Mixing
Divergence of probabilities for a position eigenstate in an energy basis
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If we have an electron in a one-dimensional infinite potential well, and we have measured its position and found it to be let's say at $x=0$ at the center of the well. The state vector after measurement becomes $|x=0>.$ If we calculate the probability then to get any eigen-value of the energy we will find it to be $1/a$ where $a$ is the width of the well. But that means the total probability (the sum of infinite terms each equals $1/a$) is infinite which is absurd. So, what exactly is happening here please?
quantum-mechanics hilbert-space wavefunction schroedinger-equation
edited 22 mins ago
Qmechanic♦
98.2k121731066
asked 1 hour ago
Arthur
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Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
1
down vote
favorite
If we have an electron in a one-dimensional infinite potential well, and we have measured its position and found it to be let's say at $x=0$ at the center of the well. The state vector after measurement becomes $|x=0>.$ If we calculate the probability then to get any eigen-value of the energy we will find it to be $1/a$ where $a$ is the width of the well. But that means the total probability (the sum of infinite terms each equals $1/a$) is infinite which is absurd. So, what exactly is happening here please?
quantum-mechanics hilbert-space wavefunction schroedinger-equation
edited 22 mins ago
Qmechanic♦
98.2k121731066
asked 1 hour ago
Arthur
62
New contributor
Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
The lab you've described has a position detector with infinite precision. Those sound nice - where did you find it? I'd like to get one for mine. (Joking aside: as you've just proved, position eigenstates are not physical, and cannot be achieved in real experiments.)
– Emilio Pisanty
1 hour ago
I liked the joke friend, but the problem persists. From totally mathematical point of view, this kind of result shouldn't occur if the axioms of QM represent a consistent mathematical theory.
– Arthur
1 hour ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If we have an electron in a one-dimensional infinite potential well, and we have measured its position and found it to be let's say at $x=0$ at the center of the well. The state vector after measurement becomes $|x=0>.$ If we calculate the probability then to get any eigen-value of the energy we will find it to be $1/a$ where $a$ is the width of the well. But that means the total probability (the sum of infinite terms each equals $1/a$) is infinite which is absurd. So, what exactly is happening here please?
quantum-mechanics hilbert-space wavefunction schroedinger-equation
edited 22 mins ago
Qmechanic♦
98.2k121731066
asked 1 hour ago
Arthur
62
New contributor
Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
If we have an electron in a one-dimensional infinite potential well, and we have measured its position and found it to be let's say at $x=0$ at the center of the well. The state vector after measurement becomes $|x=0>.$ If we calculate the probability then to get any eigen-value of the energy we will find it to be $1/a$ where $a$ is the width of the well. But that means the total probability (the sum of infinite terms each equals $1/a$) is infinite which is absurd. So, what exactly is happening here please?
quantum-mechanics hilbert-space wavefunction schroedinger-equation
quantum-mechanics hilbert-space wavefunction schroedinger-equation
edited 22 mins ago
Qmechanic♦
98.2k121731066
asked 1 hour ago
Arthur
62
New contributor
Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 22 mins ago
Qmechanic♦
98.2k121731066
asked 1 hour ago
Arthur
62
New contributor
Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 22 mins ago
Qmechanic♦
98.2k121731066
edited 22 mins ago
Qmechanic♦
98.2k121731066
edited 22 mins ago
Qmechanic♦
98.2k121731066
98.2k121731066
asked 1 hour ago
Arthur
62
New contributor
Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Arthur
62
asked 1 hour ago
Arthur
62
62
New contributor
Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Arthur is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
4
The lab you've described has a position detector with infinite precision. Those sound nice - where did you find it? I'd like to get one for mine. (Joking aside: as you've just proved, position eigenstates are not physical, and cannot be achieved in real experiments.)
– Emilio Pisanty
1 hour ago
I liked the joke friend, but the problem persists. From totally mathematical point of view, this kind of result shouldn't occur if the axioms of QM represent a consistent mathematical theory.
– Arthur
1 hour ago
add a comment |Â
4
The lab you've described has a position detector with infinite precision. Those sound nice - where did you find it? I'd like to get one for mine. (Joking aside: as you've just proved, position eigenstates are not physical, and cannot be achieved in real experiments.)
– Emilio Pisanty
1 hour ago
I liked the joke friend, but the problem persists. From totally mathematical point of view, this kind of result shouldn't occur if the axioms of QM represent a consistent mathematical theory.
– Arthur
1 hour ago
4
4
The lab you've described has a position detector with infinite precision. Those sound nice - where did you find it? I'd like to get one for mine. (Joking aside: as you've just proved, position eigenstates are not physical, and cannot be achieved in real experiments.)
– Emilio Pisanty
1 hour ago
The lab you've described has a position detector with infinite precision. Those sound nice - where did you find it? I'd like to get one for mine. (Joking aside: as you've just proved, position eigenstates are not physical, and cannot be achieved in real experiments.)
– Emilio Pisanty
1 hour ago
I liked the joke friend, but the problem persists. From totally mathematical point of view, this kind of result shouldn't occur if the axioms of QM represent a consistent mathematical theory.
– Arthur
1 hour ago
I liked the joke friend, but the problem persists. From totally mathematical point of view, this kind of result shouldn't occur if the axioms of QM represent a consistent mathematical theory.
– Arthur
1 hour ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
When the distribution of probabilities is continuous, the probability of finding an object exactly at one specific position is zero. You can calculate what is the probability of finding the object at a certain interval, you have a probability density. The right way to treat the probability is $dP(x)=dx/a$, and the total probability now becomes 1, because $int dP(x)=int^a_0 dx/a=1$. For more details see https://en.wikipedia.org/wiki/Particle_in_a_box
answered 1 hour ago
Wolphram jonny
10k22451
But the spectrum of energy is discrete, and it is this spectrum that I am trying to calculate the total probability for. The problem is that the individual probabilities are the same and are all 1/a.
– Arthur
1 hour ago
The spectrum of the energy operator is discrete, but the spectrum of the position operator is continuum. An eigenstate of energy is not an eigenstate of position, the operators do not commute
– Wolphram jonny
1 hour ago
you are correct in that a continuum spectrum is incompatible with a Hilbert space, that is why there are extensions to deal with this, although in practice this is almost never mentioned. See physics.stackexchange.com/questions/210300/…
– Wolphram jonny
1 hour ago
The fact that one observable has a continuous spectrum and the other doesn't, shouldn't cause problems if the axioms of quantum mechanics are true. I appreciate it if you know how they deal with such a problem my friend.
– Arthur
58 mins ago
Well, that is the problem, the formalism is incomplete if you want to use continumm spectrum operators. There is a whole bunch of literature to deal with it in a rigurous way, but that is complex and there is a lot of literarture on the web. In practice we just use probabilty densities and assume that everything works well
– Wolphram jonny
54 mins ago
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
When the distribution of probabilities is continuous, the probability of finding an object exactly at one specific position is zero. You can calculate what is the probability of finding the object at a certain interval, you have a probability density. The right way to treat the probability is $dP(x)=dx/a$, and the total probability now becomes 1, because $int dP(x)=int^a_0 dx/a=1$. For more details see https://en.wikipedia.org/wiki/Particle_in_a_box
answered 1 hour ago
Wolphram jonny
10k22451
But the spectrum of energy is discrete, and it is this spectrum that I am trying to calculate the total probability for. The problem is that the individual probabilities are the same and are all 1/a.
– Arthur
1 hour ago
The spectrum of the energy operator is discrete, but the spectrum of the position operator is continuum. An eigenstate of energy is not an eigenstate of position, the operators do not commute
– Wolphram jonny
1 hour ago
you are correct in that a continuum spectrum is incompatible with a Hilbert space, that is why there are extensions to deal with this, although in practice this is almost never mentioned. See physics.stackexchange.com/questions/210300/…
– Wolphram jonny
1 hour ago
The fact that one observable has a continuous spectrum and the other doesn't, shouldn't cause problems if the axioms of quantum mechanics are true. I appreciate it if you know how they deal with such a problem my friend.
– Arthur
58 mins ago
Well, that is the problem, the formalism is incomplete if you want to use continumm spectrum operators. There is a whole bunch of literature to deal with it in a rigurous way, but that is complex and there is a lot of literarture on the web. In practice we just use probabilty densities and assume that everything works well
– Wolphram jonny
54 mins ago
 |Â
show 3 more comments
up vote
2
down vote
When the distribution of probabilities is continuous, the probability of finding an object exactly at one specific position is zero. You can calculate what is the probability of finding the object at a certain interval, you have a probability density. The right way to treat the probability is $dP(x)=dx/a$, and the total probability now becomes 1, because $int dP(x)=int^a_0 dx/a=1$. For more details see https://en.wikipedia.org/wiki/Particle_in_a_box
answered 1 hour ago
Wolphram jonny
10k22451
But the spectrum of energy is discrete, and it is this spectrum that I am trying to calculate the total probability for. The problem is that the individual probabilities are the same and are all 1/a.
– Arthur
1 hour ago
The spectrum of the energy operator is discrete, but the spectrum of the position operator is continuum. An eigenstate of energy is not an eigenstate of position, the operators do not commute
– Wolphram jonny
1 hour ago
you are correct in that a continuum spectrum is incompatible with a Hilbert space, that is why there are extensions to deal with this, although in practice this is almost never mentioned. See physics.stackexchange.com/questions/210300/…
– Wolphram jonny
1 hour ago
The fact that one observable has a continuous spectrum and the other doesn't, shouldn't cause problems if the axioms of quantum mechanics are true. I appreciate it if you know how they deal with such a problem my friend.
– Arthur
58 mins ago
Well, that is the problem, the formalism is incomplete if you want to use continumm spectrum operators. There is a whole bunch of literature to deal with it in a rigurous way, but that is complex and there is a lot of literarture on the web. In practice we just use probabilty densities and assume that everything works well
– Wolphram jonny
54 mins ago
 |Â
show 3 more comments
up vote
2
down vote
up vote
2
down vote
When the distribution of probabilities is continuous, the probability of finding an object exactly at one specific position is zero. You can calculate what is the probability of finding the object at a certain interval, you have a probability density. The right way to treat the probability is $dP(x)=dx/a$, and the total probability now becomes 1, because $int dP(x)=int^a_0 dx/a=1$. For more details see https://en.wikipedia.org/wiki/Particle_in_a_box
answered 1 hour ago
Wolphram jonny
10k22451
When the distribution of probabilities is continuous, the probability of finding an object exactly at one specific position is zero. You can calculate what is the probability of finding the object at a certain interval, you have a probability density. The right way to treat the probability is $dP(x)=dx/a$, and the total probability now becomes 1, because $int dP(x)=int^a_0 dx/a=1$. For more details see https://en.wikipedia.org/wiki/Particle_in_a_box
answered 1 hour ago
Wolphram jonny
10k22451
answered 1 hour ago
Wolphram jonny
10k22451
answered 1 hour ago
Wolphram jonny
10k22451
answered 1 hour ago
Wolphram jonny
10k22451
10k22451
But the spectrum of energy is discrete, and it is this spectrum that I am trying to calculate the total probability for. The problem is that the individual probabilities are the same and are all 1/a.
– Arthur
1 hour ago
The spectrum of the energy operator is discrete, but the spectrum of the position operator is continuum. An eigenstate of energy is not an eigenstate of position, the operators do not commute
– Wolphram jonny
1 hour ago
you are correct in that a continuum spectrum is incompatible with a Hilbert space, that is why there are extensions to deal with this, although in practice this is almost never mentioned. See physics.stackexchange.com/questions/210300/…
– Wolphram jonny
1 hour ago
The fact that one observable has a continuous spectrum and the other doesn't, shouldn't cause problems if the axioms of quantum mechanics are true. I appreciate it if you know how they deal with such a problem my friend.
– Arthur
58 mins ago
Well, that is the problem, the formalism is incomplete if you want to use continumm spectrum operators. There is a whole bunch of literature to deal with it in a rigurous way, but that is complex and there is a lot of literarture on the web. In practice we just use probabilty densities and assume that everything works well
– Wolphram jonny
54 mins ago
 |Â
show 3 more comments
But the spectrum of energy is discrete, and it is this spectrum that I am trying to calculate the total probability for. The problem is that the individual probabilities are the same and are all 1/a.
– Arthur
1 hour ago
The spectrum of the energy operator is discrete, but the spectrum of the position operator is continuum. An eigenstate of energy is not an eigenstate of position, the operators do not commute
– Wolphram jonny
1 hour ago
you are correct in that a continuum spectrum is incompatible with a Hilbert space, that is why there are extensions to deal with this, although in practice this is almost never mentioned. See physics.stackexchange.com/questions/210300/…
– Wolphram jonny
1 hour ago
The fact that one observable has a continuous spectrum and the other doesn't, shouldn't cause problems if the axioms of quantum mechanics are true. I appreciate it if you know how they deal with such a problem my friend.
– Arthur
58 mins ago
Well, that is the problem, the formalism is incomplete if you want to use continumm spectrum operators. There is a whole bunch of literature to deal with it in a rigurous way, but that is complex and there is a lot of literarture on the web. In practice we just use probabilty densities and assume that everything works well
– Wolphram jonny
54 mins ago
But the spectrum of energy is discrete, and it is this spectrum that I am trying to calculate the total probability for. The problem is that the individual probabilities are the same and are all 1/a.
– Arthur
1 hour ago
But the spectrum of energy is discrete, and it is this spectrum that I am trying to calculate the total probability for. The problem is that the individual probabilities are the same and are all 1/a.
– Arthur
1 hour ago
The spectrum of the energy operator is discrete, but the spectrum of the position operator is continuum. An eigenstate of energy is not an eigenstate of position, the operators do not commute
– Wolphram jonny
1 hour ago
The spectrum of the energy operator is discrete, but the spectrum of the position operator is continuum. An eigenstate of energy is not an eigenstate of position, the operators do not commute
– Wolphram jonny
1 hour ago
you are correct in that a continuum spectrum is incompatible with a Hilbert space, that is why there are extensions to deal with this, although in practice this is almost never mentioned. See physics.stackexchange.com/questions/210300/…
– Wolphram jonny
1 hour ago
you are correct in that a continuum spectrum is incompatible with a Hilbert space, that is why there are extensions to deal with this, although in practice this is almost never mentioned. See physics.stackexchange.com/questions/210300/…
– Wolphram jonny
1 hour ago
The fact that one observable has a continuous spectrum and the other doesn't, shouldn't cause problems if the axioms of quantum mechanics are true. I appreciate it if you know how they deal with such a problem my friend.
– Arthur
58 mins ago
The fact that one observable has a continuous spectrum and the other doesn't, shouldn't cause problems if the axioms of quantum mechanics are true. I appreciate it if you know how they deal with such a problem my friend.
– Arthur
58 mins ago
Well, that is the problem, the formalism is incomplete if you want to use continumm spectrum operators. There is a whole bunch of literature to deal with it in a rigurous way, but that is complex and there is a lot of literarture on the web. In practice we just use probabilty densities and assume that everything works well
– Wolphram jonny
54 mins ago
Well, that is the problem, the formalism is incomplete if you want to use continumm spectrum operators. There is a whole bunch of literature to deal with it in a rigurous way, but that is complex and there is a lot of literarture on the web. In practice we just use probabilty densities and assume that everything works well
– Wolphram jonny
54 mins ago
 |Â
show 3 more comments
Arthur is a new contributor. Be nice, and check out our Code of Conduct.
Arthur is a new contributor. Be nice, and check out our Code of Conduct.
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4
The lab you've described has a position detector with infinite precision. Those sound nice - where did you find it? I'd like to get one for mine. (Joking aside: as you've just proved, position eigenstates are not physical, and cannot be achieved in real experiments.)
– Emilio Pisanty
1 hour ago
I liked the joke friend, but the problem persists. From totally mathematical point of view, this kind of result shouldn't occur if the axioms of QM represent a consistent mathematical theory.
– Arthur
1 hour ago