Is a bound state a stationary state?












2












$begingroup$


In Shankar's discussion on the 1D infinite square well in Principles of Quantum Mechanics (2nd edition), he made the following statement:




Now $langle P rangle = 0$ in any bound state for the following reason. Since a bound state is a stationary state, $langle P rangle$ is time independent. If this $langle Prangle ne 0$, the particle must (in the average sense) drift either to the right or to the left and eventually escape to infinity, which cannot happen in a bound state.




The final sentence makes sense to me, but his reasoning in the second sentence does not. Aren't bound states and stationary states entirely different things? Does the one in fact imply the other?






quantum-mechanics hilbert-space terminology definition quantum-states







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I find that puzzling too because I would think that a state moving around in a potential is still a bound state. I guess Shankar is just using the words in a particular way.
    $endgroup$
    – DanielSank
    3 hours ago
















2












$begingroup$


In Shankar's discussion on the 1D infinite square well in Principles of Quantum Mechanics (2nd edition), he made the following statement:




Now $langle P rangle = 0$ in any bound state for the following reason. Since a bound state is a stationary state, $langle P rangle$ is time independent. If this $langle Prangle ne 0$, the particle must (in the average sense) drift either to the right or to the left and eventually escape to infinity, which cannot happen in a bound state.




The final sentence makes sense to me, but his reasoning in the second sentence does not. Aren't bound states and stationary states entirely different things? Does the one in fact imply the other?






quantum-mechanics hilbert-space terminology definition quantum-states







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    I find that puzzling too because I would think that a state moving around in a potential is still a bound state. I guess Shankar is just using the words in a particular way.
    $endgroup$
    – DanielSank
    3 hours ago














2












2








2





$begingroup$


In Shankar's discussion on the 1D infinite square well in Principles of Quantum Mechanics (2nd edition), he made the following statement:




Now $langle P rangle = 0$ in any bound state for the following reason. Since a bound state is a stationary state, $langle P rangle$ is time independent. If this $langle Prangle ne 0$, the particle must (in the average sense) drift either to the right or to the left and eventually escape to infinity, which cannot happen in a bound state.




The final sentence makes sense to me, but his reasoning in the second sentence does not. Aren't bound states and stationary states entirely different things? Does the one in fact imply the other?






quantum-mechanics hilbert-space terminology definition quantum-states







share|cite|improve this question











$endgroup$




In Shankar's discussion on the 1D infinite square well in Principles of Quantum Mechanics (2nd edition), he made the following statement:




Now $langle P rangle = 0$ in any bound state for the following reason. Since a bound state is a stationary state, $langle P rangle$ is time independent. If this $langle Prangle ne 0$, the particle must (in the average sense) drift either to the right or to the left and eventually escape to infinity, which cannot happen in a bound state.




The final sentence makes sense to me, but his reasoning in the second sentence does not. Aren't bound states and stationary states entirely different things? Does the one in fact imply the other?






quantum-mechanics hilbert-space terminology definition quantum-states




quantum-mechanics hilbert-space terminology definition quantum-states






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Qmechanic♦

106k121961226




106k121961226










asked 3 hours ago









J-JJ-J

586




586








  • 2




    $begingroup$
    I find that puzzling too because I would think that a state moving around in a potential is still a bound state. I guess Shankar is just using the words in a particular way.
    $endgroup$
    – DanielSank
    3 hours ago














  • 2




    $begingroup$
    I find that puzzling too because I would think that a state moving around in a potential is still a bound state. I guess Shankar is just using the words in a particular way.
    $endgroup$
    – DanielSank
    3 hours ago








2




2




$begingroup$
I find that puzzling too because I would think that a state moving around in a potential is still a bound state. I guess Shankar is just using the words in a particular way.
$endgroup$
– DanielSank
3 hours ago




$begingroup$
I find that puzzling too because I would think that a state moving around in a potential is still a bound state. I guess Shankar is just using the words in a particular way.
$endgroup$
– DanielSank
3 hours ago










1 Answer
1






active

oldest

votes


















2












$begingroup$

I think most of us would agree that superposition of bound states — say, of an electron in an atom — still deserves to be called a bound state, even though most such superpositions are time-dependent. The electron is still bound to the atom.



Based on the context from which the excerpt shown in the OP was extracted, it looks like Shankar is specifically talking about the ground state. The paragraph begins with




Let us now ... discuss the fact that the lowest energy is not zero...




(emphasis added by me), and the following paragraph ends with




The uncertainty principle is often used in this fashion to provide a quick order-of-magnitude estimate for the ground-state energy.




So although Shankar doesn't say it directly, the whole derivation seems to be focused on a particular stationary state, not a generic bound state. This inference is consistent with the fact that, just a few paragraphs earlier, Shankar writes




Bound states are thus characterized by $psi(x)to 0$ [as $|x|toinfty$] ... The energy levels of bound states are always quantized.




Shankar doesn't say that bound states always have sharply-defined energies, so none of this contradicts the usual convention that a superposition of bound states is still called a bound state, whether or not it happens to be stationary.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    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
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f468307%2fis-a-bound-state-a-stationary-state%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









    2












    $begingroup$

    I think most of us would agree that superposition of bound states — say, of an electron in an atom — still deserves to be called a bound state, even though most such superpositions are time-dependent. The electron is still bound to the atom.



    Based on the context from which the excerpt shown in the OP was extracted, it looks like Shankar is specifically talking about the ground state. The paragraph begins with




    Let us now ... discuss the fact that the lowest energy is not zero...




    (emphasis added by me), and the following paragraph ends with




    The uncertainty principle is often used in this fashion to provide a quick order-of-magnitude estimate for the ground-state energy.




    So although Shankar doesn't say it directly, the whole derivation seems to be focused on a particular stationary state, not a generic bound state. This inference is consistent with the fact that, just a few paragraphs earlier, Shankar writes




    Bound states are thus characterized by $psi(x)to 0$ [as $|x|toinfty$] ... The energy levels of bound states are always quantized.




    Shankar doesn't say that bound states always have sharply-defined energies, so none of this contradicts the usual convention that a superposition of bound states is still called a bound state, whether or not it happens to be stationary.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      I think most of us would agree that superposition of bound states — say, of an electron in an atom — still deserves to be called a bound state, even though most such superpositions are time-dependent. The electron is still bound to the atom.



      Based on the context from which the excerpt shown in the OP was extracted, it looks like Shankar is specifically talking about the ground state. The paragraph begins with




      Let us now ... discuss the fact that the lowest energy is not zero...




      (emphasis added by me), and the following paragraph ends with




      The uncertainty principle is often used in this fashion to provide a quick order-of-magnitude estimate for the ground-state energy.




      So although Shankar doesn't say it directly, the whole derivation seems to be focused on a particular stationary state, not a generic bound state. This inference is consistent with the fact that, just a few paragraphs earlier, Shankar writes




      Bound states are thus characterized by $psi(x)to 0$ [as $|x|toinfty$] ... The energy levels of bound states are always quantized.




      Shankar doesn't say that bound states always have sharply-defined energies, so none of this contradicts the usual convention that a superposition of bound states is still called a bound state, whether or not it happens to be stationary.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I think most of us would agree that superposition of bound states — say, of an electron in an atom — still deserves to be called a bound state, even though most such superpositions are time-dependent. The electron is still bound to the atom.



        Based on the context from which the excerpt shown in the OP was extracted, it looks like Shankar is specifically talking about the ground state. The paragraph begins with




        Let us now ... discuss the fact that the lowest energy is not zero...




        (emphasis added by me), and the following paragraph ends with




        The uncertainty principle is often used in this fashion to provide a quick order-of-magnitude estimate for the ground-state energy.




        So although Shankar doesn't say it directly, the whole derivation seems to be focused on a particular stationary state, not a generic bound state. This inference is consistent with the fact that, just a few paragraphs earlier, Shankar writes




        Bound states are thus characterized by $psi(x)to 0$ [as $|x|toinfty$] ... The energy levels of bound states are always quantized.




        Shankar doesn't say that bound states always have sharply-defined energies, so none of this contradicts the usual convention that a superposition of bound states is still called a bound state, whether or not it happens to be stationary.






        share|cite|improve this answer









        $endgroup$



        I think most of us would agree that superposition of bound states — say, of an electron in an atom — still deserves to be called a bound state, even though most such superpositions are time-dependent. The electron is still bound to the atom.



        Based on the context from which the excerpt shown in the OP was extracted, it looks like Shankar is specifically talking about the ground state. The paragraph begins with




        Let us now ... discuss the fact that the lowest energy is not zero...




        (emphasis added by me), and the following paragraph ends with




        The uncertainty principle is often used in this fashion to provide a quick order-of-magnitude estimate for the ground-state energy.




        So although Shankar doesn't say it directly, the whole derivation seems to be focused on a particular stationary state, not a generic bound state. This inference is consistent with the fact that, just a few paragraphs earlier, Shankar writes




        Bound states are thus characterized by $psi(x)to 0$ [as $|x|toinfty$] ... The energy levels of bound states are always quantized.




        Shankar doesn't say that bound states always have sharply-defined energies, so none of this contradicts the usual convention that a superposition of bound states is still called a bound state, whether or not it happens to be stationary.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 2 hours ago









        Chiral AnomalyChiral Anomaly

        12.4k21541




        12.4k21541






























            draft saved

            draft discarded




















































            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%2f468307%2fis-a-bound-state-a-stationary-state%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

            Knooppunt Holsloot

            Image
            Knooppunt Holsloot Kenmerken Wegen A37 × N34 Type knooppunt klaverblad Portaal     Verkeer & Vervoer Knooppunt Holsloot is een verkeersknooppunt op de kruising van de autosnelweg A37 en de N34 in Drenthe, tussen Emmen en Coevorden. Oorspronkelijk was deze aansluiting een afritconstructie van het type Haarlemmermeeraansluiting, zoals bij Knooppunt Hooipolder. Sinds 2002 is het knooppunt omgebouwd tot een volledig klaverblad. Als noviteit werd hier voor het eerst gebruikgemaakt van dimbare verlichting in plaats van de normale vaste tijden waarop de verlichting aan en uit gaat, wat een grote kostenbesparing opleverde. [1] Dat Holsloot als volwaardig knooppunt is uitgevoerd, waarbij alle verkeersstromen en afbuigingen ononderbroken kunnen doorgaan, is uitzonderlijk. Het betreft hier namelijk een aansluiting tussen een autosnelweg en niet-snelweg. Een vergelijkbare situatie treft men nabij Hoensbroek bij het knooppunt Ten Esschen.     Aa
            Read more

            Altaar (religie)

            Image
            Deel van de serie over kerkelijk gerei waaronder paramenten en liturgisch vaatwerk gebruikt in de liturgie Liturgisch vaatwerk Vasa sacra Miskelk · Pateen Kelklepeltje Ciborie · Monstrans Pyxis · Custodiale Vasa non sacra Ampullen · Wijwatervat Olievaatje Wierookvat · Ablutievat Paramenten Amict · Albe · Baarkleed · Cingel Tuniek · Dalmatiek · Fanon Kazuifel · Manipel · Stola Gremiale · Benedictievelum Mijter Koorkledij Rochet · Superplie Koorkap · Cappa magna Kovel Kelkgerei Bursa · Kelkvelum Ciborievelum Kerklinnen Corporale · Kelkdoekje Lavabodoekje · Palla Altaardwaal Kerkinterieur Altaar · Ambo Biechtstoel · Communiebank Doksaal · Doopvont Faldistorium Godslamp · Hoogaltaar Heilig Kruisaltaar Katheder · Preekstoel Sedilia Tabernakel · Volksaltaar Liturgische boeken Altaarmissaal Benedictionale · Brevier Evangeliarium · Evangelistarium Graduale · Kyriale Lectionarium Psalter · Rituaal Sacramentarium · Volksmissaal Overige Flambou
            Read more

            Gregoriusmis

            Image
            Een afbeelding van de Gregoriusmis uit een getijdenboek gemaakt in de zuidelijke Nederlanden tussen 1500 en 1525 Gregoriusmis, Meester van het Akense Altaar, ca. 1500-1525. Collectie Museum Catharijneconvent De Gregoriusmis of de Mis van Sint Gregorius is een motief in katholieke kunst, dat het eerst verscheen in de late Middeleeuwen en dat nog steeds werd gebruikt tijdens de contrareformatie. Paus Gregorius I (ca. 540-604) is afgebeeld, de H. Mis opdragend, terwijl Christus als de Man van Smarten verschijnt op het altaar dat voor Sint Gregorius staat. Volgens de legende had St. Gregorius gebeden om een teken voor een betwijfelaar van de transsubstantiatieleer, waarop de lijdende Christus tijdens het opdragen van de Mis verscheen om hem te overtuigen en waarbij zijn bloed in de altaarkelk liep. Vaak zijn ook de passiewerktuigen afgebeeld. Zie de categorie Mass of Saint Gregory van Wikimedia Commons voor mediabestanden over dit onderwerp. This page is onl
            Read more