Mixing
Heisenberg energy uncertainty
Clash Royale CLAN TAG#URR8PPP
up vote
2
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?
quantum-field-theory heisenberg-uncertainty-principle virtual-particles
edited 3 hours ago
Qmechanic♦
99.1k121781096
asked 3 hours ago
Kosala
316111
add a comment |
up vote
2
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?
quantum-field-theory heisenberg-uncertainty-principle virtual-particles
edited 3 hours ago
Qmechanic♦
99.1k121781096
asked 3 hours ago
Kosala
316111
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
2 hours ago
add a comment |
up vote
2
down vote
favorite
up vote
2
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?
quantum-field-theory heisenberg-uncertainty-principle virtual-particles
edited 3 hours ago
Qmechanic♦
99.1k121781096
asked 3 hours ago
Kosala
316111
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
quantum-field-theory heisenberg-uncertainty-principle virtual-particles
edited 3 hours ago
Qmechanic♦
99.1k121781096
asked 3 hours ago
Kosala
316111
edited 3 hours ago
Qmechanic♦
99.1k121781096
asked 3 hours ago
Kosala
316111
edited 3 hours ago
Qmechanic♦
99.1k121781096
edited 3 hours ago
Qmechanic♦
99.1k121781096
edited 3 hours ago
Qmechanic♦
99.1k121781096
99.1k121781096
asked 3 hours ago
Kosala
316111
asked 3 hours ago
Kosala
316111
asked 3 hours ago
Kosala
316111
316111
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
2 hours ago
add a comment |
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
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
2 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
2 hours ago
add a comment |
3 Answers
3
active
oldest
votes
up vote
5
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.
answered 2 hours ago
Gonenc Mogol
2,71111331
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
43 mins 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
26 mins ago
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
26 mins ago
add a comment |
up vote
1
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.
answered 2 hours ago
anna v
154k7147441
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
2 hours ago
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
1 hour 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
17 mins ago
add a comment |
up vote
0
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.
answered 3 hours ago
Deeply Confused
244
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
11 mins ago
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
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.
answered 2 hours ago
Gonenc Mogol
2,71111331
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
43 mins 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
26 mins ago
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
26 mins ago
add a comment |
up vote
5
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.
answered 2 hours ago
Gonenc Mogol
2,71111331
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
43 mins 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
26 mins ago
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
26 mins ago
add a comment |
up vote
5
down vote
up vote
5
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.
answered 2 hours ago
Gonenc Mogol
2,71111331
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.
answered 2 hours ago
Gonenc Mogol
2,71111331
answered 2 hours ago
Gonenc Mogol
2,71111331
answered 2 hours ago
Gonenc Mogol
2,71111331
answered 2 hours ago
Gonenc Mogol
2,71111331
2,71111331
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
43 mins 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
26 mins ago
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
26 mins ago
add a 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
43 mins 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
26 mins ago
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
26 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
43 mins 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
43 mins 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
26 mins 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
26 mins ago
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
26 mins ago
I was actually challenging my own idea since I am not a big physics brain. Anyway thanks for the explanation.
– Kosala
26 mins ago
add a comment |
up vote
1
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.
answered 2 hours ago
anna v
154k7147441
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
2 hours ago
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
1 hour 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
17 mins ago
add a comment |
up vote
1
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.
answered 2 hours ago
anna v
154k7147441
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
2 hours ago
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
1 hour 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
17 mins ago
add a comment |
up vote
1
down vote
up vote
1
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.
answered 2 hours ago
anna v
154k7147441
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.
answered 2 hours ago
anna v
154k7147441
edited 1 hour ago
answered 2 hours ago
anna v
154k7147441
answered 2 hours ago
anna v
154k7147441
answered 2 hours ago
anna v
154k7147441
154k7147441
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
2 hours ago
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
1 hour 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
17 mins ago
add a comment |
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
2 hours ago
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
1 hour 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
17 mins ago
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
2 hours ago
Thanks for the comment. Did you mean internal lines or inclined lines?
– Kosala
2 hours ago
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
1 hour ago
internal, as the γ in the diagram. it connects to external legs through a vertex
– anna v
1 hour 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
17 mins 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
17 mins ago
add a comment |
up vote
0
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.
answered 3 hours ago
Deeply Confused
244
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
11 mins ago
add a comment |
up vote
0
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.
answered 3 hours ago
Deeply Confused
244
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
11 mins ago
add a comment |
up vote
0
down vote
up vote
0
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.
answered 3 hours ago
Deeply Confused
244
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.
answered 3 hours ago
Deeply Confused
244
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 3 hours ago
Deeply Confused
244
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 3 hours ago
Deeply Confused
244
answered 3 hours ago
Deeply Confused
244
244
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
11 mins ago
add a comment |
Thanks for the answer.
– Kosala
11 mins ago
Thanks for the answer.
– Kosala
11 mins ago
Thanks for the answer.
– Kosala
11 mins ago
add a comment |
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f439946%2fheisenberg-energy-uncertainty%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
2 hours ago