Heisenberg energy uncertainty

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I am biologist and just new to the field of quantum mechanics and trying to understand the subject by reading. It is said that virtual particles are created due to the uncertainty of energy when consider a time, energy conjugate pair. Which leads to violate the energy conservation principle temporarily. But how a huge uncertainty in energy creates virtual particles? How it could violate the energy conservation principle?










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  • 2




    Virtual particles are just tools to carry out calculations in perturbation theory. They're not states, and are not 'particles' in the usual sense. 'Explanations' based on the energy-time uncertainty relation are heuristic and not rigorous, as far as I can tell. Also, energy-time do not form a conjugate pair like $x$ and $p$ do. The $E-t$ uncertainty relation is of a different character. See also: physics.stackexchange.com/q/103724/133418, physics.stackexchange.com/q/109229/133418
    – Avantgarde
    5 hours ago











  • other possible duplicates: Virtual particles with almost infinite energy?, Why are there e.m. vacuüm fluctuations? and links therein.
    – AccidentalFourierTransform
    2 hours ago














up vote
4
down vote

favorite












I am biologist and just new to the field of quantum mechanics and trying to understand the subject by reading. It is said that virtual particles are created due to the uncertainty of energy when consider a time, energy conjugate pair. Which leads to violate the energy conservation principle temporarily. But how a huge uncertainty in energy creates virtual particles? How it could violate the energy conservation principle?










share|cite|improve this question



















  • 2




    Virtual particles are just tools to carry out calculations in perturbation theory. They're not states, and are not 'particles' in the usual sense. 'Explanations' based on the energy-time uncertainty relation are heuristic and not rigorous, as far as I can tell. Also, energy-time do not form a conjugate pair like $x$ and $p$ do. The $E-t$ uncertainty relation is of a different character. See also: physics.stackexchange.com/q/103724/133418, physics.stackexchange.com/q/109229/133418
    – Avantgarde
    5 hours ago











  • other possible duplicates: Virtual particles with almost infinite energy?, Why are there e.m. vacuüm fluctuations? and links therein.
    – AccidentalFourierTransform
    2 hours ago












up vote
4
down vote

favorite









up vote
4
down vote

favorite











I am biologist and just new to the field of quantum mechanics and trying to understand the subject by reading. It is said that virtual particles are created due to the uncertainty of energy when consider a time, energy conjugate pair. Which leads to violate the energy conservation principle temporarily. But how a huge uncertainty in energy creates virtual particles? How it could violate the energy conservation principle?










share|cite|improve this question















I am biologist and just new to the field of quantum mechanics and trying to understand the subject by reading. It is said that virtual particles are created due to the uncertainty of energy when consider a time, energy conjugate pair. Which leads to violate the energy conservation principle temporarily. But how a huge uncertainty in energy creates virtual particles? How it could violate the energy conservation principle?







quantum-field-theory heisenberg-uncertainty-principle virtual-particles






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edited 6 hours ago









Qmechanic

99.1k121781097




99.1k121781097










asked 6 hours ago









Kosala

326111




326111







  • 2




    Virtual particles are just tools to carry out calculations in perturbation theory. They're not states, and are not 'particles' in the usual sense. 'Explanations' based on the energy-time uncertainty relation are heuristic and not rigorous, as far as I can tell. Also, energy-time do not form a conjugate pair like $x$ and $p$ do. The $E-t$ uncertainty relation is of a different character. See also: physics.stackexchange.com/q/103724/133418, physics.stackexchange.com/q/109229/133418
    – Avantgarde
    5 hours ago











  • other possible duplicates: Virtual particles with almost infinite energy?, Why are there e.m. vacuüm fluctuations? and links therein.
    – AccidentalFourierTransform
    2 hours ago












  • 2




    Virtual particles are just tools to carry out calculations in perturbation theory. They're not states, and are not 'particles' in the usual sense. 'Explanations' based on the energy-time uncertainty relation are heuristic and not rigorous, as far as I can tell. Also, energy-time do not form a conjugate pair like $x$ and $p$ do. The $E-t$ uncertainty relation is of a different character. See also: physics.stackexchange.com/q/103724/133418, physics.stackexchange.com/q/109229/133418
    – Avantgarde
    5 hours ago











  • other possible duplicates: Virtual particles with almost infinite energy?, Why are there e.m. vacuüm fluctuations? and links therein.
    – AccidentalFourierTransform
    2 hours ago







2




2




Virtual particles are just tools to carry out calculations in perturbation theory. They're not states, and are not 'particles' in the usual sense. 'Explanations' based on the energy-time uncertainty relation are heuristic and not rigorous, as far as I can tell. Also, energy-time do not form a conjugate pair like $x$ and $p$ do. The $E-t$ uncertainty relation is of a different character. See also: physics.stackexchange.com/q/103724/133418, physics.stackexchange.com/q/109229/133418
– Avantgarde
5 hours ago





Virtual particles are just tools to carry out calculations in perturbation theory. They're not states, and are not 'particles' in the usual sense. 'Explanations' based on the energy-time uncertainty relation are heuristic and not rigorous, as far as I can tell. Also, energy-time do not form a conjugate pair like $x$ and $p$ do. The $E-t$ uncertainty relation is of a different character. See also: physics.stackexchange.com/q/103724/133418, physics.stackexchange.com/q/109229/133418
– Avantgarde
5 hours ago













other possible duplicates: Virtual particles with almost infinite energy?, Why are there e.m. vacuüm fluctuations? and links therein.
– AccidentalFourierTransform
2 hours ago




other possible duplicates: Virtual particles with almost infinite energy?, Why are there e.m. vacuüm fluctuations? and links therein.
– AccidentalFourierTransform
2 hours ago










3 Answers
3






active

oldest

votes

















up vote
8
down vote













This is an excellent question, which clearly demonstrated the lie perpetuated by a lot of people, that the Energy time uncertainty relationship has the same meaning as the momentum and position uncertainty relationship.



First of all, there is no energy and time uncertainty. Recall that the uncertainty principle for operators $hat A$ and $hat B$ have the form



$$ sigma _ A sigma _ B geq left| frac 1 2 i langle [ hat A , hat B ] rangle right| = frac 1 2 | langle [ hat A , hat B ] rangle | $$



Now there is no "time" operator in quantum mechanics thus, it doesn't make any sense to talk about "time" uncertainty. The best you can do is the following for a time independent operator $hat Q$:



$$Delta hat H Delta t geq frac hbar 2 , quad Delta t equiv frac Delta hat Q $$



and this is what people mean, when they talk about Energy-time uncertainty. As you can see $Delta t$ really has nothing to do with the "uncertainty in time", but rather is telling you roughly speaking, how big the uncertainty of an opeartor is compared to the variation of its expectation value.



Second thing you should know is that virtual particles also do not exist. They are essentially calculational tools that one uses for perturbation theory. The idea that two electrons see each other, "throw" a "virtual photon" and therefore repulse each other is atrocious. This view comes from overinterpreting Feynman diagrams, which are essentially pictures that simplify combinatoric calculations in perturbation theory. There is a beautiful epsiode by BPS Spacetime about virtual particles, if you want to learn more about virtual particles.



However, the upshot is that the statement that "virtual particles break conservation of energy for a very short time because of energy-time uncertainty principle" is devoid of meaning, as there is no virtual particles nor energy-time uncertainty principle.






share|cite|improve this answer
















  • 1




    "This is an excellent question, which clearly demonstrated the lie perpetuated by a lot of people, that the Energy time uncertainty relationship has the same meaning as the momentum and position uncertainty relationship." Thank you for this. One of my physics pet peeves.
    – Aaron Stevens
    3 hours ago










  • This is a great answer. @Gonenc, thanks for taking time to answer my question. Honestly, when I first read about time energy uncertainty pair, that didn't make any sense to me. I understood the concept of momentum position uncertainty but relating the time energy uncertainty didn't make any sense to me. My first impression was that well, the uncertainty of time should be how much we are far from the expectation value. I believe what I understood was correct even though further reading confused me.
    – Kosala
    3 hours ago










  • I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
    – Kosala
    3 hours ago






  • 1




    @AaronStevens Yeah if I have to rank the things in physics I find annoying, "E-t uncertainty relationship = x-p uncertainty relationship" would clearly come near the top. Sadly, I don't blame anyone (especially laypeople, physicists should know it better though), who believe in this statement, since even in a physics class for particle physics this statement was made.
    – Gonenc Mogol
    2 hours ago










  • there is of course the interpretation of $Delta t$ as being the uncertainty of the lifetime of the state that is being used to take the average $langle Q rangle$ and standard deviations.
    – hyportnex
    39 mins ago


















up vote
4
down vote













Virtual particles , as Gonenc Mogol says, are not particles, they are place holders for quantum numbers in the integrals necessary to calculate probabilities of interaction for quantum mechanical entities.



Here is a practical example :



eemumu



The virtual particle is labeled as a photon, but in effect it is a place holder for the photon quantum numbers within an integral over the variables shown in the graph, it means that the photon is off mass shell, and the mass varies within the integration which will give real numbers ( note the imaginary numbers) to be compared with experiment. The link gives details . What can be measured are the input particle four vectors and the output particles four vectors, the internal lines, as the $γ$ here, are not accessible to measurement.



The popularization of this mathematical concept creates the confusion about uncertainties and energy conservation. In order to have virtual "particles" there must always be real energy input by incoming real particles, and real energy output by outgoing real particles. In between it is mathematics and hand waving misinterpretations.






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  • Thanks for the comment. Did you mean internal lines or inclined lines?
    – Kosala
    5 hours ago










  • internal, as the γ in the diagram. it connects to external legs through a vertex
    – anna v
    5 hours ago










  • Thanks for taking time to answer my question. But I am bit confused now. According to this theory, the photon is the virtual particle. And how the positron comes into play? why does it appear? So why we call then virtual particles are countervailed?
    – Kosala
    3 hours ago










  • @Kosala How the Feynamn diagrams arise from QFT formalism is really esoteric for most of the people and uses relatively heavy machinery. So just think of "virtual particles" as a mathematical trick and not as a part of reality.
    – Gonenc Mogol
    2 hours ago

















up vote
-1
down vote













It really depends on what you mean by conservation of energy. The classical idea is that the total energy is constant at all times. However, once quantum mechanics was discovered, people realized that our classical idea of conversation of energy is really just an approximation of the quantum conservation principle. In quantum mechanics, conservation of energy is a statement about the expectation value (think the average value) of energy; it says that the expectation value of energy does not change. Quantum fluctations in the energy cancel out over time, so the average energy of an isolated system never changes.



Effectively, physicists reformulated their idea about conservation of energy to agree with quantum mechanics, because they found that the classical notion breaks down.



As for how virtual particles are created, I'm not exactly sure. Somebody who is more familiar with QFT can probably answer.






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  • Thanks for the answer.
    – Kosala
    3 hours ago










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3 Answers
3






active

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3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
8
down vote













This is an excellent question, which clearly demonstrated the lie perpetuated by a lot of people, that the Energy time uncertainty relationship has the same meaning as the momentum and position uncertainty relationship.



First of all, there is no energy and time uncertainty. Recall that the uncertainty principle for operators $hat A$ and $hat B$ have the form



$$ sigma _ A sigma _ B geq left| frac 1 2 i langle [ hat A , hat B ] rangle right| = frac 1 2 | langle [ hat A , hat B ] rangle | $$



Now there is no "time" operator in quantum mechanics thus, it doesn't make any sense to talk about "time" uncertainty. The best you can do is the following for a time independent operator $hat Q$:



$$Delta hat H Delta t geq frac hbar 2 , quad Delta t equiv frac Delta hat Q $$



and this is what people mean, when they talk about Energy-time uncertainty. As you can see $Delta t$ really has nothing to do with the "uncertainty in time", but rather is telling you roughly speaking, how big the uncertainty of an opeartor is compared to the variation of its expectation value.



Second thing you should know is that virtual particles also do not exist. They are essentially calculational tools that one uses for perturbation theory. The idea that two electrons see each other, "throw" a "virtual photon" and therefore repulse each other is atrocious. This view comes from overinterpreting Feynman diagrams, which are essentially pictures that simplify combinatoric calculations in perturbation theory. There is a beautiful epsiode by BPS Spacetime about virtual particles, if you want to learn more about virtual particles.



However, the upshot is that the statement that "virtual particles break conservation of energy for a very short time because of energy-time uncertainty principle" is devoid of meaning, as there is no virtual particles nor energy-time uncertainty principle.






share|cite|improve this answer
















  • 1




    "This is an excellent question, which clearly demonstrated the lie perpetuated by a lot of people, that the Energy time uncertainty relationship has the same meaning as the momentum and position uncertainty relationship." Thank you for this. One of my physics pet peeves.
    – Aaron Stevens
    3 hours ago










  • This is a great answer. @Gonenc, thanks for taking time to answer my question. Honestly, when I first read about time energy uncertainty pair, that didn't make any sense to me. I understood the concept of momentum position uncertainty but relating the time energy uncertainty didn't make any sense to me. My first impression was that well, the uncertainty of time should be how much we are far from the expectation value. I believe what I understood was correct even though further reading confused me.
    – Kosala
    3 hours ago










  • I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
    – Kosala
    3 hours ago






  • 1




    @AaronStevens Yeah if I have to rank the things in physics I find annoying, "E-t uncertainty relationship = x-p uncertainty relationship" would clearly come near the top. Sadly, I don't blame anyone (especially laypeople, physicists should know it better though), who believe in this statement, since even in a physics class for particle physics this statement was made.
    – Gonenc Mogol
    2 hours ago










  • there is of course the interpretation of $Delta t$ as being the uncertainty of the lifetime of the state that is being used to take the average $langle Q rangle$ and standard deviations.
    – hyportnex
    39 mins ago















up vote
8
down vote













This is an excellent question, which clearly demonstrated the lie perpetuated by a lot of people, that the Energy time uncertainty relationship has the same meaning as the momentum and position uncertainty relationship.



First of all, there is no energy and time uncertainty. Recall that the uncertainty principle for operators $hat A$ and $hat B$ have the form



$$ sigma _ A sigma _ B geq left| frac 1 2 i langle [ hat A , hat B ] rangle right| = frac 1 2 | langle [ hat A , hat B ] rangle | $$



Now there is no "time" operator in quantum mechanics thus, it doesn't make any sense to talk about "time" uncertainty. The best you can do is the following for a time independent operator $hat Q$:



$$Delta hat H Delta t geq frac hbar 2 , quad Delta t equiv frac Delta hat Q $$



and this is what people mean, when they talk about Energy-time uncertainty. As you can see $Delta t$ really has nothing to do with the "uncertainty in time", but rather is telling you roughly speaking, how big the uncertainty of an opeartor is compared to the variation of its expectation value.



Second thing you should know is that virtual particles also do not exist. They are essentially calculational tools that one uses for perturbation theory. The idea that two electrons see each other, "throw" a "virtual photon" and therefore repulse each other is atrocious. This view comes from overinterpreting Feynman diagrams, which are essentially pictures that simplify combinatoric calculations in perturbation theory. There is a beautiful epsiode by BPS Spacetime about virtual particles, if you want to learn more about virtual particles.



However, the upshot is that the statement that "virtual particles break conservation of energy for a very short time because of energy-time uncertainty principle" is devoid of meaning, as there is no virtual particles nor energy-time uncertainty principle.






share|cite|improve this answer
















  • 1




    "This is an excellent question, which clearly demonstrated the lie perpetuated by a lot of people, that the Energy time uncertainty relationship has the same meaning as the momentum and position uncertainty relationship." Thank you for this. One of my physics pet peeves.
    – Aaron Stevens
    3 hours ago










  • This is a great answer. @Gonenc, thanks for taking time to answer my question. Honestly, when I first read about time energy uncertainty pair, that didn't make any sense to me. I understood the concept of momentum position uncertainty but relating the time energy uncertainty didn't make any sense to me. My first impression was that well, the uncertainty of time should be how much we are far from the expectation value. I believe what I understood was correct even though further reading confused me.
    – Kosala
    3 hours ago










  • I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
    – Kosala
    3 hours ago






  • 1




    @AaronStevens Yeah if I have to rank the things in physics I find annoying, "E-t uncertainty relationship = x-p uncertainty relationship" would clearly come near the top. Sadly, I don't blame anyone (especially laypeople, physicists should know it better though), who believe in this statement, since even in a physics class for particle physics this statement was made.
    – Gonenc Mogol
    2 hours ago










  • there is of course the interpretation of $Delta t$ as being the uncertainty of the lifetime of the state that is being used to take the average $langle Q rangle$ and standard deviations.
    – hyportnex
    39 mins ago













up vote
8
down vote










up vote
8
down vote









This is an excellent question, which clearly demonstrated the lie perpetuated by a lot of people, that the Energy time uncertainty relationship has the same meaning as the momentum and position uncertainty relationship.



First of all, there is no energy and time uncertainty. Recall that the uncertainty principle for operators $hat A$ and $hat B$ have the form



$$ sigma _ A sigma _ B geq left| frac 1 2 i langle [ hat A , hat B ] rangle right| = frac 1 2 | langle [ hat A , hat B ] rangle | $$



Now there is no "time" operator in quantum mechanics thus, it doesn't make any sense to talk about "time" uncertainty. The best you can do is the following for a time independent operator $hat Q$:



$$Delta hat H Delta t geq frac hbar 2 , quad Delta t equiv frac Delta hat Q $$



and this is what people mean, when they talk about Energy-time uncertainty. As you can see $Delta t$ really has nothing to do with the "uncertainty in time", but rather is telling you roughly speaking, how big the uncertainty of an opeartor is compared to the variation of its expectation value.



Second thing you should know is that virtual particles also do not exist. They are essentially calculational tools that one uses for perturbation theory. The idea that two electrons see each other, "throw" a "virtual photon" and therefore repulse each other is atrocious. This view comes from overinterpreting Feynman diagrams, which are essentially pictures that simplify combinatoric calculations in perturbation theory. There is a beautiful epsiode by BPS Spacetime about virtual particles, if you want to learn more about virtual particles.



However, the upshot is that the statement that "virtual particles break conservation of energy for a very short time because of energy-time uncertainty principle" is devoid of meaning, as there is no virtual particles nor energy-time uncertainty principle.






share|cite|improve this answer












This is an excellent question, which clearly demonstrated the lie perpetuated by a lot of people, that the Energy time uncertainty relationship has the same meaning as the momentum and position uncertainty relationship.



First of all, there is no energy and time uncertainty. Recall that the uncertainty principle for operators $hat A$ and $hat B$ have the form



$$ sigma _ A sigma _ B geq left| frac 1 2 i langle [ hat A , hat B ] rangle right| = frac 1 2 | langle [ hat A , hat B ] rangle | $$



Now there is no "time" operator in quantum mechanics thus, it doesn't make any sense to talk about "time" uncertainty. The best you can do is the following for a time independent operator $hat Q$:



$$Delta hat H Delta t geq frac hbar 2 , quad Delta t equiv frac Delta hat Q $$



and this is what people mean, when they talk about Energy-time uncertainty. As you can see $Delta t$ really has nothing to do with the "uncertainty in time", but rather is telling you roughly speaking, how big the uncertainty of an opeartor is compared to the variation of its expectation value.



Second thing you should know is that virtual particles also do not exist. They are essentially calculational tools that one uses for perturbation theory. The idea that two electrons see each other, "throw" a "virtual photon" and therefore repulse each other is atrocious. This view comes from overinterpreting Feynman diagrams, which are essentially pictures that simplify combinatoric calculations in perturbation theory. There is a beautiful epsiode by BPS Spacetime about virtual particles, if you want to learn more about virtual particles.



However, the upshot is that the statement that "virtual particles break conservation of energy for a very short time because of energy-time uncertainty principle" is devoid of meaning, as there is no virtual particles nor energy-time uncertainty principle.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 5 hours ago









Gonenc Mogol

2,74611331




2,74611331







  • 1




    "This is an excellent question, which clearly demonstrated the lie perpetuated by a lot of people, that the Energy time uncertainty relationship has the same meaning as the momentum and position uncertainty relationship." Thank you for this. One of my physics pet peeves.
    – Aaron Stevens
    3 hours ago










  • This is a great answer. @Gonenc, thanks for taking time to answer my question. Honestly, when I first read about time energy uncertainty pair, that didn't make any sense to me. I understood the concept of momentum position uncertainty but relating the time energy uncertainty didn't make any sense to me. My first impression was that well, the uncertainty of time should be how much we are far from the expectation value. I believe what I understood was correct even though further reading confused me.
    – Kosala
    3 hours ago










  • I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
    – Kosala
    3 hours ago






  • 1




    @AaronStevens Yeah if I have to rank the things in physics I find annoying, "E-t uncertainty relationship = x-p uncertainty relationship" would clearly come near the top. Sadly, I don't blame anyone (especially laypeople, physicists should know it better though), who believe in this statement, since even in a physics class for particle physics this statement was made.
    – Gonenc Mogol
    2 hours ago










  • there is of course the interpretation of $Delta t$ as being the uncertainty of the lifetime of the state that is being used to take the average $langle Q rangle$ and standard deviations.
    – hyportnex
    39 mins ago













  • 1




    "This is an excellent question, which clearly demonstrated the lie perpetuated by a lot of people, that the Energy time uncertainty relationship has the same meaning as the momentum and position uncertainty relationship." Thank you for this. One of my physics pet peeves.
    – Aaron Stevens
    3 hours ago










  • This is a great answer. @Gonenc, thanks for taking time to answer my question. Honestly, when I first read about time energy uncertainty pair, that didn't make any sense to me. I understood the concept of momentum position uncertainty but relating the time energy uncertainty didn't make any sense to me. My first impression was that well, the uncertainty of time should be how much we are far from the expectation value. I believe what I understood was correct even though further reading confused me.
    – Kosala
    3 hours ago










  • I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
    – Kosala
    3 hours ago






  • 1




    @AaronStevens Yeah if I have to rank the things in physics I find annoying, "E-t uncertainty relationship = x-p uncertainty relationship" would clearly come near the top. Sadly, I don't blame anyone (especially laypeople, physicists should know it better though), who believe in this statement, since even in a physics class for particle physics this statement was made.
    – Gonenc Mogol
    2 hours ago










  • there is of course the interpretation of $Delta t$ as being the uncertainty of the lifetime of the state that is being used to take the average $langle Q rangle$ and standard deviations.
    – hyportnex
    39 mins ago








1




1




"This is an excellent question, which clearly demonstrated the lie perpetuated by a lot of people, that the Energy time uncertainty relationship has the same meaning as the momentum and position uncertainty relationship." Thank you for this. One of my physics pet peeves.
– Aaron Stevens
3 hours ago




"This is an excellent question, which clearly demonstrated the lie perpetuated by a lot of people, that the Energy time uncertainty relationship has the same meaning as the momentum and position uncertainty relationship." Thank you for this. One of my physics pet peeves.
– Aaron Stevens
3 hours ago












This is a great answer. @Gonenc, thanks for taking time to answer my question. Honestly, when I first read about time energy uncertainty pair, that didn't make any sense to me. I understood the concept of momentum position uncertainty but relating the time energy uncertainty didn't make any sense to me. My first impression was that well, the uncertainty of time should be how much we are far from the expectation value. I believe what I understood was correct even though further reading confused me.
– Kosala
3 hours ago




This is a great answer. @Gonenc, thanks for taking time to answer my question. Honestly, when I first read about time energy uncertainty pair, that didn't make any sense to me. I understood the concept of momentum position uncertainty but relating the time energy uncertainty didn't make any sense to me. My first impression was that well, the uncertainty of time should be how much we are far from the expectation value. I believe what I understood was correct even though further reading confused me.
– Kosala
3 hours ago












I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
3 hours ago




I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
3 hours ago




1




1




@AaronStevens Yeah if I have to rank the things in physics I find annoying, "E-t uncertainty relationship = x-p uncertainty relationship" would clearly come near the top. Sadly, I don't blame anyone (especially laypeople, physicists should know it better though), who believe in this statement, since even in a physics class for particle physics this statement was made.
– Gonenc Mogol
2 hours ago




@AaronStevens Yeah if I have to rank the things in physics I find annoying, "E-t uncertainty relationship = x-p uncertainty relationship" would clearly come near the top. Sadly, I don't blame anyone (especially laypeople, physicists should know it better though), who believe in this statement, since even in a physics class for particle physics this statement was made.
– Gonenc Mogol
2 hours ago












there is of course the interpretation of $Delta t$ as being the uncertainty of the lifetime of the state that is being used to take the average $langle Q rangle$ and standard deviations.
– hyportnex
39 mins ago





there is of course the interpretation of $Delta t$ as being the uncertainty of the lifetime of the state that is being used to take the average $langle Q rangle$ and standard deviations.
– hyportnex
39 mins ago











up vote
4
down vote













Virtual particles , as Gonenc Mogol says, are not particles, they are place holders for quantum numbers in the integrals necessary to calculate probabilities of interaction for quantum mechanical entities.



Here is a practical example :



eemumu



The virtual particle is labeled as a photon, but in effect it is a place holder for the photon quantum numbers within an integral over the variables shown in the graph, it means that the photon is off mass shell, and the mass varies within the integration which will give real numbers ( note the imaginary numbers) to be compared with experiment. The link gives details . What can be measured are the input particle four vectors and the output particles four vectors, the internal lines, as the $γ$ here, are not accessible to measurement.



The popularization of this mathematical concept creates the confusion about uncertainties and energy conservation. In order to have virtual "particles" there must always be real energy input by incoming real particles, and real energy output by outgoing real particles. In between it is mathematics and hand waving misinterpretations.






share|cite|improve this answer






















  • Thanks for the comment. Did you mean internal lines or inclined lines?
    – Kosala
    5 hours ago










  • internal, as the γ in the diagram. it connects to external legs through a vertex
    – anna v
    5 hours ago










  • Thanks for taking time to answer my question. But I am bit confused now. According to this theory, the photon is the virtual particle. And how the positron comes into play? why does it appear? So why we call then virtual particles are countervailed?
    – Kosala
    3 hours ago










  • @Kosala How the Feynamn diagrams arise from QFT formalism is really esoteric for most of the people and uses relatively heavy machinery. So just think of "virtual particles" as a mathematical trick and not as a part of reality.
    – Gonenc Mogol
    2 hours ago














up vote
4
down vote













Virtual particles , as Gonenc Mogol says, are not particles, they are place holders for quantum numbers in the integrals necessary to calculate probabilities of interaction for quantum mechanical entities.



Here is a practical example :



eemumu



The virtual particle is labeled as a photon, but in effect it is a place holder for the photon quantum numbers within an integral over the variables shown in the graph, it means that the photon is off mass shell, and the mass varies within the integration which will give real numbers ( note the imaginary numbers) to be compared with experiment. The link gives details . What can be measured are the input particle four vectors and the output particles four vectors, the internal lines, as the $γ$ here, are not accessible to measurement.



The popularization of this mathematical concept creates the confusion about uncertainties and energy conservation. In order to have virtual "particles" there must always be real energy input by incoming real particles, and real energy output by outgoing real particles. In between it is mathematics and hand waving misinterpretations.






share|cite|improve this answer






















  • Thanks for the comment. Did you mean internal lines or inclined lines?
    – Kosala
    5 hours ago










  • internal, as the γ in the diagram. it connects to external legs through a vertex
    – anna v
    5 hours ago










  • Thanks for taking time to answer my question. But I am bit confused now. According to this theory, the photon is the virtual particle. And how the positron comes into play? why does it appear? So why we call then virtual particles are countervailed?
    – Kosala
    3 hours ago










  • @Kosala How the Feynamn diagrams arise from QFT formalism is really esoteric for most of the people and uses relatively heavy machinery. So just think of "virtual particles" as a mathematical trick and not as a part of reality.
    – Gonenc Mogol
    2 hours ago












up vote
4
down vote










up vote
4
down vote









Virtual particles , as Gonenc Mogol says, are not particles, they are place holders for quantum numbers in the integrals necessary to calculate probabilities of interaction for quantum mechanical entities.



Here is a practical example :



eemumu



The virtual particle is labeled as a photon, but in effect it is a place holder for the photon quantum numbers within an integral over the variables shown in the graph, it means that the photon is off mass shell, and the mass varies within the integration which will give real numbers ( note the imaginary numbers) to be compared with experiment. The link gives details . What can be measured are the input particle four vectors and the output particles four vectors, the internal lines, as the $γ$ here, are not accessible to measurement.



The popularization of this mathematical concept creates the confusion about uncertainties and energy conservation. In order to have virtual "particles" there must always be real energy input by incoming real particles, and real energy output by outgoing real particles. In between it is mathematics and hand waving misinterpretations.






share|cite|improve this answer














Virtual particles , as Gonenc Mogol says, are not particles, they are place holders for quantum numbers in the integrals necessary to calculate probabilities of interaction for quantum mechanical entities.



Here is a practical example :



eemumu



The virtual particle is labeled as a photon, but in effect it is a place holder for the photon quantum numbers within an integral over the variables shown in the graph, it means that the photon is off mass shell, and the mass varies within the integration which will give real numbers ( note the imaginary numbers) to be compared with experiment. The link gives details . What can be measured are the input particle four vectors and the output particles four vectors, the internal lines, as the $γ$ here, are not accessible to measurement.



The popularization of this mathematical concept creates the confusion about uncertainties and energy conservation. In order to have virtual "particles" there must always be real energy input by incoming real particles, and real energy output by outgoing real particles. In between it is mathematics and hand waving misinterpretations.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 5 hours ago

























answered 5 hours ago









anna v

154k7147441




154k7147441











  • Thanks for the comment. Did you mean internal lines or inclined lines?
    – Kosala
    5 hours ago










  • internal, as the γ in the diagram. it connects to external legs through a vertex
    – anna v
    5 hours ago










  • Thanks for taking time to answer my question. But I am bit confused now. According to this theory, the photon is the virtual particle. And how the positron comes into play? why does it appear? So why we call then virtual particles are countervailed?
    – Kosala
    3 hours ago










  • @Kosala How the Feynamn diagrams arise from QFT formalism is really esoteric for most of the people and uses relatively heavy machinery. So just think of "virtual particles" as a mathematical trick and not as a part of reality.
    – Gonenc Mogol
    2 hours ago
















  • Thanks for the comment. Did you mean internal lines or inclined lines?
    – Kosala
    5 hours ago










  • internal, as the γ in the diagram. it connects to external legs through a vertex
    – anna v
    5 hours ago










  • Thanks for taking time to answer my question. But I am bit confused now. According to this theory, the photon is the virtual particle. And how the positron comes into play? why does it appear? So why we call then virtual particles are countervailed?
    – Kosala
    3 hours ago










  • @Kosala How the Feynamn diagrams arise from QFT formalism is really esoteric for most of the people and uses relatively heavy machinery. So just think of "virtual particles" as a mathematical trick and not as a part of reality.
    – Gonenc Mogol
    2 hours ago















Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
5 hours ago




Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
5 hours ago












internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
5 hours ago




internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
5 hours ago












Thanks for taking time to answer my question. But I am bit confused now. According to this theory, the photon is the virtual particle. And how the positron comes into play? why does it appear? So why we call then virtual particles are countervailed?
– Kosala
3 hours ago




Thanks for taking time to answer my question. But I am bit confused now. According to this theory, the photon is the virtual particle. And how the positron comes into play? why does it appear? So why we call then virtual particles are countervailed?
– Kosala
3 hours ago












@Kosala How the Feynamn diagrams arise from QFT formalism is really esoteric for most of the people and uses relatively heavy machinery. So just think of "virtual particles" as a mathematical trick and not as a part of reality.
– Gonenc Mogol
2 hours ago




@Kosala How the Feynamn diagrams arise from QFT formalism is really esoteric for most of the people and uses relatively heavy machinery. So just think of "virtual particles" as a mathematical trick and not as a part of reality.
– Gonenc Mogol
2 hours ago










up vote
-1
down vote













It really depends on what you mean by conservation of energy. The classical idea is that the total energy is constant at all times. However, once quantum mechanics was discovered, people realized that our classical idea of conversation of energy is really just an approximation of the quantum conservation principle. In quantum mechanics, conservation of energy is a statement about the expectation value (think the average value) of energy; it says that the expectation value of energy does not change. Quantum fluctations in the energy cancel out over time, so the average energy of an isolated system never changes.



Effectively, physicists reformulated their idea about conservation of energy to agree with quantum mechanics, because they found that the classical notion breaks down.



As for how virtual particles are created, I'm not exactly sure. Somebody who is more familiar with QFT can probably answer.






share|cite|improve this answer








New contributor




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

















  • Thanks for the answer.
    – Kosala
    3 hours ago














up vote
-1
down vote













It really depends on what you mean by conservation of energy. The classical idea is that the total energy is constant at all times. However, once quantum mechanics was discovered, people realized that our classical idea of conversation of energy is really just an approximation of the quantum conservation principle. In quantum mechanics, conservation of energy is a statement about the expectation value (think the average value) of energy; it says that the expectation value of energy does not change. Quantum fluctations in the energy cancel out over time, so the average energy of an isolated system never changes.



Effectively, physicists reformulated their idea about conservation of energy to agree with quantum mechanics, because they found that the classical notion breaks down.



As for how virtual particles are created, I'm not exactly sure. Somebody who is more familiar with QFT can probably answer.






share|cite|improve this answer








New contributor




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

















  • Thanks for the answer.
    – Kosala
    3 hours ago












up vote
-1
down vote










up vote
-1
down vote









It really depends on what you mean by conservation of energy. The classical idea is that the total energy is constant at all times. However, once quantum mechanics was discovered, people realized that our classical idea of conversation of energy is really just an approximation of the quantum conservation principle. In quantum mechanics, conservation of energy is a statement about the expectation value (think the average value) of energy; it says that the expectation value of energy does not change. Quantum fluctations in the energy cancel out over time, so the average energy of an isolated system never changes.



Effectively, physicists reformulated their idea about conservation of energy to agree with quantum mechanics, because they found that the classical notion breaks down.



As for how virtual particles are created, I'm not exactly sure. Somebody who is more familiar with QFT can probably answer.






share|cite|improve this answer








New contributor




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









It really depends on what you mean by conservation of energy. The classical idea is that the total energy is constant at all times. However, once quantum mechanics was discovered, people realized that our classical idea of conversation of energy is really just an approximation of the quantum conservation principle. In quantum mechanics, conservation of energy is a statement about the expectation value (think the average value) of energy; it says that the expectation value of energy does not change. Quantum fluctations in the energy cancel out over time, so the average energy of an isolated system never changes.



Effectively, physicists reformulated their idea about conservation of energy to agree with quantum mechanics, because they found that the classical notion breaks down.



As for how virtual particles are created, I'm not exactly sure. Somebody who is more familiar with QFT can probably answer.







share|cite|improve this answer








New contributor




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









share|cite|improve this answer



share|cite|improve this answer






New contributor




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









answered 6 hours ago









Deeply Confused

224




224




New contributor




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





New contributor





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






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











  • Thanks for the answer.
    – Kosala
    3 hours ago
















  • Thanks for the answer.
    – Kosala
    3 hours ago















Thanks for the answer.
– Kosala
3 hours ago




Thanks for the answer.
– Kosala
3 hours ago

















 

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