Successive amplitudes in quantum mechanicsWhat is the relationship between distinguishability and probability in Quantum mechanics?How is the classical EM field modeled in quantum mechanics?Propagators, Green’s functions, path integrals and transition amplitudes in quantum mechanics and quantum field theoryIs there a charge density in quantum mechanics?What is the etymology of the term “amplitude” as used in quantum mechanics?
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Successive amplitudes in quantum mechanics
What is the relationship between distinguishability and probability in Quantum mechanics?How is the classical EM field modeled in quantum mechanics?Propagators, Green’s functions, path integrals and transition amplitudes in quantum mechanics and quantum field theoryIs there a charge density in quantum mechanics?What is the etymology of the term “amplitude” as used in quantum mechanics?
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$begingroup$
In quantum mechanics we define amplitudes for events, like propagation from one point to some other point. Lets say that from a source to detector we have some amplitude (D/S).
But, lets now say that we have one mid point. Now we can say that the amplitude from S to M (midpoint) and then to D is (D/M)(M/S). Why multiply?
quantum-mechanics newtonian-mechanics
edited Sep 29 at 13:17
DanielSank
19.9k4 gold badges58 silver badges84 bronze badges
asked Sep 29 at 12:47
Žarko TomičićŽarko Tomičić
1,4247 silver badges12 bronze badges
$endgroup$
add a comment
|
$begingroup$
In quantum mechanics we define amplitudes for events, like propagation from one point to some other point. Lets say that from a source to detector we have some amplitude (D/S).
But, lets now say that we have one mid point. Now we can say that the amplitude from S to M (midpoint) and then to D is (D/M)(M/S). Why multiply?
quantum-mechanics newtonian-mechanics
edited Sep 29 at 13:17
DanielSank
19.9k4 gold badges58 silver badges84 bronze badges
asked Sep 29 at 12:47
Žarko TomičićŽarko Tomičić
1,4247 silver badges12 bronze badges
$endgroup$
$begingroup$
What would you do instead? Add? Why add?
$endgroup$
– infinitezero
Sep 29 at 13:01
$begingroup$
Didnt say that. But i was just wondering.
$endgroup$
– Žarko Tomičić
Sep 29 at 17:05
add a comment
|
$begingroup$
In quantum mechanics we define amplitudes for events, like propagation from one point to some other point. Lets say that from a source to detector we have some amplitude (D/S).
But, lets now say that we have one mid point. Now we can say that the amplitude from S to M (midpoint) and then to D is (D/M)(M/S). Why multiply?
quantum-mechanics newtonian-mechanics
edited Sep 29 at 13:17
DanielSank
19.9k4 gold badges58 silver badges84 bronze badges
asked Sep 29 at 12:47
Žarko TomičićŽarko Tomičić
1,4247 silver badges12 bronze badges
$endgroup$
In quantum mechanics we define amplitudes for events, like propagation from one point to some other point. Lets say that from a source to detector we have some amplitude (D/S).
But, lets now say that we have one mid point. Now we can say that the amplitude from S to M (midpoint) and then to D is (D/M)(M/S). Why multiply?
quantum-mechanics newtonian-mechanics
quantum-mechanics newtonian-mechanics
edited Sep 29 at 13:17
DanielSank
19.9k4 gold badges58 silver badges84 bronze badges
asked Sep 29 at 12:47
Žarko TomičićŽarko Tomičić
1,4247 silver badges12 bronze badges
edited Sep 29 at 13:17
DanielSank
19.9k4 gold badges58 silver badges84 bronze badges
asked Sep 29 at 12:47
Žarko TomičićŽarko Tomičić
1,4247 silver badges12 bronze badges
edited Sep 29 at 13:17
DanielSank
19.9k4 gold badges58 silver badges84 bronze badges
edited Sep 29 at 13:17
DanielSank
19.9k4 gold badges58 silver badges84 bronze badges
edited Sep 29 at 13:17
DanielSank
19.9k4 gold badges58 silver badges84 bronze badges
19.9k4 gold badges58 silver badges84 bronze badges
asked Sep 29 at 12:47
Žarko TomičićŽarko Tomičić
1,4247 silver badges12 bronze badges
asked Sep 29 at 12:47
Žarko TomičićŽarko Tomičić
1,4247 silver badges12 bronze badges
asked Sep 29 at 12:47
Žarko TomičićŽarko Tomičić
1,4247 silver badges12 bronze badges
1,4247 silver badges12 bronze badges
$begingroup$
What would you do instead? Add? Why add?
$endgroup$
– infinitezero
Sep 29 at 13:01
$begingroup$
Didnt say that. But i was just wondering.
$endgroup$
– Žarko Tomičić
Sep 29 at 17:05
add a comment
|
$begingroup$
What would you do instead? Add? Why add?
$endgroup$
– infinitezero
Sep 29 at 13:01
$begingroup$
Didnt say that. But i was just wondering.
$endgroup$
– Žarko Tomičić
Sep 29 at 17:05
$begingroup$
What would you do instead? Add? Why add?
$endgroup$
– infinitezero
Sep 29 at 13:01
$begingroup$
What would you do instead? Add? Why add?
$endgroup$
– infinitezero
Sep 29 at 13:01
$begingroup$
Didnt say that. But i was just wondering.
$endgroup$
– Žarko Tomičić
Sep 29 at 17:05
$begingroup$
Didnt say that. But i was just wondering.
$endgroup$
– Žarko Tomičić
Sep 29 at 17:05
add a comment
|
1 Answer
1
active
oldest
votes
$begingroup$
Because amplitudes are related to probabilities, and that's how the laws of probability work.
A fair die has a probability of $1/6$ of landing on any side. If you roll the die twice, the probability of rolling a 6 the first time and rolling a 6 the second time is $1/6times 1/6=1/36$. Likewise, for a single roll, the probability of rolling a 5 or rolling a 6 is $1/6+1/6=1/3$. So you see that the laws of probability dictate the following:
The probability of one event happening and another mutually-exclusive event happening is the product of the probabilities of the two events happening.
The probability of one event happening or another mutually-exclusive event happening is the sum of the probabilities of the two events happening.
So the probability of the particle traveling from S to M to D is the probability of traveling from S to M and traveling from M to D, hence is equal to the product of the two individual probabilities. Since the squared magnitude of the amplitude is the probability, it should be straightforward to see that amplitudes should follow the same rule, since $|a||b|=|ab|$.
answered Sep 29 at 13:36
probably_someoneprobably_someone
25.1k1 gold badge37 silver badges76 bronze badges
$endgroup$
$begingroup$
Tnx man. I thought so my self but wasnt sure.
$endgroup$
– Žarko Tomičić
Sep 29 at 17:04
$begingroup$
Also, to add, I see this law for successive probabilities as consequence of the fact that in two dice rolls you have total of 36 permutations , 6 times 6. So, if you define probability as number of desireable events over the number of total events you get for example 1 over 36. Now what I was worried was the generalization of this to events like those in QM where we do not have a discrete situation. I guess generalization is ok. But also would be nice to see the proof. In QM book also. Just to be sure.
$endgroup$
– Žarko Tomičić
Sep 30 at 6:33
add a comment
|
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$begingroup$
Because amplitudes are related to probabilities, and that's how the laws of probability work.
A fair die has a probability of $1/6$ of landing on any side. If you roll the die twice, the probability of rolling a 6 the first time and rolling a 6 the second time is $1/6times 1/6=1/36$. Likewise, for a single roll, the probability of rolling a 5 or rolling a 6 is $1/6+1/6=1/3$. So you see that the laws of probability dictate the following:
The probability of one event happening and another mutually-exclusive event happening is the product of the probabilities of the two events happening.
The probability of one event happening or another mutually-exclusive event happening is the sum of the probabilities of the two events happening.
So the probability of the particle traveling from S to M to D is the probability of traveling from S to M and traveling from M to D, hence is equal to the product of the two individual probabilities. Since the squared magnitude of the amplitude is the probability, it should be straightforward to see that amplitudes should follow the same rule, since $|a||b|=|ab|$.
answered Sep 29 at 13:36
probably_someoneprobably_someone
25.1k1 gold badge37 silver badges76 bronze badges
$endgroup$
$begingroup$
Tnx man. I thought so my self but wasnt sure.
$endgroup$
– Žarko Tomičić
Sep 29 at 17:04
$begingroup$
Also, to add, I see this law for successive probabilities as consequence of the fact that in two dice rolls you have total of 36 permutations , 6 times 6. So, if you define probability as number of desireable events over the number of total events you get for example 1 over 36. Now what I was worried was the generalization of this to events like those in QM where we do not have a discrete situation. I guess generalization is ok. But also would be nice to see the proof. In QM book also. Just to be sure.
$endgroup$
– Žarko Tomičić
Sep 30 at 6:33
add a comment
|
$begingroup$
Because amplitudes are related to probabilities, and that's how the laws of probability work.
A fair die has a probability of $1/6$ of landing on any side. If you roll the die twice, the probability of rolling a 6 the first time and rolling a 6 the second time is $1/6times 1/6=1/36$. Likewise, for a single roll, the probability of rolling a 5 or rolling a 6 is $1/6+1/6=1/3$. So you see that the laws of probability dictate the following:
The probability of one event happening and another mutually-exclusive event happening is the product of the probabilities of the two events happening.
The probability of one event happening or another mutually-exclusive event happening is the sum of the probabilities of the two events happening.
So the probability of the particle traveling from S to M to D is the probability of traveling from S to M and traveling from M to D, hence is equal to the product of the two individual probabilities. Since the squared magnitude of the amplitude is the probability, it should be straightforward to see that amplitudes should follow the same rule, since $|a||b|=|ab|$.
answered Sep 29 at 13:36
probably_someoneprobably_someone
25.1k1 gold badge37 silver badges76 bronze badges
$endgroup$
$begingroup$
Tnx man. I thought so my self but wasnt sure.
$endgroup$
– Žarko Tomičić
Sep 29 at 17:04
$begingroup$
Also, to add, I see this law for successive probabilities as consequence of the fact that in two dice rolls you have total of 36 permutations , 6 times 6. So, if you define probability as number of desireable events over the number of total events you get for example 1 over 36. Now what I was worried was the generalization of this to events like those in QM where we do not have a discrete situation. I guess generalization is ok. But also would be nice to see the proof. In QM book also. Just to be sure.
$endgroup$
– Žarko Tomičić
Sep 30 at 6:33
add a comment
|
$begingroup$
Because amplitudes are related to probabilities, and that's how the laws of probability work.
A fair die has a probability of $1/6$ of landing on any side. If you roll the die twice, the probability of rolling a 6 the first time and rolling a 6 the second time is $1/6times 1/6=1/36$. Likewise, for a single roll, the probability of rolling a 5 or rolling a 6 is $1/6+1/6=1/3$. So you see that the laws of probability dictate the following:
The probability of one event happening and another mutually-exclusive event happening is the product of the probabilities of the two events happening.
The probability of one event happening or another mutually-exclusive event happening is the sum of the probabilities of the two events happening.
So the probability of the particle traveling from S to M to D is the probability of traveling from S to M and traveling from M to D, hence is equal to the product of the two individual probabilities. Since the squared magnitude of the amplitude is the probability, it should be straightforward to see that amplitudes should follow the same rule, since $|a||b|=|ab|$.
answered Sep 29 at 13:36
probably_someoneprobably_someone
25.1k1 gold badge37 silver badges76 bronze badges
$endgroup$
Because amplitudes are related to probabilities, and that's how the laws of probability work.
A fair die has a probability of $1/6$ of landing on any side. If you roll the die twice, the probability of rolling a 6 the first time and rolling a 6 the second time is $1/6times 1/6=1/36$. Likewise, for a single roll, the probability of rolling a 5 or rolling a 6 is $1/6+1/6=1/3$. So you see that the laws of probability dictate the following:
The probability of one event happening and another mutually-exclusive event happening is the product of the probabilities of the two events happening.
The probability of one event happening or another mutually-exclusive event happening is the sum of the probabilities of the two events happening.
So the probability of the particle traveling from S to M to D is the probability of traveling from S to M and traveling from M to D, hence is equal to the product of the two individual probabilities. Since the squared magnitude of the amplitude is the probability, it should be straightforward to see that amplitudes should follow the same rule, since $|a||b|=|ab|$.
answered Sep 29 at 13:36
probably_someoneprobably_someone
25.1k1 gold badge37 silver badges76 bronze badges
answered Sep 29 at 13:36
probably_someoneprobably_someone
25.1k1 gold badge37 silver badges76 bronze badges
answered Sep 29 at 13:36
probably_someoneprobably_someone
25.1k1 gold badge37 silver badges76 bronze badges
answered Sep 29 at 13:36
probably_someoneprobably_someone
25.1k1 gold badge37 silver badges76 bronze badges
25.1k1 gold badge37 silver badges76 bronze badges
$begingroup$
Tnx man. I thought so my self but wasnt sure.
$endgroup$
– Žarko Tomičić
Sep 29 at 17:04
$begingroup$
Also, to add, I see this law for successive probabilities as consequence of the fact that in two dice rolls you have total of 36 permutations , 6 times 6. So, if you define probability as number of desireable events over the number of total events you get for example 1 over 36. Now what I was worried was the generalization of this to events like those in QM where we do not have a discrete situation. I guess generalization is ok. But also would be nice to see the proof. In QM book also. Just to be sure.
$endgroup$
– Žarko Tomičić
Sep 30 at 6:33
add a comment
|
$begingroup$
Tnx man. I thought so my self but wasnt sure.
$endgroup$
– Žarko Tomičić
Sep 29 at 17:04
$begingroup$
Also, to add, I see this law for successive probabilities as consequence of the fact that in two dice rolls you have total of 36 permutations , 6 times 6. So, if you define probability as number of desireable events over the number of total events you get for example 1 over 36. Now what I was worried was the generalization of this to events like those in QM where we do not have a discrete situation. I guess generalization is ok. But also would be nice to see the proof. In QM book also. Just to be sure.
$endgroup$
– Žarko Tomičić
Sep 30 at 6:33
$begingroup$
Tnx man. I thought so my self but wasnt sure.
$endgroup$
– Žarko Tomičić
Sep 29 at 17:04
$begingroup$
Tnx man. I thought so my self but wasnt sure.
$endgroup$
– Žarko Tomičić
Sep 29 at 17:04
$begingroup$
Also, to add, I see this law for successive probabilities as consequence of the fact that in two dice rolls you have total of 36 permutations , 6 times 6. So, if you define probability as number of desireable events over the number of total events you get for example 1 over 36. Now what I was worried was the generalization of this to events like those in QM where we do not have a discrete situation. I guess generalization is ok. But also would be nice to see the proof. In QM book also. Just to be sure.
$endgroup$
– Žarko Tomičić
Sep 30 at 6:33
$begingroup$
Also, to add, I see this law for successive probabilities as consequence of the fact that in two dice rolls you have total of 36 permutations , 6 times 6. So, if you define probability as number of desireable events over the number of total events you get for example 1 over 36. Now what I was worried was the generalization of this to events like those in QM where we do not have a discrete situation. I guess generalization is ok. But also would be nice to see the proof. In QM book also. Just to be sure.
$endgroup$
– Žarko Tomičić
Sep 30 at 6:33
add a comment
|
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$begingroup$
What would you do instead? Add? Why add?
$endgroup$
– infinitezero
Sep 29 at 13:01
$begingroup$
Didnt say that. But i was just wondering.
$endgroup$
– Žarko Tomičić
Sep 29 at 17:05