Could quantum fluctuations spawn real matter?











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Would it be plausible for fluctuations in the QED vacuum to spawn actual matter (such as quarks, electrons the constituents of a hydrgen atom) given enough time and space?










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    What is the meaning of your question? Real particle/antiparticle pairs are created all the time, but it requires a real source of energy. Are you asking if real particles can be created without consuming energy? (Obviously not.) Or are you asking if energy applied to a vacuum can create particle pairs? (In which case there is no vacuum, because it contains energy and therefore particles). Could you please clarify?
    – safesphere
    Nov 28 at 15:15










  • Please keep in mind that comments are to be used only for suggesting improvements to, or requesting clarification on, the question. I've deleted some comments that don't fit that purpose.
    – David Z
    Nov 28 at 22:21















up vote
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Would it be plausible for fluctuations in the QED vacuum to spawn actual matter (such as quarks, electrons the constituents of a hydrgen atom) given enough time and space?










share|cite|improve this question


















  • 3




    What is the meaning of your question? Real particle/antiparticle pairs are created all the time, but it requires a real source of energy. Are you asking if real particles can be created without consuming energy? (Obviously not.) Or are you asking if energy applied to a vacuum can create particle pairs? (In which case there is no vacuum, because it contains energy and therefore particles). Could you please clarify?
    – safesphere
    Nov 28 at 15:15










  • Please keep in mind that comments are to be used only for suggesting improvements to, or requesting clarification on, the question. I've deleted some comments that don't fit that purpose.
    – David Z
    Nov 28 at 22:21













up vote
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Would it be plausible for fluctuations in the QED vacuum to spawn actual matter (such as quarks, electrons the constituents of a hydrgen atom) given enough time and space?










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Would it be plausible for fluctuations in the QED vacuum to spawn actual matter (such as quarks, electrons the constituents of a hydrgen atom) given enough time and space?







quantum-field-theory big-bang cosmological-inflation virtual-particles matter






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asked Nov 28 at 11:27









Justin

37049




37049








  • 3




    What is the meaning of your question? Real particle/antiparticle pairs are created all the time, but it requires a real source of energy. Are you asking if real particles can be created without consuming energy? (Obviously not.) Or are you asking if energy applied to a vacuum can create particle pairs? (In which case there is no vacuum, because it contains energy and therefore particles). Could you please clarify?
    – safesphere
    Nov 28 at 15:15










  • Please keep in mind that comments are to be used only for suggesting improvements to, or requesting clarification on, the question. I've deleted some comments that don't fit that purpose.
    – David Z
    Nov 28 at 22:21














  • 3




    What is the meaning of your question? Real particle/antiparticle pairs are created all the time, but it requires a real source of energy. Are you asking if real particles can be created without consuming energy? (Obviously not.) Or are you asking if energy applied to a vacuum can create particle pairs? (In which case there is no vacuum, because it contains energy and therefore particles). Could you please clarify?
    – safesphere
    Nov 28 at 15:15










  • Please keep in mind that comments are to be used only for suggesting improvements to, or requesting clarification on, the question. I've deleted some comments that don't fit that purpose.
    – David Z
    Nov 28 at 22:21








3




3




What is the meaning of your question? Real particle/antiparticle pairs are created all the time, but it requires a real source of energy. Are you asking if real particles can be created without consuming energy? (Obviously not.) Or are you asking if energy applied to a vacuum can create particle pairs? (In which case there is no vacuum, because it contains energy and therefore particles). Could you please clarify?
– safesphere
Nov 28 at 15:15




What is the meaning of your question? Real particle/antiparticle pairs are created all the time, but it requires a real source of energy. Are you asking if real particles can be created without consuming energy? (Obviously not.) Or are you asking if energy applied to a vacuum can create particle pairs? (In which case there is no vacuum, because it contains energy and therefore particles). Could you please clarify?
– safesphere
Nov 28 at 15:15












Please keep in mind that comments are to be used only for suggesting improvements to, or requesting clarification on, the question. I've deleted some comments that don't fit that purpose.
– David Z
Nov 28 at 22:21




Please keep in mind that comments are to be used only for suggesting improvements to, or requesting clarification on, the question. I've deleted some comments that don't fit that purpose.
– David Z
Nov 28 at 22:21










4 Answers
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The answer is no. And to be clear about this: the set of quantum fields in their least energy state, which we call the vacuum, when left to its own devices, in the absence of stuff (including gravitating stuff) does not fluctuate at all.



In this context the term 'fluctuation' was introduced by well-meaning physicists in an attempt to draw an analogy between quantum uncertainty (i.e. standard deviation in wavefunctions and operator measurements) and thermal (or other) fluctuation. The analogy does not always work very well. See here Is the term "quantum fluctuation" an aide to understanding? for some further discussion.



Added item



A brief comment on Hawking and Unruh radiation. The latter (Unruh) is a statement about what happens when a physical system is accelerated through the vacuum. Something has to provide the force which causes this acceleration! Hawking radiation is similar in some respects, in that, as I understand it, an apparatus in free-fall near the horizon does not get its internal degrees of freedom excited (except by the classical tidal forces). But in any case my statement above is about vacuum in the absence of extreme curvature. I would be happy to admit that gravitational tidal forces can be causal, just as other forces are.



And yet ... after writing the above I added a comment to the effect that a particle detector in free fall in otherwise empty flat space will not click. Then I was pushed to justify that and had to admit I'm not completely sure on this specific issue, so I have asked a question here Particles, fluctuations and the quantum vacuum: is this right? to get clarity. (Also, you can see in the comments that the original questioner had in mind not the quantum vacuum but rather a space filled with a scalar field not in its ground state (as a model of dark energy). Such a field can in principle excite other fields, i.e. form particles.)






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  • Thanks for your insight
    – Justin
    Nov 28 at 13:46










  • Not true. The interaction Hamiltonian in QED connects states with and without particles. Perhaps "oscillation" is a better term? A state with no electrons and no photons cannot be the vacuum of QED, because it only an eigenstate of the free Hamiltonian. A state with no electrons and no photons will "oscillate" between this state and states with electrons and photons.
    – Eric David Kramer
    Nov 28 at 14:30








  • 1




    What I intend to say is that whatever the ground state is, that is the one I am referring to when I say it doesn't fluctuate. I see that this is not an eigenstate of the free Hamiltonian, but as far as I know if you put a particle detector in free fall in otherwise empty space it will not click. Would be happy to learn I am wrong!
    – Andrew Steane
    Nov 28 at 16:35










  • @Andrew Im a mere muggle so i realise my question may sound nieve but i am a theoretical physics lecture junkie.
    – Justin
    Nov 28 at 19:18










  • I liked your question and I note answer from L. B. Crowell which adds something useful I think.
    – Andrew Steane
    Nov 28 at 19:25


















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The answers above are correct. Kramer and Steane make reference to gravitation. This is where things get strange, for a quantum vacuum is only defined in a local region. A black hole for instance has local internal frames patched together by transition functions that under derivatives give connections terms that further with covariant derivatives give curvatures. Because of this you have not a vacuum but a set of vacua. For this reason a transformation on one vacuum can result in a vacuum plus particles or bosons.



Coleman showed how a potential function that with a hump, say a quartic function, that is also asymmetric may generate a bubble of spacetime. Quantum states in the right can tunnel into the left, and where there is an energy gap. This means there might be a transition that produces bosons or particles. This radiation may in general be bubble cosmologies.



enter image description here



So in general to generate particles you either need a set of vacua, such as those in local regions of spacetime containing an event horizon, or some potential function with different vacuum states. You can't get particles generated out of a global vacuum state. That is unless there is some sort of instability that causes that vacuum to transition into another vacuum at lower energy.






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  • Thanks for this concise and pertinant extra info. But for anyone looking in, note that the use of the term "vacuum" for something with huge gravitational or other fields, at energies well above their ground state, is a technical usage far removed from everyday language.
    – Andrew Steane
    Nov 28 at 19:28










  • @AndrewSteane It's actually completely standard. For example, the Higgs field in our universe might not be at its global minimum right now. That is, seemingly empty space might contain much more Higgs field energy than the very lowest it could; this doesn't have any effects because we can just shift the origin of energy. If you insist on only defining a vacuum as the unique global minimum of energy, then we don't know if we live in one, and probably won't know for at least a thousand years.
    – knzhou
    2 days ago












  • @knzhou I see what you mean; the terminology is partly a matter of taste. But during the period of time where such a Higgs field has no effects, then clearly it is not transferring energy to other fields. As soon as its effects become apparent owing to the appearance of an accessible further minimum, it becomes questionable whether it should be called a "vacuum" or a vacuum state. I mean questionable in the context of being helpful to non-experts eager to learn. Better to say "it's like a metastable state" or "it's like opening a flood-gate at the base of a dam."
    – Andrew Steane
    2 days ago


















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Note: People seem to think that an empty universe is an eigenstate of the full QED Hamiltonian. This is not true. The Hamiltonian is



$$H= sum_{e^+e^-} omega_p (b^dagger_p b_p +c^dagger_p c_p)+ sum_{rm photons}!k,a^dagger_k a_k,+, sum_{k+p+p'=0}(a^dagger_k + a_k)(b_p c_{p'} + c^dagger_p b^dagger_{p'})$$



If we define the vacuum as



$$a_k|0rangle = b_p|0rangle = c_p|0rangle =0,$$



Then



$$H|0rangle = sum_{k+p+p'=0}a^dagger_kc^dagger_p b^dagger_{p'}|0rangle neq 0$$



That means that the true vacuum



$H|Omegarangle =0$



has



$$a_kb_kc_k|Omegarangle neq 0.$$



The number of particles is not conserved. Neither is energy with respect to the free Hamiltonian. Even for a single particle.



====================================



Normally, a process that violates energy conservation (of the free Hamiltonian!) by an amount $Delta E$ can only survive for a time $hbar/Delta E$. Energy is only strictly conserved (in scattering for instance) for processes where the initial and final states are at $t=pm infty$. For example, you could create the state with two positrons and two electrons through the process (time going up):



enter image description here



but it would only last a time $hbar/4m_e sim 10^{-22}$ seconds. The reason in that in the interaction picture for scattering from $t=-T$ to $t=+T$, the scattering probability will look something like



$$P sim left|int_{-T}^T!dt,e^{i(E_f-E_i)t},langle f | V|irangleright|^2sim frac{sin^2(E_f-E_i)T}{(E_f-E_i)^2}left|langle f | V|irangleright|^2$$



You can check that this is non-zero for times $Tlesssim 1/(E_f-E_i)$.



However, when gravity is included, time can be stretched, making short-lived states very long-lived. This happens when you have a horizon, like when there is a black hole or in an accelerating frame, in which case time can be infinitely stretched, producing matter. I think this also happens in inflation but there the particles get pulled away from each other. Not sure how to show this in terms of quantum mechanics.






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  • In your last sentence you mention inflation. Could you refer to my comment on my own question. Do you think inflation could create long lived matter?
    – Justin
    Nov 28 at 14:12






  • 1




    @Justin I do not know much about it, but yes, apparently inflation does create long-lived matter. Most texts on inflation describe how when you define the vacuum at two different times in an accelerating universe, the matter content is different between the two. I don't know how to show it in terms of what I explained in my answer, but you could imagine the short-lived state being created and then the particles being pulled away very far from each other by inflation, creating a long-lived state.
    – Eric David Kramer
    Nov 28 at 14:27










  • The idea that quantum mechanics allows you to energy conservation for short times is an annoyingly persistent myth -- one of the first things you do in every intro QM class is prove energy is conserved! This idea is one of those things that's been being passed down from popsci book to popsci book unchanged for 100 years, despite having no relation to the truth.
    – knzhou
    2 days ago










  • @knzhou Yes you are absolutely right, energy is conserved. An energy eigenstate will stay in that state forever. But a state with no matter cannot be an energy eigenstate when interactions are included. An energy eigenstate of the free Hamiltonian (i.e. a state with no matter) will, over time, change its energy eigenvalue with respect to this free Hamiltonian (i.e. particles will be created). No myth here.
    – Eric David Kramer
    2 days ago






  • 1




    @EricDavidKramer Similarly, the only sensible definition of "empty universe" is the vacuum state of the full theory, not the free theory. While it is certainly true that the vacuum state of the free theory is not an eigenstate of the full Hamiltonian, this state has no physical meaning anyway, and is not what anybody is ever referring to when they speak of "the vacuum".
    – knzhou
    2 days ago




















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Also in QED, total Energy is conserved for every time! A difference between QED and classical electrodynamics is that the Expression for total Energy is slightly altered. Classical electrodynamics has kinetic Energy $T$ and a potential Energy $V$ arising from electromagnetic fields. In classical Theory, total Energy is conserved, i.e.



$T+V = const.$



But in Quantum electrodynamics and other Quantum field theories, you have also additional zero-Point Energy $hbar omega$, which is responsible for e.g. the Casimir effect (when non-charged plates attrackt each other when These have extremely small distance). The frequency $omega$ can be interpreted as how fast significant changes in the System takes place. For a many-particle System with extremely high collision frequency, the value $omega$ will be also high, altering the Energy balance to



$T+V + hbar omega = const.$



The last term in this equation on the left Hand side is also called "self-energy" in Quantum kinetic Theory. This self-energy is a complex-valued quantity, where the real part describes the



zero-Point Energy



, while the imaginary part is antiproportional to



the lifetime of the excited state



Hence, higher zero-Point Energy imply Shorter Lifetimes of excited states.



Another example of the Change of effective Energy/Hamiltonian due to Quantum effects is shown e.g. in this paper:



https://arxiv.org/abs/0706.1090






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    4 Answers
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    The answer is no. And to be clear about this: the set of quantum fields in their least energy state, which we call the vacuum, when left to its own devices, in the absence of stuff (including gravitating stuff) does not fluctuate at all.



    In this context the term 'fluctuation' was introduced by well-meaning physicists in an attempt to draw an analogy between quantum uncertainty (i.e. standard deviation in wavefunctions and operator measurements) and thermal (or other) fluctuation. The analogy does not always work very well. See here Is the term "quantum fluctuation" an aide to understanding? for some further discussion.



    Added item



    A brief comment on Hawking and Unruh radiation. The latter (Unruh) is a statement about what happens when a physical system is accelerated through the vacuum. Something has to provide the force which causes this acceleration! Hawking radiation is similar in some respects, in that, as I understand it, an apparatus in free-fall near the horizon does not get its internal degrees of freedom excited (except by the classical tidal forces). But in any case my statement above is about vacuum in the absence of extreme curvature. I would be happy to admit that gravitational tidal forces can be causal, just as other forces are.



    And yet ... after writing the above I added a comment to the effect that a particle detector in free fall in otherwise empty flat space will not click. Then I was pushed to justify that and had to admit I'm not completely sure on this specific issue, so I have asked a question here Particles, fluctuations and the quantum vacuum: is this right? to get clarity. (Also, you can see in the comments that the original questioner had in mind not the quantum vacuum but rather a space filled with a scalar field not in its ground state (as a model of dark energy). Such a field can in principle excite other fields, i.e. form particles.)






    share|cite|improve this answer























    • Thanks for your insight
      – Justin
      Nov 28 at 13:46










    • Not true. The interaction Hamiltonian in QED connects states with and without particles. Perhaps "oscillation" is a better term? A state with no electrons and no photons cannot be the vacuum of QED, because it only an eigenstate of the free Hamiltonian. A state with no electrons and no photons will "oscillate" between this state and states with electrons and photons.
      – Eric David Kramer
      Nov 28 at 14:30








    • 1




      What I intend to say is that whatever the ground state is, that is the one I am referring to when I say it doesn't fluctuate. I see that this is not an eigenstate of the free Hamiltonian, but as far as I know if you put a particle detector in free fall in otherwise empty space it will not click. Would be happy to learn I am wrong!
      – Andrew Steane
      Nov 28 at 16:35










    • @Andrew Im a mere muggle so i realise my question may sound nieve but i am a theoretical physics lecture junkie.
      – Justin
      Nov 28 at 19:18










    • I liked your question and I note answer from L. B. Crowell which adds something useful I think.
      – Andrew Steane
      Nov 28 at 19:25















    up vote
    13
    down vote













    The answer is no. And to be clear about this: the set of quantum fields in their least energy state, which we call the vacuum, when left to its own devices, in the absence of stuff (including gravitating stuff) does not fluctuate at all.



    In this context the term 'fluctuation' was introduced by well-meaning physicists in an attempt to draw an analogy between quantum uncertainty (i.e. standard deviation in wavefunctions and operator measurements) and thermal (or other) fluctuation. The analogy does not always work very well. See here Is the term "quantum fluctuation" an aide to understanding? for some further discussion.



    Added item



    A brief comment on Hawking and Unruh radiation. The latter (Unruh) is a statement about what happens when a physical system is accelerated through the vacuum. Something has to provide the force which causes this acceleration! Hawking radiation is similar in some respects, in that, as I understand it, an apparatus in free-fall near the horizon does not get its internal degrees of freedom excited (except by the classical tidal forces). But in any case my statement above is about vacuum in the absence of extreme curvature. I would be happy to admit that gravitational tidal forces can be causal, just as other forces are.



    And yet ... after writing the above I added a comment to the effect that a particle detector in free fall in otherwise empty flat space will not click. Then I was pushed to justify that and had to admit I'm not completely sure on this specific issue, so I have asked a question here Particles, fluctuations and the quantum vacuum: is this right? to get clarity. (Also, you can see in the comments that the original questioner had in mind not the quantum vacuum but rather a space filled with a scalar field not in its ground state (as a model of dark energy). Such a field can in principle excite other fields, i.e. form particles.)






    share|cite|improve this answer























    • Thanks for your insight
      – Justin
      Nov 28 at 13:46










    • Not true. The interaction Hamiltonian in QED connects states with and without particles. Perhaps "oscillation" is a better term? A state with no electrons and no photons cannot be the vacuum of QED, because it only an eigenstate of the free Hamiltonian. A state with no electrons and no photons will "oscillate" between this state and states with electrons and photons.
      – Eric David Kramer
      Nov 28 at 14:30








    • 1




      What I intend to say is that whatever the ground state is, that is the one I am referring to when I say it doesn't fluctuate. I see that this is not an eigenstate of the free Hamiltonian, but as far as I know if you put a particle detector in free fall in otherwise empty space it will not click. Would be happy to learn I am wrong!
      – Andrew Steane
      Nov 28 at 16:35










    • @Andrew Im a mere muggle so i realise my question may sound nieve but i am a theoretical physics lecture junkie.
      – Justin
      Nov 28 at 19:18










    • I liked your question and I note answer from L. B. Crowell which adds something useful I think.
      – Andrew Steane
      Nov 28 at 19:25













    up vote
    13
    down vote










    up vote
    13
    down vote









    The answer is no. And to be clear about this: the set of quantum fields in their least energy state, which we call the vacuum, when left to its own devices, in the absence of stuff (including gravitating stuff) does not fluctuate at all.



    In this context the term 'fluctuation' was introduced by well-meaning physicists in an attempt to draw an analogy between quantum uncertainty (i.e. standard deviation in wavefunctions and operator measurements) and thermal (or other) fluctuation. The analogy does not always work very well. See here Is the term "quantum fluctuation" an aide to understanding? for some further discussion.



    Added item



    A brief comment on Hawking and Unruh radiation. The latter (Unruh) is a statement about what happens when a physical system is accelerated through the vacuum. Something has to provide the force which causes this acceleration! Hawking radiation is similar in some respects, in that, as I understand it, an apparatus in free-fall near the horizon does not get its internal degrees of freedom excited (except by the classical tidal forces). But in any case my statement above is about vacuum in the absence of extreme curvature. I would be happy to admit that gravitational tidal forces can be causal, just as other forces are.



    And yet ... after writing the above I added a comment to the effect that a particle detector in free fall in otherwise empty flat space will not click. Then I was pushed to justify that and had to admit I'm not completely sure on this specific issue, so I have asked a question here Particles, fluctuations and the quantum vacuum: is this right? to get clarity. (Also, you can see in the comments that the original questioner had in mind not the quantum vacuum but rather a space filled with a scalar field not in its ground state (as a model of dark energy). Such a field can in principle excite other fields, i.e. form particles.)






    share|cite|improve this answer














    The answer is no. And to be clear about this: the set of quantum fields in their least energy state, which we call the vacuum, when left to its own devices, in the absence of stuff (including gravitating stuff) does not fluctuate at all.



    In this context the term 'fluctuation' was introduced by well-meaning physicists in an attempt to draw an analogy between quantum uncertainty (i.e. standard deviation in wavefunctions and operator measurements) and thermal (or other) fluctuation. The analogy does not always work very well. See here Is the term "quantum fluctuation" an aide to understanding? for some further discussion.



    Added item



    A brief comment on Hawking and Unruh radiation. The latter (Unruh) is a statement about what happens when a physical system is accelerated through the vacuum. Something has to provide the force which causes this acceleration! Hawking radiation is similar in some respects, in that, as I understand it, an apparatus in free-fall near the horizon does not get its internal degrees of freedom excited (except by the classical tidal forces). But in any case my statement above is about vacuum in the absence of extreme curvature. I would be happy to admit that gravitational tidal forces can be causal, just as other forces are.



    And yet ... after writing the above I added a comment to the effect that a particle detector in free fall in otherwise empty flat space will not click. Then I was pushed to justify that and had to admit I'm not completely sure on this specific issue, so I have asked a question here Particles, fluctuations and the quantum vacuum: is this right? to get clarity. (Also, you can see in the comments that the original questioner had in mind not the quantum vacuum but rather a space filled with a scalar field not in its ground state (as a model of dark energy). Such a field can in principle excite other fields, i.e. form particles.)







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited yesterday

























    answered Nov 28 at 12:30









    Andrew Steane

    2,190522




    2,190522












    • Thanks for your insight
      – Justin
      Nov 28 at 13:46










    • Not true. The interaction Hamiltonian in QED connects states with and without particles. Perhaps "oscillation" is a better term? A state with no electrons and no photons cannot be the vacuum of QED, because it only an eigenstate of the free Hamiltonian. A state with no electrons and no photons will "oscillate" between this state and states with electrons and photons.
      – Eric David Kramer
      Nov 28 at 14:30








    • 1




      What I intend to say is that whatever the ground state is, that is the one I am referring to when I say it doesn't fluctuate. I see that this is not an eigenstate of the free Hamiltonian, but as far as I know if you put a particle detector in free fall in otherwise empty space it will not click. Would be happy to learn I am wrong!
      – Andrew Steane
      Nov 28 at 16:35










    • @Andrew Im a mere muggle so i realise my question may sound nieve but i am a theoretical physics lecture junkie.
      – Justin
      Nov 28 at 19:18










    • I liked your question and I note answer from L. B. Crowell which adds something useful I think.
      – Andrew Steane
      Nov 28 at 19:25


















    • Thanks for your insight
      – Justin
      Nov 28 at 13:46










    • Not true. The interaction Hamiltonian in QED connects states with and without particles. Perhaps "oscillation" is a better term? A state with no electrons and no photons cannot be the vacuum of QED, because it only an eigenstate of the free Hamiltonian. A state with no electrons and no photons will "oscillate" between this state and states with electrons and photons.
      – Eric David Kramer
      Nov 28 at 14:30








    • 1




      What I intend to say is that whatever the ground state is, that is the one I am referring to when I say it doesn't fluctuate. I see that this is not an eigenstate of the free Hamiltonian, but as far as I know if you put a particle detector in free fall in otherwise empty space it will not click. Would be happy to learn I am wrong!
      – Andrew Steane
      Nov 28 at 16:35










    • @Andrew Im a mere muggle so i realise my question may sound nieve but i am a theoretical physics lecture junkie.
      – Justin
      Nov 28 at 19:18










    • I liked your question and I note answer from L. B. Crowell which adds something useful I think.
      – Andrew Steane
      Nov 28 at 19:25
















    Thanks for your insight
    – Justin
    Nov 28 at 13:46




    Thanks for your insight
    – Justin
    Nov 28 at 13:46












    Not true. The interaction Hamiltonian in QED connects states with and without particles. Perhaps "oscillation" is a better term? A state with no electrons and no photons cannot be the vacuum of QED, because it only an eigenstate of the free Hamiltonian. A state with no electrons and no photons will "oscillate" between this state and states with electrons and photons.
    – Eric David Kramer
    Nov 28 at 14:30






    Not true. The interaction Hamiltonian in QED connects states with and without particles. Perhaps "oscillation" is a better term? A state with no electrons and no photons cannot be the vacuum of QED, because it only an eigenstate of the free Hamiltonian. A state with no electrons and no photons will "oscillate" between this state and states with electrons and photons.
    – Eric David Kramer
    Nov 28 at 14:30






    1




    1




    What I intend to say is that whatever the ground state is, that is the one I am referring to when I say it doesn't fluctuate. I see that this is not an eigenstate of the free Hamiltonian, but as far as I know if you put a particle detector in free fall in otherwise empty space it will not click. Would be happy to learn I am wrong!
    – Andrew Steane
    Nov 28 at 16:35




    What I intend to say is that whatever the ground state is, that is the one I am referring to when I say it doesn't fluctuate. I see that this is not an eigenstate of the free Hamiltonian, but as far as I know if you put a particle detector in free fall in otherwise empty space it will not click. Would be happy to learn I am wrong!
    – Andrew Steane
    Nov 28 at 16:35












    @Andrew Im a mere muggle so i realise my question may sound nieve but i am a theoretical physics lecture junkie.
    – Justin
    Nov 28 at 19:18




    @Andrew Im a mere muggle so i realise my question may sound nieve but i am a theoretical physics lecture junkie.
    – Justin
    Nov 28 at 19:18












    I liked your question and I note answer from L. B. Crowell which adds something useful I think.
    – Andrew Steane
    Nov 28 at 19:25




    I liked your question and I note answer from L. B. Crowell which adds something useful I think.
    – Andrew Steane
    Nov 28 at 19:25










    up vote
    5
    down vote













    The answers above are correct. Kramer and Steane make reference to gravitation. This is where things get strange, for a quantum vacuum is only defined in a local region. A black hole for instance has local internal frames patched together by transition functions that under derivatives give connections terms that further with covariant derivatives give curvatures. Because of this you have not a vacuum but a set of vacua. For this reason a transformation on one vacuum can result in a vacuum plus particles or bosons.



    Coleman showed how a potential function that with a hump, say a quartic function, that is also asymmetric may generate a bubble of spacetime. Quantum states in the right can tunnel into the left, and where there is an energy gap. This means there might be a transition that produces bosons or particles. This radiation may in general be bubble cosmologies.



    enter image description here



    So in general to generate particles you either need a set of vacua, such as those in local regions of spacetime containing an event horizon, or some potential function with different vacuum states. You can't get particles generated out of a global vacuum state. That is unless there is some sort of instability that causes that vacuum to transition into another vacuum at lower energy.






    share|cite|improve this answer





















    • Thanks for this concise and pertinant extra info. But for anyone looking in, note that the use of the term "vacuum" for something with huge gravitational or other fields, at energies well above their ground state, is a technical usage far removed from everyday language.
      – Andrew Steane
      Nov 28 at 19:28










    • @AndrewSteane It's actually completely standard. For example, the Higgs field in our universe might not be at its global minimum right now. That is, seemingly empty space might contain much more Higgs field energy than the very lowest it could; this doesn't have any effects because we can just shift the origin of energy. If you insist on only defining a vacuum as the unique global minimum of energy, then we don't know if we live in one, and probably won't know for at least a thousand years.
      – knzhou
      2 days ago












    • @knzhou I see what you mean; the terminology is partly a matter of taste. But during the period of time where such a Higgs field has no effects, then clearly it is not transferring energy to other fields. As soon as its effects become apparent owing to the appearance of an accessible further minimum, it becomes questionable whether it should be called a "vacuum" or a vacuum state. I mean questionable in the context of being helpful to non-experts eager to learn. Better to say "it's like a metastable state" or "it's like opening a flood-gate at the base of a dam."
      – Andrew Steane
      2 days ago















    up vote
    5
    down vote













    The answers above are correct. Kramer and Steane make reference to gravitation. This is where things get strange, for a quantum vacuum is only defined in a local region. A black hole for instance has local internal frames patched together by transition functions that under derivatives give connections terms that further with covariant derivatives give curvatures. Because of this you have not a vacuum but a set of vacua. For this reason a transformation on one vacuum can result in a vacuum plus particles or bosons.



    Coleman showed how a potential function that with a hump, say a quartic function, that is also asymmetric may generate a bubble of spacetime. Quantum states in the right can tunnel into the left, and where there is an energy gap. This means there might be a transition that produces bosons or particles. This radiation may in general be bubble cosmologies.



    enter image description here



    So in general to generate particles you either need a set of vacua, such as those in local regions of spacetime containing an event horizon, or some potential function with different vacuum states. You can't get particles generated out of a global vacuum state. That is unless there is some sort of instability that causes that vacuum to transition into another vacuum at lower energy.






    share|cite|improve this answer





















    • Thanks for this concise and pertinant extra info. But for anyone looking in, note that the use of the term "vacuum" for something with huge gravitational or other fields, at energies well above their ground state, is a technical usage far removed from everyday language.
      – Andrew Steane
      Nov 28 at 19:28










    • @AndrewSteane It's actually completely standard. For example, the Higgs field in our universe might not be at its global minimum right now. That is, seemingly empty space might contain much more Higgs field energy than the very lowest it could; this doesn't have any effects because we can just shift the origin of energy. If you insist on only defining a vacuum as the unique global minimum of energy, then we don't know if we live in one, and probably won't know for at least a thousand years.
      – knzhou
      2 days ago












    • @knzhou I see what you mean; the terminology is partly a matter of taste. But during the period of time where such a Higgs field has no effects, then clearly it is not transferring energy to other fields. As soon as its effects become apparent owing to the appearance of an accessible further minimum, it becomes questionable whether it should be called a "vacuum" or a vacuum state. I mean questionable in the context of being helpful to non-experts eager to learn. Better to say "it's like a metastable state" or "it's like opening a flood-gate at the base of a dam."
      – Andrew Steane
      2 days ago













    up vote
    5
    down vote










    up vote
    5
    down vote









    The answers above are correct. Kramer and Steane make reference to gravitation. This is where things get strange, for a quantum vacuum is only defined in a local region. A black hole for instance has local internal frames patched together by transition functions that under derivatives give connections terms that further with covariant derivatives give curvatures. Because of this you have not a vacuum but a set of vacua. For this reason a transformation on one vacuum can result in a vacuum plus particles or bosons.



    Coleman showed how a potential function that with a hump, say a quartic function, that is also asymmetric may generate a bubble of spacetime. Quantum states in the right can tunnel into the left, and where there is an energy gap. This means there might be a transition that produces bosons or particles. This radiation may in general be bubble cosmologies.



    enter image description here



    So in general to generate particles you either need a set of vacua, such as those in local regions of spacetime containing an event horizon, or some potential function with different vacuum states. You can't get particles generated out of a global vacuum state. That is unless there is some sort of instability that causes that vacuum to transition into another vacuum at lower energy.






    share|cite|improve this answer












    The answers above are correct. Kramer and Steane make reference to gravitation. This is where things get strange, for a quantum vacuum is only defined in a local region. A black hole for instance has local internal frames patched together by transition functions that under derivatives give connections terms that further with covariant derivatives give curvatures. Because of this you have not a vacuum but a set of vacua. For this reason a transformation on one vacuum can result in a vacuum plus particles or bosons.



    Coleman showed how a potential function that with a hump, say a quartic function, that is also asymmetric may generate a bubble of spacetime. Quantum states in the right can tunnel into the left, and where there is an energy gap. This means there might be a transition that produces bosons or particles. This radiation may in general be bubble cosmologies.



    enter image description here



    So in general to generate particles you either need a set of vacua, such as those in local regions of spacetime containing an event horizon, or some potential function with different vacuum states. You can't get particles generated out of a global vacuum state. That is unless there is some sort of instability that causes that vacuum to transition into another vacuum at lower energy.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 28 at 15:39









    Lawrence B. Crowell

    10.8k11023




    10.8k11023












    • Thanks for this concise and pertinant extra info. But for anyone looking in, note that the use of the term "vacuum" for something with huge gravitational or other fields, at energies well above their ground state, is a technical usage far removed from everyday language.
      – Andrew Steane
      Nov 28 at 19:28










    • @AndrewSteane It's actually completely standard. For example, the Higgs field in our universe might not be at its global minimum right now. That is, seemingly empty space might contain much more Higgs field energy than the very lowest it could; this doesn't have any effects because we can just shift the origin of energy. If you insist on only defining a vacuum as the unique global minimum of energy, then we don't know if we live in one, and probably won't know for at least a thousand years.
      – knzhou
      2 days ago












    • @knzhou I see what you mean; the terminology is partly a matter of taste. But during the period of time where such a Higgs field has no effects, then clearly it is not transferring energy to other fields. As soon as its effects become apparent owing to the appearance of an accessible further minimum, it becomes questionable whether it should be called a "vacuum" or a vacuum state. I mean questionable in the context of being helpful to non-experts eager to learn. Better to say "it's like a metastable state" or "it's like opening a flood-gate at the base of a dam."
      – Andrew Steane
      2 days ago


















    • Thanks for this concise and pertinant extra info. But for anyone looking in, note that the use of the term "vacuum" for something with huge gravitational or other fields, at energies well above their ground state, is a technical usage far removed from everyday language.
      – Andrew Steane
      Nov 28 at 19:28










    • @AndrewSteane It's actually completely standard. For example, the Higgs field in our universe might not be at its global minimum right now. That is, seemingly empty space might contain much more Higgs field energy than the very lowest it could; this doesn't have any effects because we can just shift the origin of energy. If you insist on only defining a vacuum as the unique global minimum of energy, then we don't know if we live in one, and probably won't know for at least a thousand years.
      – knzhou
      2 days ago












    • @knzhou I see what you mean; the terminology is partly a matter of taste. But during the period of time where such a Higgs field has no effects, then clearly it is not transferring energy to other fields. As soon as its effects become apparent owing to the appearance of an accessible further minimum, it becomes questionable whether it should be called a "vacuum" or a vacuum state. I mean questionable in the context of being helpful to non-experts eager to learn. Better to say "it's like a metastable state" or "it's like opening a flood-gate at the base of a dam."
      – Andrew Steane
      2 days ago
















    Thanks for this concise and pertinant extra info. But for anyone looking in, note that the use of the term "vacuum" for something with huge gravitational or other fields, at energies well above their ground state, is a technical usage far removed from everyday language.
    – Andrew Steane
    Nov 28 at 19:28




    Thanks for this concise and pertinant extra info. But for anyone looking in, note that the use of the term "vacuum" for something with huge gravitational or other fields, at energies well above their ground state, is a technical usage far removed from everyday language.
    – Andrew Steane
    Nov 28 at 19:28












    @AndrewSteane It's actually completely standard. For example, the Higgs field in our universe might not be at its global minimum right now. That is, seemingly empty space might contain much more Higgs field energy than the very lowest it could; this doesn't have any effects because we can just shift the origin of energy. If you insist on only defining a vacuum as the unique global minimum of energy, then we don't know if we live in one, and probably won't know for at least a thousand years.
    – knzhou
    2 days ago






    @AndrewSteane It's actually completely standard. For example, the Higgs field in our universe might not be at its global minimum right now. That is, seemingly empty space might contain much more Higgs field energy than the very lowest it could; this doesn't have any effects because we can just shift the origin of energy. If you insist on only defining a vacuum as the unique global minimum of energy, then we don't know if we live in one, and probably won't know for at least a thousand years.
    – knzhou
    2 days ago














    @knzhou I see what you mean; the terminology is partly a matter of taste. But during the period of time where such a Higgs field has no effects, then clearly it is not transferring energy to other fields. As soon as its effects become apparent owing to the appearance of an accessible further minimum, it becomes questionable whether it should be called a "vacuum" or a vacuum state. I mean questionable in the context of being helpful to non-experts eager to learn. Better to say "it's like a metastable state" or "it's like opening a flood-gate at the base of a dam."
    – Andrew Steane
    2 days ago




    @knzhou I see what you mean; the terminology is partly a matter of taste. But during the period of time where such a Higgs field has no effects, then clearly it is not transferring energy to other fields. As soon as its effects become apparent owing to the appearance of an accessible further minimum, it becomes questionable whether it should be called a "vacuum" or a vacuum state. I mean questionable in the context of being helpful to non-experts eager to learn. Better to say "it's like a metastable state" or "it's like opening a flood-gate at the base of a dam."
    – Andrew Steane
    2 days ago










    up vote
    2
    down vote













    Note: People seem to think that an empty universe is an eigenstate of the full QED Hamiltonian. This is not true. The Hamiltonian is



    $$H= sum_{e^+e^-} omega_p (b^dagger_p b_p +c^dagger_p c_p)+ sum_{rm photons}!k,a^dagger_k a_k,+, sum_{k+p+p'=0}(a^dagger_k + a_k)(b_p c_{p'} + c^dagger_p b^dagger_{p'})$$



    If we define the vacuum as



    $$a_k|0rangle = b_p|0rangle = c_p|0rangle =0,$$



    Then



    $$H|0rangle = sum_{k+p+p'=0}a^dagger_kc^dagger_p b^dagger_{p'}|0rangle neq 0$$



    That means that the true vacuum



    $H|Omegarangle =0$



    has



    $$a_kb_kc_k|Omegarangle neq 0.$$



    The number of particles is not conserved. Neither is energy with respect to the free Hamiltonian. Even for a single particle.



    ====================================



    Normally, a process that violates energy conservation (of the free Hamiltonian!) by an amount $Delta E$ can only survive for a time $hbar/Delta E$. Energy is only strictly conserved (in scattering for instance) for processes where the initial and final states are at $t=pm infty$. For example, you could create the state with two positrons and two electrons through the process (time going up):



    enter image description here



    but it would only last a time $hbar/4m_e sim 10^{-22}$ seconds. The reason in that in the interaction picture for scattering from $t=-T$ to $t=+T$, the scattering probability will look something like



    $$P sim left|int_{-T}^T!dt,e^{i(E_f-E_i)t},langle f | V|irangleright|^2sim frac{sin^2(E_f-E_i)T}{(E_f-E_i)^2}left|langle f | V|irangleright|^2$$



    You can check that this is non-zero for times $Tlesssim 1/(E_f-E_i)$.



    However, when gravity is included, time can be stretched, making short-lived states very long-lived. This happens when you have a horizon, like when there is a black hole or in an accelerating frame, in which case time can be infinitely stretched, producing matter. I think this also happens in inflation but there the particles get pulled away from each other. Not sure how to show this in terms of quantum mechanics.






    share|cite|improve this answer























    • In your last sentence you mention inflation. Could you refer to my comment on my own question. Do you think inflation could create long lived matter?
      – Justin
      Nov 28 at 14:12






    • 1




      @Justin I do not know much about it, but yes, apparently inflation does create long-lived matter. Most texts on inflation describe how when you define the vacuum at two different times in an accelerating universe, the matter content is different between the two. I don't know how to show it in terms of what I explained in my answer, but you could imagine the short-lived state being created and then the particles being pulled away very far from each other by inflation, creating a long-lived state.
      – Eric David Kramer
      Nov 28 at 14:27










    • The idea that quantum mechanics allows you to energy conservation for short times is an annoyingly persistent myth -- one of the first things you do in every intro QM class is prove energy is conserved! This idea is one of those things that's been being passed down from popsci book to popsci book unchanged for 100 years, despite having no relation to the truth.
      – knzhou
      2 days ago










    • @knzhou Yes you are absolutely right, energy is conserved. An energy eigenstate will stay in that state forever. But a state with no matter cannot be an energy eigenstate when interactions are included. An energy eigenstate of the free Hamiltonian (i.e. a state with no matter) will, over time, change its energy eigenvalue with respect to this free Hamiltonian (i.e. particles will be created). No myth here.
      – Eric David Kramer
      2 days ago






    • 1




      @EricDavidKramer Similarly, the only sensible definition of "empty universe" is the vacuum state of the full theory, not the free theory. While it is certainly true that the vacuum state of the free theory is not an eigenstate of the full Hamiltonian, this state has no physical meaning anyway, and is not what anybody is ever referring to when they speak of "the vacuum".
      – knzhou
      2 days ago

















    up vote
    2
    down vote













    Note: People seem to think that an empty universe is an eigenstate of the full QED Hamiltonian. This is not true. The Hamiltonian is



    $$H= sum_{e^+e^-} omega_p (b^dagger_p b_p +c^dagger_p c_p)+ sum_{rm photons}!k,a^dagger_k a_k,+, sum_{k+p+p'=0}(a^dagger_k + a_k)(b_p c_{p'} + c^dagger_p b^dagger_{p'})$$



    If we define the vacuum as



    $$a_k|0rangle = b_p|0rangle = c_p|0rangle =0,$$



    Then



    $$H|0rangle = sum_{k+p+p'=0}a^dagger_kc^dagger_p b^dagger_{p'}|0rangle neq 0$$



    That means that the true vacuum



    $H|Omegarangle =0$



    has



    $$a_kb_kc_k|Omegarangle neq 0.$$



    The number of particles is not conserved. Neither is energy with respect to the free Hamiltonian. Even for a single particle.



    ====================================



    Normally, a process that violates energy conservation (of the free Hamiltonian!) by an amount $Delta E$ can only survive for a time $hbar/Delta E$. Energy is only strictly conserved (in scattering for instance) for processes where the initial and final states are at $t=pm infty$. For example, you could create the state with two positrons and two electrons through the process (time going up):



    enter image description here



    but it would only last a time $hbar/4m_e sim 10^{-22}$ seconds. The reason in that in the interaction picture for scattering from $t=-T$ to $t=+T$, the scattering probability will look something like



    $$P sim left|int_{-T}^T!dt,e^{i(E_f-E_i)t},langle f | V|irangleright|^2sim frac{sin^2(E_f-E_i)T}{(E_f-E_i)^2}left|langle f | V|irangleright|^2$$



    You can check that this is non-zero for times $Tlesssim 1/(E_f-E_i)$.



    However, when gravity is included, time can be stretched, making short-lived states very long-lived. This happens when you have a horizon, like when there is a black hole or in an accelerating frame, in which case time can be infinitely stretched, producing matter. I think this also happens in inflation but there the particles get pulled away from each other. Not sure how to show this in terms of quantum mechanics.






    share|cite|improve this answer























    • In your last sentence you mention inflation. Could you refer to my comment on my own question. Do you think inflation could create long lived matter?
      – Justin
      Nov 28 at 14:12






    • 1




      @Justin I do not know much about it, but yes, apparently inflation does create long-lived matter. Most texts on inflation describe how when you define the vacuum at two different times in an accelerating universe, the matter content is different between the two. I don't know how to show it in terms of what I explained in my answer, but you could imagine the short-lived state being created and then the particles being pulled away very far from each other by inflation, creating a long-lived state.
      – Eric David Kramer
      Nov 28 at 14:27










    • The idea that quantum mechanics allows you to energy conservation for short times is an annoyingly persistent myth -- one of the first things you do in every intro QM class is prove energy is conserved! This idea is one of those things that's been being passed down from popsci book to popsci book unchanged for 100 years, despite having no relation to the truth.
      – knzhou
      2 days ago










    • @knzhou Yes you are absolutely right, energy is conserved. An energy eigenstate will stay in that state forever. But a state with no matter cannot be an energy eigenstate when interactions are included. An energy eigenstate of the free Hamiltonian (i.e. a state with no matter) will, over time, change its energy eigenvalue with respect to this free Hamiltonian (i.e. particles will be created). No myth here.
      – Eric David Kramer
      2 days ago






    • 1




      @EricDavidKramer Similarly, the only sensible definition of "empty universe" is the vacuum state of the full theory, not the free theory. While it is certainly true that the vacuum state of the free theory is not an eigenstate of the full Hamiltonian, this state has no physical meaning anyway, and is not what anybody is ever referring to when they speak of "the vacuum".
      – knzhou
      2 days ago















    up vote
    2
    down vote










    up vote
    2
    down vote









    Note: People seem to think that an empty universe is an eigenstate of the full QED Hamiltonian. This is not true. The Hamiltonian is



    $$H= sum_{e^+e^-} omega_p (b^dagger_p b_p +c^dagger_p c_p)+ sum_{rm photons}!k,a^dagger_k a_k,+, sum_{k+p+p'=0}(a^dagger_k + a_k)(b_p c_{p'} + c^dagger_p b^dagger_{p'})$$



    If we define the vacuum as



    $$a_k|0rangle = b_p|0rangle = c_p|0rangle =0,$$



    Then



    $$H|0rangle = sum_{k+p+p'=0}a^dagger_kc^dagger_p b^dagger_{p'}|0rangle neq 0$$



    That means that the true vacuum



    $H|Omegarangle =0$



    has



    $$a_kb_kc_k|Omegarangle neq 0.$$



    The number of particles is not conserved. Neither is energy with respect to the free Hamiltonian. Even for a single particle.



    ====================================



    Normally, a process that violates energy conservation (of the free Hamiltonian!) by an amount $Delta E$ can only survive for a time $hbar/Delta E$. Energy is only strictly conserved (in scattering for instance) for processes where the initial and final states are at $t=pm infty$. For example, you could create the state with two positrons and two electrons through the process (time going up):



    enter image description here



    but it would only last a time $hbar/4m_e sim 10^{-22}$ seconds. The reason in that in the interaction picture for scattering from $t=-T$ to $t=+T$, the scattering probability will look something like



    $$P sim left|int_{-T}^T!dt,e^{i(E_f-E_i)t},langle f | V|irangleright|^2sim frac{sin^2(E_f-E_i)T}{(E_f-E_i)^2}left|langle f | V|irangleright|^2$$



    You can check that this is non-zero for times $Tlesssim 1/(E_f-E_i)$.



    However, when gravity is included, time can be stretched, making short-lived states very long-lived. This happens when you have a horizon, like when there is a black hole or in an accelerating frame, in which case time can be infinitely stretched, producing matter. I think this also happens in inflation but there the particles get pulled away from each other. Not sure how to show this in terms of quantum mechanics.






    share|cite|improve this answer














    Note: People seem to think that an empty universe is an eigenstate of the full QED Hamiltonian. This is not true. The Hamiltonian is



    $$H= sum_{e^+e^-} omega_p (b^dagger_p b_p +c^dagger_p c_p)+ sum_{rm photons}!k,a^dagger_k a_k,+, sum_{k+p+p'=0}(a^dagger_k + a_k)(b_p c_{p'} + c^dagger_p b^dagger_{p'})$$



    If we define the vacuum as



    $$a_k|0rangle = b_p|0rangle = c_p|0rangle =0,$$



    Then



    $$H|0rangle = sum_{k+p+p'=0}a^dagger_kc^dagger_p b^dagger_{p'}|0rangle neq 0$$



    That means that the true vacuum



    $H|Omegarangle =0$



    has



    $$a_kb_kc_k|Omegarangle neq 0.$$



    The number of particles is not conserved. Neither is energy with respect to the free Hamiltonian. Even for a single particle.



    ====================================



    Normally, a process that violates energy conservation (of the free Hamiltonian!) by an amount $Delta E$ can only survive for a time $hbar/Delta E$. Energy is only strictly conserved (in scattering for instance) for processes where the initial and final states are at $t=pm infty$. For example, you could create the state with two positrons and two electrons through the process (time going up):



    enter image description here



    but it would only last a time $hbar/4m_e sim 10^{-22}$ seconds. The reason in that in the interaction picture for scattering from $t=-T$ to $t=+T$, the scattering probability will look something like



    $$P sim left|int_{-T}^T!dt,e^{i(E_f-E_i)t},langle f | V|irangleright|^2sim frac{sin^2(E_f-E_i)T}{(E_f-E_i)^2}left|langle f | V|irangleright|^2$$



    You can check that this is non-zero for times $Tlesssim 1/(E_f-E_i)$.



    However, when gravity is included, time can be stretched, making short-lived states very long-lived. This happens when you have a horizon, like when there is a black hole or in an accelerating frame, in which case time can be infinitely stretched, producing matter. I think this also happens in inflation but there the particles get pulled away from each other. Not sure how to show this in terms of quantum mechanics.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago

























    answered Nov 28 at 12:53









    Eric David Kramer

    93216




    93216












    • In your last sentence you mention inflation. Could you refer to my comment on my own question. Do you think inflation could create long lived matter?
      – Justin
      Nov 28 at 14:12






    • 1




      @Justin I do not know much about it, but yes, apparently inflation does create long-lived matter. Most texts on inflation describe how when you define the vacuum at two different times in an accelerating universe, the matter content is different between the two. I don't know how to show it in terms of what I explained in my answer, but you could imagine the short-lived state being created and then the particles being pulled away very far from each other by inflation, creating a long-lived state.
      – Eric David Kramer
      Nov 28 at 14:27










    • The idea that quantum mechanics allows you to energy conservation for short times is an annoyingly persistent myth -- one of the first things you do in every intro QM class is prove energy is conserved! This idea is one of those things that's been being passed down from popsci book to popsci book unchanged for 100 years, despite having no relation to the truth.
      – knzhou
      2 days ago










    • @knzhou Yes you are absolutely right, energy is conserved. An energy eigenstate will stay in that state forever. But a state with no matter cannot be an energy eigenstate when interactions are included. An energy eigenstate of the free Hamiltonian (i.e. a state with no matter) will, over time, change its energy eigenvalue with respect to this free Hamiltonian (i.e. particles will be created). No myth here.
      – Eric David Kramer
      2 days ago






    • 1




      @EricDavidKramer Similarly, the only sensible definition of "empty universe" is the vacuum state of the full theory, not the free theory. While it is certainly true that the vacuum state of the free theory is not an eigenstate of the full Hamiltonian, this state has no physical meaning anyway, and is not what anybody is ever referring to when they speak of "the vacuum".
      – knzhou
      2 days ago




















    • In your last sentence you mention inflation. Could you refer to my comment on my own question. Do you think inflation could create long lived matter?
      – Justin
      Nov 28 at 14:12






    • 1




      @Justin I do not know much about it, but yes, apparently inflation does create long-lived matter. Most texts on inflation describe how when you define the vacuum at two different times in an accelerating universe, the matter content is different between the two. I don't know how to show it in terms of what I explained in my answer, but you could imagine the short-lived state being created and then the particles being pulled away very far from each other by inflation, creating a long-lived state.
      – Eric David Kramer
      Nov 28 at 14:27










    • The idea that quantum mechanics allows you to energy conservation for short times is an annoyingly persistent myth -- one of the first things you do in every intro QM class is prove energy is conserved! This idea is one of those things that's been being passed down from popsci book to popsci book unchanged for 100 years, despite having no relation to the truth.
      – knzhou
      2 days ago










    • @knzhou Yes you are absolutely right, energy is conserved. An energy eigenstate will stay in that state forever. But a state with no matter cannot be an energy eigenstate when interactions are included. An energy eigenstate of the free Hamiltonian (i.e. a state with no matter) will, over time, change its energy eigenvalue with respect to this free Hamiltonian (i.e. particles will be created). No myth here.
      – Eric David Kramer
      2 days ago






    • 1




      @EricDavidKramer Similarly, the only sensible definition of "empty universe" is the vacuum state of the full theory, not the free theory. While it is certainly true that the vacuum state of the free theory is not an eigenstate of the full Hamiltonian, this state has no physical meaning anyway, and is not what anybody is ever referring to when they speak of "the vacuum".
      – knzhou
      2 days ago


















    In your last sentence you mention inflation. Could you refer to my comment on my own question. Do you think inflation could create long lived matter?
    – Justin
    Nov 28 at 14:12




    In your last sentence you mention inflation. Could you refer to my comment on my own question. Do you think inflation could create long lived matter?
    – Justin
    Nov 28 at 14:12




    1




    1




    @Justin I do not know much about it, but yes, apparently inflation does create long-lived matter. Most texts on inflation describe how when you define the vacuum at two different times in an accelerating universe, the matter content is different between the two. I don't know how to show it in terms of what I explained in my answer, but you could imagine the short-lived state being created and then the particles being pulled away very far from each other by inflation, creating a long-lived state.
    – Eric David Kramer
    Nov 28 at 14:27




    @Justin I do not know much about it, but yes, apparently inflation does create long-lived matter. Most texts on inflation describe how when you define the vacuum at two different times in an accelerating universe, the matter content is different between the two. I don't know how to show it in terms of what I explained in my answer, but you could imagine the short-lived state being created and then the particles being pulled away very far from each other by inflation, creating a long-lived state.
    – Eric David Kramer
    Nov 28 at 14:27












    The idea that quantum mechanics allows you to energy conservation for short times is an annoyingly persistent myth -- one of the first things you do in every intro QM class is prove energy is conserved! This idea is one of those things that's been being passed down from popsci book to popsci book unchanged for 100 years, despite having no relation to the truth.
    – knzhou
    2 days ago




    The idea that quantum mechanics allows you to energy conservation for short times is an annoyingly persistent myth -- one of the first things you do in every intro QM class is prove energy is conserved! This idea is one of those things that's been being passed down from popsci book to popsci book unchanged for 100 years, despite having no relation to the truth.
    – knzhou
    2 days ago












    @knzhou Yes you are absolutely right, energy is conserved. An energy eigenstate will stay in that state forever. But a state with no matter cannot be an energy eigenstate when interactions are included. An energy eigenstate of the free Hamiltonian (i.e. a state with no matter) will, over time, change its energy eigenvalue with respect to this free Hamiltonian (i.e. particles will be created). No myth here.
    – Eric David Kramer
    2 days ago




    @knzhou Yes you are absolutely right, energy is conserved. An energy eigenstate will stay in that state forever. But a state with no matter cannot be an energy eigenstate when interactions are included. An energy eigenstate of the free Hamiltonian (i.e. a state with no matter) will, over time, change its energy eigenvalue with respect to this free Hamiltonian (i.e. particles will be created). No myth here.
    – Eric David Kramer
    2 days ago




    1




    1




    @EricDavidKramer Similarly, the only sensible definition of "empty universe" is the vacuum state of the full theory, not the free theory. While it is certainly true that the vacuum state of the free theory is not an eigenstate of the full Hamiltonian, this state has no physical meaning anyway, and is not what anybody is ever referring to when they speak of "the vacuum".
    – knzhou
    2 days ago






    @EricDavidKramer Similarly, the only sensible definition of "empty universe" is the vacuum state of the full theory, not the free theory. While it is certainly true that the vacuum state of the free theory is not an eigenstate of the full Hamiltonian, this state has no physical meaning anyway, and is not what anybody is ever referring to when they speak of "the vacuum".
    – knzhou
    2 days ago












    up vote
    0
    down vote













    Also in QED, total Energy is conserved for every time! A difference between QED and classical electrodynamics is that the Expression for total Energy is slightly altered. Classical electrodynamics has kinetic Energy $T$ and a potential Energy $V$ arising from electromagnetic fields. In classical Theory, total Energy is conserved, i.e.



    $T+V = const.$



    But in Quantum electrodynamics and other Quantum field theories, you have also additional zero-Point Energy $hbar omega$, which is responsible for e.g. the Casimir effect (when non-charged plates attrackt each other when These have extremely small distance). The frequency $omega$ can be interpreted as how fast significant changes in the System takes place. For a many-particle System with extremely high collision frequency, the value $omega$ will be also high, altering the Energy balance to



    $T+V + hbar omega = const.$



    The last term in this equation on the left Hand side is also called "self-energy" in Quantum kinetic Theory. This self-energy is a complex-valued quantity, where the real part describes the



    zero-Point Energy



    , while the imaginary part is antiproportional to



    the lifetime of the excited state



    Hence, higher zero-Point Energy imply Shorter Lifetimes of excited states.



    Another example of the Change of effective Energy/Hamiltonian due to Quantum effects is shown e.g. in this paper:



    https://arxiv.org/abs/0706.1090






    share|cite|improve this answer



























      up vote
      0
      down vote













      Also in QED, total Energy is conserved for every time! A difference between QED and classical electrodynamics is that the Expression for total Energy is slightly altered. Classical electrodynamics has kinetic Energy $T$ and a potential Energy $V$ arising from electromagnetic fields. In classical Theory, total Energy is conserved, i.e.



      $T+V = const.$



      But in Quantum electrodynamics and other Quantum field theories, you have also additional zero-Point Energy $hbar omega$, which is responsible for e.g. the Casimir effect (when non-charged plates attrackt each other when These have extremely small distance). The frequency $omega$ can be interpreted as how fast significant changes in the System takes place. For a many-particle System with extremely high collision frequency, the value $omega$ will be also high, altering the Energy balance to



      $T+V + hbar omega = const.$



      The last term in this equation on the left Hand side is also called "self-energy" in Quantum kinetic Theory. This self-energy is a complex-valued quantity, where the real part describes the



      zero-Point Energy



      , while the imaginary part is antiproportional to



      the lifetime of the excited state



      Hence, higher zero-Point Energy imply Shorter Lifetimes of excited states.



      Another example of the Change of effective Energy/Hamiltonian due to Quantum effects is shown e.g. in this paper:



      https://arxiv.org/abs/0706.1090






      share|cite|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        Also in QED, total Energy is conserved for every time! A difference between QED and classical electrodynamics is that the Expression for total Energy is slightly altered. Classical electrodynamics has kinetic Energy $T$ and a potential Energy $V$ arising from electromagnetic fields. In classical Theory, total Energy is conserved, i.e.



        $T+V = const.$



        But in Quantum electrodynamics and other Quantum field theories, you have also additional zero-Point Energy $hbar omega$, which is responsible for e.g. the Casimir effect (when non-charged plates attrackt each other when These have extremely small distance). The frequency $omega$ can be interpreted as how fast significant changes in the System takes place. For a many-particle System with extremely high collision frequency, the value $omega$ will be also high, altering the Energy balance to



        $T+V + hbar omega = const.$



        The last term in this equation on the left Hand side is also called "self-energy" in Quantum kinetic Theory. This self-energy is a complex-valued quantity, where the real part describes the



        zero-Point Energy



        , while the imaginary part is antiproportional to



        the lifetime of the excited state



        Hence, higher zero-Point Energy imply Shorter Lifetimes of excited states.



        Another example of the Change of effective Energy/Hamiltonian due to Quantum effects is shown e.g. in this paper:



        https://arxiv.org/abs/0706.1090






        share|cite|improve this answer














        Also in QED, total Energy is conserved for every time! A difference between QED and classical electrodynamics is that the Expression for total Energy is slightly altered. Classical electrodynamics has kinetic Energy $T$ and a potential Energy $V$ arising from electromagnetic fields. In classical Theory, total Energy is conserved, i.e.



        $T+V = const.$



        But in Quantum electrodynamics and other Quantum field theories, you have also additional zero-Point Energy $hbar omega$, which is responsible for e.g. the Casimir effect (when non-charged plates attrackt each other when These have extremely small distance). The frequency $omega$ can be interpreted as how fast significant changes in the System takes place. For a many-particle System with extremely high collision frequency, the value $omega$ will be also high, altering the Energy balance to



        $T+V + hbar omega = const.$



        The last term in this equation on the left Hand side is also called "self-energy" in Quantum kinetic Theory. This self-energy is a complex-valued quantity, where the real part describes the



        zero-Point Energy



        , while the imaginary part is antiproportional to



        the lifetime of the excited state



        Hence, higher zero-Point Energy imply Shorter Lifetimes of excited states.



        Another example of the Change of effective Energy/Hamiltonian due to Quantum effects is shown e.g. in this paper:



        https://arxiv.org/abs/0706.1090







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 28 at 13:17









        Chair

        3,59572034




        3,59572034










        answered Nov 28 at 12:47









        kryomaxim

        1,597620




        1,597620






























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