Heisenberg energy uncertainty [duplicate]
up vote
10
down vote
favorite
This question already has an answer here:
Why do many people say that virtual particles do not conserve energy?
5 answers
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
marked as duplicate by knzhou, AccidentalFourierTransform, John Rennie, Jon Custer, peterh Nov 11 at 19:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
up vote
10
down vote
favorite
This question already has an answer here:
Why do many people say that virtual particles do not conserve energy?
5 answers
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
marked as duplicate by knzhou, AccidentalFourierTransform, John Rennie, Jon Custer, peterh Nov 11 at 19:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
other possible duplicates: Virtual particles with almost infinite energy?, Why are there e.m. vacuüm fluctuations? and links therein.
– AccidentalFourierTransform
Nov 9 at 22:22
add a comment |
up vote
10
down vote
favorite
up vote
10
down vote
favorite
This question already has an answer here:
Why do many people say that virtual particles do not conserve energy?
5 answers
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
This question already has an answer here:
Why do many people say that virtual particles do not conserve energy?
5 answers
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?
This question already has an answer here:
Why do many people say that virtual particles do not conserve energy?
5 answers
quantum-field-theory heisenberg-uncertainty-principle virtual-particles
quantum-field-theory heisenberg-uncertainty-principle virtual-particles
edited Nov 9 at 17:59
Qmechanic♦
100k121791122
100k121791122
asked Nov 9 at 17:54
Kosala
356112
356112
marked as duplicate by knzhou, AccidentalFourierTransform, John Rennie, Jon Custer, peterh Nov 11 at 19:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by knzhou, AccidentalFourierTransform, John Rennie, Jon Custer, peterh Nov 11 at 19:22
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
other possible duplicates: Virtual particles with almost infinite energy?, Why are there e.m. vacuüm fluctuations? and links therein.
– AccidentalFourierTransform
Nov 9 at 22:22
add a comment |
other possible duplicates: Virtual particles with almost infinite energy?, Why are there e.m. vacuüm fluctuations? and links therein.
– AccidentalFourierTransform
Nov 9 at 22:22
other possible duplicates: Virtual particles with almost infinite energy?, Why are there e.m. vacuüm fluctuations? and links therein.
– AccidentalFourierTransform
Nov 9 at 22:22
other possible duplicates: Virtual particles with almost infinite energy?, Why are there e.m. vacuüm fluctuations? and links therein.
– AccidentalFourierTransform
Nov 9 at 22:22
add a comment |
3 Answers
3
active
oldest
votes
up vote
13
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 $hat Q$:
$$Delta hat H Delta t geq frac { hbar } { 2 } , quad Delta t equiv frac { Delta hat Q } { left| frac { d left< hat Q right> } { d t } right| } $$
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.
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
Nov 9 at 20:48
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
Nov 9 at 21:04
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
Nov 9 at 21:04
2
@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
Nov 9 at 22:29
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
Nov 10 at 0:03
|
show 1 more comment
up vote
8
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 :
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.
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
Nov 9 at 19:29
1
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
Nov 9 at 19:33
2
@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
Nov 9 at 22:35
1
Kosala this is an example of how we calculate real numbers to compare with exeriment. The experiment that is described by this particular diagram is the scattering of e+ on e- , two beams in a particle collider, to give a $μ^+$ $μ^-$ the muon energy and momentum measured in the detectors of the experiment. The theory is quantum field theory and gives a recipe for getting the numbers to compare to experiments using Feynman diagrams as a tool for easy mathematical organization
– anna v
Nov 10 at 4:45
1
. Virtual internal lines appear due to the need to conserve quantum numbers and other conservation laws, labeling functions that affect the integration by the term "virtual" to make clear that they are off mass shell, in the case above the mass not zero but varying according to the function shown.
– anna v
Nov 10 at 4:50
|
show 2 more comments
up vote
2
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.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
13
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 $hat Q$:
$$Delta hat H Delta t geq frac { hbar } { 2 } , quad Delta t equiv frac { Delta hat Q } { left| frac { d left< hat Q right> } { d t } right| } $$
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.
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
Nov 9 at 20:48
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
Nov 9 at 21:04
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
Nov 9 at 21:04
2
@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
Nov 9 at 22:29
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
Nov 10 at 0:03
|
show 1 more comment
up vote
13
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 $hat Q$:
$$Delta hat H Delta t geq frac { hbar } { 2 } , quad Delta t equiv frac { Delta hat Q } { left| frac { d left< hat Q right> } { d t } right| } $$
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.
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
Nov 9 at 20:48
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
Nov 9 at 21:04
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
Nov 9 at 21:04
2
@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
Nov 9 at 22:29
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
Nov 10 at 0:03
|
show 1 more comment
up vote
13
down vote
up vote
13
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 $hat Q$:
$$Delta hat H Delta t geq frac { hbar } { 2 } , quad Delta t equiv frac { Delta hat Q } { left| frac { d left< hat Q right> } { d t } right| } $$
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.
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 $hat Q$:
$$Delta hat H Delta t geq frac { hbar } { 2 } , quad Delta t equiv frac { Delta hat Q } { left| frac { d left< hat Q right> } { d t } right| } $$
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.
answered Nov 9 at 18:59
Gonenc Mogol
2,83111332
2,83111332
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
Nov 9 at 20:48
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
Nov 9 at 21:04
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
Nov 9 at 21:04
2
@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
Nov 9 at 22:29
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
Nov 10 at 0:03
|
show 1 more comment
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
Nov 9 at 20:48
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
Nov 9 at 21:04
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
Nov 9 at 21:04
2
@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
Nov 9 at 22:29
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
Nov 10 at 0:03
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
Nov 9 at 20:48
"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
Nov 9 at 20:48
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
Nov 9 at 21:04
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
Nov 9 at 21:04
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
Nov 9 at 21:04
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
Nov 9 at 21:04
2
2
@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
Nov 9 at 22:29
@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
Nov 9 at 22:29
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
Nov 10 at 0:03
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
Nov 10 at 0:03
|
show 1 more comment
up vote
8
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 :
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.
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
Nov 9 at 19:29
1
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
Nov 9 at 19:33
2
@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
Nov 9 at 22:35
1
Kosala this is an example of how we calculate real numbers to compare with exeriment. The experiment that is described by this particular diagram is the scattering of e+ on e- , two beams in a particle collider, to give a $μ^+$ $μ^-$ the muon energy and momentum measured in the detectors of the experiment. The theory is quantum field theory and gives a recipe for getting the numbers to compare to experiments using Feynman diagrams as a tool for easy mathematical organization
– anna v
Nov 10 at 4:45
1
. Virtual internal lines appear due to the need to conserve quantum numbers and other conservation laws, labeling functions that affect the integration by the term "virtual" to make clear that they are off mass shell, in the case above the mass not zero but varying according to the function shown.
– anna v
Nov 10 at 4:50
|
show 2 more comments
up vote
8
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 :
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.
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
Nov 9 at 19:29
1
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
Nov 9 at 19:33
2
@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
Nov 9 at 22:35
1
Kosala this is an example of how we calculate real numbers to compare with exeriment. The experiment that is described by this particular diagram is the scattering of e+ on e- , two beams in a particle collider, to give a $μ^+$ $μ^-$ the muon energy and momentum measured in the detectors of the experiment. The theory is quantum field theory and gives a recipe for getting the numbers to compare to experiments using Feynman diagrams as a tool for easy mathematical organization
– anna v
Nov 10 at 4:45
1
. Virtual internal lines appear due to the need to conserve quantum numbers and other conservation laws, labeling functions that affect the integration by the term "virtual" to make clear that they are off mass shell, in the case above the mass not zero but varying according to the function shown.
– anna v
Nov 10 at 4:50
|
show 2 more comments
up vote
8
down vote
up vote
8
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 :
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.
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 :
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.
edited Nov 9 at 19:34
answered Nov 9 at 19:21
anna v
155k7148443
155k7148443
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
Nov 9 at 19:29
1
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
Nov 9 at 19:33
2
@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
Nov 9 at 22:35
1
Kosala this is an example of how we calculate real numbers to compare with exeriment. The experiment that is described by this particular diagram is the scattering of e+ on e- , two beams in a particle collider, to give a $μ^+$ $μ^-$ the muon energy and momentum measured in the detectors of the experiment. The theory is quantum field theory and gives a recipe for getting the numbers to compare to experiments using Feynman diagrams as a tool for easy mathematical organization
– anna v
Nov 10 at 4:45
1
. Virtual internal lines appear due to the need to conserve quantum numbers and other conservation laws, labeling functions that affect the integration by the term "virtual" to make clear that they are off mass shell, in the case above the mass not zero but varying according to the function shown.
– anna v
Nov 10 at 4:50
|
show 2 more comments
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
Nov 9 at 19:29
1
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
Nov 9 at 19:33
2
@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
Nov 9 at 22:35
1
Kosala this is an example of how we calculate real numbers to compare with exeriment. The experiment that is described by this particular diagram is the scattering of e+ on e- , two beams in a particle collider, to give a $μ^+$ $μ^-$ the muon energy and momentum measured in the detectors of the experiment. The theory is quantum field theory and gives a recipe for getting the numbers to compare to experiments using Feynman diagrams as a tool for easy mathematical organization
– anna v
Nov 10 at 4:45
1
. Virtual internal lines appear due to the need to conserve quantum numbers and other conservation laws, labeling functions that affect the integration by the term "virtual" to make clear that they are off mass shell, in the case above the mass not zero but varying according to the function shown.
– anna v
Nov 10 at 4:50
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
Nov 9 at 19:29
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
Nov 9 at 19:29
1
1
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
Nov 9 at 19:33
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
Nov 9 at 19:33
2
2
@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
Nov 9 at 22:35
@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
Nov 9 at 22:35
1
1
Kosala this is an example of how we calculate real numbers to compare with exeriment. The experiment that is described by this particular diagram is the scattering of e+ on e- , two beams in a particle collider, to give a $μ^+$ $μ^-$ the muon energy and momentum measured in the detectors of the experiment. The theory is quantum field theory and gives a recipe for getting the numbers to compare to experiments using Feynman diagrams as a tool for easy mathematical organization
– anna v
Nov 10 at 4:45
Kosala this is an example of how we calculate real numbers to compare with exeriment. The experiment that is described by this particular diagram is the scattering of e+ on e- , two beams in a particle collider, to give a $μ^+$ $μ^-$ the muon energy and momentum measured in the detectors of the experiment. The theory is quantum field theory and gives a recipe for getting the numbers to compare to experiments using Feynman diagrams as a tool for easy mathematical organization
– anna v
Nov 10 at 4:45
1
1
. Virtual internal lines appear due to the need to conserve quantum numbers and other conservation laws, labeling functions that affect the integration by the term "virtual" to make clear that they are off mass shell, in the case above the mass not zero but varying according to the function shown.
– anna v
Nov 10 at 4:50
. Virtual internal lines appear due to the need to conserve quantum numbers and other conservation laws, labeling functions that affect the integration by the term "virtual" to make clear that they are off mass shell, in the case above the mass not zero but varying according to the function shown.
– anna v
Nov 10 at 4:50
|
show 2 more comments
up vote
2
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.
add a comment |
up vote
2
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.
add a comment |
up vote
2
down vote
up vote
2
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.
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.
answered Nov 9 at 18:22
Ricky P
574
574
add a comment |
add a comment |
other possible duplicates: Virtual particles with almost infinite energy?, Why are there e.m. vacuüm fluctuations? and links therein.
– AccidentalFourierTransform
Nov 9 at 22:22