty^inftyfracdk~psi[k]2pi~phi_k(x).$$



        And we know how to Schrödinger-ize such a sum, we just add $e^-iomega_k t$ terms to each of the eigenfunctions, turning $phi_k$ into $Phi_k.$ So we get,
        $$Psi(x, t) = frac12piint_-infty^infty dk ~e^i (k x - omega_k t) ~psi[k].$$
        You do not have to do it this way, you can try to do some sort of $$expleft[-i frachbar t2m fracpartial^2~partial x^2right] e^-a x^2$$
        monstrosity, expanding the operator in a power series and then seeing whether there are patterns you can use among the $n^textth$ derivatives of Gaussians to simplify. But the operator expansion way looks really pretty difficult, while the eigenvector way is really easy.



        The reason it is really easy is that both $hat H$ and $ihbar partial_t$ are linear operators: they distribute over sums. So if you are still feeling queasy about this procedure, convince yourself by just writing it out: calculate this value $$0 = left(ihbar fracpartial~partial t + frachbar^22mfracpartial^2~partial x^2right) frac12pi int_-infty^infty dk~psi[k] ~e^i (k x - omega_k t).$$ Notice that it holds with pretty much no restriction on the actual form of $psi[k]$ so that you only need to choose coefficients $psi[k]$ such that $Psi(x, 0) = psi(x).$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 54 mins ago

























        answered 2 hours ago









        CR DrostCR Drost

        23.2k11964




        23.2k11964




















            M-B is a new contributor. Be nice, and check out our Code of Conduct.









            draft saved

            draft discarded


















            M-B is a new contributor. Be nice, and check out our Code of Conduct.












            M-B is a new contributor. Be nice, and check out our Code of Conduct.











            M-B is a new contributor. Be nice, and check out our Code of Conduct.














            Thanks for contributing an answer to Physics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid


            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.

            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f473865%2ftime-evolution-of-a-gaussian-wave-packet-why-convert-to-k-space%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Are there any AGPL-style licences that require source code modifications to be public? Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Force derivative works to be publicAre there any GPL like licenses for Apple App Store?Do you violate the GPL if you provide source code that cannot be compiled?GPL - is it distribution to use libraries in an appliance loaned to customers?Distributing App for free which uses GPL'ed codeModifications of server software under GPL, with web/CLI interfaceDoes using an AGPLv3-licensed library prevent me from dual-licensing my own source code?Can I publish only select code under GPLv3 from a private project?Is there published precedent regarding the scope of covered work that uses AGPL software?If MIT licensed code links to GPL licensed code what should be the license of the resulting binary program?If I use a public API endpoint that has its source code licensed under AGPL in my app, do I need to disclose my source?

            2013 GY136 Descoberta | Órbita | Referências Menu de navegação«List Of Centaurs and Scattered-Disk Objects»«List of Known Trans-Neptunian Objects»

            Mortes em março de 2019 Referências Menu de navegação«Zhores Alferov, Nobel de Física bielorrusso, morre aos 88 anos - Ciência»«Fallece Rafael Torija, o bispo emérito de Ciudad Real»«Peter Hurford dies at 88»«Keith Flint, vocalista do The Prodigy, morre aos 49 anos»«Luke Perry, ator de 'Barrados no baile' e 'Riverdale', morre aos 52 anos»«Former Rangers and Scotland captain Eric Caldow dies, aged 84»«Morreu, aos 61 anos, a antiga lenda do wrestling King Kong Bundy»«Fallece el actor y director teatral Abraham Stavans»«In Memoriam Guillaume Faye»«Sidney Sheinberg, a Force Behind Universal and Spielberg, Is Dead at 84»«Carmine Persico, Colombo Crime Family Boss, Is Dead at 85»«Dirigent Michael Gielen gestorben»«Ciclista tricampeã mundial e prata na Rio 2016 é encontrada morta em casa aos 23 anos»«Pagan Community Notes: Raven Grimassi dies, Indianapolis pop-up event cancelled, Circle Sanctuary announces new podcast, and more!»«Hal Blaine, Wrecking Crew Drummer, Dies at 90»«Morre Coutinho, que editou dupla lendária com Pelé no Santos»«Cantor Demétrius, ídolo da Jovem Guarda, morre em SP»«Ex-presidente do Vasco, Eurico Miranda morre no Rio de Janeiro»«Bronze no Mundial de basquete de 1971, Laís Elena morre aos 76 anos»«Diretor de Corridas da F1, Charlie Whiting morre aos 66 anos às vésperas do GP da Austrália»«Morreu o cardeal Danneels, da Bélgica»«Morreu o cartoonista Augusto Cid»«Morreu a atriz Maria Isabel de Lizandra, de "Vale Tudo" e novelas da Tupi»«WS Merwin, prize-winning poet of nature, dies at 91»«Atriz Márcia Real morre em São Paulo aos 88 anos»«Mauritanie: décès de l'ancien président Mohamed Mahmoud ould Louly»«Morreu Dick Dale, o rei da surf guitar e de "Pulp Fiction"»«Falleció Víctor Genes»«João Carlos Marinho, autor de 'O Gênio do Crime', morre em SP»«Legendary Horror Director and SFX Artist John Carl Buechler Dies at 66»«Morre em Salvador a religiosa Makota Valdina»«مرگ بازیکن‌ سابق نساجی بر اثر سقوط سنگ در مازندران»«Domingos Oliveira morre no Rio»«Morre Airton Ravagniani, ex-São Paulo, Fla, Vasco, Grêmio e Sport - Notícias»«Morre o escritor Flavio Moreira da Costa»«Larry Cohen, Writer-Director of 'It's Alive' and 'Hell Up in Harlem,' Dies at 77»«Scott Walker, experimental singer-songwriter, dead at 76»«Joseph Pilato, Day of the Dead Star and Horror Favorite, Dies at 70»«Sheffield United set to pay tribute to legendary goalkeeper Ted Burgin who has died at 91»«Morre Rafael Henzel, sobrevivente de acidente aéreo da Chapecoense»«Morre Valery Bykovsky, um dos primeiros cosmonautas da União Soviética»«Agnès Varda, cineasta da Nouvelle Vague, morre aos 90 anos»«Agnès Varda, cineasta francesa, morre aos 90 anos»«Tania Mallet, James Bond Actress and Helen Mirren's Cousin, Dies at 77»e