Which basis does the wavefunction collapse to?What are the differences between a $psi$-epistemic ontological model and a $psi$-ontic model of quantum mechanics, exactly?wavefunction collapse and uncertainty principleWhat happens after the collapse of a wavefunction?How does wave function collapse when I measure position?Is something beyond the material needed to solve the Von Neumann Chain?Collapse of wave functionDoes measurement of momentum always collapse the wave function into a plane wave?Wave function. Measurement of the absence

Unconventional examples of mathematical modelling

Vegetarian dishes on Russian trains (European part)

How do I cope with haze for the photos containing sky and trees at a distance?

Designing a prison for a telekinetic race

Best model for precedence constraints within scheduling problem

Existence of a certain set of 0/1-sequences without the Axiom of Choice

Which basis does the wavefunction collapse to?

Subgroup generated by a subgroup and a conjugate of it

How to train a replacement without them knowing?

Why don't modern jet engines use forced exhaust mixing?

My new Acer Aspire 7 doesn't have a Legacy Boot option, what can I do to get it?

Combinatorial Argument for Exponential and Logarithmic Function Being Inverse

How could Tony Stark wield the Infinity Nano Gauntlet - at all?

Will some rockets really collapse under their own weight?

Meaning and structure of headline "Hair it is: A List of ..."

The anatomy of an organic infrared generator

Can I submit a paper computer science conference using an alias if using my real name can cause legal trouble in my original country

Representing an indicator function: binary variables and "indicator constraints"

Why should I pay for an SSL certificate?

How to render "have ideas above his station" into German

Trying to understand how Digital Certificates and CA are indeed secure

How to use the passive form to say "This flower was watered."

What are some tips and tricks for finding the cheapest flight when luggage and other fees are not revealed until far into the booking process?

Build a mob of suspiciously happy lenny faces ( ͡° ͜ʖ ͡°)



Which basis does the wavefunction collapse to?


What are the differences between a $psi$-epistemic ontological model and a $psi$-ontic model of quantum mechanics, exactly?wavefunction collapse and uncertainty principleWhat happens after the collapse of a wavefunction?How does wave function collapse when I measure position?Is something beyond the material needed to solve the Von Neumann Chain?Collapse of wave functionDoes measurement of momentum always collapse the wave function into a plane wave?Wave function. Measurement of the absence






.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








4












$begingroup$


When we measure position for example, how does the system "know" that we're measuring position in order to collapse to a position eigenvector? Does the wave function always evolve from the state that it collapsed to? For example, if we measure the position (whatever that means) does the wave evolve from a delta function?










share|cite|improve this question











$endgroup$




















    4












    $begingroup$


    When we measure position for example, how does the system "know" that we're measuring position in order to collapse to a position eigenvector? Does the wave function always evolve from the state that it collapsed to? For example, if we measure the position (whatever that means) does the wave evolve from a delta function?










    share|cite|improve this question











    $endgroup$
















      4












      4








      4





      $begingroup$


      When we measure position for example, how does the system "know" that we're measuring position in order to collapse to a position eigenvector? Does the wave function always evolve from the state that it collapsed to? For example, if we measure the position (whatever that means) does the wave evolve from a delta function?










      share|cite|improve this question











      $endgroup$




      When we measure position for example, how does the system "know" that we're measuring position in order to collapse to a position eigenvector? Does the wave function always evolve from the state that it collapsed to? For example, if we measure the position (whatever that means) does the wave evolve from a delta function?







      quantum-mechanics wavefunction wavefunction-collapse






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 8 hours ago









      Qmechanic

      112k13 gold badges219 silver badges1331 bronze badges




      112k13 gold badges219 silver badges1331 bronze badges










      asked 8 hours ago









      Jeff BassJeff Bass

      464 bronze badges




      464 bronze badges























          3 Answers
          3






          active

          oldest

          votes


















          7












          $begingroup$

          The system doesn't "know" anything.



          The only uncontroversial statement one can make about the (strong) measurement of a quantum system is that you will make the correct predictions if you assume that the state after the measurement was the eigenstate corresponding to the measured value of the observable (so, for position, indeed a $delta$-function, if we ignore issues with that not being a real function, which would be a distraction here). But what we mean by "state" in the first place - i.e. what ontology, if any, corresponds to the statement "the system is in the quantum state $lvert psirangle$" - is ambiguous to begin with:



          Whether the original state "collapsed" to this new state, whether the "state" is just an imperfect representation of our knowledge and the "collapse" is just updating our information (cf. "$psi$-ontic" vs "$psi$-epistemic", see e.g. this answer by Emilio Pisanty) instead of an actual physical process, or something else entirely, is a matter of quantum interpretation. In some interpretations, there is collapse, in others there isn't, but in any case, the formalism of quantum mechanics itself does not provide a single "correct" interpretation.



          That is, your question is essentially unanswerable unless you specify the interpretation within which it is to be answered. But none of the predictions of quantum mechanics depends on it anyway - you do not need to have a concept of "how" collapse works to compute the outcome of measurements.






          share|cite|improve this answer









          $endgroup$














          • $begingroup$
            Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
            $endgroup$
            – Jeff Bass
            1 hour ago


















          2












          $begingroup$

          The collapse happens in all bases. What I mean by that is that the wavefunction can be expressed in any basis you want to. It's just that the easiest basis to look at right after measurement is the one corresponding to what you measured, since the state is the eigenstate corresponding to your measurement.



          Always remember the wavefunction isn't physical. It's an abstract thing that we can only describe and "look at" as shadows from their projections. We can choose any projection we want to, but that choice doesn't change the wavefunction






          share|cite|improve this answer











          $endgroup$














          • $begingroup$
            This does not answer the question. The question is not about representing a wavefunction in different bases.
            $endgroup$
            – eigenvalue
            4 hours ago











          • $begingroup$
            @eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
            $endgroup$
            – Aaron Stevens
            3 hours ago



















          1












          $begingroup$

          I'm just posting a quick answer, mainly to say this question is about what is called the "preferred basis problem" and it is a well-studied aspect of quantum measurement theory. The main thing to say is that it can happen that for one basis an off-diagonal density matrix element such as $langle phi_i |psirangle langle psi | phi_j rangle$ (where $phi_i$ are states of the basis) will either evolve very quickly or can be sensitive to very small disturbances, whereas for another basis this may not be so. In this case the off-diagonal elements of the density matrix average to zero over any practical timescale, so we have decoherence between states of such a basis. It is called a pointer basis.






          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
            );



            );













            draft saved

            draft discarded


















            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f497374%2fwhich-basis-does-the-wavefunction-collapse-to%23new-answer', 'question_page');

            );

            Post as a guest















            Required, but never shown

























            3 Answers
            3






            active

            oldest

            votes








            3 Answers
            3






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            7












            $begingroup$

            The system doesn't "know" anything.



            The only uncontroversial statement one can make about the (strong) measurement of a quantum system is that you will make the correct predictions if you assume that the state after the measurement was the eigenstate corresponding to the measured value of the observable (so, for position, indeed a $delta$-function, if we ignore issues with that not being a real function, which would be a distraction here). But what we mean by "state" in the first place - i.e. what ontology, if any, corresponds to the statement "the system is in the quantum state $lvert psirangle$" - is ambiguous to begin with:



            Whether the original state "collapsed" to this new state, whether the "state" is just an imperfect representation of our knowledge and the "collapse" is just updating our information (cf. "$psi$-ontic" vs "$psi$-epistemic", see e.g. this answer by Emilio Pisanty) instead of an actual physical process, or something else entirely, is a matter of quantum interpretation. In some interpretations, there is collapse, in others there isn't, but in any case, the formalism of quantum mechanics itself does not provide a single "correct" interpretation.



            That is, your question is essentially unanswerable unless you specify the interpretation within which it is to be answered. But none of the predictions of quantum mechanics depends on it anyway - you do not need to have a concept of "how" collapse works to compute the outcome of measurements.






            share|cite|improve this answer









            $endgroup$














            • $begingroup$
              Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
              $endgroup$
              – Jeff Bass
              1 hour ago















            7












            $begingroup$

            The system doesn't "know" anything.



            The only uncontroversial statement one can make about the (strong) measurement of a quantum system is that you will make the correct predictions if you assume that the state after the measurement was the eigenstate corresponding to the measured value of the observable (so, for position, indeed a $delta$-function, if we ignore issues with that not being a real function, which would be a distraction here). But what we mean by "state" in the first place - i.e. what ontology, if any, corresponds to the statement "the system is in the quantum state $lvert psirangle$" - is ambiguous to begin with:



            Whether the original state "collapsed" to this new state, whether the "state" is just an imperfect representation of our knowledge and the "collapse" is just updating our information (cf. "$psi$-ontic" vs "$psi$-epistemic", see e.g. this answer by Emilio Pisanty) instead of an actual physical process, or something else entirely, is a matter of quantum interpretation. In some interpretations, there is collapse, in others there isn't, but in any case, the formalism of quantum mechanics itself does not provide a single "correct" interpretation.



            That is, your question is essentially unanswerable unless you specify the interpretation within which it is to be answered. But none of the predictions of quantum mechanics depends on it anyway - you do not need to have a concept of "how" collapse works to compute the outcome of measurements.






            share|cite|improve this answer









            $endgroup$














            • $begingroup$
              Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
              $endgroup$
              – Jeff Bass
              1 hour ago













            7












            7








            7





            $begingroup$

            The system doesn't "know" anything.



            The only uncontroversial statement one can make about the (strong) measurement of a quantum system is that you will make the correct predictions if you assume that the state after the measurement was the eigenstate corresponding to the measured value of the observable (so, for position, indeed a $delta$-function, if we ignore issues with that not being a real function, which would be a distraction here). But what we mean by "state" in the first place - i.e. what ontology, if any, corresponds to the statement "the system is in the quantum state $lvert psirangle$" - is ambiguous to begin with:



            Whether the original state "collapsed" to this new state, whether the "state" is just an imperfect representation of our knowledge and the "collapse" is just updating our information (cf. "$psi$-ontic" vs "$psi$-epistemic", see e.g. this answer by Emilio Pisanty) instead of an actual physical process, or something else entirely, is a matter of quantum interpretation. In some interpretations, there is collapse, in others there isn't, but in any case, the formalism of quantum mechanics itself does not provide a single "correct" interpretation.



            That is, your question is essentially unanswerable unless you specify the interpretation within which it is to be answered. But none of the predictions of quantum mechanics depends on it anyway - you do not need to have a concept of "how" collapse works to compute the outcome of measurements.






            share|cite|improve this answer









            $endgroup$



            The system doesn't "know" anything.



            The only uncontroversial statement one can make about the (strong) measurement of a quantum system is that you will make the correct predictions if you assume that the state after the measurement was the eigenstate corresponding to the measured value of the observable (so, for position, indeed a $delta$-function, if we ignore issues with that not being a real function, which would be a distraction here). But what we mean by "state" in the first place - i.e. what ontology, if any, corresponds to the statement "the system is in the quantum state $lvert psirangle$" - is ambiguous to begin with:



            Whether the original state "collapsed" to this new state, whether the "state" is just an imperfect representation of our knowledge and the "collapse" is just updating our information (cf. "$psi$-ontic" vs "$psi$-epistemic", see e.g. this answer by Emilio Pisanty) instead of an actual physical process, or something else entirely, is a matter of quantum interpretation. In some interpretations, there is collapse, in others there isn't, but in any case, the formalism of quantum mechanics itself does not provide a single "correct" interpretation.



            That is, your question is essentially unanswerable unless you specify the interpretation within which it is to be answered. But none of the predictions of quantum mechanics depends on it anyway - you do not need to have a concept of "how" collapse works to compute the outcome of measurements.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 8 hours ago









            ACuriousMindACuriousMind

            75.5k18 gold badges139 silver badges352 bronze badges




            75.5k18 gold badges139 silver badges352 bronze badges














            • $begingroup$
              Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
              $endgroup$
              – Jeff Bass
              1 hour ago
















            • $begingroup$
              Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
              $endgroup$
              – Jeff Bass
              1 hour ago















            $begingroup$
            Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
            $endgroup$
            – Jeff Bass
            1 hour ago




            $begingroup$
            Interesting I've never heard it explained that way. So the precision of your observation must play a part then, right? If you don't carefully pin something down to exactly one location how could the wave localize to that point? However, I thought that the wave needed to collapse to a basis vector, so which basis does it use? I assume I'm misunderstanding what constitutes a basis vector. Is it more like the wave "constricts" to match the updated possibilities for the state?
            $endgroup$
            – Jeff Bass
            1 hour ago













            2












            $begingroup$

            The collapse happens in all bases. What I mean by that is that the wavefunction can be expressed in any basis you want to. It's just that the easiest basis to look at right after measurement is the one corresponding to what you measured, since the state is the eigenstate corresponding to your measurement.



            Always remember the wavefunction isn't physical. It's an abstract thing that we can only describe and "look at" as shadows from their projections. We can choose any projection we want to, but that choice doesn't change the wavefunction






            share|cite|improve this answer











            $endgroup$














            • $begingroup$
              This does not answer the question. The question is not about representing a wavefunction in different bases.
              $endgroup$
              – eigenvalue
              4 hours ago











            • $begingroup$
              @eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
              $endgroup$
              – Aaron Stevens
              3 hours ago
















            2












            $begingroup$

            The collapse happens in all bases. What I mean by that is that the wavefunction can be expressed in any basis you want to. It's just that the easiest basis to look at right after measurement is the one corresponding to what you measured, since the state is the eigenstate corresponding to your measurement.



            Always remember the wavefunction isn't physical. It's an abstract thing that we can only describe and "look at" as shadows from their projections. We can choose any projection we want to, but that choice doesn't change the wavefunction






            share|cite|improve this answer











            $endgroup$














            • $begingroup$
              This does not answer the question. The question is not about representing a wavefunction in different bases.
              $endgroup$
              – eigenvalue
              4 hours ago











            • $begingroup$
              @eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
              $endgroup$
              – Aaron Stevens
              3 hours ago














            2












            2








            2





            $begingroup$

            The collapse happens in all bases. What I mean by that is that the wavefunction can be expressed in any basis you want to. It's just that the easiest basis to look at right after measurement is the one corresponding to what you measured, since the state is the eigenstate corresponding to your measurement.



            Always remember the wavefunction isn't physical. It's an abstract thing that we can only describe and "look at" as shadows from their projections. We can choose any projection we want to, but that choice doesn't change the wavefunction






            share|cite|improve this answer











            $endgroup$



            The collapse happens in all bases. What I mean by that is that the wavefunction can be expressed in any basis you want to. It's just that the easiest basis to look at right after measurement is the one corresponding to what you measured, since the state is the eigenstate corresponding to your measurement.



            Always remember the wavefunction isn't physical. It's an abstract thing that we can only describe and "look at" as shadows from their projections. We can choose any projection we want to, but that choice doesn't change the wavefunction







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 8 hours ago

























            answered 8 hours ago









            Aaron StevensAaron Stevens

            21k4 gold badges36 silver badges75 bronze badges




            21k4 gold badges36 silver badges75 bronze badges














            • $begingroup$
              This does not answer the question. The question is not about representing a wavefunction in different bases.
              $endgroup$
              – eigenvalue
              4 hours ago











            • $begingroup$
              @eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
              $endgroup$
              – Aaron Stevens
              3 hours ago

















            • $begingroup$
              This does not answer the question. The question is not about representing a wavefunction in different bases.
              $endgroup$
              – eigenvalue
              4 hours ago











            • $begingroup$
              @eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
              $endgroup$
              – Aaron Stevens
              3 hours ago
















            $begingroup$
            This does not answer the question. The question is not about representing a wavefunction in different bases.
            $endgroup$
            – eigenvalue
            4 hours ago





            $begingroup$
            This does not answer the question. The question is not about representing a wavefunction in different bases.
            $endgroup$
            – eigenvalue
            4 hours ago













            $begingroup$
            @eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
            $endgroup$
            – Aaron Stevens
            3 hours ago





            $begingroup$
            @eigenvalue I answer the question at the beginning and then qualify and explain my answer. The point is that there isn't a basis that the wavefunction collapses to. You can choose any basis your want. It's just how you represent it. It doesn't go into some basis.
            $endgroup$
            – Aaron Stevens
            3 hours ago












            1












            $begingroup$

            I'm just posting a quick answer, mainly to say this question is about what is called the "preferred basis problem" and it is a well-studied aspect of quantum measurement theory. The main thing to say is that it can happen that for one basis an off-diagonal density matrix element such as $langle phi_i |psirangle langle psi | phi_j rangle$ (where $phi_i$ are states of the basis) will either evolve very quickly or can be sensitive to very small disturbances, whereas for another basis this may not be so. In this case the off-diagonal elements of the density matrix average to zero over any practical timescale, so we have decoherence between states of such a basis. It is called a pointer basis.






            share|cite|improve this answer









            $endgroup$



















              1












              $begingroup$

              I'm just posting a quick answer, mainly to say this question is about what is called the "preferred basis problem" and it is a well-studied aspect of quantum measurement theory. The main thing to say is that it can happen that for one basis an off-diagonal density matrix element such as $langle phi_i |psirangle langle psi | phi_j rangle$ (where $phi_i$ are states of the basis) will either evolve very quickly or can be sensitive to very small disturbances, whereas for another basis this may not be so. In this case the off-diagonal elements of the density matrix average to zero over any practical timescale, so we have decoherence between states of such a basis. It is called a pointer basis.






              share|cite|improve this answer









              $endgroup$

















                1












                1








                1





                $begingroup$

                I'm just posting a quick answer, mainly to say this question is about what is called the "preferred basis problem" and it is a well-studied aspect of quantum measurement theory. The main thing to say is that it can happen that for one basis an off-diagonal density matrix element such as $langle phi_i |psirangle langle psi | phi_j rangle$ (where $phi_i$ are states of the basis) will either evolve very quickly or can be sensitive to very small disturbances, whereas for another basis this may not be so. In this case the off-diagonal elements of the density matrix average to zero over any practical timescale, so we have decoherence between states of such a basis. It is called a pointer basis.






                share|cite|improve this answer









                $endgroup$



                I'm just posting a quick answer, mainly to say this question is about what is called the "preferred basis problem" and it is a well-studied aspect of quantum measurement theory. The main thing to say is that it can happen that for one basis an off-diagonal density matrix element such as $langle phi_i |psirangle langle psi | phi_j rangle$ (where $phi_i$ are states of the basis) will either evolve very quickly or can be sensitive to very small disturbances, whereas for another basis this may not be so. In this case the off-diagonal elements of the density matrix average to zero over any practical timescale, so we have decoherence between states of such a basis. It is called a pointer basis.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 8 hours ago









                Andrew SteaneAndrew Steane

                9,8461 gold badge12 silver badges52 bronze badges




                9,8461 gold badge12 silver badges52 bronze badges






























                    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%2f497374%2fwhich-basis-does-the-wavefunction-collapse-to%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

                    Canceling a color specificationRandomly assigning color to Graphics3D objects?Default color for Filling in Mathematica 9Coloring specific elements of sets with a prime modified order in an array plotHow to pick a color differing significantly from the colors already in a given color list?Detection of the text colorColor numbers based on their valueCan color schemes for use with ColorData include opacity specification?My dynamic color schemes

                    Invision Community Contents History See also References External links Navigation menuProprietaryinvisioncommunity.comIPS Community ForumsIPS Community Forumsthis blog entry"License Changes, IP.Board 3.4, and the Future""Interview -- Matt Mecham of Ibforums""CEO Invision Power Board, Matt Mecham Is a Liar, Thief!"IPB License Explanation 1.3, 1.3.1, 2.0, and 2.1ArchivedSecurity Fixes, Updates And Enhancements For IPB 1.3.1Archived"New Demo Accounts - Invision Power Services"the original"New Default Skin"the original"Invision Power Board 3.0.0 and Applications Released"the original"Archived copy"the original"Perpetual licenses being done away with""Release Notes - Invision Power Services""Introducing: IPS Community Suite 4!"Invision Community Release Notes

                    199年 目錄 大件事 到箇年出世嗰人 到箇年死嗰人 節慶、風俗習慣 導覽選單